# Nonisomorphic central products on the same pair of groups?

A central product of two groups $$G$$ and $$H$$ is determined as follows. The groups $$G$$ and $$H$$ have respective central subgroups $$A$$ and $$B$$ which are isomorphic, let $$\delta:A\rightarrow B$$ be such an isomorphism. In this scenario, the diagonal subgroup $$\Delta=\{(a^{-1},a^\delta)\mid a\in A\}$$ is normal in $$G\times H$$ and the quotient $$(G\times H)/\Delta$$ is a central product. The idea of course is that we can "pinch" the "two" copies of $$A$$ together into a single copy in the quotient.

Places that define central products (e.g., the Wikipedia article "Central product") are quick to point out that central products are not just dependent on the groups $$G,H,A,B$$ but also on the specific isomorphism $$\delta$$, that distinct isomorphisms $$A\rightarrow B$$ can indeed yield different central products. However, the most prominent application of central products is (or at least appears to be) the classification of extraspecial $$p$$-groups, a scenario where it is known that any pair of distinct isomorphisms define isomorphic quotients. As many writings introduce central products as a means towards the extraspecial classification, they do not provide details on nonisomorphic examples.

The Question: What are some groups $$G$$ and $$H$$ with respective central isomorphic subgroups $$A$$ and $$B$$ where distinct isomorphisms $$A\rightarrow B$$ yield distinct central products?

Bonus Question: Are there examples with $$G=H$$ and $$A=B$$, that a pair of distinct automorphisms on $$A$$ yield distinct central products?

My main interest is finite groups but am willing to entertain infinite examples as well.

• One starting point might be to take a group with center of order divisible by 3 — the first example that comes to mind is $Q_8\times C_3$ — and try the two different automorphisms on the $C_3$ portion of the center. Feb 12 at 18:27

The smallest example: $$G = H = \mathbb{Z}/4 \times \mathbb{Z}/2$$, generated by say $$x$$ of order 4 and $$y$$ of order $$2$$, and $$A = B = \langle x^2, y \rangle \cong \mathbb{Z} / 2 \times \mathbb{Z} / 2$$. Then the identity automorphism gives central product $$\mathbb{Z}/4 \times \mathbb{Z}/2 \times \mathbb{Z}/2$$ whereas the automorphism swapping $$x^2$$ and $$y$$ gives $$\mathbb{Z}/4 \times \mathbb{Z}/4$$.