Locally inner automorphism-balanced not implies central factor

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., locally inner automorphism) need not satisfy the second subgroup property (i.e., central factor)
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Statement

It is possible to have a group $G$ and a locally inner automorphism-balanced subgroup $H$of $G$ (i.e., every inner automorphism of $G$ restricts to a locally inner automorphism of $H$) such that $H$ is not a central factor of $G$.

Here, central factor means that every inner automorphism of $G$ restricts to an inner automorphism of $H$. This is equivalent to the condition that $HC_{G}(H)=G$.

Note that we cannot have an example where $H$ is a finitely generated group (and even more generally, we cannot have an example where $H$ is a group with finitely generated inner automorphism group), because finitely generated inner automorphism group implies every locally inner automorphism is inner.

Facts used

1. Finitary symmetric group is locally inner automorphism-balanced in symmetric group
2. Finitary symmetric group is centralizer-free in symmetric group
3. Restricted direct product is locally inner automorphism-balanced in unrestricted direct product

Proof

Proof using the finitary symmetric group

Suppose $S$ is an infinite set. Let $G=\operatorname {Sym} (S)$ be the symmetric group on $S$ and let $H=\operatorname {FSym} (S)$ be the finitary symmetric group on $S$.

• $H$ is locally inner automorphism-balanced in $G$: This follows from Fact (1).
• $H$ is not a central factor of $G$: To see this, note that by fact (2), $C_{G}(H)$ is trivial, so $HC_{G}(H)=H\neq G$.

Proof using direct products

Suppose $G_{i},i\in I$ is an infinite collection of (possibly repeated) centerless groups. Let $G=\prod _{i\in I}G_{i}$ be the external direct product and let $H$ be the subgroup of $G$ given as the unrestricted direct product.

• $H$ is locally inner automorphism-balanced in $G$: This follows from Fact (3).
• $H$ is not a central factor of $G$: We can compute that $C_{G}(H)$ is trivial, and therefore that $HC_{G}(H)=H\neq G$.