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:



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> Is a subgroup of $F\_2 \times F\_2$ not containing a copy of $F\_2 \times F\_2$ coherent?
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> 
> 


That is to say, is containing $F\_2 \times F\_2$ the only obstruction to being coherent?



Coherent subgroups of $F_2 \times F_2$
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.


