Metabelian Group

Question

Is there any infinite metabelian group or it is naturally defined to be finite? In fact, I am investigating for finding another solvable group around (apart of $\mathbb{Z}_{p^{\infty}}$ or additive group $\mathbb{Z}$).

Answer 1

A group $G$ is metabelian if and only if $[G,G]$ is abelian. In particular, every abelian group is metabelian.

Less trivially so: a group $G$ is metabelian if and only if there exists an abelian normal subgroup $N\triangleleft G$ such that $G/N$ is abelian. In particular, any extension of an abelian group by an abelian group is abelian. In particular, any metacyclic group (extension of a cyclic group by a cyclic group) is metabelian.

An example of an infinite metacyclic group is given by the holomorph of $\mathbb{Z}$, $\mathbb{Z}\rtimes\mathrm{Aut}(\mathbb{Z})$, also known as the infinite dihedral group $D_{\infty}$: $$D_{\infty} = \bigl\langle a,b\,\bigm| ba = a^{-1}b,\quad b^2=1\bigr\rangle.$$

Another "big" example: if $F$ is the free group on two generators, and $[F,F]$ is the commutator subgroup of $F$, then $[F,F]$ is free of infinite rank. So $[F,F]/[[F,F],[F,F]]$ is free abelian of infinite rank. Hence $$F_2^{\mathfrak{A}^2} = \frac{F}{\bigl[ [F,F],[F,F]\bigr]}$$ is a metabelian group (the commutator subgroup is $[F,F]/[[F,F],[F,F]]$), and infinite (the commutator subgroup is infinite).

In fact, $F_2^{\mathfrak{A}^2}$ is the relatively free metabelian group of rank $2$.

You can repeat the operation with any free group of rank $r\gt 1$ to get the relatively free metabelian groups of finite rank. Any finitely generated metabelian group is a quotient of one of these.