Suppose $G$ is a finitely generated Hausdorff topological group. Must $G$ be first countable (or perhaps a sequential space)? What if we restrict to the abelian case?


I wonder if this is even true for the additive group of integers $\mathbb{Z}$. There certainly are non-discrete, Hausdorff group topologies on $\mathbb{Z}$ where a basis at $0$ consists of subgroups (such as that used in Furstenberg's proof of the infinitude of primes). On the other hand, determining if there is a Hausdorff group topology that makes a given sequence converge to $0$ is non-trivial. For instance, it is known that the sequence of squares $n^2$ can't converge to $0$ in any Hausdorff group topology and that if there is a Hausdorff group topology on $\mathbb{Z}$ such that the sequence of primes $2,3,5,...,p,..$ converges to $0$, then the twin prime conjecture is false.



Hausdorff group topologies on finitely generated groups
No. The Bohr topology on $\mathbb{Z}$ is not first countable, in fact the least size of a local base at $0$ is $2^{\aleph\_0}$. It is also known that this topology is not sequential (because there are no non-trivial convergent sequences).


