Simplicity of the Alternating group

Denote by S_n the set of all permutations on a set with n elements, which we take without loss to be $\{1,2,\ldots ,n\}.$ Under the composition of permutations S_n is a group, called the Symmetric group of degree n.

Let S_n act on the set of polynomials with integral coefficients; specifically, $g\in S_n$ sends $Q(x_1,\cdots ,x_n)$ to $Q(x_{ g(1)},\ldots ,x_{g(n)})$. The orbit of $P=\prod_{i<j}(x_i-x_j)$ under this action is clearly $\{ P,-P\}.$ Thus the stabilizer of P in S_n is a subgroup of index 2; call it A_n. Note that the transposition (ij) sends P to -P, for all $i\neq j ;$ all transpositions lie therefore outside A_n.

Represent S_n on the left cosets of A_n by left translation. Denote the cosets by $A_n=\bar {1}$ and $\bar {A}_n=\bar {2}.$ The kernel of this representation is

$\{g\in S_n;gA_n=A_n,g\bar {A}_n=\bar {A}_n\}=A_ n.$

We conclude that A_n is normal in S_n. To summarize, this representation sends $g\in A_n$ into $(\bar {1})(\bar {2})$ and $g\in\bar {A}_n$ into $(\bar {1}\bar {2}).$

If $g_1,\ldots ,g_m\in\bar {A}_n$, then $g_1\cdots g_m\rightarrow (\bar {1}\bar {2})\cdots (\bar {1}\bar {2})=(\bar {1}\bar {2} )^m.$ Clearly $(\bar {1}\bar {2})^m=(\bar {1})(\bar {2})$ if and only if m is even. Thus $g_1\cdots g_m\in A_n$ if and only if m is even. A cycle $(12\cdots k)=(1k)\cdots (12)$ decomposes into k-1 transpositions. Therefore a cycle of length k is in A_n if and only if k is odd. By the decomposition into disjoint cycles of any element of S_ n we conclude as follows:

* A permutation belong to A_n if and only if it has an even number of cycles of even length.

Elements of A_n are called even permutations. Elements of $\bar {A}_n$ are called odd permutations. We call A_n the alternat ing group of degree n.

A group is simple if it has no normal subgroups other that itself and 1.

Theorem (Simplicity of ${\bf A}_n)$ The alternating group of degree n is simple; $n\neq 4.$

Proof: The group A₃ is easily seen to be simple. When n=4 we have the normal subgroup 1, (12)(34), (13)(24), (14)(23) in A₄, so A₄ is not simple. Assume $n\geq 5.$

Step 1 A_n is generated by the 3-cycles. [In fact A_n is generated by the n-2 3-cycles $\{ (12k):k\geq 3\}.]$

Indeed, any element of A_n is a product of transpositions of the form (ab)(cd) or (ab)(ac). Since (ab)(cd)=(acb) (acd) and (ab)(ac)=(acb) we conclude that A_n is generated by the 3-cycles. Furthermore, (1a2)=(12a)^-1, (1ab)=(12b)(12a)^-1, (2ab)=(12b)^{- 1}(12a), and (abc)=(12a)^-1(12c)(12b)^-1(12a), which shows that every 3-cycle is generated by a cycle of the form (12k).

Step 2 If H is a normal subgroup of A_n and H contains a 3-cycle, then H=A_n.

Without loss $(123)\in H.$ Then $(12k)=((12)(3k))(123)^{-1}((12)(3k))^{-1}=((123)^{-1})^{(12 )(3k)}\in H,$ by normality. Thus A_n=<(12k): all $k\geq 3>\leq H,$ and H=A_n.

We now assume that $H(\neq 1$) is normal in A_n. The idea is to show that H contains a 3-cycle, necessarily. We then invoke Step 2 to conclude the proof. Consider the exhausting possibilities examined below.

Case 1. Without loss assume that H contains $\sigma =(12\cdots r)\tau ,$ where $r\geq 4,$ and $\tau$ is disjoint of $\{1,2,\ldots ,r\}.$ Then, by the normality of H, H contains $\sigma^{-1}\sigma^{(123)}=(12r).$ We are done by Step 2.

Case 2. Assume without loss that $\sigma =(123)(456 )\tau\in H,$ where $\tau$ is a product of disjoint transpositions. Then H contains $\sigma^{ -1}\sigma^{(124)}=(14263).$ We are done by Step 1.

Case 3. Assume without loss that $(123)\tau\in H,$ with $\tau$ a product of disjoint transpositions. Then H contains $(123)\tau (123) \tau =(132),$ and we are done by Step 2.

Case 4. Assume that H contains elements that are products of disjoint transpositions. Without loss let $(12)(34)\tau$ be such an element of H. Then $(12)(34)\tau ((12)(34)\tau )^{(123)}=(13)(24)\in H.$ Since $n\geq 5,$ consider $(123)\in A_n.$ By normality H contains (13)(24)((13)(24 ))⁽¹³⁵⁾=(135), and we are done by Step 2. This ends the proof.