11-1100-Pgadey-Lect5 - Revision history

Pgadey at 01:14, 5 October 2011

2011-10-05T01:14:16Z

← Older revision		Revision as of 21:14, 4 October 2011
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	* <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math>		* <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math>
	* 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>		* 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>
	@@color:green ~~; //~~[Where does the shirt come into the business?! ]/~~/ @@~~		<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>
	* 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>.		* 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>.
	* <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.		* <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.

Pgadey at 01:12, 5 October 2011

2011-10-05T01:12:31Z

← Older revision		Revision as of 21:12, 4 October 2011
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	the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.		the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.

	;
	Claim		;Claim
	: 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>.		: 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>.

Pgadey at 01:11, 5 October 2011

2011-10-05T01:11:53Z

← Older revision		Revision as of 21:11, 4 October 2011
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	The claim then follows by Case 1.		The claim then follows by Case 1.

	<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>		<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>

	== Throwback to <math>S_4</math> ==		== Throwback to <math>S_4</math> ==

Pgadey at 01:10, 5 October 2011

2011-10-05T01:10:42Z

← Older revision		Revision as of 21:10, 4 October 2011
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	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.		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.

	For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism </span>		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>

	<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>		<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>

Pgadey at 01:06, 5 October 2011

2011-10-05T01:06:25Z

← Older revision		Revision as of 21:06, 4 October 2011
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	<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>		<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>

	We proceed with some unmotivated computations,		We proceed with some unmotivated computations:
		⚫	<math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math>

			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>
	Some computations:
⚫	</math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math>
	These are the main ingredients of the proof

	; Lemma 1		; Lemma 1
	: <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>.		: <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>.

Pgadey at 01:02, 5 October 2011

2011-10-05T01:02:37Z

← Older revision		Revision as of 21:02, 4 October 2011
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	The claim then follows by Case 1.		The claim then follows by Case 1.

	<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>		<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>

	== Throwback to <math>S_4</math> ==		== Throwback to <math>S_4</math> ==
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	:<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.		:<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.

	</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.		<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.

	[ Suppose that <math>(123) \in H </math>, then		[ Suppose that <math>(123) \in H </math>, then

Pgadey at 01:01, 5 October 2011

2011-10-05T01:01:56Z

← Older revision		Revision as of 21:01, 4 October 2011
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	!! Simplicity of <math>A_n</math>.		== Simplicity of <math>A_n</math>. ==

	;Claim		;Claim
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	<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>		<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>

	!! Throwback: <math>S_4 </math> ~~contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.~~		== Throwback to <math>S_4</math> ==
			;Claim
			:<math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.

	</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.		</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.

	//[ Suppose that <math>(123) \in H </math>, then		[ Suppose that <math>(123) \in H </math>, then

	<math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math>		<math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math>
	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.]//		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.]

	!!Group Actions.		== Group Actions ==

	; A group <math>G </math> acting on a set <math>X </math>		; A group <math>G </math> acting on a set <math>X </math>
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	* [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.]		* [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.]

	!! Examples of group actions		=== Examples of group actions ===
	* <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math>		* <math>G </math> acting on itself by conjugation (a right action). <math>(g,g') \mapsto g^{g'} </math>
	* 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>		* 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>

Pgadey at 00:59, 5 October 2011

2011-10-05T00:59:43Z

← Older revision		Revision as of 20:59, 4 October 2011
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	For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism </span>		For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism </span>

		⚫	<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>

	We proceed with some unmotivated computations,		We proceed with some unmotivated computations,
⚫	<~~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~~>

	Some computations:		Some computations:
	</math> ~~</math>~~ (12)(23) = (123) \quad \quad (12)(34) = (123)(234) ~~<math>~~ </math>		</math>(12)(23) = (123) \quad \quad (12)(34) = (123)(234)</math>
	These are the main ingredients of the proof		These are the main ingredients of the proof

	; Lemma 1		; Lemma 1
	: <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>.		: <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>.

	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.		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.

	; Lemma 2		; Lemma 2
	: If <math>N \triangleleft A_n </math> contains a 3-cycle then <math>N=A_n </math>.		: If <math>N \triangleleft A_n</math> contains a 3-cycle then <math>N=A_n</math>.
	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		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
	</math> ~~</math>~~(123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N ~~<math>~~ </math>		<math> (123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N</math>
	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>.		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>.

	;//Case I//		;''Case I''
	: <math>N </math> contains a cycle of length <math>\geq 4 </math>.		: <math>N</math> contains a cycle of length <math>\geq 4</math>.
	<~~/math> </~~math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N ~~<math>~~ </math>		<math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N </math>
	The claim then follows by Lemma 2.		The claim then follows by Lemma 2.

	;//Case II//		;''Case II''
	: If <math>N </math> contains an with two cycles of length 3.		: If <math>N </math> contains an with two cycles of length 3.
	<~~/math> </~~math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N ~~</math>~~ </math>		<math> \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N </math>
	The claim then follows by //Case I//.		The claim then follows by ''Case I''.

	;//Case III//		;''Case III''
	: If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math>		: If <math>N </math> contains <math>\sigma = (123)(\textrm{a product of disjoint 2-cycles}) </math>
	We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.		We have that <math>\sigma^2 = (132) \in N </math>. The claim then follows by Lemma 1.

	;//Case IV//		;''Case IV''
	: If every element of <math>N </math> is a product of disjoint 2-cycles.		: If every element of <math>N </math> is a product of disjoint 2-cycles.
	We have that		We have that
	</math> ~~</math>~~\sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N ~~</math>~~ </math>		<math> \sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math>
	But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.		But then <math>\tau^{-1}(125)\tau(125)^{-1} = (13452) \in N </math>.
	The claim then follows by Case 1.		The claim then follows by Case 1.

	@@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 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>

	!! Throwback: <math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.		!! Throwback: <math>S_4 </math> contains no normal <math>H </math> such that <math>H \simeq S_3 </math>.
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	//[ Suppose that <math>(123) \in H </math>, then		//[ Suppose that <math>(123) \in H </math>, then

	<~~/math> </~~math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H ~~<math>~~ </math>		<math> S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H </math>
	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.]//		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.]//

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	: 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.		: 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.

	@@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 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>

	; Claim		; Claim
	: Left <math>G </math>-sets form a category.		: Left <math>G </math>-sets form a category.

	@@color:green ; //[Dror: I'm being a little bit biased. I prefer the left over the right. Parker: Propaganda? ]//@@

	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>.		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>.

Pgadey at 00:53, 5 October 2011

2011-10-05T00:53:46Z

← Older revision		Revision as of 20:53, 4 October 2011
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	!! Simplicity of <math>A_n </math>.		!! Simplicity of <math>A_n</math>.

	;Claim		;Claim
	: <math>A_n </math> is simple for <math>n \neq 4 </math>.		: <math>A_n</math> is simple for <math>n \neq 4</math>.
	For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.		For <math>n = 1 </math> we have that <math>A_n = \{e\} </math> which is simple.
	For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.		For <math>n=2 </math> we have that <math>S_n = \{(12), e\} </math>, and once again <math>A_n = \{e\} </math>.
	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.		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.

	For <math>n = 4 </math> we have @@color: red ; Dror's Favourite Homomorphism @@		For <math>n = 4 </math> we have <span style="color:red">Dror's Favourite Homomorphism </span>

	We proceed with some unmotivated computations,		We proceed with some unmotivated computations,
	~~@@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.]// @@		<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>

	Some computations:		Some computations:

Pgadey at 00:49, 5 October 2011

2011-10-05T00:49:43Z

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