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.
Best Answer
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)?