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

## Statement

Let $S$ be a set, $K = \operatorname{Sym}(S)$ be the Symmetric group (?) on $S$, and $G = \operatorname{FSym}(S)$ be the subgroup of $G$ comprising the finitary permutations, i.e., $H$ is the Finitary symmetric group (?) on $S$. Suppose $\sigma$ is an automorphism of $H$ that sends transpositions to transpositions. Then, $\sigma$ is induced by conjugation by some element of $G$.

## Proof

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

To prove: $\sigma$ is induced by conjugation by a permutation in $K$, the whole symmetric group on $S$.

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 $b \ne a$: For any $b \ne c$ 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 $g(a) \ne g(b)$, so $\sigma((a,b)) = (g(a),g(b))$. Thus, $\sigma$ is induced by conjugation by the permutation $g$.