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Let $G$ be a group, and consider a fibre product of the form $H=G\times_{D,\phi,\psi}G$, i.e. the set 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 naive 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 $|G|<12$, 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. Thank you!

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.

Let $G$ be a group, and consider a fibre product of the form $H=G\times_{D,\phi,\psi}G$, i.e. the set 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 naive 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 $|G|<12$, 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 groups of automorphisms of $\mathbb P^1\times\mathbb P^1$), but I would like to know if there is a more general answer. Thank you!

Let $G$ be a group, and consider a fibre product of the form $H=G\times_{D,\phi,\psi}G$, i.e. the set 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 naive 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 $|G|<12$, 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. Thank you!

Let $G$ be a group, and consider a fibre product of the form $H=G\times_{D,\phi,\psi}G$, i.e. the set 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 naive 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 $|G|<12$, 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 groups of automorphisms of $\mathbb P^1\times\mathbb P^1$), but I would like to know if there is a more general answer. Thank you!

Let $G$ be a group, and consider a fibre product of the form $H=G\times_{D,\phi,\psi}G$, i.e. the set 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 naive 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 $|G|<12$, 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 groups of automorphisms of $\mathbb P^1\times\mathbb P^1$), but I would like to know if there is a more general answer. Thank you!