On proving that $\operatorname{Aut} A_n \cong \operatorname{Aut} S_n$?

Question

I have long been thinking about the automorphism group of $A_n$ since I'd learned the structure of $\operatorname{Aut} S_n$. I went through several pages on the web, each of which asserts that $\operatorname{Aut} A_n \cong \operatorname{Aut} S_n \; (n\geq 4)$ or an equivalent statement without proof, and many of them seem to regard it as a trivial fact. I remember a Math SE post asking what $\operatorname{Aut} A_n$ looks like, which got marked as a duplicate of a question on $\operatorname{Aut} S_n$. My question is, is this fact so easy to prove?

Of course the simpler part is that $\operatorname{Aut} S_n$ embeds to $\operatorname{Aut} A_n$ (even this part requires some arguments). To prove this part, it suffices to show the natural homomorphism $\operatorname{Aut} S_n\to \operatorname{Aut} A_n$ is injective. My proof goes as follows: Suppose $\phi$ is an automorphism of $S_n$ that fixes every element of $A_n$. Noticing that every odd permutation can be written as $\alpha(12)$ where $\alpha\in A_n$, we deduce that $\phi$ sends every odd permutation $\sigma$ to $\sigma (12)(12)^\phi$. Let $g=(12)(12)^\phi$. For each $\alpha\in A_n$ we have $$ (12) g \alpha = (12)^\phi\alpha = (12)^\phi \alpha^\phi = ((12)\alpha)^\phi = (12)\alpha g, $$ so that $g$ commutes with every group member of $A_n$. But when $n\geq 4$, the group $A_n$ is centerless (for if $A_n$ has nontrivial center $Z$, then $Z \operatorname{char} A_n \lhd S_n \Rightarrow Z\lhd S_n$, so that $Z=A_n$, but $A_n$ is nonabelian for $n\geq 4$, a contradiction). We arrive at the conclusion that $g=1$, and therefore $\phi$ is the identity.

What remains is to prove surjectivity of natural homomorphism $\operatorname{Aut} S_n\to \operatorname{Aut} A_n$. It's equivalent to "every automorphism of $A_n$ extends to $S_n$". I did some attempt finally to conclude that it's too hard for me. Time to state my precise question.

How can you prove "for $n\geq 4$, $\operatorname{Aut} A_n \cong \operatorname{Aut} S_n$" (best in an easy manner, if possible)? Or how do you prove that "every automorphism of $A_n$ extends to $S_n$"?

Any help is appreciated.

Answer 1

This answer starts in the spirit of this proof that $S_n$ has only inner automorphisms for $n\ne6$ (that is, any automorphism is conjugation by some element of $S_n$). That proof involves two ideas:

• The conjugacy class $C(2^1)$ of transpositions is the unique conjugacy class of involutions with the cardinality that it has ($n\ne6$), therefore it is fixed (setwise) by any automorphism.
• $\mathrm{Aut}(S_n)$'s induced action on $C(2^1)$ in turn induces an action on $[n]:=\{1,\cdots,n\}$, and the action on $C(2^1)$ coincides with the conjugation action induced from the one on $[n]$, hence so too for the action on all of $S_n$ since $C(2^1)$ is a generating set.

We will modify this to work for $3$-cycles instead of $2$-cycles (assuming $n\ge5$ and $n\ne6$). We can show $\mathrm{Aut}(A_n)$ acts on $C(3^1)$, which induces an action on "edges" corresponding to ordered pairs $(i,j)$, which induces an action on "vertices" that we identify with $[n]$, and the action on $C(3^1)$ coincides with the conjugation action induced from said action on $[n]$.

This latter construction process is reminiscent (at least to me) of some I've seen used for exceptional isomorphisms such as $\mathrm{PGL}(3,2)\cong\mathrm{PSL}(2,7)$.

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Lemma. $C(3^1)$ is the unique conjugacy class in $A_n$ of order $3$ elements with the cardinality that it has ($n\ne6$), therefore $\mathrm{Aut}(A_n)$ acts on it.

Proof. Any element of order $3$ in $S_n$ is a product of $3$-cycles, and $3$-cycles are even, so the same is true of $A_n$. The size of the conjugacy class of a product of $k$ disjoint $3$-cycles in $S_n$ is

$$ |C(3^k)|= \frac{2^k}{k!}\binom{n}{3}\binom{n-3}{3}\cdots\binom{n-3(k-1)}{3} \tag{1}$$

Why? To construct a product of $k$ $3$-cycles:

• Pick $3$ out of $n$ elements, then pick another $3$ out of $n-3$ remaining, and so on;
• For each of the $k$ sets of $3$ choose one of $2$ orientations to make it a $3$-cycle;
• Notice this overcounts by a factor of $k!$ since we can permute the $k$ $3$-cycles arbitrarily.

Let's show $C(3^k)$ is too large to be $C(3^1)$. (Note $C(3^k)$ does not split in $A_n$.) For contradiction's sake, set it equal to $|C(3^1)|=2\binom{n}{3}$, multiply by $3^{k-1}k!/(3k-3)!$ and simplify to

$$ \binom{n-3}{3k-3}=\frac{3^{k-1}k!}{(3k-3)!} \tag{2}$$

The RHS is $<1$ if $k>2$ by induction, and $k=2$ would force $n=6$ which we're ruling out.

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Incidence. Observe the products of two $3$-cycles have the following possible shapes:

• $(i\,j\,k)\,(\ell\,m\,n)$ (order $3$)
• $(i\,j\,k)\,(k\,\ell\,m)=(i\,j\,k\,\ell\,m)$ (order $5$)
• $(i\,j\,k)\,(j\,k\,\ell)=(i\,j)\,(k\,\ell)$ (order $2$)
• $(i\,j\,k)\,(k\,j\,\ell)=(i\,j\,\ell)$ (order $3$)

Thus, two distinct $3$-cycles send $j\mapsto k$ iff their product is an involution; call them incident.

Edges. Define an "edge" to be any maximal collection of pairwise incident $3$-cycles; edges are in natural bijection with ordered pairs $(j,k)$ if $n\ge5$. Since "product is an involution" is a property preserved by automorphisms, $\mathrm{Aut}(A_n)$ respects incidence and acts on the set of edges.

For any edge $e$, we can define the inverse edge $e^{-1}=\{\Delta^{-1}\mid \Delta\in e\}$: if $e$ corresponds to $(j,k)$ then $e^{-1}$ corresponds to $(k,j)$. Any two edges intersect in either a singleton $\{\Delta\}$ or the empty set $\varnothing$ when their corresponding ordered pairs are either part of a common $3$-cycle $\Delta$ or not (resp).

Coterminality. Edges $e_1,e_2$ are coterminal (i.e. their corresponding ordered pairs have identical second component) iff $\Delta\, e_1\,\Delta^{-1}=e_2^{-1}$ where $e_1\cap e_2^{-1}=\{\Delta\}$. Indeed, $e_1\cap e_2^{-1}=\{(i\,j\,k)\}$ is WLOG equivalent to $e_1,e_2$ corresponding to $(j\,k),(i\,k)$, in which case we can check directly $(i\,j\,k)\,(j\,k\,\ell)\,(i\,j\,k)^{-1}=(k\,i\,\ell)\in e_2^{-1}$ for all $(j\,k\,\ell)\in e_1$.

Vertices. From its definition we see coterminality is preserved by automorphisms. From the ordered pair perspective, we see coterminality is an equivalence relation and the equivalence classes, which we'll dub "vertices," are in natural bijection with $[n]:=\{1,\cdots,n\}$ (which we'll identify). This induces an action of $\mathrm{Aut}(A_n)$ on $[n]$.

Theorem. The induced action $\mathrm{Aut}(A_n)\to S_n$ is an isomorphism.

Proof. The composition $S_n\hookrightarrow\mathrm{Aut}(A_n)\to S_n$ is the identity, so it remains to see the latter map is one-to-one, i.e. the action on $[n]$ determines the one on $A_n$. Since $A_n$ is generated by $C(3^1)$, it suffices to show the action on $[n]$ determines the one on $C(3^1)$.

Given an arbitrary $3$-cycle $\Delta$, there are exactly three edges with $\Delta\in e_1,e_2,e_3$ (cyclically ordered so $e_i,e_{i+1}^{-1}$ are coterminal for each $i$ mod $3$), which in turn determine three vertices $v_1,v_2,v_3$ characterized by $e_i,e_{i+1}^{-1}\in v_i$, and then $\Delta=(v_1\,v_2\,v_3)$. In other words, $(i\,j\,k)$ corresponds to $(i,j),(j,k),(k,i)$ which correspond to $i,j,k$ (respectively).

If $\Delta=(v_1\,v_2\,v_3)$, then the three edges $\Phi(\Delta)\in\Phi(e_1),\Phi(e_2),\Phi(e_3)$ determine the three vertices $\Phi(v_1),\Phi(v_2),\Phi(v_3)$, and therefore $\Phi(\Delta)=(\Phi(v_1)\,\Phi(v_2)\,\Phi(v_3))$. In conclusion, $\Phi$'s action on $[n]$ determines its action on $C(3^1)$, hence on $A_n$.