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Dec 19, 2014 at 11:07 comment added Sean Eberhard This is a nice lemma. The heart of the matter seems to be in proving that a compact connected group is divisible, which uses a little structure theory. Once you know this then you know that any map $G\to\mathbf{Z}$ factors through the profinite group $G/G_0$, and then by the universal property of profinite completion you need only check there is no nontrivial homomorphism $\widehat{\mathbf{Z}}\to\mathbf{Z}$, which is straightforward.
Nov 17, 2011 at 7:49 vote accept Florent MARTIN
Nov 17, 2011 at 7:49
Nov 15, 2011 at 12:45 history edited Alain Valette CC BY-SA 3.0
added 52 characters in body; deleted 19 characters in body
Nov 15, 2011 at 12:33 history answered Alain Valette CC BY-SA 3.0