# Transposition-preserving automorphism of symmetric group is inner

From Groupprops

## Statement

Let be a finite set and be the Symmetric group (?) on . Suppose is such that sends Transposition (?)s to transpositions; in other words, preserves the conjugacy class of transpositions. Then, .

## Related facts

### Infinitary version

### Stronger facts

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

## Facts used

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

## Proof

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

**To prove**: is inner.

**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 .