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

## Statement

Let be a set, be the Symmetric group (?) on , and be the subgroup of comprising the finitary permutations, i.e., is the Finitary symmetric group (?) on . Suppose is an automorphism of that sends transpositions to transpositions. Then, is induced by conjugation by some element of .

## Related facts

- Transposition-preserving automorphism of symmetric group is inner: This is the version for finite sets.

### Stronger facts

- Symmetric groups on finite sets are complete: This holds except when the set has size two or six.
- Symmetric groups on infinite sets are complete

## Facts used

## Proof

**Given**: A finite set . is the finitary symmetric group on . is an automorphism of that sends transpositions to transpositions.

**To prove**: is induced by conjugation by a permutation in , the whole symmetric group on .

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

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

- Pick distinct elements . Then, both and are transpositions. We claim that these transpositions have exactly one element in common: If and have no element in common, then they commute, and hence, commutes with , which is not true. Thus, and have exactly one element in commmon.
- Pick distinct elements . Then, there are elements such that , and : This follows directly from the previous step.
- For any element , there is a unique element that is involved in the transposition for every : For any distinct from , there exist as described in the previous step. We now show that is also a transposition involving .
- Since commutes with , commutes with . In particular, it cannot be a transposition involving or .
- Since does not commute with , does not commute with , so they must have an element in common. Thus, either or is involved in .
- These together force to be involved in .
- We can thus define as the unique element involved in for
*every*transposition involving .

- The previous step yields a map such that for any transposition , is a transposition involving both and . Step (2) makes it clear that , so . Thus, is induced by conjugation by the permutation .