Symmetric Group $S_n$ is complete

I am just checking some properties of the symmetric group and found on Wikipedia the statement that "Conversely, for $n \neq 6$, $S_n$ has no outer automorphisms, and for $n \neq 2$ it has no center, so for $n \neq 2, 6$ it is a complete group, as discussed in automorphism group, below."

I could show that $Z(S_n)=\epsilon$ but do you know an easy proof for $\mathrm{Aut}(S_n)=\mathrm{Inn}(S_n)$ ?

• Coincidentally I looked up this proof in Hupperts book about finite groups (I) this week since I also stumbled over this fact. The proof in Hupperts book is in my opinion rather technical (which is also the reason that I cannot recall it). – Hans Giebenrath Dec 14 '12 at 18:17
• It is proved in Dummit and Foote's "Abstract Algebra" page 138 exercise 18. – ougao Dec 14 '12 at 18:31
• Thank you ougao for the information. I was able to show part c and d, but could you tell me how a and b works? – Alexander Dec 14 '12 at 19:07
• For part a, it is clear when you notice that if a conjugacy class $\mathcal{K}$ is generated by $\tau$, then $\sigma{\mathcal{K}}$ is generated by $\sigma(\tau)$. For part b, note that any element of order 2 in $S_n$ is a product of two cycles (i.e., transpositions) and one cycles, so the conjugacy class it generated must have smaller size. and note that $\tau(i_1, \cdots, i_k)\tau^{-1}=(\tau(i_1), \cdots, \tau(i_k)).$ – ougao Dec 14 '12 at 20:05

Lemma. Let $\mathrm{Aut}(G)$ be the automorphism group of a group $G$ and $\mathrm{cl}(G)$ the conjugacy classes of elements of $G$. Then $\mathrm{Aut}(G)$ acts on $\mathrm{cl}(G)$.

Explanation. Let $\alpha$ be an automorphism of $G$ and $\mathrm{cl}(a)=\{gag^{-1}:g\in G\}$ the conjugacy class of $a$. Then $\alpha(\mathrm{cl}(a))=\{\alpha(g)\alpha(a)\alpha(g)^{-1}:g\in G\}=\{t\alpha(a)t^{-1}:t\in G\}=\mathrm{cl}(\alpha(a))$ is also a class.

Lemma. If $\alpha\in\mathrm{Aut}(S_n)$ stabilizes the conjugacy class of transpositions then $\alpha\in\mathrm{Inn}(S_n)$ is inner.

Proof. We want to exhibit an element $\sigma\in S_n$ such that $\alpha(g)=\sigma g\sigma^{-1}$ for every $g$. It suffices to find a $\sigma$ for which $\alpha(\tau)=\sigma \tau\sigma^{-1}$ holds for every transposition $\tau$, since every permutation is a product of transpositions. It further suffices to find a $\sigma$ for which $\alpha(1k)=\sigma(1k)\sigma^{-1}$ for each $1< k\le n$, since these $n-1$ transpositions generate all the others.

Pick distinct $1<\ell_1,\ell_2\le n$. Then $\rho=(1\ell_2\ell_1)=(1\ell_1)(1\ell_2)$ has order $3$, in which case $\alpha(1\ell_1)\alpha(1\ell_2)$ has order three as well. Without loss of generality, $\alpha(1\ell_1)=(ab)$ and $\alpha(1\ell_2)=(ac)$ for some $a,b,c$; since this holds for each distinct pair $\ell_1,\ell_2$, we find that for every $k$, $\alpha(1k)=(af(k))$ for some $f(k)$.

Let $\sigma$ send $1$ to $a$ and $k$ to $f(k)$ for each $k>1$. This permutation satisfies our requirements.

Lemma. For $n\ne6$, the class of transpositions is the unique class of involutions with ${n\choose2}$ elements.

Proof. Consider the class of involutions with cycle type a product of $k>1$ disjoint cycles. Suppose

$$\frac{1}{k!}{n\choose 2}{n-2\choose2}\cdots{n-2(k-1)\choose2}={n\choose2}.$$

The LHS is the size of the new conjugacy class. Setting $m=n-2$ and $k=r+1$, cancelling ${n\choose2}$s,

$${m\choose2}\cdots{m-2(r-1)\choose2}=(r+1)!$$

The LHS is telescoping. Multiply by $2^r/(2r)!$ and get

$${m\choose2r}=\frac{(r+1)!2^r}{(2r)!}=\frac{r+1}{(2r-1)!!}$$

For $r>2$ the RHS is not even an integer (it is $<1$). The case $r=1$ gives $m(m-1)=4$ with no solution in $m$. Finally $r=2$ gives $m(m-1)(m-2)(m-3)=24$ which has positive solution $4$.

Theorem. For $n\ne6$, $\mathrm{Out}(S_n)=1$.

Proof. Combine the three lemmas. When $n\ne6$, automorphisms permute conjugacy classes and in particular stabilize the transposition class (since the size of this class is unique among the classes), hence every automorphism is inner.

Remark. This proof is adapted from this sketch from Wikipedia and these notes.

• Presumably I am downvoted for adapting a proof from other places? Or perhaps is it for having a proof that is not easy, as desired? – anon Dec 15 '12 at 18:21
• This is quite elegant. By far, my favorite proof of this fact I've seen yet. – Robert Wolfe Jul 19 '14 at 3:31