# Every group of given order is a permutable complement for symmetric groups

## Statement

### Symbolic statement

Let be a finite group of order . Then the following are true:

- Via the regular representation, we can view as a subgroup of the symmetric group on elements (this is the idea behind Cayley's theorem).
- Under this representation, is a permutable complement to the symmetric group on any of the elements

In particular this means that every isomorphism class of a group of order occurs as a permutable complement of in . Since there can exist non-isomorphic groups of the same order, this shows that the isomorphism type of the permutable complement of in is not uniquely determined.

## Analysis

In fact, define the following equivalence relation on groups:

Two groups and are equivalent if there is a group with embeddings of and in , such that both and have a common permutable complement.

*A priori*, it is clear that two finite groups can be equivalent only if they have the same order. The above result shows that the converse is also true: any two finite groups that have the same order are equivalent.

## Related facts

- Complement to normal subgroup is isomorphic to quotient: This rules out a variant of the result where the
*other*subgroup is normal. - Complements to abelian normal subgroup are automorphic
- Complements to normal subgroup need not be automorphic
- Retract not implies normal complements are isomorphic
- Direct product is cancellative for finite groups
- Semidirect product is not left-cancellative for finite groups

## Facts used

- Cayley's theorem
- Group equals product of transitive subgroup and isotropy of a point (get more related information at proving product of subgroups)

## Proof

**Given**: A group of order .

**To prove**: is isomorphic to a permutable complement to the symmetric group on (which is the isotropy of the point , in the symmetric group on .

**Proof**:

- By Cayley's theorem, embeds inside -- the group of permutations on the underlying set of , via the action by left multiplication.
- Under this embedding, every non-identity element of acts as a fixed point-free permutation. In particular, the intersection of with the isotropy subgroup of any point is trivial.
- Also, the action of on itself by left multiplication is transitive. Thus, by fact (2), is the product of the subgroup and the isotropy of any point.
- Combining steps (2) and (3), we see that is a permutable complement to the isotropy of any point in .
- A set-theoretic bijection between and the set thus shows that is isomorphic to a permutable complement to the symmetric group on in .