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?
Finite index subgroup with free abelianization
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group-theory
1 Answers
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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.
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0@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. – 2011-08-25