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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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    $\begingroup$ 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. $\endgroup$
    – Julien Melleray
    Commented Nov 15, 2011 at 9:11
  • $\begingroup$ 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 $\endgroup$
    – Florent MARTIN
    Commented Nov 15, 2011 at 9:53

3 Answers 3

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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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    $\begingroup$ 1+. It's hard to believe that such a result was only proven recently, in 2011. $\endgroup$
    – Martin Brandenburg
    Commented Nov 15, 2011 at 11:43
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    $\begingroup$ this is a 2nd generation proof -- the first one is from 2005 :) $\endgroup$
    – kassabov
    Commented Nov 15, 2011 at 16:43
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Jun 5, 2012 at 21:48
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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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    $\begingroup$ 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. $\endgroup$
    – Sean Eberhard
    Commented Dec 19, 2014 at 11:07
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Sorry for resurrecting such an old question, but I think we can give a much simpler proof here. We'll reduce the problem from $G$ to the Bohr compactification $B\mathbf{Z}$ of $\mathbf{Z}$, then from $B\mathbf{Z}$ to the profinite completion $\hat{\mathbf{Z}}=\prod_p\mathbf{Z}_p$ of $\mathbf{Z}$, and then we'll argue directly.

Let $\phi:G\to\mathbf{Z}$ be a homomorphism and fix $x\in G$. The map $\mathbf{Z}\to G$ extending $1\mapsto x$ induces a map $B\mathbf{Z}\to G$ such that $1\mapsto x$, and thus we obtain a map $\phi':B\mathbf{Z}\to \mathbf{Z}$ such that $\phi'(1)=\phi(x)$.

Recall that to construct $B\mathbf{Z}$ one takes the dual of $\mathbf{Z}$, namely $\mathbf{R}/\mathbf{Z}$, strips the topology to get the discrete group $\mathbf{R}_d/\mathbf{Z}\cong\mathbf{R}_d\times\mathbf{Q}/\mathbf{Z}$, then takes the dual again. The result is that $B\mathbf{Z} \cong B\mathbf{R}\times\hat{\mathbf{Z}}$. Since $B\mathbf{R}$ is divisible $\phi'$ must vanish on $B\mathbf{R}$. Since $\prod_{p\neq 2}\mathbf{Z}_p$ is infinitely $2$-divisibile and $\mathbf{Z}_2$ is infinitely $3$-divisible, $\phi'$ vanishes on $\hat{\mathbf{Z}}$. Thus $\phi'$ is identically $0$, so $\phi(x)=\phi'(1)=0$.

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  • $\begingroup$ Very nice. Let me slightly restate your argument: 1) replacing $G$ by the closure of some $x\in\phi^{-1}(\{1\})$, we can suppose that $G$ is abelian with a dense cyclic subgroup. 2) if $G$ is any connected compact abelian group then the result holds because $G$ is divisible (as a projective limit of tori) 3) now supposing $G$ has dense cyclic subgroup, $\phi$ vanishes on its unit connected component because of (2), hence we can suppose $G$ totally disconnected, hence a quotient of the profinite completion $\hat{Z}\simeq\prod_p\mathbf{Z}_p$, and your last argument finishes the job. $\endgroup$
    – YCor
    Commented Jan 22, 2015 at 22:43
  • $\begingroup$ @YCor, I think you mean $x \not\in \phi^{-1}(\{0\})$, right? $\endgroup$
    – LSpice
    Commented Jan 22, 2015 at 23:39
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    $\begingroup$ You do not need $\phi(x) \not= 0$, as your argument directly shows every homomorphism $\phi \colon G \rightarrow \mathbf Z$ is identically $0$. Pick any $x \in G$ and your argument gives a homomorphism $\phi' \colon B\mathbf Z \rightarrow \mathbf Z$ where $\phi'(1) = \phi(x)$. You show $B\mathbf Z = B\mathbf R \times \widehat{\mathbf Z}$, and you give an argument that any homomorphism $B\mathbf R \rightarrow \mathbf Z$ is $0$ and any homomorphism $\widehat{\mathbf Z} \rightarrow \mathbf Z$ is $0$. Thus $\phi'$ is $0$, so $\phi(x) = \phi'(1) = 0$. Since $x \in G$ was arbitrary we're done. $\endgroup$
    – KConrad
    Commented Jan 23, 2015 at 0:18
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    $\begingroup$ I'm extremely happy to see an old question resurrected because of a new and informative answer, rather than merely bumped by reformatting or tweaking of existing answers $\endgroup$
    – Yemon Choi
    Commented Jan 23, 2015 at 12:54
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    $\begingroup$ This is really a proof from the Book. Thanks very much. $\endgroup$
    – Todd Trimble
    Commented Jan 23, 2015 at 14:10

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