Transposition-preserving automorphism of symmetric group is inner

Statement

Let $S$ be a finite set and $G=\operatorname {Sym} (S)$ be the Symmetric group (?) on $S$. Suppose $\sigma \in \operatorname {Aut} (G)$ is such that $\sigma$ sends Transposition (?)s to transpositions; in other words, $\sigma$ preserves the conjugacy class of transpositions. Then, $\sigma \in \operatorname {Inn} (G)$.

Related facts

Infinitary version

• Transposition-preserving automorphism of finitary symmetric group is induced by conjugation by a permutation

Stronger facts

• Symmetric groups on finite sets are complete: The symmetric group on a finite set of size other than $2$ or $6$ is complete. Thus, for almost all symmetric groups on finite sets, every automorphism is inner.

Facts used

1. Transpositions generate the finitary symmetric group: In particular, for a finite set, transpositions generate the symmetric group on the set.
2. Transpositions commute iff they are disjoint

Proof

Given: A finite set $S$. $G$ is the symmetric group on $S$. $\sigma$ is an automorphism of $G$ that sends transpositions to transpositions.

To prove: $\sigma$ is inner.

Proof: By fact (1), it suffices to find $g\in G$ such that conjugation by $g$ agrees with $\sigma$ on the set of transpositions. Further, we can assume that $S$ has at least three elements (The statement is obviously true for $S$ having zero, one, or two elements). We now describe how the permutation $g$ can be constructed explicitly.

Also, we use fact (2) at many steps, without explicitly acknowledging it.

1. Pick distinct elements $a,b,c\in S$. Then, both $\sigma ((a,b))$ and $\sigma ((a,c))$ are transpositions. We claim that these transpositions have exactly one element in common: If $\sigma ((a,b))$ and $\sigma ((a,c))$ have no element in common, then they commute, and hence, $(a,b)$ commutes with $(a,c)$, which is not true. Thus, $\sigma ((a,b))$ and $\sigma ((a,c))$ have exactly one element in commmon.
2. Pick distinct elements $a,b,c\in S$. Then, there are elements $a',b',c'\in S$ such that $\sigma ((a,b))=(a',b')$, $\sigma ((b,c))=(b',c')$ and $\sigma ((c,a))=(c',a')$: This follows directly from the previous step.
3. For any element $a\in S$, there is a unique element $g(a)$ that is involved in the transposition $\sigma ((a,b))$ for every $b\neq a$: For any $b\neq c$ distinct from $a$, there exist $a',b',c'$ as described in the previous step. We now show that $\sigma ((a,d))$ is also a transposition involving $a'$.
   • Since $(a,d)$ commutes with $(b,c)$, $\sigma ((a,d))$ commutes with $(b',c')$. In particular, it cannot be a transposition involving $b'$ or $c'$.
   • Since $(a,b)$ does not commute with $(a,d)$, $\sigma ((a,b))$ does not commute with $\sigma ((a,d))$, so they must have an element in common. Thus, either $a'$ or $b'$ is involved in $\sigma ((a,d))$.
   • These together force $a'$ to be involved in $\sigma ((a,d))$.
   • We can thus define $g(a)$ as the unique element involved in $\sigma ((a,x))$ for every transposition involving $a$.
4. The previous step yields a map $g:S\to S$ such that for any transposition $(a,b)$, $\sigma ((a,b))$ is a transposition involving both $g(a)$ and $g(b)$. Step (2) makes it clear that $g(a)\neq g(b)$, so $\sigma ((a,b))=(g(a),g(b))$. Thus, $\sigma$ is induced by conjugation by the permutation $g$.