Outer Automorphisms of $S_n$

Question

There are a lots of questions (and several good explanations) on understanding the non-trivial outer automorphism of $S_6$. What I haven't seen, though, is a good explanation of why there are no such outer automorphisms for $n\neq 6$. Wikipedia makes a passing mention of this fact here.

In particular, they have this cryptic claim:

For every symmetric group other than $S_6$, there is no other conjugacy class of elements of order 2 with the same number of elements as the class of transpositions.

This seems like the natural thing to prove, given the knowledge that you can construct an outer automorphism of $S_6$ which takes transpositions to products of three transpositions (i.e. things like $(12)\mapsto (12)(34)(56)$.) Also, the second reason Wikipedia gives doesn't feel very intuitive to me.

My question, then, is how does one justify the original (quoted) statement above?

Here is the naive computation of sizes of conjugacy classes: We need to show that when $n, k\neq 6,3$ $$\frac{n!}{2^k k! (n-2k)!}\neq\frac{n!}{2(n-2)!},$$which is equivalent to $$2^k k!(n-2k)!\neq 2(n-2)!$$

What I want to say is something like, if $n{-}2$ is bigger than $k$, then there is always some factor on the RHS that doesn't appear in the LHS (except when it is a power of two). Will this type of analysis get me anywhere, or is there a better way to think about this?

Answer 1

Yes, as you mention, $S_n$ has $n\choose2$ transpositions and the conjugacy class whose elements consist of $2\le k\le \frac n2$ disjoint transposition has $\frac{n!}{(n-2k)!2^kk!}$ elements. As you say, the condition that these numbers are the same is equivalent to $$\tag1 (n-2k)!2^{k-1}k!= (n-2)!$$

For $k=2$ this becomes $4(n-4)!=(n-2)!$ or equivalently $(n-2)(n-3)=4$, which has no integer solution.

For $k=3$, we obtain $ 24(n-6)!=(n-2)!$ or equivalently $(n-5)(n-4)(n-3)(n-2)=24$. This has integer solutions $n=6$ and $n=1$, the first leading to the exceptional situation with $S_6$, the other being invalid because it has $n<2k$.

For the rest of the argument, we may assume $k>3$. Then Bertrand's postulate states that there is a prime $p$ with $k<p<2k-2$. This $p$ does not divide $2^{k-1}k!$ but $n-2\ge2k-2$ implies that $p$ divides $(n-2)!$ so that we conclude that $p$ divides $(n-2k)!$ and finally that $p\le n-2k$. Rewrite (1) as $$\underbrace{2\cdot2\cdots 2}_{k-1}\cdot \underbrace{k\cdot (k-1)\cdots 2}_{k-1} = \underbrace{(n-2)(n-3)\cdots(n-2k+1)}_{2k-2}$$ Each of the $2k-2$ factors on the left is $\le k$, each of the $2k-2$ factors on the right is $>p>k$, hence equality cannot hold (as $2k-2\ne0$).

Answer 2

It takes a little work, but first you want to show that for $n \not= 6$, any automorphism of $S_n$ will take transpositions to transpositions. Next, note that $S_n$ is generated by $\{(12),(13),...,(1n)\}$. Now show that any automorphism will take this set to a set of the form $\{(ab_1),(ab_2),...,(ab_{n-1})\}$, which also generates $S_n$. Since each automorphism is uniquely determined by how it acts on the generators of the group, this gives at most $n!$ choices for automorphisms. That is, $|Aut(S_n)|\leq n!$. This implies that $Aut(S_n)=Inn(S_n)$, thus $Out(S_n)=\{1\}$.

Answer 3

Just for interest, I'll treat the case $k >3$, trying to avoid the use of Bertrand's Postulate. Note that for $k >1$, $(n-2)!$ is divisible by $(n-2k)!(2k-2)!$. It suffices when $k > 3$ to prove that $2^{k-1}k! < (2k-2)!$, or equivalently that $(2k-2) \ldots (k+1) > 2^{k-1}$. There are $k-2$ terms in the left product, all at least $4.$ Also, we have $4^{k-2} > 2^{k-1}$ as $k >3.$

Answer 4

As factorials are not defined for negative integers, we also require $n \ge 2$ and $n \ge 2k$. Next, assuming we have equality instead, means that

$$\begin{equation}\begin{aligned} 2(n-2)! &= 2^{k}k!(n-2k)! \\ \frac{(n-2)!}{k!(n-2k)!} & = 2^{k-1} \\ \frac{\color{green}{(n-2)(n-3)\cdots(n-k+1)}(n-k)!}{k!(n-2k)!} & = 2^{k-1} \\ \frac{(n-k)!}{k!(n-2k)!} & = \frac{2^{k-1}}{\color{green}{(n-2)(n-3)\cdots(n-k+1)}} \\ \binom{n-k}{k} & = \frac{2^{k-1}}{(n-2)(n-3)\cdots(n-k+1)} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

First, for $k = 2$, we get

$$\begin{equation}\begin{aligned} \binom{n-2}{2} & = 2^{2-1} \\ \frac{(n-2)(n-3)}{2} & = 2 \\ (n-2)(n-3) & = 4 \\ n^2 - 5n + 2 & = 0 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

This is a quadratic equation in $n$. However, the discriminant of $5^2 - 4(2) = 17$ is not a perfect square, so $n$ can't be an integer.

Next, for $k = 3$, the RHS of \eqref{eq1A} becomes $\frac{2^{3-1}}{n-2} = \frac{4}{n-2}$. Since the LHS is always an integer, $n - 2 \mid 4 \; \to \; n-2 \le 4 \; \to \; n \le 6$. However, $n \ge 2k \; \to \; n \ge 6$, so this means $n = 6$ is the only possibility, but that's excluded.

Finally, for $k \ge 4$ (so $n \ge 8$), there are at least $2$ consecutive integers in the denominator on the RHS of \eqref{eq1A}, so at least one of them is an odd integer $\gt 1$. This odd integer can't divide into the numerator of $2^{k-1}$, so the fraction is not an integer, contradicting that the LHS is always an integer.

This means \eqref{eq1A} is not true for $n \ge 2$, with $n \neq 6$, for all $2 \le k \le \left\lfloor \frac{n}{2} \right\rfloor$.