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I wonder if it there exists a topological compact group $G$ (by compact, I mean Hausdorff and quasi-compact) and a non-zero group morphism $\phi : G \to \mathbb{Z}$ (without assuming any topological condition on this morphism).

For compact Lie groups, using the exponential map, the answers is no, but in general I don't know.

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If G is metrizable then the answer is no - by Dudley's theorem, $\phi$ must be continuous hence is zero everywhere. I don't know what happens in general. –  Julien Melleray Nov 15 '11 at 9:11
Thanks. However, if I got it well I can use Dudley's theorem has, if G is metric and complete, a morphism from G to Z is continuous? For Dudley's theorem I found this math.univ-lyon1.fr/~melleray/rapport-LeMaitre.pdf , part 3. Otherwise, there is the following (part 3 too), but I didn't understand the proof : math.univ-lyon1.fr/~melleray/Rosendal.pdf –  Florent MARTIN Nov 15 '11 at 9:53

2 Answers 2

up vote 37 down vote accepted

The answer is no in general, but this is a rather deep fact.

Theorem: (Nikolov, Segal) If $G$ is any compact Hausdorff topological group, then every finitely generated (abstract) quotient of $G$ is finite.

N. Nikolov and D. Segal, Generators and commutators in finite groups; abstract quotients of compact groups, arXiv, http://arxiv.org/abs/1102.3037

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1+. It's hard to believe that such a result was only proven recently, in 2011. –  Martin Brandenburg Nov 15 '11 at 11:43
this is a 2nd generation proof -- the first one is from 2005 :) –  kassabov Nov 15 '11 at 16:43
As Alain mentions, the first proof (with target $\mathbf{Z}$) is from R. Alperin in the early 80's; actually the 1982 paper refers to the older [Compact groups acting on trees. Houston J. Math. 6 (1980), no. 4, 439--441.] The proof is subtle but quite elementary, this is considerably simpler than the Nikolov-Segal result. –  YCor Jun 5 '12 at 21:48

Andreas shot first, but I still encourage everybody to have a look at the lemma on p.263 of R. Alperin,Locally compact groups acting on trees and property $T$. Monatsh. Math. 93 (1982), no. 4, 261–265: any homomorphism from a locally compact group to $\mathbb{Z}$, is continuous. This answers Florent's question.

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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. –  Sean Eberhard yesterday

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