# Why do meta-abelian groups contain no free subgroup of rank two?

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?

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I think that in the definition of metabelian you must require $N$ to also be abelian... – DonAntonio May 24 '12 at 16:06
yes. i edited it. – Peter May 24 '12 at 19:44

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.

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Nice. Thank you very much. – Peter May 24 '12 at 19:46

The group $\,G:=F_2 \times A\,$ seems to contradict this, with $F_2=$ the free group in two generators, $A=$ any abelian group

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Even $F_2$ itself violates this - set $N$ to be the commutator subgroup, and the quotient is $\mathbb{Z} \times \mathbb{Z}$. Peter must have intended for $N$ to be abelian as well? – MartianInvader May 24 '12 at 16:11
Either that or he didn't actually mean "metabelian" as this requires N to be abelian...can't say for sure.. – DonAntonio May 24 '12 at 16:24

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.)

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Nice. Thank you. – Peter May 29 '12 at 17:50