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Let $G$ be a finitely presented group. Does there exist a finite index subgroup $H$ such that its abelianisation $H^{\text{ab}} = H/[H, H]$ is free abelian?

Note, if $G^{\text{ab}}$ is not already free abelian, then it is non-zero and surjects onto a non-trivial finite group, so $G$ admits a finite index subgroup.

My motivation comes from compact manifolds. If $M$ is a compact manifold, then $\pi_1(M)$ is finitely presented and its abelianisation is $H_1(M; \mathbb{Z})$ which is a finitely generated abelian group which may have torsion. A finite index subgroup of $\pi_1(M)$ corresponds to a finite cover $N \to M$. So my question above is asking whether I can take a finite cover $N$ of $M$ such that $H_1(N; \mathbb{Z})$ is torsion-free.

If $M = k\mathbb{RP}^2$ is a non-orientable surface, then $H_1(M; \mathbb{Z}) \cong \mathbb{Z}^{k-1}\oplus\mathbb{Z}_2$. Its orientable double cover $N = \Sigma_{k-1}$ has $H_1(N; \mathbb{Z}) \cong \mathbb{Z}^{2k-2}$ which is torsion-free. This example shows that the rank of $H_1$ can change by passing to covers (in terms of groups, the rank of the abelianisation can change by passing to finite index subgroups).

If $G$ is virtually free, then of course the answer is yes. Nilpotent groups also have the desired finite index subgroups, see this MO question. In this case, more is true. Namely, the finite index subgroup $H$ can be chosen so that $H^{\text{ab}}$ has the same rank as $G^{\text{ab}}$.

Michael Albanese
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1 Answers1

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The Baumslag-Solitar group $\mathrm{BS}(1,n)$ is the group with presentation $\langle a, t \mid tat^{-1} = a^n \rangle$. It is easy to check that $\mathrm{BS}(1,n)$ has torsion in its Abelianisation when $n > 2$. Moreover, it is known that all the finite index subgroups of $\mathrm{BS}(1,n)$ are isomorphic to $\mathrm{BS}(1,m)$ for some value $m \geqslant n$. It follows that all of the finite index subgroups of $\mathrm{BS}(1,n)$ have torsion in their Abelianisation.

It might be worth mentioning that $\mathrm{BS}(1,n)$ is solvable, so the fact about nilpotent groups that you mentioned in your question does not extend to solvable groups.

SFSH
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  • Do you have a reference for the fact that a finite index subgroup of $BS(1, n)$ is isomorphic to $BS(1, m)$ for some $m \geq n$? I did a quick google search and was unsuccessful. – Michael Albanese Jan 31 '24 at 16:04
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    @MichaelAlbanese: A fairly simple case is when $n=p$ is prime, so that the normal closure $A=\langle\langle a\rangle\rangle$ of $a$ is isomorphic to $\mathbb{Z}[1/p]$. Any finite index subgroup $H$ will intersect this in a finite index subgroup, so that $H\cap A\cong m\mathbb{Z}[1/p]=\langle\langle a^m\rangle\rangle$ for some $m$ coprime to $p$. Since the image of $H$ has finite index in $G/A$, $H$ also contains some $t^ka^r$. These end up generating $H$ and it's not hard to see what the action of $t^ka^r$ does to $a^m$. – Steve D Jan 31 '24 at 18:13
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    One way to do the general case is with Bass-Serre theory. Realise BS$(1,n)$ as an HNN extension $\mathbb Z*_{\mathbb Z}$ and look at the action on the Bass-Serre tree. It is a directed tree with $n$ incoming edges and one outgoing edge at each vertex. The quotient graph by a finite index subgroup will be finite, and you can show that because the action is orientation preserving it must be a loop representing some BS$(1,n^k)$. @Steve D's comment suffices though, without using Bass-Serre theory. – SFSH Jan 31 '24 at 20:43