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.

If $G$ is free on $k$ generators, and $H \subset G$ has index $n$, then $H$ is free on $m=1+(k1)n$ generators (NielsenSchreier 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 $[G:H]^k$. 


No. Consider $\langle a, baba^{1}b^{1}, ba^2b^{1}, b^2 \rangle$ of index three in $\langle a,b\rangle$. (I'm assuming you mean the image of $H^{ab}$ in $G^{ab}$.) 


If $H$ is normal in $G$ (and $G$ is free as in your case), then you get by explicit computation (or the the LyndonHochschildSerre spectral sequence) an exact sequence $$0 \to H_2(G/H, \mathbb Z) \to H_{ab} \otimes_{\mathbb Z[G/H]} \mathbb Z \to G_{ab} \to (G/H)_{ab} \to 0.$$ Here, the second term is involving the natural action of $G/H$ on $H_{ab}$. Hence, the index of the image of $H_{ab}$ in $G_{ab}$ is precisely the cardinality of the abelianization of $G/H$. Hence, if $G/H$ is perfect, then $H_{ab}$ will map onto $G_{ab}$. 

