# Alternating groups are simple

## Statement

For $n \ge 5$, the Alternating group (?) $A_n$, i.e., the group of all even permutations on $n$ letters, is a simple non-Abelian group.

## 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. For full proof, refer: A5 is simple

### 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.