Alternating groups are simple

Statement

For $n\geq 5$, the 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. For full proof, refer: A5 is simple

Induction step

We prove that for $n\geq 5$, 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 $n\geq 5$, then $n+1\geq 6$. 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 $n+1\geq 6$). 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.