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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.

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Well, H will also be free. The abelianization of both will be free on the number of generators. If life works out nicely it seems like the index should be the same. But this is just idle speculation- hence a comment not an answer. – Dylan Wilson Dec 4 '10 at 4:07
@Phi Le Well, have you thought about the case when $G$ is free on one generator, hence abelian? – Alex B. Dec 4 '10 at 4:15
up vote 1 down vote accepted

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 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 $[G:H]^k$.

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I thought the question is about the index of the image of $H_{ab}$ in $G_{ab}$, but your are right: this is not so clear. – Andreas Thom Dec 4 '10 at 12:46
Thank you so much, Johannes. I think your answer is the one I want to know. – Phi Le Dec 4 '10 at 18:18

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 $G^{ab}$.)

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Yes. In fact, if $G$ is any group and $H\subset G$ is a non-normal subgroup of index 3 then $H^{ab}$ maps onto $G^{ab}$. – Tom Goodwillie Dec 4 '10 at 4:52
Good point, Tom. In free groups, you can arrange most situations (perhaps all) using (the proof of) Hall's theorem that any finitely generated subgroup $H$ of a finitely generated free group $F$ is a free factor in some finite index subgroup $H * K$ of $F$. – Richard Kent Dec 4 '10 at 5:04

If $H$ is normal in $G$ (and $G$ is free as in your case), then you get by explicit computation (or the the Lyndon-Hochschild-Serre 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 $G_{ab}$.

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This is very nice! A small typo: The first term in the short exact sequence should involve $G/H$, not $G/N$. – Theo Buehler Dec 4 '10 at 13:09
Thanks Theo. Good to hear from you. – Andreas Thom Dec 4 '10 at 19:13

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