This problem is problem 5.1.12.b of Dummit and Foote and Ive been tearing my hair out. I managed to do 5.1.12.a and 5.1.13 which are related and were pretty simple but I can't for the life of me figure this one out. I tried to look up a solution online in the end but the solution was wrong so that didn't help either. Here is the problem:
Let $Z_4=\left<x\right>$. Let $D_8 = \left<r,s\right>$ and $Q_8=\left<i,j\right>$ be given by their usual generators and relations. Let $Z_4 * D_8$ be the central product of $Z_4$ and $D_8$ which identifies $x^2$ and $r^2$ (ie $Z_1=\left<x^2\right>, Z_2=\left<r^2\right>$ and the isomorphism is $x^2 \mapsto r^2$) and let $Z_4 * Q_8$ be the central product of $Z_4$ and $Q_8$ which identifies $x^2$ and $-1$. Prove that $Z_4 * D_8 \cong Z_4 * Q_8$.
For reference of what the problem means by $Z_1$ and $Z_2$ here is the problem's definition of a central product:
Let $A$ and $B$ be groups. Assume $Z(A)$ contains a subgroup $Z_1$ and $Z(B)$ contains a subgroup $Z_2$ with $Z_1\cong Z_2$. Let this isomorphism be given 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
$$\begin{equation}
(A \times B)/Z \text{ where } Z=\{(x_i,y_i^{-1}) : x_i\in Z_1\}
\end{equation}$$
and is denoted by $A * B$.
I realize I could technically list out the elements of each set since it's only a total of 32 elements and then construct the first isomorphism that jumps out to me but that would not be interesting and very unsatisfying. Would love to hear if anyone has an elegant way to solve this.
Best Answer
You can verify that the following mappings on the generators define homomorphisms and are inverses to each other: $$Z_4*D_8\to Z_4*Q_8: \\ \quad\quad x\mapsto x,\ r\mapsto i,\ s\mapsto xj\\ Z_4*Q_8\to Z_4*D_8:\\ \quad\quad x\mapsto x,\ i\mapsto r,\ j\mapsto x^{-1}s\,.$$ Note that, by the direct product construction, $x$ and thus its image needs to commute with everything, which narrows down the possibilities, and also you can quickly spot the orders to find candidates of correspondence.
As a consequence of this and Birkhoff's HSP theorem, we obtain that these two $8$ element groups $D_8$ and $Q_8$ satisfy exactly the same universally quantified identities, because $Q_8\in \Bbb H\Bbb S\Bbb P(D_8)$ and $D_8\in\Bbb H\Bbb S\Bbb P(Q_8)$.