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

Solution 1:

Lemma: The centralizer of $A_n$ in $S_n$ is trivial for $n \geq 4$.

Proof: Since the centralizer is normal, it has to be trivial for $n \geq 5$. For $n = 4$, the only normal subgroup is $V_4$, but $(123)\,(12)(34) = (134) \neq (12)(34)\,(123)$, so it is trivial for $n \geq 4$.

Proof of the problem: For $n = 3$, notice that $A_3 = \{e, (123), (132)\}$ is abelian, so it has no inner automorphisms. For $n \geq 4$, suppose there is an odd $\sigma \in S_n$ such that conjugation by $\sigma$ is an inner automorphism in $A_n$, i.e. there is $\rho \in A_n$ such that $\sigma \tau\sigma^{-1} = \rho \tau \rho^{-1}$ for all $\tau \in A_n$, then $$\rho^{-1} \sigma \tau \sigma^{-1} \rho = \tau.$$ Hence $\rho^{-1} \sigma$ is a centralizer of $A_n$, so $\rho^{-1} \sigma = e$ by the lemma, but $\sigma$ is odd and $\rho$ is even, which is a contradiction.

Solution 2:

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.