There are two propositions I'm trying to prove that seem fairly intuitive to me, but I haven't been able to prove:

Let $G = A \ast B$ with $A,B$ infinite. No subgroup of $G$ with finite index is isomorphic to $\mathbb{Z}$

If we did have $\mathbb{Z}$ as a finite index subgroup, then we'd be able to write any $g \in G$ as $r_i z$ for $\{r_1, \ldots, r_n\}$ the coset representatives of $\mathbb{Z}$ and $z \in \mathbb{Z}$. I want to think about this geometrically. It seems to me that $G$ being differing in a "finite way" from $\mathbb{Z}$ means it is quasi-isometric to $\mathbb{Z}$, which seems like it shouldn't be true. But we don't know that $A$ and $B$ are finitely presented or even generated, so I don't know where I can go from here.

Let $G$ be a torsion-free finitely presented group with a finite index subgroup isomorphic to $\mathbb{Z}$. Then $G$ is 2-ended and in fact isomorphic to $\mathbb{Z}$

I can kind of see the first part - finitely presented groups are quasi-isometric to finite index subgroups. $\mathbb{Z}$ is 2-ended, so $G$ is 2-ended also, since this is preserved under quasi-isometry. But why must this be isomorphic to $\mathbb{Z}$? Stallings' theorem applies here, since we have $\geq 2$ ends, we can split $G$ (as an HNN-extension or amalgamated product) over a finite subgroup. But already I see a problem, $\mathbb{Z}$ isn't an amalgamated free product, it's $\{e\} \ast _{\{e\}}$ right?

I guess intuitively if you have a torsion free 2-ended group, looking at the Cayley graph I can see on a large scale that if you take some element close to $e$ and keep multiplying by it, you just keep moving away from $e$, this is similar to what happens in $\mathbb{Z}$. But again I don't really know what to do here.

  • $\begingroup$ Using Stallings' theorem in such a situation is a bit senseless, since Stallings' theorem is far more involved and actually makes use of these facts as prerequisites in the 2-ended case. There is a theorem by Wall (1967 I think) that every group $G$ with a finite index isomorphic to $\mathbf{Z}$ has a (unique) finite normal subgroup $W=W(G)$ such that $G/W$ is either infinite cyclic, or infinite dihedral. $\endgroup$ – YCor Apr 23 '19 at 18:42
  • $\begingroup$ It is easy to see that any group with a finite index subgroup isomorphic to ${\mathbb Z}$ is finitely presented, so there is no need to make that assumption in the second question. $\endgroup$ – Derek Holt Apr 23 '19 at 19:08

This is a hint, although you basically already have it. For the first part how many ends does $A*B$, and finite index subgroups have it have? For the second part you have that $G$ is two ended, how can it split over a finite subgroup and be torsion free(you have it written down)?

  • $\begingroup$ (as YCor mentions it is overkill to use Stallings) $\endgroup$ – user29123 Apr 23 '19 at 18:49
  • $\begingroup$ Thanks for the hint, I got the second part, the torsion-free property means the splittings either lead to a group without $\mathbb{Z}$ as finite index subgroup (using the first part) or $\mathbb{Z}$. But for the first part I am still confused: if we have $\mathbb{Z}$ as finite index subgroup of $A \ast B$, then it's finitely generated, is quasi-isometric to $\mathbb{Z}$ so has 2 ends. But I think it's really supposed to have an infinite number of ends, by analogy with $\mathbb{Z} \ast \mathbb{Z}$. Is this correct? How do I prove this? $\endgroup$ – pizzaroll Apr 24 '19 at 14:42
  • $\begingroup$ I got it eventually, and this hint is what got me there, so I'm accepting this. For anyone reading, $A \ast B$ having infinitely many ends is obvious if you e.g. try to draw the Cayley graph. $\endgroup$ – pizzaroll May 12 '19 at 20:10

Let $G$ be a torsion-free group with a finite index, infinite cyclic subgroup; let's show that $G$ itself is infinite cyclic. We can choose this subgroup, say $N$, to be normal in $G$.

For every $g\in G$, the action of $G$ on $N$ is by multiplication by $s(g)\in\{-1,1\}$.

For every $g\neq 1$ in $G$, $\langle g\rangle$ is infinite cyclic (because $G$ is torsion-free), hence its intersection with $N$ has finite index in $\langle g\rangle$. So there exists $n\ge 1$ such that $g^n\in G$. Since $g$ commutes with $g^n$, we deduce that $s(g)=1$. This shows that $N$ is central in $G$.

So the commutator map $G^2\to G$ factors through a map $(G/N)^2\to G$; hence $G$ has finitely many commutators. A classical result then implies that $G$ has a finite derived subgroup. Since $G$ is torsion-free, this implies that $G$ has a trivial derived subgroup. So $G$ is abelian, and in this case the conclusion immediately follows.


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