Every group of given order is a permutable complement for symmetric groups
- 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.
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.
- 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
- Cayley's theorem
- Group equals product of transitive subgroup and isotropy of a point (get more related information at proving product of subgroups)
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 .
- 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 .