Does a fibre product of a group $G$ with itself have a subgroup isomorphic to $G$?

Question

Let $G$ be a group, and consider a fibre product of the form $H=G\times_{D,\phi,\psi}G$, i.e. the group of pairs $(g,g'),\phi(g)=\psi(g')$, for some surjective group morphisms $\phi:G\rightarrow D$ and $\psi:G\rightarrow D$.

Is it always true that $H$ has a subgroup isomorphic to $G$?

My naïve guess is that the answer is no, because I don't see any natural pattern related to the occurrence of such subgroup in general, and because otherwise such result would probably be classical. There is no counter example for $\lvert G\rvert<12$, unless I am mistaken. I proved a positive answer when $G$ splits in a nice way:

Suppose that $G=\ker(\psi)D'$, with $\ker(\psi)\cap D'=\{e\}$ and $\ker(\phi)=\ker(\psi)$. Then $H$ has a subgroup isomorphic to $G$.

This case is enough for my geometric problems (related to subgroups of automorphisms of $\mathbb P^1\times\mathbb P^1$), but I would like to know if there is a more general answer.

Answer 1

There is a counterexample with $G$ equal to $2^3 \cdot L_3(2)$, the nonsplit extension of the natural module for $L_3(2)$ by $L_3(2)$, and $D = L_3(2)$.

We let $\phi$ be the natural projection $G \to D$, and $\psi$ be $\phi$ composed with an outer automorphism of $L_3(2)$ (which maps the natural module to its dual).

Then the resulting fibre product $H=G \times_{D,\phi,\psi} G$ of order $2^6 \times 168 = 10752$ has no subgroup isomorphic to $G$.

If you want to check this computationally, you can define $H$ by

H := PerfectGroup(10752,7);

in GAP or by

D := PerfectGroupDatabase();
H := PermutationGroup(D,10752,7);

in Magma.

Answer 2

By mistake, I did not find any counterexample for the case of $G=D_4$. I have done the computations again, and this actually gives the smallest possible counterexample.

Let $r$ and $s$ be the usual generators of $D_4$, i.e. such that $D_4=\left<r,s\vert r^4=s^2=1,srs=r^{-1}\right>$. Let $\phi,\psi:D_4\rightarrow\mathbb Z_2^2$ be the surjective group morphisms given by: \begin{align} \phi:r&\mapsto(0,1)\nonumber\\ s&\mapsto(1,0)\nonumber\\ \psi:r&\mapsto(1,0)\nonumber\\ s&\mapsto(1,1).\nonumber \end{align}

Then $D_4\times_{D,\phi,\psi}D_4$ is isomorphic to $\mathbb Z_2^2\rtimes\mathbb Z_4$, which does not have any subgroup isomorphic to $D_4$.