Completely unaware of the Bohr topology, I recently [asked](https://mathoverflow.net/questions/114816/hausdorff-group-topologies-on-finitely-generated-groups) whether or not there was a Hausdorff group topology on the integers $\mathbb{Z}$ which made the group fail to be first countable. For me, this topological group is a bit extreme since there are no non-trivial convergent sequences. I'm very interested to know if there is a sequential example.


If $\mathbb{Z}$ is given a Hausdorff group topology which makes it a sequential space, must it be first countable?



The integers as a sequential but non-first countable topological group
The answer is no. 


It is proved in *Topologies on Abelian Groups* (E.G. Zelenyuk and I.V. Protasov, Math. USSR Izvestiya, 1991), that on every infinite abelian group there exists a sequential Hausdorff group topology which is not first-countable.


