The Problem: Let $A$ and $B$ be groups. Assume $Z(A)$ contains a subgroup $Z_{1}$ and $Z(B)$ contains a subgroup $Z_{2}$. Suppose $Z_{1}$ is isomorphic to $Z_{2}$ by the map $x_{i} \mapsto y_{i}$ for all $x_{i} \in Z_{1}$. A central product of $A$ and $B$ is a quotient $(A \times B) / Z$ where $Z = \{(x_{i}, y_{i}^{-1})\mid x_{i} \in Z_{1}\}$ and is denoted by $A*B$. Find $|A*B|$.
My Question: Clearly $|A*B|=|(A\times B)/Z|=|(A\times B)|/|Z|$ IF $A\times B$ is finite. But what if $|A\times B|=\infty$? We can't necessarily conclude that $|A*B|=\infty$ in this case.
Any help would be greatly appreciated.
I think we can say that $|A * B| = \infty$ if $A \times B$ is infinite.
If $Z_1$ is finite, then $A * B$ is factor of an infinite group by finite, thus infinite.
Otherwise, because elements of form $(a, e_B)$ belong to distinct cosets in $Z$ for different $a\in Z_1$, thus they already give us infinite number of elements in $Z_1 * Z_2$. And as $Z_1 * Z_2$ is subgroup of $A * B$, this implies that $A * B$ is also infinite.