A group $G$ is called meta-abelian if there exists a normal abelian subgroup $N\unlhd G$, such that $G/N$ is abelian.
If $G$ is meta-abelian, then why does $G$ contain no free group of rank two?
Thanks for your help.
A group $G$ is called meta-abelian if there exists a normal abelian subgroup $N\unlhd G$, such that $G/N$ is abelian.
If $G$ is meta-abelian, then why does $G$ contain no free group of rank two?
Thanks for your help.
As others have pointed out, a group is meta-abelian if there exists an abelian normal subgroup $N\trianglelefteq G$ such that $G/N$ is abelian.
To prove that a meta-abelian group contains no free group of rank 2, suppose the $G$ is meta-abelian, and let $F\leq G$ be a free subgroup of rank $2$. Let $N\trianglelefteq G$ be an abelian normal subgroup such that $G/N$ is abelian. Then $N\cap F$ is abelian and normal in $F$. However, the only abelian subgroups of $F_2$ are cyclic, so the only abelian normal subgroup of $F_2$ is the identity. Then $N\cap F = 1$, which means that $F$ maps isomorphically to $G/N$ via the quotient map. This is a contradiction, since $G/N$ is abelian.
The group $\,G:=F_2 \times A\,$ seems to contradict this, with $F_2=$ the free group in two generators, $A=$ any abelian group
The metabelian groups form a variety: it is a class of groups closed under subgroups (subgroup of metabelian is metabelian), quotients (quotient of metabelian is metabelian), and arbitrary direct products (an arbitrary direct product of metabelian groups is metabelian).
In particular, if $K$ is not metabelian, then $K$ cannot be a subgroup of a metabelian group. The free group of rank $2$ is not metabelian: its commutator subgroup is free of countable rank. So no metabelian group can contain the free group of rank $2$.
(Metabelian is equivalent to "solvable of solvability length at most two"; that is, a group $G$ is metabelian if and only if $[G,G]$ is abelian.)