# Finite presentability of semi-direct product of free group and its commutator subgroup

Let $$F_n$$ be a free group of rank $$n \geq 2$$. The group $$F_n$$ acts on its commutator subgroup $$[F_n,\, F_n]$$ by conjugation. Let $$G = [F_n,\, F_n] \rtimes F_n$$. It's not hard to see that $$G$$ is finitely generated.

Question: Is $$G$$ finitely presentable? Presumably not, but I can't seem to prove it.

• Look at this question: math.stackexchange.com/questions/788307/… Oct 14 at 4:04
• @markvs: It's an interesting question, but I can't figure out how to extract an answer to my question from any of the answers or comments. Oct 14 at 4:15
• Just a remark: The retraction $F_n \to F_2$ (which kills all the generators but two) induces a retraction $[F_n,F_n] \rtimes F_n \to [F_2,F_2] \rtimes F_2$. As a consequence, if you are able to prove that the group is not finitely presented for $n=2$, then it automatically implies that it is not finitely presented for any $n \geq 2$. Oct 14 at 5:52
• I think Mark's point is that $G$ is an $n$-times iterated HNN-extension, each time having base $[F_n,F_n]$. Like, toss in the generators of $F_n$ one at a time. So the base $[F_n,F_n]$ being non-finitely generated should do it. (Except... to literally apply the result in the link you'd need $[F_n,F_n]\rtimes F_{n-1}$ to be finitely presented, which it's not, so I don't think the result in the link does it immediately....) Oct 14 at 11:08

This group is not finitely presentable. Indeed, write $$F$$ for the given free group and $$F'$$ for its derived subgroup. The map $$F\ltimes F'\to F\times F,\quad (f,g)\mapsto (f,fg)$$ is an injective group homomorphism, and its image is the fibre product of two copies of $$F$$ over the abelianization map $$F\to F/F'$$. (Alternatively, start from this fibre product, and observe that it is the semidirect product $$H\ltimes K$$, where $$H\simeq F$$ is the diagonal and $$K=F'\times\{1\}$$.)
PS: the same argument works equally if $$N$$ is an arbitrary nontrivial normal subgroup of infinite index in $$F$$: then $$F\ltimes N$$ is not finitely presentable.