Find perfect finite group whose quotient by center equals the same quotient for two other groups and has both as a quotient

Question

Let $G_1$ and $G_2$ be two finite perfect groups such that $G_1 / Z(G_1) \cong G_2 / Z(G_2)$. Then there exists a finite perfect group $G$ and subgroups $Z_1, Z_2 \le Z(G)$ with $$ G / Z(G) \cong G_i / Z(G_i) \quad \mbox{and} \quad G / Z_i \cong G_i, \quad i = 1,2. $$

I want to construct the group $G$. I tried to start from the direct product $G_1 \times G_2$ and factor out an appropriate normal subgroup, but everything I try does not give the desired properties. And also I have no idea how to incorporate the assumption that $G_1$ and $G_2$ are perfect. So any hints on this exercise?

This exercise is taken from the book Theory of finite groups by H. Kurzweil and B. Stellmacher, and appears there on page 36.

Answer 1

This is just an outline argument, not a full solution.

Let $X= G_1/Z(G_1) \cong G_2/Z(G_2)$ and let $\rho_1$, $\rho_2$ be the associated epimorphisms $\rho_1:G_1 \to X$ and $\rho_2:G_2 \to X$.

Now let $H = \{ (g_1,g_2) \in G_1 \times G_2 \mid \rho_1(g_1) = \rho_2(g_2) \}$ and put $G = [H,H]$. Note that $G$ is perfect by 1.5.3 in the book, and $G$ projects onto both $G_1$ and $G_2$.

Answer 2

Here is a more elaborated sketch of the answer to this question.

1. Write $\bar G_i=G_i/Z(G_i)$ and let $\theta\colon \bar G_1\to\bar G_2$ be an isomorphism. Consider the following commutative diagram

2. Define $$ H = \{\phi_1\circ\rho_1=\varphi_2\circ\rho_2\}\leqslant G_1\times G_2. $$

3. Observe that $$ H = \{(x_1,x_2)\mid \theta(\bar x_1)=\bar x_2\}. $$

4. Note that $$ [(\bar x,\theta(x)),(\bar y,\theta(\bar y))] = ([\bar x,\bar y],\theta([\bar x,\bar y])). $$

5. Put $Z=Z(G_1\times G_2)$. Then $H/Z$ (the graph of $\theta$) is perfect.

6. $G=H'$ is perfect.

7. $G\cap Z=Z(G)$.

8. The (restriction of the) projections $G\to G_i$ are epi.

9. Define $Z_i=\text{ker}(G\to G_i)$. Then $Z_i\subseteq Z(G)$.