View source for 11-1100-Pgadey-Lect5

== Simplicity of <math>A_n</math>. ==

;Claim
: <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=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 = 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>

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>
; 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>.

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
: 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 
	 <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>.

;''Case I''
: <math>N</math> contains a cycle of length <math>\geq 4</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.

;''Case II''
: If <math>N </math> contains an with two cycles of length 3.
 <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''.

;''Case III''
: 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.

;''Case IV''
: If every element of <math>N </math> is a product of disjoint 2-cycles.
We have that
	 <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>.
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>

== 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.

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

 <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.]

== Group Actions ==

; A group <math>G </math> acting on a set <math>X </math>
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:
* <math>ex = x </math> (resp. <math>xe = x </math>)
* <math>(g_1g_2)x = g_1(g_2x) </math> (resp <math>(xg_1)g_2 </math>)
* [The above implies <math>ex = x </math> and <math>gy = x \Rightarrow g^{-1}y=x </math>.] 

=== Examples of group actions ===
* <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>
@@color:green ; //[Where does the shirt come into the business?! ]// @@
* 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. 
* 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>. 
* We have <math>S_{n-1} \leq S_n </math> and <math>|S_n / S_{n-1}| = n!/(n-1)! = n </math>. 
; Exercise
: 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.

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

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>. 

; Isomorphism of <math>G </math>-sets
: An isomorphism of <math>G </math>-sets is a morphism which is bijective.

; Silly fact
: 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.

the next statement combines the silly observation above, with the construction of an action of <math>G </math> on <math>G/H </math>.
;
 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>.