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Suppose $G$ is a nilpotent, finitely generated group such that it's abelianization has rank $r$. How does one go on proving that $G$ has a finite-index subgroup $H$ with free abelianization of rank $r$?

Martin Sleziak
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Matthew Clayton
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1 Answers1

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The abelianization gives you a surjection $G\twoheadrightarrow G^{ab}\cong\mathbb Z^r\oplus T$, where $T$ is a finite torsion group. (Finite because the abelianization is finitely generated.) Now let $H$ be the inverse image of $\mathbb Z^r$ under this surjection.

Edit: As pointed out in the comments, $H^{ab}$ could be $\mathbb Z^r\oplus T_1$. So you repeat the process by taking a finite index subgroup of $H$, say $H_1$. Now it's not hard to show that $G,H,H_1,\ldots$ is a descending central series. That means each term is contained in the corresponding term of the lower central series and the quotients $H_i/H_{i+1}$ are contained in the quotients $\Gamma_i/\Gamma_{i+1}$ of the lower central series, which are eventually trivial.

Cheerful Parsnip
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    That's how I tried to do it, but I'm not able to prove that abelianization of H is free. I mean, could it be that H' is a proper subgroup of G', hence abelianization of the embedding H -> G is not monic? – Matthew Clayton Aug 24 '11 at 23:24
  • $H$ maps onto $\mathbb Z^r$ so its abelianization is at least as big as $\mathbb Z^r$, since any homomorphism to an abelian group factors through its abelianization. Perhaps this is where nilpotency is used, since you can only repeat this process finitely many times. – Cheerful Parsnip Aug 24 '11 at 23:54
  • I agree, but how do you prove that you can repeat this process only finitely many times> – Matthew Clayton Aug 25 '11 at 00:00
  • You're building a series of nested subgroups whose successive quotients are abelian so that smells like the lower central series, but I admit I don't see the argument at the moment. – Cheerful Parsnip Aug 25 '11 at 00:05
  • @Jim: Are you using rank in the sense of "minimum number of generators", or in some other sense? If you are, then the rank of $\mathbb{Z}^{r} \oplus T$ is greater than $r$ if the torsion part is non-zero. – Geoff Robinson Aug 25 '11 at 13:40
  • @Geoff: I'm using "rank of an abelian group" to mean the largest copy of $\mathbb Z^r$ inside the group. This is equivalent to the dimension of the group tensored with the rationals as a vector spaces. This is a common usage. – Cheerful Parsnip Aug 25 '11 at 14:45