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 @@color: red ; Dror's Favourite Homomorphism @@

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.]// @@

Some computations: $$ (12)(23) = (123) \quad \quad (12)(34) = (123)(234) $$ 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 $$(123)^\sigma = \left( (123)^{(12)} \right)^{\sigma'} \in N $$ 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 $\geq 4$.

$$ \sigma= (123456)\sigma' \in N \Rightarrow \sigma^{-1} (123) \sigma (123)^{-1} = (136) \in N $$ The claim then follows by Lemma 2.

//Case II//

If $N$ contains an with two cycles of length 3.

$$ \sigma = (123)(456) \sigma' \in N \Rightarrow \sigma^{-1}(124)\sigma(124)^{-1} = (14263) \in N$$ The claim then follows by //Case I//.

//Case III//

If $N$ contains $\sigma = (123)(\textrm{a product of disjoint 2-cycles})$

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 $$\sigma = (12)(34)\sigma' \Rightarrow \sigma^{-1}(123)\sigma(123)^{-1} = (13)(24) = \tau \in N$$ 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 $H \simeq S_3$.

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

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

$$ S = \{ e, (123), (132), (124), (142), (134), (143), (234), (243)\} \subset H $$ 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. $xe = x$)
• $(g_1g_2)x = g_1(g_2x)$ (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 $H \leq G$.