Coherent subgroups of $F_2 \times F_2$ A group is coherent if its finitely generated subgroups are finitely presented. For instance, $F_2 \times F_2$ is a well-known example of incoherent group. My question is:

Is a subgroup of $F_2 \times F_2$ not containing a copy of $F_2 \times F_2$ coherent?

That is to say, is containing $F_2 \times F_2$ the only obstruction to being coherent?
 A: This is true. I change the notation for convenience: let $F_1,F_2$ be two free groups (on any number of generators). Let a subgroup $H$ of $F_1\times F_2$ not contain any direct product of two non-abelian free groups. Then $H$ is coherent.
Proof: let $P_i$ be the projection of $H$ on $F_i$, and $J_i=H\cap F_i$. Note that $J_i$ is normal in $P_i$, and $H$ induces an isomorphism $J_1/P_1\simeq J_2/P_2$. Then $H$ contains $J_1\times J_2$, hence at least one of $J_1$ and $J_2$, say $J_2$, is abelian (hence cyclic). We have two cases:


*

*if $J_2=1$, then $H$ is isomorphic to $P_1$, hence is free.

*if $J_2$ is infinite cyclic, then since $P_2$ normalizes $J_2$ and $F_2$ is free, $P_2$ has finite index in $J_2$ (which is the centralizer of $P_2$), and hence $P_1$ also has finite index in $J_1$ as well. Thus $H$ admits $J_1\times J_2$ as a subgroup of finite index (and $P_1\times P_2$ as an over group of finite index).


Thus we have proved that if $H$ does not contain the product of two non-abelian free subgroups, then it is either free or is contained with finite index in the product of a cyclic group and a free group, and hence the same holds for all subgroups of $H$. In particular $H$ is coherent.
