Alternating groups are simple

Statement
For $$n \ge 5$$, the fact about::alternating group $$A_n$$, i.e., the group of all even permutations on $$n$$ letters, is a simple non-Abelian group.

Related facts

 * Finitary alternating groups are simple: The finitary alternating group on an infinite set is simple.

Facts used

 * 1) A5 is simple
 * 2) Normality satisfies transfer condition: The intersection of a normal subgroup of the whole group, with any subgroup, is normal in the subgroup.
 * 3) Even permutation IAPS is padding-contranormal: $$A_n$$ is a contranormal subgroup inside $$A_{n+1}$$. More specifically, the even permutations fixing any particular letter form a contranormal subgroup of the group of all even permutations.

Proof
The proof is by induction on $$n$$.

Base case
The case $$n = 5$$ is dealt with separately, by a direct argument.

Induction step
We prove that for $$n \ge 5$$, if $$A_n$$ is simple, then $$A_{n+1}$$ is simple.

Given: $$A_{n+1}$$ is the group of even permutations on $$\{ 1,2,3,\dots,n+1 \}$$. $$N$$ is a normal subgroup of $$A_{n+1}$$.

To prove: $$N = A_{n+1}$$ or $$N$$ is trivial.

Proof: Let $$H_i$$ denote the subgroup of $$A_{n+1}$$ that stabilizes the letter $$i$$. Then, each $$H_i$$ consists of the even permutations on $$n$$ letters (the letters excluding $$i$$) and is hence isomorphic to $$A_n$$. Thus, each $$H_i$$ is simple.

Now, since normality satisfies transfer condition, $$N \cap H_i$$ is normal in $$H_i$$ for every $$i$$. By simplicity of $$H_i$$, either $$N$$ contains $$H_i$$, or $$N$$ intersects $$H_i$$ trivially.

Suppose there exists $$i$$ for which $$N$$ contains $$H_i$$. Then, by fact (3) stated above, $$H_i$$ is contranormal inside $$A_{n+1}$$, i.e., its normal closure is $$A_{n+1}$$. Since $$N$$ is normal, this forces $$N = A_{n+1}$$, and we are done.

Otherwise, $$N \cap H_i$$ is trivial for every $$i$$. Thus, no nontrivial element of $$N$$ fixes any letter. Let's use this to show that $$N$$ can have no nontrivial elements.

Suppose $$\sigma \in N$$ is nontrivial. Then, first observe that in the cycle decomposition of $$\sigma$$, every element must be in a cycle of the same length $$k$$ (otherwise, some power of $$\sigma$$ would fix a letter). Thus, $$\sigma$$ has $$r$$ cycles each of size $$k$$, where $$kr = n + 1$$.

Now, if $$n \ge 5$$, then $$n + 1 \ge 6$$. Choose an $$i$$ and a double transposition $$\tau \in A_{n+1}$$ such that $$\tau$$ fixes both $$i$$ and $$\sigma(i)$$, but such that $$\tau$$ does not commute with $$\sigma$$ (this is possible because $$n+1 \ge 6$$). Then, $$\sigma^{-1}\tau\sigma\tau^{-1} \in N$$ is a nontrivial element of $$N$$ fixing both $$i$$ and $$\sigma(i)$$, giving the required contradiction.