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For $n\ge 3$. Let $A_n$ be the alternating group of degree $n$ acting on $\{1,2,\ldots,n\}$, and let $H_i$ be the isotropy subgroup of $i\in J_n$. If $\phi:A_n\to A_n$ is automorphism, show that $\phi$ is induced by an inner automorphism of $S_n$ if and only if $\phi(H_1)=H_i$ for some $1\le i\le n$.

Showing that $\phi$ induced by an inner automorphism of $S_n$ implies $\phi(H_1)=H_i$ is easy, but I'm stuck on the other direction.

I've thought about looking at how the automorphisms act on $3$-cycles, since they generate $A_n$, but this seems messy and inefficient.

Tim Raczkowski
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  • For $n\neq 6$, every automorphism of $A_n$ is induced by an inner automorphism of $S_n$. – Nishant Apr 30 '15 at 01:00
  • Yes, I'm trying to prove the statement without that fact. This is an Exercise 40(b) in Lang's Algebra. The exercises are leading up to proving that $A_6$ has an automorphism which is not induced by an inner automorphism of $S_n$. – Tim Raczkowski Apr 30 '15 at 01:06
  • It couldn't be as simple as exhibiting some inner-automorphism $\phi$ of $S_n$ that sends $H_1$ to $H_i$, could it? Because you're definitely guaranteed that those exist. – pjs36 Apr 30 '15 at 02:53
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    @pjs36 Yes, I've already show that. What I'm not getting is how to show that the only automorphisms which send $H_1$ to an $H_i$ are the ones induced by inner automorphisms of $S_n$. – Tim Raczkowski Apr 30 '15 at 03:28
  • You could look at http://math.stackexchange.com/questions/1049663 – Derek Holt Apr 30 '15 at 07:28

1 Answers1

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For $n\le 3$, $H_1=1$ and the result has to be checked by hand (easy since $A_2$ is trivial and $A_3$ is cyclic fo order 3). Now assume $n\ge 4$.

The set $C=\{H_1,\dots,H_n\}$ is a single $G$-conjugacy class of subgroups of $A_n$. Hence if $\phi(H_1)=H_i$, then $\phi(C)=C$. The set of $\phi$ such that $\phi(C)=C$ form a subgroup $I_n$ of $\mathrm{Aut}(A_n)$, containing automorphisms induced by elements of $S_n$.

In particular, for $\phi\in I_n$ there is a permutation $s=s(\phi)$ of $\{1,\dots,n\}$ such that $\phi(H_j)=H_{s(j)}$ for all $j$. Then $\phi\mapsto s(\phi)$ is a homomorphism $I_n\mapsto S_n$. If $\phi\in S_n$ then $s(\phi)=\phi$. Hence to show that $I_n=S_n$ it is enough to show that any $\psi$ in the kernel of $s$ is trivial.

By assumption, $\psi(H_j)=H_j$ for all $j$ and let us show $\psi$ is the identity. In particular, $\psi$ preserves the intersections of any $n-3$ of the $H_j$; thus it maps any 3-cycle to either itself or its inverse. Since all cyclic subgroups generated by 3-cycles are conjugate (because $A_n$ acts transitively on 3-element subsets), if $\psi$ is not the identity, then $\psi$ maps all 3-cycles to their inverse. Now (using the right action, so reading permutations from the left to compute compositions), but this yields a contradiction since it would imply $$(134)=\psi((143))=\psi((123)(243))=\psi((123))\psi((243))=(321)(342)=(142).$$

Tim Raczkowski
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YCor
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  • Just been looking over your argument. I'm having some trouble with the third paragraph. When you say "If $\phi\in S_n$ are you saying that $\phi$ induced by an inner automorphism of $S_n$. It's also not clear to me what $\psi$ is. – Tim Raczkowski May 01 '15 at 02:51
  • @TimRaczkowski: I introduced $\psi$ end of the 3rd paragraph (the word "in" missing: I mean "any $\psi$ in the kernel". For the previous sentence, you're right I'm making a slight abuse of notation: I mean: if $\phi$ is induced by an element $\sigma$ of $S_n$ then $s(\phi)=\sigma$. – YCor May 01 '15 at 07:41