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

Statement

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. Cayley's theorem
2. 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 $\{ 1,2,3,\dots,n \}$.