# Commutator subgroup of a finitely generated nilpotent group.

Let $G$ be a finitely generated nilpotent group.
Is the commutator subgroup $[G,G]$ finitely presented?

Edit: I am also interested in the weaker question: is $[G,G]$ finitely generated?

• Yes, all subgroups of $G$ are finitely presented. – Derek Holt Jan 19 '14 at 18:13
• Hmmm....how does "finitely generated" ==> finitely presented in this case, @DerekHolt ? – DonAntonio Jan 19 '14 at 18:23
• Well you would probably need to read this in a group theory textbook to understand it properly, but I have written a brief summary of how to prove these things below. – Derek Holt Jan 19 '14 at 19:49
• @Lepanais your weaker question is identical to your original question! – Derek Holt Jan 19 '14 at 19:50
• Indeed, I changed it. – Lepanais Jan 19 '14 at 20:31

Let $G=\gamma_1(G) \ge \gamma_2(G) \ge \cdots \ge \gamma_{k+1}(G)=1$ be the lower central series of the finitely generated nilpotent $G$.

Suppose that $G$ is generated by $x_{11},x_{12},\ldots,x_{1n_1}$. Then $\gamma_2(G)/\gamma_3(G)$ is generated by the images of the commutators $x_{21},\ldots,x_{2n_2}$, say, of the elements $x_{1i}$.

Since $\gamma_{m}(G)=[G,\gamma_{m-1}(G)]$, you can show by induction on $m$ that $\gamma_m(G)/\gamma_{m+1}(G)$ is generated by the commutators $x_{m1},\ldots,x_{mn_m}$ of the $x_{1i}$ with $x_{m-1,j}$.

So each subgroup $\gamma_m(G)$ is finitely generated by the elements $x_{li}$ with $l \ge m$, and every element of $G$ can be written in normal form as a product of powers of all these generators. You can get a finite presentation of $G$ by evaluating the normal form words of all commutators of pairs of generators, and making those the relations. For generators $x_{mi}$ that have finite order in $\gamma_m(G)/\gamma_{m+1}(G)$, you also need power relations that express that power in normal form. These realtions enable you to put an arbitrary word in the generators into normal form, which shows that they constitute a presentation of the group.

The easiest nonabelian example is the Heisenberg group ($3 \times 3$ upper unitriangular integer matrices) which, in this notation, has the presentation

$\langle x_{11},x_{12},x_{21} \mid [x_{11},x_{12}]=x_{21}, [x_{11},x_{21}]=[x_{12},x_{21}]=1 \rangle$.

You can also show that any subgroup of $H \le G$ is finitely generated, and hence finitely presented. To do that, assume inductively that $H \gamma_k(G)/\gamma_k(G) \cong H/(H \cap \gamma_k(G))$ is finitely generated, and then observe that $H \cap \gamma(G)$ is a subgroup of the finitely generated abelian group $\gamma_k(G)$ and hence is itself fintiely generated.