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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 $[G^\text{ab}:H^\text{ab}]>1$ or can we have any property of $[G^\text{ab}:H^\text{ab}]$? Where $G^\text{ab}$ is the abelianization of $G$.


Edit (YCor Sep. 2023) as it has been pointed out, the question is not meaningful since $H^\text{ab}$ is not a subgroup of $G^\text{ab}$; more precisely, the canonical map $f:H^\text{ab}\to G^\text{ab}$ is not injective. However, the questions about the index of $f(H^\text{ab})$ in $G^\text{ab}$ make sense and have been addressed in the answers.

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  • $\begingroup$ 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. $\endgroup$ Dec 4, 2010 at 4:07
  • $\begingroup$ @Phi Le Well, have you thought about the case when $G$ is free on one generator, hence abelian? $\endgroup$
    – Alex B.
    Dec 4, 2010 at 4:15

3 Answers 3

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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 PlanetMath). 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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  • $\begingroup$ 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. $\endgroup$ Dec 4, 2010 at 12:46
  • $\begingroup$ Thank you so much, Johannes. I think your answer is the one I want to know. $\endgroup$
    – Phi Le
    Dec 4, 2010 at 18:18
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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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  • $\begingroup$ 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 $G^{ab}$. $\endgroup$ Dec 4, 2010 at 4:52
  • $\begingroup$ 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 $F$. $\endgroup$ Dec 4, 2010 at 5:04
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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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  • $\begingroup$ This is very nice! A small typo: The first term in the short exact sequence should involve $G/H$, not $G/N$. $\endgroup$ Dec 4, 2010 at 13:09

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