Locally inner automorphism-balanced not implies central factor

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) uses::Finitary symmetric group is locally inner automorphism-balanced in symmetric group
 * 2) uses::Finitary symmetric group is centralizer-free in symmetric group
 * 3) uses::Restricted direct product is locally inner automorphism-balanced in unrestricted direct product

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 $$HC_G(H) = H \ne G$$.

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 $$HC_G(H) = H \ne G$$.