Is it true that if G is a free group and H is a subgroup of G such that $[G:H]=n$ where $n>1$ then $[G^{ab}:H^{ab}]>1$ or can we have any property of $[G^{ab}:H^{ab}]$? Where $G^{ab}$ is the abelianization of $G$.Thank you.



Question on the index of abelianization of groups.
If $G$ is free on $k$ generators, and $H \subset G$ has index $n$, then $H$ is free on $m=1+(k-1)n$ generators (Nielsen-Schreier theorem, see <http://planetmath.org/?op=getobj&from=objects&id=4693>). If $n$ grows, then the rank of $H$ also grows, and you see that the induced map $H^{ab} \to G^{ab}$ is not injective and hence you cannot even talk about $[G^{ab}:H^{ab}]$.


EDIT: But of course you can talk about the image of the map $H^{ab} \to G^{ab}$, and I can offer you at least an estimate of this. Take a $[G:H]$-sheeted cover of graphs $p:Y \to X$ which realizes the map $H \to G$ on fundamental group; consider the transfer $p^{!}:H\_1 (X) \to H\_1 (Y)$ and use the equation $p\_\* \circ p^{!} = [G:H]$. Thus the image of $H^{ab} \to G^{ab}$ contains the subgroup of elements divisible by $[G:H]$, which has index $[G:H]^k$.


