# Groups containing finite index subgroups isomorphic to $\mathbb{Z}$

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.

• 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. – YCor Apr 23 '19 at 18:42
• 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. – 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)?

• (as YCor mentions it is overkill to use Stallings) – user29123 Apr 23 '19 at 18:49
• 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? – pizzaroll Apr 24 '19 at 14:42
• 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. – 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.