Why is conjugation by an odd permutation in $S_n$ not an inner automorphism on $A_n$?

Question

I was reading about the outer automorphism group on wikipedia, and it mentions that conjugation by an odd permutation is an outer automorphism on the alternating group $A_n$. This suggests the automorphism defined on $A_n$ by $$ \varphi: A_n\to A_n:\varphi(\tau)=\sigma\tau\sigma^{-1} $$ where $\sigma\in S_n$ is odd it not an inner automorphism of $A_n$. Is there a way to see explicitly that $\varphi$ isn't just $\tau\mapsto\rho\tau\rho^{-1}$ for $\rho\in A_n$ in disguise? To avoid trivial cases I guess we can assume $n>2$. Thanks.

Answer 1

Note that if $\sigma$ is an odd permutation then $(12)\sigma$ is an even permutation, thus $\tau\mapsto (12)\sigma\tau\sigma^{-1}(12)$ is an inner automorphism, and so conjugation by $\sigma$ is an inner automorphism iff conjugation by $(12)$ is, as conjugation by $\sigma$ is the composition of conjugation by $(12)$ with conjugation by $(12)\sigma$ and inner automorphisms form a subgroup.

Suppose conjugation by $(12)$ is equivalent to conjugation by some $\pi\in S_n$. Let $$m=\begin{cases} n&\text{if $n$ is odd}\\ n-1&\text{if $n$ is even}\end{cases}$$ and note that $(12)(123\cdots m)(12)=(213\cdots m)$, thus $\pi(123\cdots m)\pi^{-1}=(213\cdots m)$. Since $m$-cycles can be written uniquely up to cyclic permutation, we have $\pi=\pi'(12)$ for some $m$-cycle $\pi'$, and thus $\pi\notin A_n$ so conjugation by $(12)$ is outer.