Let $S$ be a finite generating set of a finitely generated group $G$. Then the set $S'$ of $[a,b]$ for $a,b \in S$ normally generates $G'$, i.e., any element of $G'$ is a product of conjugates of commutators of generators. Indeed, denoting the normal closure of $S'$ by $\langle \langle S' \rangle \rangle$ we have that $G/\langle \langle S' \rangle \rangle$ is abelian (because the generators commute), so $G' \subset \langle \langle S' \rangle \rangle$, and that $S' \subset G'$. Hence, $G' = \langle \langle S' \rangle \rangle$ is finitely normally generated. (Notice that the word "normally" is important since, e.g., $F_2'$ is not finitely generated, where $F_2$ is the free group on two generators.) See also this answer explaining what I just explained.

Does this generalize to higher order terms in the derived series? More concretely: Let $G$ be a finitely generated group. Is the $k$-th term of the derived series $G^{(k)}$ finitely normally generated in $G$ for $k \geq 2$?

My suspicion is that the group $F_2/F_2''$ is not finitely presented, which would answer my question in the negative.

  • $\begingroup$ Your terminology is a bit nonstandard: Do you mean the derived series? $\endgroup$ Jul 19, 2020 at 10:42
  • $\begingroup$ Yes. I edited my question to account for this. Thanks! $\endgroup$
    – rawbacon
    Jul 19, 2020 at 11:06
  • $\begingroup$ In general metabelian groups are not finitely presentable. A standard example is the lamplighter group. $\endgroup$
    – Derek Holt
    Jul 19, 2020 at 11:28
  • $\begingroup$ It is not clear to me that the lamplighter group is of the form $G/G^{(k)}$ for a finitely generated group $G$, am I missing something obvious? $\endgroup$
    – rawbacon
    Jul 19, 2020 at 11:36
  • $\begingroup$ It is not obvious, but is a theorem of Shmelkin that free solvable group on at least 2 generators and of class $\ge 2$ is not finitely presentable. Honestly, YCor should be writing an answer since he has a paper on this subject. $\endgroup$ Jul 19, 2020 at 15:24

2 Answers 2


Let $G$ be the group defined by the presentation $$\langle x,\, y_i\,(i \in {\mathbb Z}),\,z_i\,(i > 0) \mid y_i^2=1,\,x^{-1}y_ix=y_{i+1}\, (i \in {\mathbb Z}),\,[y_i,y_j] = z_{|i-j|}\,(i\ne j),\,z_i\ {\rm central}\,\rangle.$$ Note that $G = \langle x,y_1 \rangle$ is finitely generated.

Let $Z = \langle z_i \,(i >0) \rangle$. Then $Z =Z(G)$, and $G/Z$ is isomorphic to the Lamplighter Group.

Now $G^{(1)} = \langle y_iy_{i+1}\,(i \in {\mathbb Z}),\,z_i\, (i>0)\rangle$ and $G^{(2)}$ is an infinitely generated subgroup of $Z$. If we let $C$ be a complement of $G^{(2)}$ in $Z$ and define $\bar{G} = G/C$, then $\bar{G}$ is finitely generated and $\bar{G}^{(2)}$ is not finite normally generated, because $\bar{G}^{(2)}$ is an infinitely generated central subgroup.


@DerekHolt 's simple elegant answer makes this answer redundant, but for variety and a visual argument:

Let $F_2$ be freely generated by elements, $a,b$ and let $A=F_2/F_2'\cong \mathbb{Z}\oplus \mathbb{Z}$. Then $F_2'$ is freely generated by elements $\{e_x\}_{x\in A}$ where $e_{(i,j)}=a^ib^j[a,b]b^{-j}a^{-j}$.

Let $\mathbb{R}^A$ denote the real vector space with basis elements $\{v_x\}_{x\in A}$. This has a natural decomposition as a cubical complex $C$, with the vertices of cubes occuring at $\mathbb{Z}^A\subset \mathbb{R}^A$.

Let $C^{(1)}$, denote the 1-skeleton of $C$. Then: $$F_2''=\pi_1\left(C^{(1)}\right)$$ and killing off the conjugation action of $F_2''$ on itself we get: $$F_2''/[F_2'',F_2'']=H_1\left(C^{(1)}\right)$$

As $$H_1\left(\mathbb{R}^A\right)=0,$$ we have $H_1\left(C^{(1)}\right)$ generated (as an Abelian group) by the boundaries of squares in $C^{(2)}$.

As $$H_2\left(\mathbb{R}^A\right)=0,$$ we know that the relations between these generators are generated by the boundaries of cubes in $C^{(3)}$.

Killing off the conjugation action of $F_2'/F_2''$ on $F_2''/[F_2'',F_2'']$ we get: $$F_2''/[F_2'',F_2']=H_1\left(C^{(1)}\right)\otimes_{{C_\infty}^{\!\!A}}\mathbb{Z}$$ where ${C_\infty}^{\!\!A}\cong F_2'/F_2''$ acts on $H_1\left(C^{(1)}\right)$ by translating boundaries of squares in the natural way.

Thus $H_1\left(C^{(1)}\right)\otimes_{{C_\infty}^{\!\!A}}\mathbb{Z}$ is generated by boundaries of squares with the origin and $v_x+v_y$ as opposite vertices, which we may index $\{s_{\{x,y\}}\}_{\{x,y\}\in A^{(2)}}$.

As the boundary of a 3 dimensional cube consists of pairs of parallel squares, with ${\it opposite}$ orientations, we have $$H_1\left(C^{(1)}\right)\otimes_{{C_\infty}^{\!\!A}}\mathbb{Z}$$ freely generated by the $\{s_{\{x,y\}}\}_{\{x,y\}\in A^{(2)}}$ as an abelian group.

Finally we kill off the conjugation action of $A=F_2/F_2'$ on $F_2''/[F_2'',F_2']$. The conjugation action of $z\in A$ on $s_{\{x,y\}}$ is given by: $$zs_{\{x,y\}}=s_{\{x+z,y+z\}}$$

Thus $F_2''/[F_2'',F_2]$ is freely generated as an abelian group by $\{s_{\{0,x\}}\}_{x\in A}$, which is infinite. Any set of elements which normally generates $F_2''$ would generate this abelian group, so must be infinite.

  • $\begingroup$ The geometric part of this argument could be bypassed using $$H_2(G)=(R\cap[F,F])/[R,F]$$ for any group presentation $G=F/R$, to give: $$H_2(F_2'/F_2'')= F_2''/[F_2'',F_2']$$ $\endgroup$
    – tkf
    Jul 21, 2020 at 0:10

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