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Suppose G is a finitely generated nilpotent group with abelianization of rank r. Does G always have a subgroup H of finite index, such that H abelianized is a free abelian group of rank r?

Since this is MathOverflow, I will push the question further - under what conditions can we expect abelianization of a monic map to be monic?

Edit: We assume r to be positive.

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So the subgroup H would have to be simple non-abelian, and it can't because it's nilpotent? You're right, I'll edit the question. – Matthew Clayton Aug 25 '11 at 11:41
@Geoff: I believe rank here means the number of $\mathbb{Z}$ summands, i.e., the Betti number. – Steve D Aug 25 '11 at 14:23
@Steve: Thanks for clearing that up – Geoff Robinson Aug 25 '11 at 14:43
up vote 0 down vote accepted

The answer is "yes" because every f.g. nilpotent group has a torsion-free finite index subgroup and because the abelianization of a torsion-free nilpotent group is torsion-free. I assume that by "rank" you meant the torsion-free (${\mathbb Q}$-)rank.

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I love the answer, but could you give a sketch of proof or at least a reference? – Matthew Clayton Aug 25 '11 at 14:16
Finite index torsion-free follows from the fact the torsion subgroup of your G is finite, and G is residually finite. I don't know why the abelianization must be torsion free. – Steve D Aug 25 '11 at 14:25
The first statement follows trivially from the fact that nilpotent groups are linear. It also trivially follows from Theorem 16.2.7 in Kargapolov-Merzlyakov and the fact that f.g. nilpotent groups are residually finite. The second is Exercise 16.2.10 in Kargapolov-Merzlyakov. – Mark Sapir Aug 25 '11 at 14:30
@Mark: take a look at exercise 16.2.11 in the same book! – Steve D Aug 25 '11 at 14:34
To be specific, the group $\langle a,b,c \mid ac=ca, bc=cb, ba=abc^2 \rangle$ is torsion-free nilpotent, but its abelianization has is $\mathbb{Z}^2 \oplus \mathbb{Z}/2\mathbb{Z}$. – Derek Holt Aug 25 '11 at 20:03

This seems surprisingly difficult! Let's try and do it by induction on the nilpotency class of $G$. The result is clear for abelian groups.

Let $Z$ be the last nontrivial subgroup in the lower central series of $G$. So $Z \le G' \cap Z(G)$. By induction, $G$ has a finite index normal subgroup $H$ containing $Z$ such that the abelianization $H/ZH'$ of $H/Z$ is free abelian.

Since $G$ is nilpotent, $|G:H|$ finite implies $|G':H'|$ finite. So $|Z:Z \cap H'| = |H'Z:H'|$ is finite, and $H'Z/H'$ has a free abelian normal complement $K/H'$ in $H/H'$ with $|H:K|$ finite.

But $H = KZ$ and $Z$ is central imply $K' = H'$, and hence $K/K' = K/H'$ is free abelian.

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Does $[G:H]$ finite implying $[G':H']$ finite hold, even if $G$ is not torsion-free? – Steve D Aug 25 '11 at 22:24
Great answer, by the way! – Steve D Aug 25 '11 at 22:25
Steve: yes. The torsion subgroup is unimportant because it is finite and can be factored out. The statement is equivalent to a (f.g.) virtually abelian nilpotent group has finite derived group. You could could prove that by induction on the finite bit at the top, and reduce to the case of an extension of a free abelian group by a cyclic group of prime order. – Derek Holt Aug 26 '11 at 7:47
@DerekHolt: Could you please elaborate why $[G:H]$ finite implies $[G':H']$ finite when $G$ is nilpotent? – user0810 Aug 15 '14 at 0:44

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