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

Symbolic statement
Let $$G$$ be a finite group of order $$n$$. Then the following are true:


 * Via the regular representation, we can view $$G$$ as a subgroup of the symmetric group on $$n$$ elements (this is the idea behind Cayley's theorem).
 * Under this representation, $$G$$ is a permutable complement to the symmetric group on any $$n-1$$ of the elements

In particular this means that every isomorphism class of a group of order $$n$$ occurs as a permutable complement of $$S_{n-1}$$ in $$S_n$$. Since there can exist non-isomorphic groups of the same order, this shows that the isomorphism type of the permutable complement of $$S_{n-1}$$ in $$S_n$$ is not uniquely determined.

Analysis
In fact, define the following equivalence relation on groups:

Two groups $$K_1$$ and $$K_2$$ are equivalent if there is a group $$G$$ with embeddings of $$K_1$$ and $$K_2$$ in $$G$$, such that both $$K_1$$ and $$K_2$$ 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

 * 1) uses::Cayley's theorem
 * 2) uses::Group equals product of transitive subgroup and isotropy of a point (get more related information at proving product of subgroups)

Proof
Given: A group $$G$$ of order $$n$$.

To prove: $$G$$ is isomorphic to a permutable complement to the symmetric group on $$\{ 1,2,3, \dots, n-1 \}$$ (which is the isotropy of the point $$\{ n \}$$, in the symmetric group on $$\{ 1,2,3, \dots, n \}$$.

Proof:


 * 1) By Cayley's theorem, $$G$$ embeds inside $$\operatorname{Sym}(G)$$ -- the group of permutations on the underlying set of $$G$$, via the action by left multiplication.
 * 2) Under this embedding, every non-identity element of $$G$$ acts as a fixed point-free permutation. In particular, the intersection of $$G$$ with the isotropy subgroup of any point is trivial.
 * 3) Also, the action of $$G$$ on itself by left multiplication is transitive. Thus, by fact (2), $$\operatorname{Sym}(G)$$ is the product of the subgroup $$G$$ and the isotropy of any point.
 * 4) Combining steps (2) and (3), we see that $$G$$ is a permutable complement to the isotropy of any point in $$\operatorname{Sym}(G)$$.
 * 5) A set-theoretic bijection between $$G$$ and the set $$\{ 1,2,3,\dots,n \}$$ thus shows that $$G$$ is isomorphic to a permutable complement to the symmetric group on $$\{ 1,2,3,\dots,n-1\}$$ in $$\{ 1,2,3,\dots,n \}$$.