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Here is a question I posted some months ago in Math.SE, and t.b. mentioned to the following question by Florent MARTIN which is somehow related to my question;

Let $G$ be a compact Hausdorff topological group, and let $H$ be a torsion-free group satisfying the ascending condition, i.e. there are no infinite strictly ascending chains $H_1 < H_2<...$ of subgroups of $H.$

Prove that there is no non-trivial homomorphism of $G$ into $H.$

Note that, no topology is considered on $H$ and "homomorphism" simply means "group homomorphism."

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  • $\begingroup$ If $G$ is finite, it's false. So I guess you should take $G$ infinite. Do you want to show that there is no injective homomorphism $G \to H$? Because then you just need to show that $G$ fails to have the ascending condition. Hint: $G$ is uncountable. If that isn't what you mean, then take $G$ profinite and let $H$ be a finite quotient. Or am I missing something? $\endgroup$
    – Autumn Kent
    Commented Dec 20, 2011 at 1:22
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    $\begingroup$ Doesn't the answer to Florent's question answer yours? The acc implies finitely generated and by the answer to that question all fg images are finite. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 20, 2011 at 1:23
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    $\begingroup$ Richard, he said H is torsion-free so in particular not finite. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 20, 2011 at 1:24

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There are no non-trivial homomorphisms from a compact group to a torsion-free finitely generated group by the theorem of Nikolov and Segal quoted in the answer by Andreas to Florent's question (mentioned by the OP above). Since ascending chain condition on subgroups implies finite generation, this answers the question.

Nb. Please upvote Andreas's answer if you like this one.

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  • $\begingroup$ Hmmm. BTW is there a more down to earth approach? $\endgroup$
    – Ehsan M. Kermani
    Commented Dec 20, 2011 at 3:03
  • $\begingroup$ Most likely not. $\endgroup$
    – Benjamin Steinberg
    Commented Dec 20, 2011 at 3:30

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