$\mathbb Z^2\subset G$ implies $F\not \stackrel{f.i.}{\subset} G$

Question

Let $G$ a finitely generated group which has a subgroup isomorphic to $\mathbb Z^2$. I need to show that $G$ cannot have a free group $F$ as a finite index subgroup.

So far I noticed that then $\mathbb Z^2 \cap F$ has finite index in $\mathbb Z^2$. But how do I proceed from here??

Answer 1

Hint: show that (a) if $G$ has a free subgroup of finite index, then every subgroup of $G$ has a free subgroup of finite index. (b) $\mathbf{Z}^2$ has no free subgroup of finite index.

(Remark: this relies on the classical result that subgroups of free groups are free; nevertheless here it is enough to check that every abelian subgroup of a free group is free [and hence cyclic]; this latter fact is easier: in a free group, every nontrivial element has an infinite cyclic centralizer.)