11-1100-Pgadey-Lect5

!! Simplicity of $A_{n}$.

Claim

$A_{n}$ is simple for $n\neq 4$.

For $n=1$ we have that $A_{n}=\{e\}$ which is simple. For $n=2$ we have that $S_{n}=\{(12),e\}$, and once again $A_{n}=\{e\}$. 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.

For $n=4$ we have Dror's Favourite Homomorphism

We proceed with some unmotivated computations,

 //[ 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.]// 

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

Lemma 1

$A_{n}$ is generated by three cycles in $S_{n}$. That is, $A_{n}=\langle \{(ijk)\in S_{n}\}\rangle$.

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.

Lemma 2

If $N\triangleleft A_{n}$ contains a 3-cycle then $N=A_{n}$.

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 </math> </math>(123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N ${\displaystyle }$ Since all 3-cycles are conjugate to $(123)$ we have that all 3-cycles are in $N$. It follows by Lemma 1 that $N=A_{n}$.

//Case I//

$N$ contains a cycle of length $\geq 4$.

</math> </math> \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N ${\displaystyle }$

The claim then follows by Lemma 2.

//Case II//

If $N$ 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>

The claim then follows by //Case I//.

//Case III//

If $N$ contains $\sigma =(123)({\textrm {aproductofdisjoint2-cycles}})$

We have that $\sigma ^{2}=(132)\in N$. The claim then follows by Lemma 1.

//Case IV//

If every element of $N$ is a product of disjoint 2-cycles.

We have that </math> </math>\sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N </math> </math> But then $\tau ^{-1}(125)\tau (125)^{-1}=(13452)\in N$. 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. ]// @@

!! Throwback: $S_{4}$ contains no normal $H$ such that $H\simeq S_{3}$.

</math>S_3 </math> has an element of order three, therefore $H$ does. We then conjugate to get all the three cycles. Then $H$ is too big.

//[ Suppose that $(123)\in H$, then

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

Which implies that $|S_{3}|=6<9\leq |H|$, but since $H\simeq S_{3}$ we have $|H|=|S_{3}|$, a contradiction.]//

!!Group Actions.

A group $G$ acting on a set $X$

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:

• $ex=x$ (resp. $xe=x$)
• $(g_{1}g_{2})x=g_{1}(g_{2}x)$ (resp $(xg_{1})g_{2}$)
• [The above implies $ex=x$ and $gy=x\Rightarrow g^{-1}y=x$.]

!! Examples of group actions

• $G$ acting on itself by conjugation (a right action). $(g,g')\mapsto g^{g'}$
• 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 $x\mapsto gx:X\rightarrow X\in S(X)$

@@color:green ; //[Where does the shirt come into the business?! ]// @@

• 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 $G\rightarrow S({\mathcal {G}})$.
• $SO(n)$ 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.
• If $H\leq G$ which may not be normal, then we have an action of $G$ on $G/H$ given by $g(xH)=(gx)H$.
• We have $S_{n-1}\leq S_{n}$ and $|S_{n}/S_{n-1}|=n!/(n-1)!=n$.

Exercise

Show that $S_{n}$ acting on $\{1,2,\dots ,n\}$ and $S_{n}/S_{n-1}$ are isomorphic $S_{n}$-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 $T$ and then looking at the space of paths with identities given by staying still, and composition of paths given by concatenation.]//@@

Claim

Left $G$-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 $G\times X\rightarrow X$. 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 $\gamma (gx)=g(\gamma (x))$.

Isomorphism of $G$-sets

An isomorphism of $G$-sets is a morphism which is bijective.

Silly fact

If $X_{1}$ and $X_{2}$ are $G$-sets then so is $X_{1}\coprod X_{2}$, the disjoint union of the two.

the next statement combines the silly observation above, with the construction of an action of $G$ on $G/H$.

Claim

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 $H\leq G$.