Every group is a subgroup of a complete group

This article gives the statement, and possibly proof, of an embeddability theorem: a result that states that any group of a certain kind can be embedded in a group of a more restricted kind.
View a complete list of embeddability theorems

Statement

Let $G$ be a group. Then, there exists a complete group $H$ such that $G\leq H$.

Definitions used

Further information: Complete group

A group is termed complete if it satisfies the following two conditions:

• It is centerless: its center is the trivial group.
• Every automorphism of the group is an inner automorphism.

Related facts

Stronger facts

• Every group is a malnormal subgroup of a complete group

Other related facts

• Every finite group is a subgroup of a finite simple group
• Every finite group is a subgroup of a finite complete group: The proof is the same -- we only need to observe that when the group is finite, the complete group constructed here is also finite.

Facts used

1. Cayley's theorem: Every group is a subgroup of a symmetric group -- in fact, of the symmetric group on its underlying set.
2. Symmetric groups on finite sets are complete: The symmetric group on a finite set of size $n$ is a complete group if $n\neq 2,6$.
3. Symmetric groups on infinite sets are complete

Proof

Given: A group $G$.

To prove: $G$ is a subgroup of a complete group.

Proof: Let $K=\operatorname {Sym} (G)$. By Cayley's theorem (fact (1)), $G$ is a subgroup of $K$. We make two cases:

• The order of $G$ is not equal to $2$ or $6$: In this case facts (2) and (3) tell us that $K$ is a complete group, and we are done.
• The order of $G$ is equal to $2$ or $6$: In this case, let $H$ be the symmetric group on the set $G\sqcup \{x_{0}\}$, so $G\leq K\leq H$. Further, $H$ is the symmetric group on a set of size $3$ or $7$, which is complete, so $G$ is a subgroup of a complete group.