# Are the terms of the derived series of finitely generated groups finitely normally generated?

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.

• Your terminology is a bit nonstandard: Do you mean the derived series? Jul 19, 2020 at 10:42
• Yes. I edited my question to account for this. Thanks! Jul 19, 2020 at 11:06
• In general metabelian groups are not finitely presentable. A standard example is the lamplighter group. Jul 19, 2020 at 11:28
• 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? Jul 19, 2020 at 11:36
• 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. Jul 19, 2020 at 15:24

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.

• 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: $$H_2(F_2'/F_2'')= F_2''/[F_2'',F_2']$$
– tkf
Jul 21, 2020 at 0:10