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.

  • $\begingroup$ Look at this question: math.stackexchange.com/questions/788307/… $\endgroup$
    – markvs
    Oct 14 at 4:04
  • $\begingroup$ @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. $\endgroup$
    – Cindy
    Oct 14 at 4:15
  • 1
    $\begingroup$ 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 $n \geq 2$. $\endgroup$
    – AGenevois
    Oct 14 at 5:52
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    $\begingroup$ 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....) $\endgroup$ 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 $K=F'\times\{1\}$.)

It is a result of Baumslag and Roseblade that a subgroup of a direct product of two free groups is "almost never" finitely presented, i.e. not finitely presentable unless one of the projection is injective on the subgroup, or if it is virtually the direct product of intersection with the factors. Since this is not such an exception, the above group is not finitely presentable.

Reference: G. Baumslag and J. E. Roseblade, Subgroups of direct products of free groups, J. London Math. Soc. (2) 30 (1984), 44-52. link DOI behind paywall

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.


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