Every group is a subgroup of a complete group

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

Definitions used
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.

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) uses::Cayley's theorem: Every group is a subgroup of a symmetric group -- in fact, of the symmetric group on its underlying set.
 * 2) uses::Symmetric groups on finite sets are complete: The symmetric group on a finite set of size $$n$$ is a complete group if $$n \ne 2, 6$$.
 * 3) uses::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 $$G \le K \le H$$. 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.