Drorbn - User contributions [en]

https://drorbn.net/api.php?action=feedcontributions&feedformat=atom&user=Pgadey Drorbn - User contributions [en] 2026-05-04T14:51:56Z User contributions MediaWiki 1.39.6 https://drorbn.net/index.php?title=User:Pgadey&diff=11088 User:Pgadey 2011-11-26T20:02:10Z

<p>Pgadey: </p> <hr /> <div>I'm Parker Glynn-Adey.<br /> I've been TeXing up notes sloppily and with lots of asides.<br /> <br /> Also -- I apologize for the ugliness of the results.<br /> <br /> * [[11-1100-Pgadey-Lect5]]<br /> ** Simplicity of <math>A_n</math>, definition <math>G</math>-sets.<br /> * [[11-1100-Pgadey-Lect6]]<br /> ** Structure of <math>G</math>-sets, Sylow 1~2~3 (up to permutation), groups of order 15.<br /> <br /> I wrote a quick hack for [[Computing GCDs over the Gaussian Integers]].<br /> <br /> Also -- One can convert LaTeX to MediaWiki using <pre>pandoc -f latex -t mediawiki</pre> The future is here my friends!!<br /> <br /> <br /> ==Messages ==<br /> <br /> Hey. I noticed that you've been writing up notes. So have I. On that note, since we both have written up proofs to the alternating group, I'm trying to create a [[11-1100/The Simplicity of the Alternating Groups| redirect]] page. Wanna give me a hand? I'm making direct modifications to the [[11-1100/Navigation |navigation template]] but they're not showing up when that page is embedded. Any ideas? [[User:Tholden|Tholden]] 17:30, 6 October 2011 (EDT)<br /> <br /> :Nevermind, just had to hard-purge the cache. [[User:Tholden|Tholden]] 10:22, 7 October 2011 (EDT)</div>

Pgadey https://drorbn.net/index.php?title=User:Pgadey&diff=11087 User:Pgadey 2011-11-26T20:01:53Z

<p>Pgadey: </p> <hr /> <div>I'm Parker Glynn-Adey.<br /> I've been TeXing up notes sloppily and with lots of asides.<br /> <br /> Also -- I apologize for the ugliness of the results.<br /> <br /> * [[11-1100-Pgadey-Lect5]]<br /> ** Simplicity of <math>A_n</math>, definition <math>G</math>-sets.<br /> * [[11-1100-Pgadey-Lect6]]<br /> ** Structure of <math>G</math>-sets, Sylow 1~2~3 (up to permutation), groups of order 15.<br /> <br /> I wrote a quick hack for [[Computing GCDs over the Gaussian Integers]].<br /> <br /> Also -- One can convert LaTeX to MediaWiki using <pre>pandoc -f latex -t mediawiki</pre>. The future is here my friends!!<br /> <br /> <br /> ==Messages ==<br /> <br /> Hey. I noticed that you've been writing up notes. So have I. On that note, since we both have written up proofs to the alternating group, I'm trying to create a [[11-1100/The Simplicity of the Alternating Groups| redirect]] page. Wanna give me a hand? I'm making direct modifications to the [[11-1100/Navigation |navigation template]] but they're not showing up when that page is embedded. Any ideas? [[User:Tholden|Tholden]] 17:30, 6 October 2011 (EDT)<br /> <br /> :Nevermind, just had to hard-purge the cache. [[User:Tholden|Tholden]] 10:22, 7 October 2011 (EDT)</div>

Pgadey https://drorbn.net/index.php?title=Computing_GCDs_over_the_Gaussian_Integers&diff=11086 Computing GCDs over the Gaussian Integers 2011-11-26T19:56:52Z

<p>Pgadey: </p> <hr /> <div>I wrote up a very simple Perl script for computing GCDs over the Gaussian integers. It comes with no guarantee or proof of correctness. One can sketchily argue something like this however:<br /> <br /> ''Claim'': <math>\mathbb{Z}[i]</math> is a Euclidean domain, and hence a PID, and hence a UFD.<br /><br /> <br /> <br /> Consider the norm function <math>\nu(a + bi) = a^2 + b^2</math>. We write the standard norm on the complex plane as <math>||a+bi||_{\mathbb{C}} = \sqrt{a^2 + b^2}</math>. We show that for all <math>x \neq 0</math> and <math>y</math> we have, <math> y = qx + r </math> where <math>\nu(r) = 0</math> or <math>\nu(r) < \nu(x)</math>. Consider <math>x</math> and <math>y</math> as points on the complex plane. Since <math>x \neq 0</math> we have that <math>y/x</math> is another point in the complex plane. Consider <math>\mathbb{Z}[i]</math> as the points with integer coordinates in <math>\mathbb{C}</math>. For any point <math>\zeta \in \mathbb{C}</math> we can find a <math>\zeta' \in \mathbb{Z}[i]</math> such that <math>||\zeta - \zeta'||_{\mathbb{C}} \leq \sqrt{2}/2</math> by taking the <math>\zeta'</math> to be the nearest lattice point to <math>\zeta</math>. Since any point on the plane is contained in a square whose vertices are lattice points and whose side lengths are one, there must be a lattice point nearer than half the diagonal of the square. We then take <math>z</math> to be the nearest lattice point to <math>y/x</math>. We then have <math>y = zx + (y - zx)</math>. We compute the norm of <math>\nu(y - zx)</math>. We have: <math> \sqrt{\nu(y - zx)} = ||y - zx||_{\mathbb{C}} = ||x||_{\mathbb{C}} \cdot ||y/x - z||_{\mathbb{C}} \leq ||x||_{\mathbb{C}} \frac{\sqrt{2}}{2} < ||x||_{\mathbb{C}} = \sqrt{\nu(x)} </math> It follows that <math>\nu(y - zy) < \nu(x)</math>. We obtain that <math>\mathbb{Z}[i]</math> is a Euclidean domain. Thus <math>\mathbb{Z}[i]</math> is a PID and hence a UFD by results given in class.<br /> <br /> <br /> <br /> ----<br /> <br /> <pre><br /> #!/usr/bin/perl<br /> use Math::Complex;<br /> <br /> ## A Quick hack for computing GCDs of Gaussian integers.<br /> <br /> $z2 = 857 + i;<br /> $z1 = 255;<br /> <br /> <br /> sub gcd {<br /> # the Euclidean algorithm<br /> <br /> my $x = $_[0];<br /> my $y = $_[1];<br /> <br /> if ($x * $y == 0) {<br /> print "Done!\n";<br /> } else {<br /> $q = &approx($x/$y); <br /> $r = $x - $q*$y;<br /> print "($x) = ($q)($y) + ($r)\n";<br /> <br /> &gcd($y,$r);<br /> }<br /> }<br /> <br /> sub approx {<br /> # find the nearest Gaussian integer to a point on the complex plane<br /> <br /> my $z = $_[0];<br /> <br /> my $x = int(Re($z));<br /> my $y = int(Im($z));<br /> <br /> if (abs($z - (($x+1) + i*$y) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*$y;<br /> } elsif (abs($z - (($x) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return $x + i*($y+1);<br /> } elsif (abs($z - (($x+1) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*($y+1);<br /> } else { <br /> return $x + i*$y;<br /> } <br /> <br /> <br /> }<br /> <br /> &gcd($z1, $z2);<br /> <br /> </pre></div>

Pgadey https://drorbn.net/index.php?title=Computing_GCDs_over_the_Gaussian_Integers&diff=11085 Computing GCDs over the Gaussian Integers 2011-11-26T19:55:55Z

<p>Pgadey: </p> <hr /> <div>I wrote up a very simple Perl script for computing GCDs over the Gaussian integers. It comes with no guarantee or proof of correctness. One can sketchily argue something like this however:<br /> <br /> <br /> ''Claim'': <math>\ZZ[i]</math> is a Euclidean domain, and hence a PID, and hence a UFD.<br /><br /> <br /> <br /> Consider the norm function <math>\nu(a + bi) = a^2 + b^2</math>. We write the standard norm on the complex plane as <math>||a+bi||_{\mathbb{C}} = \sqrt{a^2 + b^2}</math>. We show that for all <math>x \neq 0</math> and <math>y</math> we have, <math> y = qx + r </math> where <math>\nu(r) = 0</math> or <math>\nu(r) < \nu(x)</math>. Consider <math>x</math> and <math>y</math> as points on the complex plane. Since <math>x \neq 0</math> we have that <math>y/x</math> is another point in the complex plane. Consider <math>\ZZ[i]</math> as the points with integer coordinates in <math>\mathbb{C}</math>. For any point <math>\zeta \in \mathbb{C}</math> we can find a <math>\zeta' \in \ZZ[i]</math> such that <math>||\zeta - \zeta'||_{\mathbb{C}} \leq \sqrt{2}/2</math> by taking the <math>\zeta'</math> to be the nearest lattice point to <math>\zeta</math>. Since any point on the plane is contained in a square whose vertices are lattice points and whose side lengths are one, there must be a lattice point nearer than half the diagonal of the square. We then take <math>z</math> to be the nearest lattice point to <math>y/x</math>. We then have <math>y = zx + (y - zx)</math>. We compute the norm of <math>\nu(y - zx)</math>. We have: <math> \sqrt{\nu(y - zx)} = ||y - zx||_{\mathbb{C}} = ||x||_{\mathbb{C}} \cdot ||y/x - z||_{\mathbb{C}} \leq ||x||_{\mathbb{C}} \frac{\sqrt{2}}{2} < ||x||_{\mathbb{C}} = \sqrt{\nu(x)} </math> It follows that <math>\nu(y - zy) < \nu(x)</math>. We obtain that <math>\ZZ[i]</math> is a Euclidean domain. Thus <math>\ZZ[i]</math> is a PID and hence a UFD by results given in class.<br /> <br /> <br /> <br /> ----<br /> <br /> <pre><br /> #!/usr/bin/perl<br /> use Math::Complex;<br /> <br /> ## A Quick hack for computing GCDs of Gaussian integers.<br /> <br /> $z2 = 857 + i;<br /> $z1 = 255;<br /> <br /> <br /> sub gcd {<br /> # the Euclidean algorithm<br /> <br /> my $x = $_[0];<br /> my $y = $_[1];<br /> <br /> if ($x * $y == 0) {<br /> print "Done!\n";<br /> } else {<br /> $q = &approx($x/$y); <br /> $r = $x - $q*$y;<br /> print "($x) = ($q)($y) + ($r)\n";<br /> <br /> &gcd($y,$r);<br /> }<br /> }<br /> <br /> sub approx {<br /> # find the nearest Gaussian integer to a point on the complex plane<br /> <br /> my $z = $_[0];<br /> <br /> my $x = int(Re($z));<br /> my $y = int(Im($z));<br /> <br /> if (abs($z - (($x+1) + i*$y) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*$y;<br /> } elsif (abs($z - (($x) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return $x + i*($y+1);<br /> } elsif (abs($z - (($x+1) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*($y+1);<br /> } else { <br /> return $x + i*$y;<br /> } <br /> <br /> <br /> }<br /> <br /> &gcd($z1, $z2);<br /> <br /> </pre></div>

Pgadey https://drorbn.net/index.php?title=11-1100&diff=11084 11-1100 2011-11-26T19:50:17Z

<p>Pgadey: </p> <hr /> <div>__NOEDITSECTION__<br /> __NOTOC__<br /> {{11-1100/Navigation}}<br /> ==Core Algebra I==<br /> ===Department of Mathematics, University of Toronto, Fall 2011===<br /> <br /> {{11-1100/Crucial Information}}<br /> <br /> ===Texts===<br /> Lang's ''Algebra'', Selick's [http://www.math.toronto.edu/mat1100/ lecture notes for this class], Dummit and Foote's ''Abstract Algebra'', Hungerford's ''Abstract Algebra''.<br /> <br /> ===Further Resources===<br /> <br /> * [http://www.math.utoronto.ca/cms/graduate-program/ Graduate Studies] at the [http://www.math.toronto.edu/ UofT Math Department]. In particular, [http://www.math.utoronto.ca/cms/tentative-2010-2011-graduate-course-descriptions/ Graduate Course Descriptions].<br /> <br /> * My {{Pensieve Link|Classes/11-1100/|11-1100 notebook}}.<br /> <br /> * My [[10-1100|2010 Class]].<br /> <br /> * Paul Selick's [http://www.math.toronto.edu/mat1100/ 2007 class] ([[11-1100/Errata_to_Prof._Selick's_Notes|Errata]]).<br /> <br /> * Some (mostly complete) notes from last year's class: [[10-1100-Notes]].<br /> <br /> * Some blackboard shots start at {{BBS Link|11_1100-111024-110239.jpg}}.<br /> <br /> {{Template:11-1100:Dror/Students Divider}}<br /> <br /> * Student solutions to homework problems: [[11-1100/Homework Solutions]].<br /> <br /> * Student notes from class: [[11-1100/Notes]].<br /> <br /> * Very cool video explaining how to visualize some concepts in group theory ([[User:jmracek]]): [http://web.bentley.edu/empl/c/ncarter/vgt/VisualizingGroupTheory-320x240.mov]<br /> <br /> * Summary of the course ([[User: Lp.thibault]]): [[Media:11-1100_Summary4.pdf|Summary of the course]].<br /> <br /> * Read Along on material covering Modules (Dummit Foote and Lang): [[User:Vanessa.foster]]<br /> <br /> * A rough and ready Perl script for [[Computing GCDs over the Gaussian Integers]]. (I provide no proof of correctness, all faults are my own, etc. -- [[User:pgadey]])<br /> <br /> [[Image:xkcdsettheory.jpg]]<br /> Link: [[http://xkcd.com/982/]]</div>

Pgadey https://drorbn.net/index.php?title=11-1100&diff=11083 11-1100 2011-11-26T19:49:35Z

<p>Pgadey: </p> <hr /> <div>__NOEDITSECTION__<br /> __NOTOC__<br /> {{11-1100/Navigation}}<br /> ==Core Algebra I==<br /> ===Department of Mathematics, University of Toronto, Fall 2011===<br /> <br /> {{11-1100/Crucial Information}}<br /> <br /> ===Texts===<br /> Lang's ''Algebra'', Selick's [http://www.math.toronto.edu/mat1100/ lecture notes for this class], Dummit and Foote's ''Abstract Algebra'', Hungerford's ''Abstract Algebra''.<br /> <br /> ===Further Resources===<br /> <br /> * [http://www.math.utoronto.ca/cms/graduate-program/ Graduate Studies] at the [http://www.math.toronto.edu/ UofT Math Department]. In particular, [http://www.math.utoronto.ca/cms/tentative-2010-2011-graduate-course-descriptions/ Graduate Course Descriptions].<br /> <br /> * My {{Pensieve Link|Classes/11-1100/|11-1100 notebook}}.<br /> <br /> * My [[10-1100|2010 Class]].<br /> <br /> * Paul Selick's [http://www.math.toronto.edu/mat1100/ 2007 class] ([[11-1100/Errata_to_Prof._Selick's_Notes|Errata]]).<br /> <br /> * Some (mostly complete) notes from last year's class: [[10-1100-Notes]].<br /> <br /> * Some blackboard shots start at {{BBS Link|11_1100-111024-110239.jpg}}.<br /> <br /> {{Template:11-1100:Dror/Students Divider}}<br /> <br /> * Student solutions to homework problems: [[11-1100/Homework Solutions]].<br /> <br /> * Student notes from class: [[11-1100/Notes]].<br /> <br /> * Very cool video explaining how to visualize some concepts in group theory ([[User:jmracek]]): [http://web.bentley.edu/empl/c/ncarter/vgt/VisualizingGroupTheory-320x240.mov]<br /> <br /> * Summary of the course ([[User: Lp.thibault]]): [[Media:11-1100_Summary4.pdf|Summary of the course]].<br /> <br /> * Read Along on material covering Modules (Dummit Foote and Lang): [[User:Vanessa.foster]]<br /> <br /> * A rough and ready Perl script for [[Computing GCDs over the Gaussian Integers]].<br /> <br /> [[Image:xkcdsettheory.jpg]]<br /> Link: [[http://xkcd.com/982/]]</div>

Pgadey https://drorbn.net/index.php?title=Computing_GCDs_over_the_Gaussian_Integers&diff=11082 Computing GCDs over the Gaussian Integers 2011-11-26T19:47:46Z

<p>Pgadey: </p> <hr /> <div>I wrote up a very simple Perl script for computing GCDs over the Gaussian integers. It comes with no guarantee or proof of correctness. <br /> <br /> ----<br /> <br /> <pre><br /> #!/usr/bin/perl<br /> use Math::Complex;<br /> <br /> ## A Quick hack for computing GCDs of Gaussian integers.<br /> <br /> $z2 = 857 + i;<br /> $z1 = 255;<br /> <br /> <br /> sub gcd {<br /> # the Euclidean algorithm<br /> <br /> my $x = $_[0];<br /> my $y = $_[1];<br /> <br /> if ($x * $y == 0) {<br /> print "Done!\n";<br /> } else {<br /> $q = &approx($x/$y); <br /> $r = $x - $q*$y;<br /> print "($x) = ($q)($y) + ($r)\n";<br /> <br /> &gcd($y,$r);<br /> }<br /> }<br /> <br /> sub approx {<br /> # find the nearest Gaussian integer to a point on the complex plane<br /> <br /> my $z = $_[0];<br /> <br /> my $x = int(Re($z));<br /> my $y = int(Im($z));<br /> <br /> if (abs($z - (($x+1) + i*$y) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*$y;<br /> } elsif (abs($z - (($x) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return $x + i*($y+1);<br /> } elsif (abs($z - (($x+1) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*($y+1);<br /> } else { <br /> return $x + i*$y;<br /> } <br /> <br /> <br /> }<br /> <br /> &gcd($z1, $z2);<br /> <br /> </pre></div>

Pgadey https://drorbn.net/index.php?title=Computing_GCDs_over_the_Gaussian_Integers&diff=11081 Computing GCDs over the Gaussian Integers 2011-11-26T19:46:11Z

<p>Pgadey: </p> <hr /> <div>I wrote up a very simple Perl script for computing GCDs over the Gaussian integers. It comes with no guarantee. <br /> <br /> ----<br /> <br /> <pre><br /> #!/usr/bin/perl<br /> use Math::Complex;<br /> <br /> ## A Quick hack for computing GCDs of Gaussian integers.<br /> <br /> $z2 = 857 + i;<br /> $z1 = 255;<br /> <br /> <br /> sub gcd {<br /> # the Euclidean algorithm<br /> <br /> my $x = $_[0];<br /> my $y = $_[1];<br /> <br /> if ($x * $y == 0) {<br /> print "Done!\n";<br /> } else {<br /> $q = &approx($x/$y); <br /> $r = $x - $q*$y;<br /> print "($x) = ($q)($y) + ($r)\n";<br /> <br /> &gcd($y,$r);<br /> }<br /> }<br /> <br /> sub approx {<br /> # find the nearest Gaussian integer to a point on the complex plane<br /> <br /> my $z = $_[0];<br /> <br /> my $x = int(Re($z));<br /> my $y = int(Im($z));<br /> <br /> if (abs($z - (($x+1) + i*$y) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*$y;<br /> } elsif (abs($z - (($x) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return $x + i*($y+1);<br /> } elsif (abs($z - (($x+1) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*($y+1);<br /> } else { <br /> return $x + i*$y;<br /> } <br /> <br /> <br /> }<br /> <br /> &gcd($z1, $z2);<br /> <br /> </pre></div>

Pgadey https://drorbn.net/index.php?title=Computing_GCDs_over_the_Gaussian_Integers&diff=11080 Computing GCDs over the Gaussian Integers 2011-11-26T19:45:52Z

<p>Pgadey: </p> <hr /> <div>I wrote up a very simple Perl script for computing GCDs over the Gaussian integers. It comes with no guarantee. <br /> <br /> ----<br /> <br /> <br /> #!/usr/bin/perl<br /> use Math::Complex;<br /> <br /> ## A Quick hack for computing GCDs of Gaussian integers.<br /> <br /> $z2 = 857 + i;<br /> $z1 = 255;<br /> <br /> <br /> sub gcd {<br /> # the Euclidean algorithm<br /> <br /> my $x = $_[0];<br /> my $y = $_[1];<br /> <br /> if ($x * $y == 0) {<br /> print "Done!\n";<br /> } else {<br /> $q = &approx($x/$y); <br /> $r = $x - $q*$y;<br /> print "($x) = ($q)($y) + ($r)\n";<br /> <br /> &gcd($y,$r);<br /> }<br /> }<br /> <br /> sub approx {<br /> # find the nearest Gaussian integer to a point on the complex plane<br /> <br /> my $z = $_[0];<br /> <br /> my $x = int(Re($z));<br /> my $y = int(Im($z));<br /> <br /> if (abs($z - (($x+1) + i*$y) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*$y;<br /> } elsif (abs($z - (($x) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return $x + i*($y+1);<br /> } elsif (abs($z - (($x+1) + i*($y+1)) ) < 1/sqrt(2)) {<br /> return ($x+1) + i*($y+1);<br /> } else { <br /> return $x + i*$y;<br /> } <br /> <br /> <br /> }<br /> <br /> &gcd($z1, $z2);</div>

Pgadey https://drorbn.net/index.php?title=User:Pgadey&diff=11079 User:Pgadey 2011-11-26T19:38:47Z

<p>Pgadey: </p> <hr /> <div>I'm Parker Glynn-Adey.<br /> I've been TeXing up notes sloppily and with lots of asides.<br /> <br /> Also -- I apologize for the ugliness of the results.<br /> <br /> * [[11-1100-Pgadey-Lect5]]<br /> ** Simplicity of <math>A_n</math>, definition <math>G</math>-sets.<br /> * [[11-1100-Pgadey-Lect6]]<br /> ** Structure of <math>G</math>-sets, Sylow 1~2~3 (up to permutation), groups of order 15.<br /> <br /> I wrote a quick hack for [[Computing GCDs over the Gaussian Integers]].<br /> <br /> <br /> ==Messages ==<br /> <br /> Hey. I noticed that you've been writing up notes. So have I. On that note, since we both have written up proofs to the alternating group, I'm trying to create a [[11-1100/The Simplicity of the Alternating Groups| redirect]] page. Wanna give me a hand? I'm making direct modifications to the [[11-1100/Navigation |navigation template]] but they're not showing up when that page is embedded. Any ideas? [[User:Tholden|Tholden]] 17:30, 6 October 2011 (EDT)<br /> <br /> :Nevermind, just had to hard-purge the cache. [[User:Tholden|Tholden]] 10:22, 7 October 2011 (EDT)</div>

Pgadey https://drorbn.net/index.php?title=User:Pgadey&diff=10760 User:Pgadey 2011-10-06T20:28:41Z

<p>Pgadey: </p> <hr /> <div>I'm Parker Glynn-Adey.<br /> I've been TeXing up notes sloppily and with lots of asides.<br /> <br /> Also -- I apologize for the ugliness of the results.<br /> <br /> * [[11-1100-Pgadey-Lect5]]<br /> ** Simplicity of <math>A_n</math>, definition <math>G</math>-sets.<br /> * [[11-1100-Pgadey-Lect6]]<br /> ** Structure of <math>G</math>-sets, Sylow 1~2~3 (up to permutation), groups of order 15.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10759 11-1100-Pgadey-Lect6 2011-10-06T20:22:35Z

<p>Pgadey: </p> <hr /> <div>==Theory of Transitive <math>G</math>-sets==<br /> ; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of <math>x</math>.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternion'' group (See [http://en.wikipedia.org/wiki/Quaternion_group], also [http://en.wikipedia.org/wiki/History_of_quaternions]).<br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> ==Sylow==<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the order of <math>p</math>. Assume the claim holds for all groups of order less than <math>|G|</math>. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>.<br /> * <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G</math> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\mod\ |G|</math> and <math>n_p \equiv 1\mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]</span><br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> ==Groups of Order 15==<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10758 11-1100-Pgadey-Lect6 2011-10-06T20:19:58Z

<p>Pgadey: </p> <hr /> <div>==Theory of Transitive <math>G</math>-sets==<br /> ; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of <math>x</math>.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternion'' group (See [http://en.wikipedia.org/wiki/Quaternion_group]).<br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> ==Sylow==<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the order of <math>p</math>. Assume the claim holds for all groups of order less than <math>|G|</math>. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>.<br /> * <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G</math> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\mod\ |G|</math> and <math>n_p \equiv 1\mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]</span><br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> ==Groups of Order 15==<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10757 11-1100-Pgadey-Lect6 2011-10-06T20:19:24Z

<p>Pgadey: </p> <hr /> <div>==Theory of Transitive <math>G</math>-sets==<br /> ; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of <math>x</math>.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternion'' group (See [http://en.wikipedia.org/wiki/Quaternion_group:WP]).<br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> ==Sylow==<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the order of <math>p</math>. Assume the claim holds for all groups of order less than <math>|G|</math>. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>.<br /> * <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G</math> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\mod\ |G|</math> and <math>n_p \equiv 1\mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]</span><br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> ==Groups of Order 15==<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10756 11-1100-Pgadey-Lect6 2011-10-06T20:16:29Z

<p>Pgadey: </p> <hr /> <div>==Theory of Transitive <math>G</math>-sets==<br /> ; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of <math>x</math>.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> ==Sylow==<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the order of <math>p</math>. Assume the claim holds for all groups of order less than <math>|G|</math>. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>.<br /> * <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G</math> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\mod\ |G|</math> and <math>n_p \equiv 1\mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]</span><br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> ==Groups of Order 15==<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10755 11-1100-Pgadey-Lect6 2011-10-06T20:15:28Z

<p>Pgadey: </p> <hr /> <div>==Theory of Transitive <math>G</math>-sets==<br /> ; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> ==Sylow==<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the order of <math>p</math>. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>.<br /> * <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G</math> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\mod\ |G|</math> and <math>n_p \equiv 1\mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]</span><br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> ==Groups of Order 15==<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10754 11-1100-Pgadey-Lect6 2011-10-06T20:14:37Z

<p>Pgadey: </p> <hr /> <div>==Theory of Transitive <math>G</math>-sets==<br /> ; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> ==Sylow==<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the order of <<p>>. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>.<br /> * <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G</math> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\mod\ |G|</math> and <math>n_p \equiv 1\mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]<br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> ==Groups of Order 15==<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10753 11-1100-Pgadey-Lect6 2011-10-06T20:13:57Z

<p>Pgadey: </p> <hr /> <div>==Theory of Transitive <math>G</math>-sets==<br /> ; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> ==Sylow==<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>.<br /> * <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G</math> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\mod\ |G|</math> and <math>n_p \equiv 1\mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]<br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> ==Groups of Order 15==<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10752 11-1100-Pgadey-Lect6 2011-10-06T20:12:43Z

<p>Pgadey: </p> <hr /> <div>==Theory of Transitive <math>G</math>-sets==<br /> ; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> ==Sylow==<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation."]</span> Since <math>|G| \equiv 0\mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\mod\ p</math>.<br /> * <math>|G| \not\equiv 0\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\mod\ |G|</math> and <math>n_p \equiv 1\mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]<br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> ==Groups of Order 15==<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\mod\ 15</math> and <math>n_3 \equiv 1\mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\mod\ 15</math> and <math>n_5 \equiv 1\mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10751 11-1100-Pgadey-Lect6 2011-10-06T20:11:27Z

<p>Pgadey: </p> <hr /> <div>==Theory of Transitive <math>G</math>-sets==<br /> ; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\ \mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\ \mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> ==Sylow==<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\ \mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation.]</span> Since <math>|G| \equiv 0\ \mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\ \mod\ p</math>.<br /> * <math>|G| \not\equiv 0\ \mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\ \mod\ |G|</math> and <math>n_p \equiv 1\ \mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]<br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> ==Groups of Order 15==<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\ \mod\ 15</math> and <math>n_3 \equiv 1\ \mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\ \mod\ 15</math> and <math>n_5 \equiv 1\ \mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10750 11-1100-Pgadey-Lect6 2011-10-06T20:09:29Z

<p>Pgadey: </p> <hr /> <div>; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\ \mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\ \mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> SYLOW<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\ \mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation.]</span> Since <math>|G| \equiv 0\ \mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\ \mod\ p</math>.<br /> * <math>|G| \equiv 0\ \not\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\ \mod\ |G|</math> and <math>n_p \equiv 1\ \mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \leq N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]<br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> GROUPS OF ORDER 15<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\ \mod\ 15</math> and <math>n_3 \equiv 1\ \mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\ \mod\ 15</math> and <math>n_5 \equiv 1\ \mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10749 11-1100-Pgadey-Lect6 2011-10-06T20:08:18Z

<p>Pgadey: </p> <hr /> <div>; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exists_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\ \mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\ \mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> SYLOW<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\ \mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation.]</span> Since <math>|G| \equiv 0\ \mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\ \mod\ p</math>.<br /> * <math>|G| \equiv 0\ \not\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\ \mod\ |G|</math> and <math>n_p \equiv 1\ \mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \lea N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]<br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> GROUPS OF ORDER 15<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\ \mod\ 15</math> and <math>n_3 \equiv 1\ \mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\ \mod\ 15</math> and <math>n_5 \equiv 1\ \mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10748 11-1100-Pgadey-Lect6 2011-10-06T20:07:53Z

<p>Pgadey: </p> <hr /> <div>; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> : If <math>X</math> is a transitive <math>G</math>-set and <math>x \in X</math> then <math>X \simeq G/Stab(x)</math> where the isomorphism an isomorphism of <math>G</math>-sets.<br /> <br /> ; Transitive <math>G</math>-set<br /> : A <math>G</math>-set <math>X</math> is transitive is <math>\forall_{x,y \in X} \exits_{g \in G}\ st.\ gx = y</math>.<br /> <br /> ; Stabilizer of a point<br /> : We write <math>Stab(x) = \{g \in G : gx = x\}</math> for the stabilizer subgroup of $x$.<br /> <br /> ''Proof'' We define an equivalence relation <math>x \sim y \iff \exists_{g \in G} gx = y</math>. This relation is reflexive since <math>x = ex</math> and thus <math>x \sim x</math>. This relation is symmetric since <math>y = gx</math> implies <math>g^{-1}y = x</math>. This relation is transitive, since if <math>x = gy</math> and <math>y = hz</math> then <math>x = ghz</math>. It follows that <math> X = \coprod_{i \in I} Gx_{i} </math> where <math>Gx_i</math> denote the orbit of a point <math>x_i</math>.<br /> <br /> We then claim that <math>Gx_i</math> is a transitive <math>G</math>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <math>Gx</math> is isomorphic to <math>G / Stab(x)</math> as a <math>G</math>-set.<br /> <br /> We produce two morphism <math>\Psi : Gx \rightarrow G/Stab(x)</math> and <math>\Phi : G/Stab(x) \rightarrow Gx</math>. <br /> <br /> To define <math>\Psi</math> there is only one thing we can do. We have <math>y \in Gx \Rightarrow y = gx</math> and then we define <math>\Psi(y) = g Stab(x)</math>. We check that this map is well defined. If <math>y = gx = g'x</math> then <math>g^{-1}g'x = x</math> and hence <math>g^{-1}g \in Stab(x)</math>. It follows that <math>gStab(x) = g'Stab(x)</math>. Thus <math>\Psi</math> is well defined.<br /> <br /> To define <math>\Phi</math> we take <math>gStab(x) \in G/Stab(x)</math> and define <math>\Phi(gStab(x)) = gx</math>. We show that this map is well defined. If <math>gStab(x) = g'Stab(x)</math> then <math>g^{-1}g' \in Stab(x)</math> and hence <math>g^{-1}g'x = x</math>. It follows that <math>gx = g'x</math> and hence <math>\Phi</math> is well defined.<br /> <br /> We need to check that <math>\Psi</math> and <math>\Phi</math> are mutually inverse and <math>G</math>-set morphisms. We quickly check that <math>\Phi</math> is a <math>G</math>-set morphism. If <math>y = gx</math> and <math>g_1 \in G</math> then <math>g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)</math>. Similarly, <math>\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)</math>. The last inequality follows since we can take any <math>g'</math> such that <math>g'y = g_1y</math>. Why not take <math>g' = g_1g</math> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <math>|X| < \infty</math> and <math>X = \coprod_{i \in I} Gx_i</math> then <math>|X| = \sum_{i} \frac{|G|}{Stab(x_i)}</math>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <math>p</math>-Group<br /> : A <math>p</math>-group is a group <math>G</math> with <math>|G| = p^k</math> for some <math>k</math>.<br /> <br /> <math>G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}</math><br /> <br /> The last group <math>Q</math> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <math>p</math>-group has a non-trivial centre.<br /> <br /> Let <math>G</math> act on itself by conjugation. Decompose <math>G = \coprod Gx_i</math>. Then,<br /> <math> |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> Observe that <math>|Gx_i|| = 1</math> iff <math>x_i \in Z(G)</math>. It follows that <br /> <math> |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} </math><br /> The formula above is called ''"the class formula"''. We have that <math>|G| / |Stab(x)| = p^k</math> for some <math> 1 < k</math> since <math>Stab(x)</math> is a subgroup. It follows that <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\ \mod\ p</math>. It follows that <math>|Z(G)| \equiv 0\ \mod\ p</math>. Since <math>e \in Z(G)</math> we have <math>1 \leq |Z(G)|</math> and thus <math>p \leq |Z(G)|</math>.<br /> <br /> SYLOW<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <math>A</math> is an abelian group and <math>p</math> divides <math>|A|</math>, then there is an element of order <math>p</math> in <math>A</math>.<br /> <br /> ''Proof''. Pick <math>x \in A</math>. If <math>p</math> divides the order of <math>x</math> then we have <math>x^{np} = e</math> for some <math>n</math>. It follows that <math>(x^n)^p = e</math>. We then have that the order of <math>x^n</math> is <math>p</math>. If <math>p</math> does not divide the order of <math>p</math>, then consider <math>A / <x> </math>. Since <math>A</math> is abelian, <math> <x> </math> is a normal subgroup. We have that <math>p</math> divides <math>|A/<x>|</math>, and <math>|A / <x>| < |A|</math>. We then induct. Let <math> y<x> </math> have order <math>p</math>, that is <math> (y<x>)^p = <x> </math>. We then have that <math> y^p = x^k </math> for some <math> k </math>. We write <math> |<y>| = np + r </math> where <math> 0 \leq p < r </math>. We then have <math> e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> </math>. It follows that <math>(y<x>)^r = <x> </math> contradicting the assumption that the order of <math> y<x> </math> is <math> p </math>.<br /> <br /> <br /> ; Sylow set<br /> : If <math>|G| = p^k m</math> for <math>m \not\equiv 0\ \mod\ p</math> then <math>Syl_p(G) = \{P \leq G : |P| = p^k</math>.<br /> <br /> ; Sylow I<br /> : <math>Syl_p(G) \neq \emptyset</math><br /> <br /> We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation.]</span> Since <math>|G| \equiv 0\ \mod\ p</math> we have either: <br /> * <math>|G| \equiv 0\ \mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \equiv 0\ \mod\ p</math>.<br /> * <math>|G| \equiv 0\ \not\mod\ p</math> and <math>\sum |G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>.<br /> <br /> If <math>|Z(G)| \not\equiv 0\ \mod p</math> then there exists <math>x_i</math> such that <math>|G|/|Stab(x_i)| \not\equiv 0\ \mod\ p</math>. Thus <math>p^k</math> divides <math>|Stab(x_i)|</math>. We have that <math>|Stab(x_i)| < |G|</math> <span style="color:green">[Why happens here?]</span> We then have that <math>p^k \leq Stab(x_i) < |G|</math> and by induction there is <math>|P| = p^k</math> such that <math>P \leq Stab(x_i)</math>. It follows <math>P \leq Stab(x_i) \leq G</math>. We've obtained the Sylow <math>p</math>-subgroup.<br /> <br /> WIf <math>|Z(G)| \equiv 0\ \mod p</math> then by Cauchy's Lemma, there is <math>x \in Z(G)</math> with <math>|<x>| = p</math>. Consider the group <math> G / <x> </math>. By the induction hypothesis there is <math> P' \leq G/<x> </math> where <math>|P'| = p^{k-1}</math>. Then, there is the canonical projection <math> \pi : G \rightarrow G/<x> </math>. By the fourth isomorphism theory <math> P = \pi^{-1}(P') \leq G </math> and <math> |\pi^{-1}(P')| = p(p^{k-1}) = p^k </math>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <math>p</math>-subgroup of <math>G> is conjugate. Moreover, every <math>p</math>-subgroup is contained in a Sylow <math>p</math>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <math>n_p(G) = |Syl_p(G)|</math>. We have <math>n_p \equiv 0\ \mod\ |G|</math> and <math>n_p \equiv 1\ \mod\ p</math>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <math>P \in Syl_p(G)</math> and <math>H \lea N(P)</math> is a <math>p</math>-group, then <math>H \leq P</math>.<br /> : If <math>x \in G</math> has <math>|<x>| = p^k</math> and <math>x \in N(P)</math> then <math>x \in P</math>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]<br /> <br /> We show the first statement. We have that <math>|P / P \cap H| = p^k</math> since <math>P</math> is a <math>p</math>-group. We then know that <math>PH / H \simeq P / P \cap H</math> by the second isomorphism theorem. It foolows that <math>|PH| = p^{k'}</math>. But since <math>P</math> is maximal, we have <math>P = PH</math> and thus <math>H \subseteq P</math>. The first statement implies the second by taking <math>H = <x> </math>.<br /> <br /> <br /> <br /> GROUPS OF ORDER 15<br /> <br /> If <math>|G| = 15</math> then <math>n_3 \equiv 0\ \mod\ 15</math> and <math>n_3 \equiv 1\ \mod\ 3</math>. These imply <math>n_3 = 1</math>. Moreover, <math>n_5 \equiv 0\ \mod\ 15</math> and <math>n_5 \equiv 1\ \mod\ 5</math>. These imply <math>n_5 = 1</math>. Thus we have <math>P_3</math> a normal <math>3</math>-subgroup. Moreover, we have <math>P_5</math> a normal <math>5</math>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect6&diff=10747 11-1100-Pgadey-Lect6 2011-10-06T20:05:44Z

<p>Pgadey: </p> <hr /> <div>; Theorem<br /> : Every <math>G</math>-set is a disjoint union of "transitive <math>G</math>-sets"<br /> <br /> ;Theorem<br /> : If <<X>> is a transitive <<G>>-set and <<x \in X>> then <<X \simeq G/Stab(x)>> where the isomorphism an isomorphism of <<G>>-sets.<br /> <br /> ; Transitive <<G>>-set<br /> : A <<G>>-set <<X>> is transitive is <<\forall_{x,y \in X} \exits_{g \in G}\ st.\ gx = y>>.<br /> <br /> ; Stabilizer of a point<br /> : We write <<Stab(x) = \{g \in G : gx = x\}>> for the stabilizer subgroup of $x$.<br /> <br /> ''Proof'' We define an equivalence relation <<x \sim y \iff \exists_{g \in G} gx = y>>. This relation is reflexive since <<x = ex>> and thus <<x \sim x>>. This relation is symmetric since <<y = gx>> implies <<g^{-1}y = x>>. This relation is transitive, since if <<x = gy>> and <<y = hz>> then <<x = ghz>>. It follows that << X = \coprod_{i \in I} Gx_{i} >> where <<Gx_i>> denote the orbit of a point <<x_i>>.<br /> <br /> We then claim that <<Gx_i>> is a transitive <<G>>-set. <span style="color:green"> [Dror: "[This fact] is too easy."] </span><br /> <br /> We show that <<Gx>> is isomorphic to <<G / Stab(x)>> as a <<G>>-set.<br /> <br /> We produce two morphism <<\Psi : Gx \rightarrow G/Stab(x)>> and <<\Phi : G/Stab(x) \rightarrow Gx>>. <br /> <br /> To define <<\Psi>> there is only one thing we can do. We have <<y \in Gx \Rightarrow y = gx>> and then we define <<\Psi(y) = g Stab(x)>>. We check that this map is well defined. If <<y = gx = g'x>> then <<g^{-1}g'x = x>> and hence <<g^{-1}g \in Stab(x)>>. It follows that <<gStab(x) = g'Stab(x)>>. Thus <<\Psi>> is well defined.<br /> <br /> To define <<\Phi>> we take <<gStab(x) \in G/Stab(x)>> and define <<\Phi(gStab(x)) = gx>>. We show that this map is well defined. If <<gStab(x) = g'Stab(x)>> then <<g^{-1}g' \in Stab(x)>> and hence <<g^{-1}g'x = x>>. It follows that <<gx = g'x>> and hence <<\Phi>> is well defined.<br /> <br /> We need to check that <<\Psi>> and <<\Phi>> are mutually inverse and <<G>>-set morphisms. We quickly check that <<\Phi>> is a <<G>>-set morphism. If <<y = gx>> and <<g_1 \in G>> then <<g_1\Psi(y) = g_1(gStab(x)) = (g_1g)Stab(x)>>. Similarly, <<\Psi(g_1y) = g'Stab(x) = (g_1g)Stab(x)>>. The last inequality follows since we can take any <<g'>> such that <<g'y = g_1y>>. Why not take <<g' = g_1g>> -- since we know that works.<br /> <br /> ; Theorem (Orbit-Stabilizer)<br /> : If <<|X| < \infty>> and <<X = \coprod_{i \in I} Gx_i>> then <<|X| = \sum_{i} \frac{|G|}{Stab(x_i)}>>.<br /> <br /> This is just a rewriting of the theorem above.<br /> <br /> ; <<p>>-Group<br /> : A <<p>>-group is a group <<G>> with <<|G| = p^k>> for some <<k>>.<br /> <br /> <<G = (Z/2Z)^3, (Z/2Z) \times (Z/4Z), Z/8Z, D_8, Q = \{\pm 1, \pm i, \pm j, \pm k : i^2 = j^2 = k^2 = -1, ij = k\}>><br /> <br /> The last group <<Q>> is the famous ''unit quaternions'' -- They need a better description here.<br /> <br /> ; Theorem <br /> : Any <<p>>-group has a non-trivial centre.<br /> <br /> Let <<G>> act on itself by conjugation. Decompose <<G = \coprod Gx_i>>. Then,<br /> << |G| = \sum_{|Gx_i| = 1} 1 + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} >><br /> Observe that <<|Gx_i|| = 1>> iff <<x_i \in Z(G)>>. It follows that <br /> << |G| = |Z(G)| + \sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x)|} >><br /> The formula above is called ''"the class formula"''. We have that <<|G| / |Stab(x)| = p^k>> for some << 1 < k>> since <<Stab(x)>> is a subgroup. It follows that <<|G| \equiv 0\ \mod\ p>> and <<\sum_{|Gx_i| > 1} \frac{|G|}{|Stab(x_i)|} \equiv 0\ \mod\ p>>. It follows that <<|Z(G)| \equiv 0\ \mod\ p>>. Since <<e \in Z(G)>> we have <<1 \leq |Z(G)|>> and thus <<p \leq |Z(G)|>>.<br /> <br /> SYLOW<br /> <br /> A prove a brief technical lemma, for fun, since we could deduce it from more high powered machinery which we don't have yet.<br /> <br /> ; Cauchy's Lemma<br /> : If <<A>> is an abelian group and <<p>> divides <<|A|>>, then there is an element of order <<p>> in <<A>>.<br /> <br /> ''Proof''. Pick <<x \in A>>. If <<p>> divides the order of <<x>> then we have <<x^{np} = e>> for some <<n>>. It follows that <<(x^n)^p = e>>. We then have that the order of <<x^n>> is <<p>>. If <<p>> does not divide the order of <<p>>, then consider <<A / <x> >>. Since <<A>> is abelian, << <x> >> is a normal subgroup. We have that <<p>> divides <<|A/<x>|>>, and <<|A / <x>| < |A|>>. We then induct. Let << y<x> >> have order <<p>>, that is << (y<x>)^p = <x> >>. We then have that << y^p = x^k >> for some << k >>. We write << |<y>| = np + r >> where << 0 \leq p < r >>. We then have << e = y^{|<y>|} = y^{np + r} = x^{nkp}y^r \Rightarrow y^r \in <x> >>. It follows that <<(y<x>)^r = <x> >> contradicting the assumption that the order of << y<x> >> is << p >>.<br /> <br /> <br /> ; Sylow set<br /> : If <<|G| = p^k m>> for <<m \not\equiv 0\ \mod\ p>> then <<Syl_p(G) = \{P \leq G : |P| = p^k>>.<br /> <br /> ; Sylow I<br /> : <<Syl_p(G) \neq \emptyset>><br /> <br /> We proceed by induction on the oder of $p$. Assume the claim holds for all groups of order less than $|G|$. <span style="color:green">[Dror: "Stare at the class equation.]</span> Since <<|G| \equiv 0\ \mod\ p>> we have either: <br /> * <<|G| \equiv 0\ \mod\ p>> and <<\sum |G|/|Stab(x_i)| \equiv 0\ \mod\ p>>.<br /> * <<|G| \equiv 0\ \not\mod\ p>> and <<\sum |G|/|Stab(x_i)| \not\equiv 0\ \mod\ p>>.<br /> <br /> If <<|Z(G)| \not\equiv 0\ \mod p>> then there exists <<x_i>> such that <<|G|/|Stab(x_i)| \not\equiv 0\ \mod\ p>>. Thus <<p^k>> divides <<|Stab(x_i)|>>. We have that <<|Stab(x_i)| < |G|>> <span style="color:green">[Why happens here?]</span> We then have that <<p^k \leq Stab(x_i) < |G|>> and by induction there is <<|P| = p^k>> such that <<P \leq Stab(x_i)>>. It follows <<P \leq Stab(x_i) \leq G>>. We've obtained the Sylow <<p>>-subgroup.<br /> <br /> WIf <<|Z(G)| \equiv 0\ \mod p>> then by Cauchy's Lemma, there is <<x \in Z(G)>> with <<|<x>| = p>>. Consider the group << G / <x> >>. By the induction hypothesis there is << P' \leq G/<x> >> where <<|P'| = p^{k-1}>>. Then, there is the canonical projection << \pi : G \rightarrow G/<x> >>. By the fourth isomorphism theory << P = \pi^{-1}(P') \leq G >> and << |\pi^{-1}(P')| = p(p^{k-1}) = p^k >>.<br /> <br /> ; Sylow 2<br /> : Every Sylow <<p>>-subgroup of <<G> is conjugate. Moreover, every <<p>>-subgroup is contained in a Sylow <<p>>-subgroup.<br /> <br /> ; Sylow 3<br /> : Let <<n_p(G) = |Syl_p(G)|>>. We have <<n_p \equiv 0\ \mod\ |G|>> and <<n_p \equiv 1\ \mod\ p>>.<br /> <br /> ; A Nearly Tautological Lemma<br /> : If <<P \in Syl_p(G)>> and <<H \lea N(P)>> is a <<p>>-group, then <<H \leq P>>.<br /> : If <<x \in G>> has <<|<x>| = p^k>> and <<x \in N(P)>> then <<x \in P>>.<br /> <br /> <span style="color:green">[Dror: "This lemma is nearly tautological but it is only nearly tautological once you understand that it is nearly tautological." Parker: "A tautology?"]<br /> <br /> We show the first statement. We have that <<|P / P \cap H| = p^k>> since <<P>> is a <<p>>-group. We then know that <<PH / H \simeq P / P \cap H>> by the second isomorphism theorem. It foolows that <<|PH| = p^{k'}>>. But since <<P>> is maximal, we have <<P = PH>> and thus <<H \subseteq P>>. The first statement implies the second by taking <<H = <x> >>.<br /> <br /> <br /> <br /> GROUPS OF ORDER 15<br /> <br /> If <<|G| = 15>> then <<n_3 \equiv 0\ \mod\ 15>> and <<n_3 \equiv 1\ \mod\ 3>>. These imply <<n_3 = 1>>. Moreover, <<n_5 \equiv 0\ \mod\ 15>> and <<n_5 \equiv 1\ \mod\ 5>>. These imply <<n_5 = 1>>. Thus we have <<P_3>> a normal <<3>>-subgroup. Moreover, we have <<P_5>> a normal <<5>>-subgroup. This tells us a lot about the group.</div>

Pgadey https://drorbn.net/index.php?title=User:Pgadey&diff=10746 User:Pgadey 2011-10-06T16:29:45Z

<p>Pgadey: </p> <hr /> <div>I'm Parker Glynn-Adey.<br /> I've been TeXing up notes sloppily and with lots of asides.<br /> <br /> * [[11-1100-Pgadey-Lect5]]<br /> ** Simplicity of <math>A_n</math>, definition <math>G</math>-sets.<br /> * [[11-1100-Pgadey-Lect6]]<br /> ** Structure of <math>G</math>-sets, Sylow 1~2~3 (up to permutation), groups of order 15.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10740 11-1100-Pgadey-Lect5 2011-10-05T01:14:16Z

<p>Pgadey: </p> <hr /> <div>== Simplicity of <math>A_n</math>. ==<br /> <br /> ;Claim<br /> : <math>A_n</math> is simple for <math>n \neq 4</math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism (the map <math>S_4 \rightarrow S_3</math> given by a coloured tetrahedron [http://katlas.math.toronto.edu/drorbn/AcademicPensieve/Classes/11-1100/ColouredTetrahedron.png (link)] </span> <br /> <br /> <span style="color:green">[This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]</span><br /> <br /> We proceed with some unmotivated computations:<br /> <math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math><br /> These are the main ingredients of the proof. The first equality says that a product of transpositions of non-disjoint pairs is a 3-cycle, the second equality says that the product of a pair of disjoint transpositions is a product of three cycles. Thus any product of two transpositions can be written a product of three cycles. <span style="color:green">[ The second equality is amusing with physical objects. ]</span><br /> ; Lemma 1<br /> : <math>A_n</math> is generated by three cycles in <math>S_n</math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle</math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions <span style="color:green">(braid diagrams, computation with polynomials, etc)</span>. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n</math> contains a 3-cycle then <math>N=A_n</math>.<br /> Up to changing notation, we have that <math>(123) \in N</math>. We show that <math>(123)^\sigma \in N</math> for any <math>\sigma \in S_n</math>. By normality, we have this for <math>\sigma \in A_n</math>. If <math>\sigma \not\in A_n</math> we can write <math>\sigma = (12)\sigma'</math> for <math>\sigma \in A_n</math>. But then <math>(123)^{(12)} = (123)^2</math> and thus <br /> <math> (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N</math><br /> Since all 3-cycles are conjugate to <math>(123)</math> we have that all 3-cycles are in <math>N</math>. It follows by Lemma 1 that <math>N = A_n</math>.<br /> <br /> ;''Case I''<br /> : <math>N</math> contains a cycle of length <math>\geq 4</math>.<br /> <math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;''Case II''<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> <math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math><br /> The claim then follows by ''Case I''.<br /> <br /> ;''Case III''<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;''Case IV''<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> <math> \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> <span style="color:green">[Note: This last case is the '''only''' place where we really use this mystical fifth element. Without it, this last step wouldn't go through.]</span><br /> <br /> == Throwback to <math>S_4</math> ==<br /> ;Claim<br /> :<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> <math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> [ Suppose that <math>(123) \in H </math>, then <br /> <br /> <math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]<br /> <br /> == Group Actions ==<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> === Examples of group actions ===<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> <span style="color:green">["Where does the shirt come into the business?!" Dror's remark after talking about the symmetry of a student's shirt.]</span><br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> <span style="color:green">[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]</span><br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> <br /> ;Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the "transitive <math>G </math>-sets". And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10739 11-1100-Pgadey-Lect5 2011-10-05T01:12:31Z

<p>Pgadey: </p> <hr /> <div>== Simplicity of <math>A_n</math>. ==<br /> <br /> ;Claim<br /> : <math>A_n</math> is simple for <math>n \neq 4</math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism (the map <math>S_4 \rightarrow S_3</math> given by a coloured tetrahedron [http://katlas.math.toronto.edu/drorbn/AcademicPensieve/Classes/11-1100/ColouredTetrahedron.png (link)] </span> <br /> <br /> <span style="color:green">[This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]</span><br /> <br /> We proceed with some unmotivated computations:<br /> <math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math><br /> These are the main ingredients of the proof. The first equality says that a product of transpositions of non-disjoint pairs is a 3-cycle, the second equality says that the product of a pair of disjoint transpositions is a product of three cycles. Thus any product of two transpositions can be written a product of three cycles. <span style="color:green">[ The second equality is amusing with physical objects. ]</span><br /> ; Lemma 1<br /> : <math>A_n</math> is generated by three cycles in <math>S_n</math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle</math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions <span style="color:green">(braid diagrams, computation with polynomials, etc)</span>. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n</math> contains a 3-cycle then <math>N=A_n</math>.<br /> Up to changing notation, we have that <math>(123) \in N</math>. We show that <math>(123)^\sigma \in N</math> for any <math>\sigma \in S_n</math>. By normality, we have this for <math>\sigma \in A_n</math>. If <math>\sigma \not\in A_n</math> we can write <math>\sigma = (12)\sigma'</math> for <math>\sigma \in A_n</math>. But then <math>(123)^{(12)} = (123)^2</math> and thus <br /> <math> (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N</math><br /> Since all 3-cycles are conjugate to <math>(123)</math> we have that all 3-cycles are in <math>N</math>. It follows by Lemma 1 that <math>N = A_n</math>.<br /> <br /> ;''Case I''<br /> : <math>N</math> contains a cycle of length <math>\geq 4</math>.<br /> <math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;''Case II''<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> <math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math><br /> The claim then follows by ''Case I''.<br /> <br /> ;''Case III''<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;''Case IV''<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> <math> \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> <span style="color:green">[Note: This last case is the '''only''' place where we really use this mystical fifth element. Without it, this last step wouldn't go through.]</span><br /> <br /> == Throwback to <math>S_4</math> ==<br /> ;Claim<br /> :<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> <math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> [ Suppose that <math>(123) \in H </math>, then <br /> <br /> <math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]<br /> <br /> == Group Actions ==<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> === Examples of group actions ===<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> <span style="color:green">[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]</span><br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> <br /> ;Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the "transitive <math>G </math>-sets". And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10738 11-1100-Pgadey-Lect5 2011-10-05T01:11:53Z

<p>Pgadey: </p> <hr /> <div>== Simplicity of <math>A_n</math>. ==<br /> <br /> ;Claim<br /> : <math>A_n</math> is simple for <math>n \neq 4</math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism (the map <math>S_4 \rightarrow S_3</math> given by a coloured tetrahedron [http://katlas.math.toronto.edu/drorbn/AcademicPensieve/Classes/11-1100/ColouredTetrahedron.png (link)] </span> <br /> <br /> <span style="color:green">[This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]</span><br /> <br /> We proceed with some unmotivated computations:<br /> <math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math><br /> These are the main ingredients of the proof. The first equality says that a product of transpositions of non-disjoint pairs is a 3-cycle, the second equality says that the product of a pair of disjoint transpositions is a product of three cycles. Thus any product of two transpositions can be written a product of three cycles. <span style="color:green">[ The second equality is amusing with physical objects. ]</span><br /> ; Lemma 1<br /> : <math>A_n</math> is generated by three cycles in <math>S_n</math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle</math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions <span style="color:green">(braid diagrams, computation with polynomials, etc)</span>. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n</math> contains a 3-cycle then <math>N=A_n</math>.<br /> Up to changing notation, we have that <math>(123) \in N</math>. We show that <math>(123)^\sigma \in N</math> for any <math>\sigma \in S_n</math>. By normality, we have this for <math>\sigma \in A_n</math>. If <math>\sigma \not\in A_n</math> we can write <math>\sigma = (12)\sigma'</math> for <math>\sigma \in A_n</math>. But then <math>(123)^{(12)} = (123)^2</math> and thus <br /> <math> (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N</math><br /> Since all 3-cycles are conjugate to <math>(123)</math> we have that all 3-cycles are in <math>N</math>. It follows by Lemma 1 that <math>N = A_n</math>.<br /> <br /> ;''Case I''<br /> : <math>N</math> contains a cycle of length <math>\geq 4</math>.<br /> <math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;''Case II''<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> <math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math><br /> The claim then follows by ''Case I''.<br /> <br /> ;''Case III''<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;''Case IV''<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> <math> \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> <span style="color:green">[Note: This last case is the '''only''' place where we really use this mystical fifth element. Without it, this last step wouldn't go through.]</span><br /> <br /> == Throwback to <math>S_4</math> ==<br /> ;Claim<br /> :<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> <math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> [ Suppose that <math>(123) \in H </math>, then <br /> <br /> <math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]<br /> <br /> == Group Actions ==<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> === Examples of group actions ===<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> <span style="color:green">[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]</span><br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> ;<br /> Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the ``transitive <math>G </math>-sets''. And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10737 11-1100-Pgadey-Lect5 2011-10-05T01:10:42Z

<p>Pgadey: </p> <hr /> <div>== Simplicity of <math>A_n</math>. ==<br /> <br /> ;Claim<br /> : <math>A_n</math> is simple for <math>n \neq 4</math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism (the map <math>S_4 \rightarrow S_3</math> given by a coloured tetrahedron [http://katlas.math.toronto.edu/drorbn/AcademicPensieve/Classes/11-1100/ColouredTetrahedron.png (link)] </span> <br /> <br /> <span style="color:green">[This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]</span><br /> <br /> We proceed with some unmotivated computations:<br /> <math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math><br /> These are the main ingredients of the proof. The first equality says that a product of transpositions of non-disjoint pairs is a 3-cycle, the second equality says that the product of a pair of disjoint transpositions is a product of three cycles. Thus any product of two transpositions can be written a product of three cycles. <span style="color:green">[ The second equality is amusing with physical objects. ]</span><br /> ; Lemma 1<br /> : <math>A_n</math> is generated by three cycles in <math>S_n</math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle</math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions <span style="color:green">(braid diagrams, computation with polynomials, etc)</span>. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n</math> contains a 3-cycle then <math>N=A_n</math>.<br /> Up to changing notation, we have that <math>(123) \in N</math>. We show that <math>(123)^\sigma \in N</math> for any <math>\sigma \in S_n</math>. By normality, we have this for <math>\sigma \in A_n</math>. If <math>\sigma \not\in A_n</math> we can write <math>\sigma = (12)\sigma'</math> for <math>\sigma \in A_n</math>. But then <math>(123)^{(12)} = (123)^2</math> and thus <br /> <math> (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N</math><br /> Since all 3-cycles are conjugate to <math>(123)</math> we have that all 3-cycles are in <math>N</math>. It follows by Lemma 1 that <math>N = A_n</math>.<br /> <br /> ;''Case I''<br /> : <math>N</math> contains a cycle of length <math>\geq 4</math>.<br /> <math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;''Case II''<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> <math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math><br /> The claim then follows by ''Case I''.<br /> <br /> ;''Case III''<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;''Case IV''<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> <math> \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> <span style="color:green">[Note: This last case is the _only_ place where we really use this mystical fifth element. Without it, this last step wouldn't go through.]</span><br /> <br /> == Throwback to <math>S_4</math> ==<br /> ;Claim<br /> :<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> <math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> [ Suppose that <math>(123) \in H </math>, then <br /> <br /> <math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]<br /> <br /> == Group Actions ==<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> === Examples of group actions ===<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> <span style="color:green">[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]</span><br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> ;<br /> Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the ``transitive <math>G </math>-sets''. And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10736 11-1100-Pgadey-Lect5 2011-10-05T01:06:25Z

<p>Pgadey: </p> <hr /> <div>== Simplicity of <math>A_n</math>. ==<br /> <br /> ;Claim<br /> : <math>A_n</math> is simple for <math>n \neq 4</math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism </span> <br /> <br /> <span style="color:green">[This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]</span><br /> <br /> We proceed with some unmotivated computations:<br /> <math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math><br /> These are the main ingredients of the proof. The first equality says that a product of transpositions of non-disjoint pairs is a 3-cycle, the second equality says that the product of a pair of disjoint transpositions is a product of three cycles. Thus any product of two transpositions can be written a product of three cycles. <span style="color:green">[ The second equality is amusing with physical objects. ]</span><br /> ; Lemma 1<br /> : <math>A_n</math> is generated by three cycles in <math>S_n</math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle</math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions <span style="color:green">(braid diagrams, computation with polynomials, etc)</span>. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n</math> contains a 3-cycle then <math>N=A_n</math>.<br /> Up to changing notation, we have that <math>(123) \in N</math>. We show that <math>(123)^\sigma \in N</math> for any <math>\sigma \in S_n</math>. By normality, we have this for <math>\sigma \in A_n</math>. If <math>\sigma \not\in A_n</math> we can write <math>\sigma = (12)\sigma'</math> for <math>\sigma \in A_n</math>. But then <math>(123)^{(12)} = (123)^2</math> and thus <br /> <math> (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N</math><br /> Since all 3-cycles are conjugate to <math>(123)</math> we have that all 3-cycles are in <math>N</math>. It follows by Lemma 1 that <math>N = A_n</math>.<br /> <br /> ;''Case I''<br /> : <math>N</math> contains a cycle of length <math>\geq 4</math>.<br /> <math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;''Case II''<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> <math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math><br /> The claim then follows by ''Case I''.<br /> <br /> ;''Case III''<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;''Case IV''<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> <math> \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> <span style="color:green">[Note: This last case is the _only_ place where we really use this mystical fifth element. Without it, this last step wouldn't go through.]</span><br /> <br /> == Throwback to <math>S_4</math> ==<br /> ;Claim<br /> :<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> <math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> [ Suppose that <math>(123) \in H </math>, then <br /> <br /> <math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]<br /> <br /> == Group Actions ==<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> === Examples of group actions ===<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> <span style="color:green">[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]</span><br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> ;<br /> Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the ``transitive <math>G </math>-sets''. And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10735 11-1100-Pgadey-Lect5 2011-10-05T01:02:37Z

<p>Pgadey: </p> <hr /> <div>== Simplicity of <math>A_n</math>. ==<br /> <br /> ;Claim<br /> : <math>A_n</math> is simple for <math>n \neq 4</math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism </span> <br /> <br /> <span style="color:green">[This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]</span><br /> <br /> We proceed with some unmotivated computations,<br /> <br /> Some computations:<br /> </math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math><br /> These are the main ingredients of the proof<br /> <br /> ; Lemma 1<br /> : <math>A_n</math> is generated by three cycles in <math>S_n</math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle</math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions <span style="color:green">(braid diagrams, computation with polynomials, etc)</span>. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n</math> contains a 3-cycle then <math>N=A_n</math>.<br /> Up to changing notation, we have that <math>(123) \in N</math>. We show that <math>(123)^\sigma \in N</math> for any <math>\sigma \in S_n</math>. By normality, we have this for <math>\sigma \in A_n</math>. If <math>\sigma \not\in A_n</math> we can write <math>\sigma = (12)\sigma'</math> for <math>\sigma \in A_n</math>. But then <math>(123)^{(12)} = (123)^2</math> and thus <br /> <math> (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N</math><br /> Since all 3-cycles are conjugate to <math>(123)</math> we have that all 3-cycles are in <math>N</math>. It follows by Lemma 1 that <math>N = A_n</math>.<br /> <br /> ;''Case I''<br /> : <math>N</math> contains a cycle of length <math>\geq 4</math>.<br /> <math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;''Case II''<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> <math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math><br /> The claim then follows by ''Case I''.<br /> <br /> ;''Case III''<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;''Case IV''<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> <math> \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> <span style="color:green">[Note: This last case is the _only_ place where we really use this mystical fifth element. Without it, this last step wouldn't go through.]</span><br /> <br /> == Throwback to <math>S_4</math> ==<br /> ;Claim<br /> :<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> <math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> [ Suppose that <math>(123) \in H </math>, then <br /> <br /> <math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]<br /> <br /> == Group Actions ==<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> === Examples of group actions ===<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> <span style="color:green">[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]</span><br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> ;<br /> Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the ``transitive <math>G </math>-sets''. And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10734 11-1100-Pgadey-Lect5 2011-10-05T01:01:56Z

<p>Pgadey: </p> <hr /> <div>== Simplicity of <math>A_n</math>. ==<br /> <br /> ;Claim<br /> : <math>A_n</math> is simple for <math>n \neq 4</math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism </span> <br /> <br /> <span style="color:green">[This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]</span><br /> <br /> We proceed with some unmotivated computations,<br /> <br /> Some computations:<br /> </math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math><br /> These are the main ingredients of the proof<br /> <br /> ; Lemma 1<br /> : <math>A_n</math> is generated by three cycles in <math>S_n</math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle</math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions <span style="color:green">(braid diagrams, computation with polynomials, etc)</span>. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n</math> contains a 3-cycle then <math>N=A_n</math>.<br /> Up to changing notation, we have that <math>(123) \in N</math>. We show that <math>(123)^\sigma \in N</math> for any <math>\sigma \in S_n</math>. By normality, we have this for <math>\sigma \in A_n</math>. If <math>\sigma \not\in A_n</math> we can write <math>\sigma = (12)\sigma'</math> for <math>\sigma \in A_n</math>. But then <math>(123)^{(12)} = (123)^2</math> and thus <br /> <math> (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N</math><br /> Since all 3-cycles are conjugate to <math>(123)</math> we have that all 3-cycles are in <math>N</math>. It follows by Lemma 1 that <math>N = A_n</math>.<br /> <br /> ;''Case I''<br /> : <math>N</math> contains a cycle of length <math>\geq 4</math>.<br /> <math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;''Case II''<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> <math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math><br /> The claim then follows by ''Case I''.<br /> <br /> ;''Case III''<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;''Case IV''<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> <math> \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> <span style="color:green">[Note: This last case is the _only_ place where we really use this mystical fifth element. Without it, this last step wouldn't go through. </span><br /> <br /> == Throwback to <math>S_4</math> ==<br /> ;Claim<br /> :<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> </math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> [ Suppose that <math>(123) \in H </math>, then <br /> <br /> <math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]<br /> <br /> == Group Actions ==<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> === Examples of group actions ===<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> <span style="color:green">[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]</span><br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> ;<br /> Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the ``transitive <math>G </math>-sets''. And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10733 11-1100-Pgadey-Lect5 2011-10-05T00:59:43Z

<p>Pgadey: </p> <hr /> <div>!! Simplicity of <math>A_n</math>.<br /> <br /> ;Claim<br /> : <math>A_n</math> is simple for <math>n \neq 4</math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism </span> <br /> <br /> <span style="color:green">[This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]</span><br /> <br /> We proceed with some unmotivated computations,<br /> <br /> Some computations:<br /> </math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math><br /> These are the main ingredients of the proof<br /> <br /> ; Lemma 1<br /> : <math>A_n</math> is generated by three cycles in <math>S_n</math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle</math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions <span style="color:green">(braid diagrams, computation with polynomials, etc)</span>. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n</math> contains a 3-cycle then <math>N=A_n</math>.<br /> Up to changing notation, we have that <math>(123) \in N</math>. We show that <math>(123)^\sigma \in N</math> for any <math>\sigma \in S_n</math>. By normality, we have this for <math>\sigma \in A_n</math>. If <math>\sigma \not\in A_n</math> we can write <math>\sigma = (12)\sigma'</math> for <math>\sigma \in A_n</math>. But then <math>(123)^{(12)} = (123)^2</math> and thus <br /> <math> (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N</math><br /> Since all 3-cycles are conjugate to <math>(123)</math> we have that all 3-cycles are in <math>N</math>. It follows by Lemma 1 that <math>N = A_n</math>.<br /> <br /> ;''Case I''<br /> : <math>N</math> contains a cycle of length <math>\geq 4</math>.<br /> <math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;''Case II''<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> <math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math><br /> The claim then follows by ''Case I''.<br /> <br /> ;''Case III''<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;''Case IV''<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> <math> \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> <span style="color:green">[Note: This last case is the _only_ place where we really use this mystical fifth element. Without it, this last step wouldn't go through. </span><br /> <br /> !! Throwback: <math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> </math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> //[ Suppose that <math>(123) \in H </math>, then <br /> <br /> <math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]//<br /> <br /> !!Group Actions.<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> !! Examples of group actions<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> <span style="color:green">[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]</span><br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> ;<br /> Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the ``transitive <math>G </math>-sets''. And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10732 11-1100-Pgadey-Lect5 2011-10-05T00:53:46Z

<p>Pgadey: </p> <hr /> <div>!! Simplicity of <math>A_n</math>.<br /> <br /> ;Claim<br /> : <math>A_n</math> is simple for <math>n \neq 4</math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism </span> <br /> <br /> We proceed with some unmotivated computations,<br /> <pre> //[ This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]// </pre><br /> <br /> Some computations:<br /> </math> </math> (12)(23) = (123) \quad \quad (12)(34) = (123)(234) <math> </math><br /> These are the main ingredients of the proof<br /> <br /> ; Lemma 1<br /> : <math>A_n </math> is generated by three cycles in <math>S_n </math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle </math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions@@color:green ; (braid diagrams, computation with polynomials, etc)@@. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n </math> contains a 3-cycle then <math>N=A_n </math>.<br /> Up to changing notation, we have that <math>(123) \in N </math>. We show that <math>(123)^\sigma \in N </math> for any <math>\sigma \in S_n </math>. By normality, we have this for <math>\sigma \in A_n </math>. If <math>\sigma \not\in A_n </math> we can write <math>\sigma = (12)\sigma' </math> for <math>\sigma \in A_n </math>. But then <math>(123)^{(12)} = (123)^2 </math> and thus <br /> </math> </math>(123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N <math> </math><br /> Since all 3-cycles are conjugate to <math>(123) </math> we have that all 3-cycles are in <math>N </math>. It follows by Lemma 1 that <math>N = A_n </math>.<br /> <br /> ;//Case I//<br /> : <math>N </math> contains a cycle of length <math>\geq 4 </math>.<br /> </math> </math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N <math> </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;//Case II//<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> </math> </math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math> </math><br /> The claim then follows by //Case I//.<br /> <br /> ;//Case III//<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;//Case IV//<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> </math> </math>\sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math> </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> @@color:green ; //[Note: This last case is the _only_ place where we really use this mystical fifth element. Without it, this last step wouldn't go through. ]// @@<br /> <br /> !! Throwback: <math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> </math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> //[ Suppose that <math>(123) \in H </math>, then <br /> <br /> </math> </math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H <math> </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]//<br /> <br /> !!Group Actions.<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> !! Examples of group actions<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> @@color:green ; //[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]//@@ <br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> @@color:green ; //[Dror: I'm being a little bit biased. I prefer the left over the right. Parker: Propaganda? ]//@@<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> ;<br /> Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the ``transitive <math>G </math>-sets''. And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10731 11-1100-Pgadey-Lect5 2011-10-05T00:49:43Z

<p>Pgadey: </p> <hr /> <div>!! Simplicity of <math>A_n </math>.<br /> <br /> ;Claim<br /> : <math>A_n </math> is simple for <math>n \neq 4 </math>.<br /> For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.<br /> For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.<br /> For <math>n = 3 </math> we have that <math>A_n = \{e, (123), (132)\} \simeq Z/3Z </math> which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For <math>n = 4 </math> we have @@color: red ; Dror's Favourite Homomorphism @@<br /> <br /> We proceed with some unmotivated computations,<br /> @@color: green ; //[ This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]// @@<br /> <br /> Some computations:<br /> </math> </math> (12)(23) = (123) \quad \quad (12)(34) = (123)(234) <math> </math><br /> These are the main ingredients of the proof<br /> <br /> ; Lemma 1<br /> : <math>A_n </math> is generated by three cycles in <math>S_n </math>. That is, <math>A_n = \langle \{ (ijk) \in S_n \} \rangle </math>.<br /> <br /> We have that each element of <math>A_n </math> is the product of an even number of transpositions@@color:green ; (braid diagrams, computation with polynomials, etc)@@. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If <math>N \triangleleft A_n </math> contains a 3-cycle then <math>N=A_n </math>.<br /> Up to changing notation, we have that <math>(123) \in N </math>. We show that <math>(123)^\sigma \in N </math> for any <math>\sigma \in S_n </math>. By normality, we have this for <math>\sigma \in A_n </math>. If <math>\sigma \not\in A_n </math> we can write <math>\sigma = (12)\sigma' </math> for <math>\sigma \in A_n </math>. But then <math>(123)^{(12)} = (123)^2 </math> and thus <br /> </math> </math>(123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N <math> </math><br /> Since all 3-cycles are conjugate to <math>(123) </math> we have that all 3-cycles are in <math>N </math>. It follows by Lemma 1 that <math>N = A_n </math>.<br /> <br /> ;//Case I//<br /> : <math>N </math> contains a cycle of length <math>\geq 4 </math>.<br /> </math> </math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N <math> </math><br /> The claim then follows by Lemma 2.<br /> <br /> ;//Case II//<br /> : If <math>N </math> contains an with two cycles of length 3.<br /> </math> </math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math> </math><br /> The claim then follows by //Case I//.<br /> <br /> ;//Case III//<br /> : If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math><br /> We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.<br /> <br /> ;//Case IV//<br /> : If every element of <math>N </math> is a product of disjoint 2-cycles.<br /> We have that<br /> </math> </math>\sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math> </math><br /> But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.<br /> The claim then follows by Case 1.<br /> <br /> @@color:green ; //[Note: This last case is the _only_ place where we really use this mystical fifth element. Without it, this last step wouldn't go through. ]// @@<br /> <br /> !! Throwback: <math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.<br /> <br /> </math>S_3 </math> has an element of order three, therefore <math>H </math> does. We then conjugate to get all the three cycles. Then <math>H </math> is too big.<br /> <br /> //[ Suppose that <math>(123) \in H </math>, then <br /> <br /> </math> </math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H <math> </math><br /> Which implies that <math>|S_3| = 6 < 9 \leq |H| </math>, but since <math>H \simeq S_3 </math> we have <math>|H| = |S_3| </math>, a contradiction.]//<br /> <br /> !!Group Actions.<br /> <br /> ; A group <math>G </math> acting on a set <math>X </math><br /> A left (resp. right) group action of <math>G </math> on <math>X </math> is a binary map <math>G \times X \rightarrow X </math> denotes by <math>(g,x) \mapsto gx </math> satisfying:<br /> * <math>ex = x </math> (resp. <math>xe = x </math>)<br /> * <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)<br /> * [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] <br /> <br /> !! Examples of group actions<br /> * <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math><br /> * Let <math>S(X) </math> be the set of bijections from <math>X </math> to <math>X </math>, with group structure given by composition. We then have an <math>S(X) </math>-action of <math>X </math> given <math>x \mapsto gx : X \rightarrow X \in S(X) </math><br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If <math>G = (\mathcal{G}, \cdot) </math> is a group where <math>\mathcal{S} </math> is the underlying set of <math>G </math> and <math>\cdot </math> is the group multiplication. We have an action: <math>(g,s) = g \cdot s </math> this gives a map <math>G \rightarrow S(\mathcal{G}) </math>.<br /> * <math>SO(n) </math> is the group of orientation preserving symmetries of the <math>(n-1) </math>-dimensional sphere. We have that <math>SO(2) \leq SO(3) </math> as the subgroup of rotations that fix the north and south pole. There is a map <math>SO(3)/SO(2) \rightarrow S^2 </math> given by looking at the image of the north pole. <br /> * If <math>H \leq G </math> which may not be normal, then we have an action of <math>G </math> on <math>G/H </math> given by <math>g(xH) = (gx)H </math>. <br /> * We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. <br /> ; Exercise<br /> : Show that <math>S_n </math> acting on <math>\{1, 2, \dots, n\} </math> and <math>S_n / S_{n-1} </math> are isomorphic <math>S_n </math>-sets.<br /> <br /> @@color:green ; //[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space <math>T </math> and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]//@@ <br /> <br /> ; Claim<br /> : Left <math>G </math>-sets form a category.<br /> <br /> @@color:green ; //[Dror: I'm being a little bit biased. I prefer the left over the right. Parker: Propaganda? ]//@@<br /> <br /> The objects of the category are actions <math>G \times X \rightarrow X </math>. The morphisms, if we have <math>X </math> and <math>Y </math> are <math>G </math>-sets, a morphism of <math>G </math>-sets is a function <math>\gamma : X \rightarrow Y </math> such that <math>\gamma(gx) = g(\gamma(x)) </math>. <br /> <br /> ; Isomorphism of <math>G </math>-sets<br /> : An isomorphism of <math>G </math>-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If <math>X_1 </math> and <math>X_2 </math> are <math>G </math>-sets then so is <math>X_1 \coprod X_2 </math>, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.<br /> ;<br /> Claim<br /> : Any <math>G </math>-set <math>X </math> is a disjoint unions of the ``transitive <math>G </math>-sets''. And If <math>Y </math> is a transitive <math>G </math>-set, then <math>Y \simeq G/H </math> for some <math>H \leq G </math>.</div>

Pgadey https://drorbn.net/index.php?title=11-1100-Pgadey-Lect5&diff=10730 11-1100-Pgadey-Lect5 2011-10-05T00:47:34Z

<p>Pgadey: </p> <hr /> <div>!! Simplicity of $A_n$.<br /> <br /> ;Claim<br /> : $A_n$ is simple for $n \neq 4$.<br /> For $n = 1$ we have that $A_n = \{e\}$ which is simple.<br /> For $n=2$ we have that $S_n = \{(12), e\}$, and once again $A_n = \{e\}$.<br /> For $n = 3$ we have that $A_n = \{e, (123), (132)\} \simeq Z/3Z$ which is of prime order, and hence has no proper subgroups (by Lagrange). It follows that it has no normal proper subgroups.<br /> <br /> For $n = 4$ we have @@color: red ; Dror's Favourite Homomorphism @@<br /> <br /> We proceed with some unmotivated computations,<br /> @@color: green ; //[ This proof is not a deep conceptual proof. It is the product of a lot of playing around with cycles, and generators. This is much like a solution to the Rubik's cube, it naturally arises from a lot of playing around -- but is not conceptually deep at all.]// @@<br /> <br /> Some computations:<br /> $$ (12)(23) = (123) \quad \quad (12)(34) = (123)(234) $$<br /> These are the main ingredients of the proof<br /> <br /> ; Lemma 1<br /> : $A_n$ is generated by three cycles in $S_n$. That is, $A_n = \langle \{ (ijk) \in S_n \} \rangle$.<br /> <br /> We have that each element of $A_n$ is the product of an even number of transpositions@@color:green ; (braid diagrams, computation with polynomials, etc)@@. But we can replace a pair of 2-cycles with one or two 3-cycles by the computation above. It follows that any element of the alternating group can be rewritten as a product of 3-cycles.<br /> <br /> ; Lemma 2<br /> : If $N \triangleleft A_n$ contains a 3-cycle then $N=A_n$.<br /> Up to changing notation, we have that $(123) \in N$. We show that $(123)^\sigma \in N$ for any $\sigma \in S_n$. By normality, we have this for $\sigma \in A_n$. If $\sigma \not\in A_n$ we can write $\sigma = (12)\sigma'$ for $\sigma \in A_n$. But then $(123)^{(12)} = (123)^2$ and thus <br /> $$(123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N $$<br /> Since all 3-cycles are conjugate to $(123)$ we have that all 3-cycles are in $N$. It follows by Lemma 1 that $N = A_n$.<br /> <br /> ;//Case I//<br /> : $N$ contains a cycle of length $\geq 4$.<br /> $$ \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N $$<br /> The claim then follows by Lemma 2.<br /> <br /> ;//Case II//<br /> : If $N$ contains an with two cycles of length 3.<br /> $$ \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N$$<br /> The claim then follows by //Case I//.<br /> <br /> ;//Case III//<br /> : If $N$ contains $\sigma = (123)(\textrm{a product of disjoint 2-cycles})$<br /> We have that $\sigma^2 = (132) \in N$. The claim then follows by Lemma 1.<br /> <br /> ;//Case IV//<br /> : If every element of $N$ is a product of disjoint 2-cycles.<br /> We have that<br /> $$\sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N$$<br /> But then $\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N$.<br /> The claim then follows by Case 1.<br /> <br /> @@color:green ; //[Note: This last case is the _only_ place where we really use this mystical fifth element. Without it, this last step wouldn't go through. ]// @@<br /> <br /> !! Throwback: $S_4$ contains no normal $H$ such that $H \simeq S_3$.<br /> <br /> $S_3$ has an element of order three, therefore $H$ does. We then conjugate to get all the three cycles. Then $H$ is too big.<br /> <br /> //[ Suppose that $(123) \in H$, then <br /> <br /> $$ S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H $$<br /> Which implies that $|S_3| = 6 < 9 \leq |H|$, but since $H \simeq S_3$ we have $|H| = |S_3|$, a contradiction.]//<br /> <br /> !!Group Actions.<br /> <br /> ; A group $G$ acting on a set $X$<br /> A left (resp. right) group action of $G$ on $X$ is a binary map $G \times X \rightarrow X$ denotes by $(g,x) \mapsto gx$ satisfying:<br /> * $ex = x$ (resp. $xe = x$)<br /> * $(g_1g_2)x = g_1(g_2x)$ (resp $(xg_1)g_2$)<br /> * [The above implies $ex = x$ and $gy = x \Rightarrow g^{-1}y=x$.] <br /> <br /> !! Examples of group actions<br /> * $G$ acting on itself by conjugation (a right action). $(g,g') \mapsto g^{g'}$<br /> * Let $S(X)$ be the set of bijections from $X$ to $X$, with group structure given by composition. We then have an $S(X)$-action of $X$ given $x \mapsto gx : X \rightarrow X \in S(X)$<br /> @@color:green ; //[Where does the shirt come into the business?! ]// @@<br /> * If $G = (\mathcal{G}, \cdot)$ is a group where $\mathcal{S}$ is the underlying set of $G$ and $\cdot$ is the group multiplication. We have an action: $(g,s) = g \cdot s$ this gives a map $G \rightarrow S(\mathcal{G})$.<br /> * $SO(n)$ is the group of orientation preserving symmetries of the $(n-1)$-dimensional sphere. We have that $SO(2) \leq SO(3)$ as the subgroup of rotations that fix the north and south pole. There is a map $SO(3)/SO(2) \rightarrow S^2$ given by looking at the image of the north pole. <br /> * If $H \leq G$ which may not be normal, then we have an action of $G$ on $G/H$ given by $g(xH) = (gx)H$. <br /> * We have $S_{n-1} \leq S_n$ and $|S_n / S_{n-1}| = n!/(n-1)! = n$. <br /> ; Exercise<br /> : Show that $S_n$ acting on $\{1, 2, \dots, n\}$ and $S_n / S_{n-1}$ are isomorphic $S_n$-sets.<br /> <br /> @@color:green ; //[Dror violently resists rigorously defining a category. Gives a little speech about "things" and "arrows". Gives an example of taking a topological space $T$ and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]//@@ <br /> <br /> ; Claim<br /> : Left $G$-sets form a category.<br /> <br /> @@color:green ; //[Dror: I'm being a little bit biased. I prefer the left over the right. Parker: Propaganda? ]//@@<br /> <br /> The objects of the category are actions $G \times X \rightarrow X$. The morphisms, if we have $X$ and $Y$ are $G$-sets, a morphism of $G$-sets is a function $\gamma : X \rightarrow Y$ such that $\gamma(gx) = g(\gamma(x))$. <br /> <br /> ; Isomorphism of $G$-sets<br /> : An isomorphism of $G$-sets is a morphism which is bijective.<br /> <br /> ; Silly fact<br /> : If $X_1$ and $X_2$ are $G$-sets then so is $X_1 \coprod X_2$, the disjoint union of the two.<br /> <br /> the next statement combines the silly observation above, with the construction of an action of $G$ on $G/H$.<br /> ;<br /> Claim<br /> : Any $G$-set $X$ is a disjoint unions of the ``transitive $G$-sets''. And If $Y$ is a transitive $G$-set, then $Y \simeq G/H$ for some $H \leq G$.</div>

Pgadey https://drorbn.net/index.php?title=User:Pgadey&diff=10729 User:Pgadey 2011-10-05T00:46:51Z

<p>Pgadey: </p> <hr /> <div>I'm Parker Glynn-Adey.<br /> I've been TeXing up notes sloppily and with lots of asides.<br /> <br /> * [[11-1100-Pgadey-Lect5]]<br /> ** Simplicity of <math>A_n</math>, definition <math>G</math>-sets.</div>

Pgadey https://drorbn.net/index.php?title=User:Pgadey&diff=10728 User:Pgadey 2011-10-05T00:46:34Z

<p>Pgadey: </p> <hr /> <div>I'm Parker Glynn-Adey.<br /> I've been TeXing up notes sloppily and with lots of asides.<br /> <br /> * [[11-1100-Pgadey-Lect5]]<br /> ** Simplicity of <math>A_n</math>, <math>G</math>-sets.</div>

Pgadey https://drorbn.net/index.php?title=User:Pgadey&diff=10727 User:Pgadey 2011-10-05T00:45:28Z

<p>Pgadey: </p> <hr /> <div><br /> I'm Parker Glynn-Adey.<br /> I've been TeXing up notes sloppily and with lots of asides.<br /> <br /> * 11-1100Pgadey-Lect5</div>

Pgadey https://drorbn.net/index.php?title=11-1100/Class_Photo&diff=10694 11-1100/Class Photo 2011-09-27T16:22:51Z

<p>Pgadey: </p> <hr /> <div>Our class on September 27, 2011:<br /> <br /> [[Image:11-1100-ClassPhoto.jpg|thumb|centre|800px|Class Photo: click to enlarge]]<br /> {{11-1100/Navigation}}<br /> <br /> Please identify yourself in this photo! There are two ways to do that:<br /> <br /> * [[Special:Userlogin|Log in]] to this Wiki and edit this page. Put your name, userid, email address and location in the picture in the alphabetical list below.<br /> * Send [[User:Drorbn|Dror]] an email message with this information.<br /> <br /> The first option is more fun but less private.<br /> <br /> ===Who We Are...===<br /> <br /> {| align=center border=1 cellspacing=0<br /> |-<br /> !First name<br /> !Last name<br /> !UserID<br /> !Email<br /> !In the photo<br /> !Comments<br /> {{Photo Entry|last=Bar-Natan|first=Dror|userid=Drorbn|email=drorbn@ math.toronto.edu|location=facing everybody, as the photographer|comments=Take this entry as a model and leave it first. Otherwise alphabetize by last name. Feel free to leave some fields blank. For better line-breaking, leave a space next to the "@" in email addresses.}}<br /> {{Photo Entry|last=Glynn-Adey|first=Parker|userid=pgadey|email=parker.glynn.adey@ math.toronto.edu|location=Fifth from the right in the back row|comments=Glowing bald guy with yellow shirt.}}<br /> <br /> |}<br /> <br /> </div>

Pgadey https://drorbn.net/index.php?title=Main_Page&diff=10693 Main Page 2011-09-27T16:18:58Z

<p>Pgadey: </p> <hr /> <div>__NOTOC__<br /> This is the wiki portion of {{Dror}} Bar-Natan's {{Home Link||web site}}. It is not a standalone page - it has no overall tree structure and there is no specific entry point. Rather it is just the common location for those of for my pages that could benefit from being editable by others and/or from having their history preserved and/or by having an associated "[[Talk:Main_Page|discussion]]" page.<br /> <br /> Ok. Here are some of the pages here anyway:<br /> <br /> {{Paperlets Navigation}}<br /> <br /> ===Cheat Sheets===<br /> [[Image:The MOY Invariant - Feb 2006.jpg|thumb|right|120px|MOY]]<br /> [[Image:Multivariable Alexander - December 2006.jpg|thumb|right|120px|Multivariable Alexander]]<br /> A ''cheat sheet'' is a very short (usually one page or less) informal note (often with more diagrams and formulas than words) containing the essence of a complicated definition, construction or proof for easy reference at later dates. A well designed cheat sheet may not be readable for an outsider - it is not a paper and not even a [[paperlet]]. But it should be readable by its author (and possibly his/her close collaborators) for at least a few years after it was written assuming normal memory and brain cell loss. It should be reasonably (and sometimes, very) accurate in as much as normalizations and conventions are concerned, as often the main purpose of a cheat sheet is to keep track of those. No originality is ever claimed yet credits may be sparse.<br /> <br /> * [[Media:The MOY Invariant - Feb 2006.jpg|The MOY Invariant]], February 2006, by {{Dror}}.<br /> * [[Media:Multivariable Alexander - December 2006.jpg|The Multivariable Alexander Polynomial]], December 2006, by Jana Archibald.<br /> * The Khovanov-Seidel Categorification of the Burau Representation, April 2007, by [http://individual.utoronto.ca/louisleung/ Louis Leung]: [[Media:Leung 070406-1.jpg|Side 1]], [[Media:Leung 070406-2.jpg|Side 2]]. (Based on {{arXiv|math.QA/0006056}}.)<br /> * [[Media:Leung on the Alexander State Sum.jpg|The Alexander State Sum Model]], by [http://individual.utoronto.ca/louisleung/ Louis Leung].<br /> <br /> ===Projects===<br /> * [[VasCalc - A Vassiliev Invariants Calculator]].<br /> * [[HoriAsso - Computing a Horizontal Associator]].<br /> <br /> === Seminars and Classes ===<br /> [[Knot at Lunch]], [[10-327]], [[10-1100]],[[11-1100]], [[09-240]], [[AKT-09]], [[08-401]], [[0708-1300]], [[07-401]], [[07-1352]], [[06-240]], [[06-1350]].<br /> <br /> ===Other===<br /> * [[dbnvp]] will be a style for video pages and a collection of programs to create and manage pages of that style.<br /> * [[My Refereeing Policy]].<br /> * [[HandoutBrowser.js]].<br /> * [[Survey of Finite Type Invariants]].<br /> * [[CMS Winter 2006 Session on Knot Homologies]].<br /> * [[2009 Knot Theory Program at the Fields Institute]].<br /> * [[An Algebraic Number]].</div>

Pgadey