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Oct 21, 2010 at 16:33 comment added Johannes Ebert There are quite a lot of familiar properties that a nontrivial group $G$ with $\pi_1 (G)\cong G$ cannot have. To give an example, $G$ cannot be locally simply connected and satisfy the second axiom of countability - because then the fundamental group is countable (exercise). I think groups with that property are either trivial or at least as weird as Todd Trimble's example.
Oct 21, 2010 at 8:01 comment added Pietro Majer So the first question is: how is the $\pi_1$ of an Abelian topological group? Is it torsion-free, or maybe even free? If so, how is the $\pi_1$ of a torsion-free Abelian TG (resp, free Abelian TG)? Is it trivial?
Oct 21, 2010 at 6:39 comment added David Roberts If G is Hausdorff, and your guess about the compactness is true, then it would have to be not locally compact, because connected locally compact Hausdorff abelian topological groups are isomorphic to R^n \times K for K compact.
Oct 21, 2010 at 6:34 history answered Denis Serre CC BY-SA 2.5