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

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.

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.

• Thanks Alex. I have some questions. How do you conclude that conjugation by $\sigma$ is inner iff conjugation by $(12)$ is? If conjugation by $\sigma$ is inner, then for all $\tau$, $\sigma\tau\sigma^{-1}=\rho\tau\rho^{-1}$ for some even permutation $\rho$, but why does that tell you anything about $(12)\tau(12)$? I'm also confused on the last sentence. If $(12)(123\cdots m)(12)=\pi(123\cdots m)\pi^{-1}$, what implies that $\pi$ and $(12)$ are conjugate by an $m$-cycle? – Adelaide Dokras Oct 29 '13 at 4:16
• @Alex You've edited my correction of $S_n$ to $A_n$ back, but then the sentence there is trivial (I mean the sentence right before the definition of $m$). – DKal Oct 29 '13 at 4:21
• Beside the typo $\pi\in S_n$ (that the answerer wants to keep it for strange reasons) this answer (which is a good one!) has a mistake: it concludes that $\pi$ is conjugated with $(1\ 2)$, that is, $\pi$ is a transposition. Actually $\pi$ is a cycle of even length and this solves the problem. – user89712 Oct 29 '13 at 8:36
• @user That's not a typo. I assume that $\pi\in S_n$ and show that $\pi\notin A_n$, which is what we need. It's not a proof by contradiction, which is perhaps what's confusing you. As for your other point: you are quite right and I am fixing that mistake. – Alex Becker Dec 9 '13 at 1:38
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.