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This question was accidentally trivial: For any countably infinite discrete group $G$ we have that $G^{\mathbb N}$ is a topological group structure on the Baire Space, as pointed in Qiaochu Yuan's answer. What about the coverse? Are all topological group structures on the Baire space of this form?

Carla only proves trivial prop
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  • Your first statement is true if $G$ is countably infinite, but "countable" includes finite group, and the other question doesn't imply the case $G$ finite. – Thomas Andrews Aug 05 '22 at 17:31
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    @Thomas: well, there are two conventions here. In one convention "countable" means "countably infinite" and "at most countable" means "countably infinite or finite." In the other convention "countably infinite" means "countably infinite" and "countable" means "countably infinite or finite." – Qiaochu Yuan Aug 05 '22 at 17:32
  • Well, for clarity, one should always say "countably infinite." Wikipedia gives my definition of countable. @QiaochuYuan – Thomas Andrews Aug 05 '22 at 17:34
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    @Thomas: if someone asked you whether a countable sum or product converged would you say "well, if it's a finite sum or product then it obviously converges"? There really are just two conventions here, otherwise the term "at most countable" would be redundant. – Qiaochu Yuan Aug 05 '22 at 17:38
  • @QiaochuYuan nobody asks if countable sums converge, because "infinite sum" means countably infinite. Unless the questioner very specifically is writing in the context of uncountable sums, in which case, I'd expect that to be made clear. Anyway, if $G$ is finite and non-trivial, $G^{\mathbb N}$ has the same cardinality as the Baire Space, so it might or might not be true that $G^{\mathbb N}$ is homeomorphic. So your example isn't really fair. – Thomas Andrews Aug 05 '22 at 17:45
  • I thought it was clear from context but its edited now. – Carla only proves trivial prop Aug 05 '22 at 17:48

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It's still pretty straightforward: we can take any countable product $\prod_{i=1}^{\infty} G_i$ where $G_i$ is a sequence of countable discrete groups.

I suppose there's some work to do to show that there exists such a group that cannot be isomorphic to a group of the form $G^{\mathbb{N}}$.

Qiaochu Yuan
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