The central product $G{\circ }_{\alpha }H$ of two groups along an isomorphism of central subgroups

Let $G$ and $H$ be groups, let $Z_1\le Z(G)$ and $Z_2\le Z(H)$ be subgroups of their centres (The center $Z(G)$ of a group, Subgroup), and let $\alpha :Z_1\to Z_2$ be an isomorphism (Group isomorphisms, automorphisms and the set $\mathrm{Aut}(G)$). Inside the external direct product $G\times H$ (The external direct product $G\times H$ with componentwise multiplication, $G\times H$ is a group with identity $(e_G,e_H)$, coordinatewise inverses, and homomorphic coordinate projections) put

$N:=\{(z,\alpha (z{)}^{-1}):z\in Z_1\}\mathrm{.}$

For groups $G,H$ with central subgroups $Z_1\le Z(G)$, $Z_2\le Z(H)$ and an isomorphism $\alpha :Z_1\to Z_2$, the central product $G{\circ }_{\alpha }H$ is the quotient of $G\times H$ by $N=\{(z,\alpha (z{)}^{-1}):z\in Z_1\}$, formed as in The quotient group $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$:

$G{\circ }_{\alpha }H:=(G\times H)\mathrm{/}N\mathrm{.}$

That $N$ is a normal subgroup, so that the quotient is defined (Normal subgroup: invariance under conjugation), is proved in The identified subgroup used to form a central product is central, hence normal ↗.

Write $\pi :G\times H\to G{\circ }_{\alpha }H$ for the quotient map and $\bar{g}:=\pi (g,e)$, $\bar{h}:=\pi (e,h)$ for the canonical images of $g\in G$ and $h\in H$.

The identification is along $\alpha$ and reverses the second coordinate: killing $(z,\alpha (z{)}^{-1})$ is exactly what makes $\bar{z}=\overline{\alpha (z)}$ hold in the quotient, so the two identified central subgroups become one. Killing $(z,\alpha (z))$ instead would identify $z$ with $\alpha (z{)}^{-1}$, which is the central product along ${\alpha }^{-1}$ composed with inversion rather than along $\alpha$.

Craven writes $G\ast H$ for a central product and van Beek writes $G\circ H$; the notation ${\circ }_{\alpha }$ is used here because the isomorphism is part of the data and different choices of $\alpha$ can give non-isomorphic quotients.

Taking $Z_1=Z_2=1$ gives $N=1$ and recovers the direct product, so the construction is a genuine generalisation and not a separate object.