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The question:

(1) Is every compact Lie group $G$ isomorphic (as a topological group) to some quotient $H/N$ where $H$ is a torsion-free compact metrizable group?

Or equivalently:

(2) Is every compact metrizable group $G$ isomorphic (as a topological group) to some quotient $H/N$ where $H$ is a torsion-free compact metrizable group?

((1)$\implies$(2) follows from the fact that every compact metrizable group is isomorphic to an inverse limit of compact Lie groups)

A related question that came up in our research:

(3) Is there a compact connected metrizable non-abelian torsion-free group?

(I asked the latter question without the assumption on connectivity here, and YCor pointed out that the Heisenberg group over the $p$-adic integers is a suitable example.

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The answer to all questions is "no" (and even without the metrizability requirements).

a) A negative answer to (3) implies a negative answer to (1).

Indeed, if $f:G\to H$ is a surjective continuous homomorphism between compact groups, then $G^\circ\to H^\circ$ is surjective as well. So a non-abelian connected compact group (such as $\mathrm{SO}(3)$) is not quotient of any torsion-free compact group.

b) The answer to (3) is negative: every connected torsion-free compact group $G$ is abelian. Indeed, let $G$ be a connected compact group. Then (see Bourbaki Lie, Chap. IX Appendix I) there exists a family $(S_i)_{i\in I}$ of connected simple Lie groups (with finite center) and a surjective homomorphism $S=\prod S_i\to [G,G]$ whose kernel is central in $S$. Since each $S_i$ contains non-central elements of finite order, it follows that $[G,G]$ is not torsion-free, unless $I$ is empty, in which case $G$ is abelian.

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  • $\begingroup$ (Side remark: if $W$ is the Pontryagin dual of $\mathbf{Q}$, the connected torsion-free compact groups are thus the powers $W^X$ of $W$. Indeed a connected torsion-free compact abelian group is the same as the Pontryagin dual of a discrete torsion-free divisible group, and this is always of the form $\mathbf{Q}^{(X)}$ for some set $X$.) $\endgroup$
    – YCor
    Commented Mar 28, 2022 at 18:19
  • $\begingroup$ Thank you! I understand it except for the reference. I assume you referred to Bourbaki Chap. IX, Appendix I, Sec. 3, Prop. 2/c (in the 2004 edition). It says that for a connected compact group $G$ there exists a family $(S_\lambda)_{\lambda\in L}$ of almost simple compact Lie groups and a surjective cont. hom. $\prod_{\lambda\in L} S_\lambda \to D(G)$, whose kernel is a totally discontinuous, compact, central subgroup. In Chapter III almost simple Lie groups are defined but the definition uses the notion of integral subgroups. I could not find what integral subgroups are. Can you help? $\endgroup$
    – chj
    Commented Mar 30, 2022 at 15:51
  • $\begingroup$ An almost simple compact Lie group means a non-abelian compact Lie group in which every proper normal (closed) subgroup is finite central. (I didn't look up in Bourbaki) $\endgroup$
    – YCor
    Commented Mar 30, 2022 at 16:40
  • $\begingroup$ With this definition everything is clear, thank you very much! $\endgroup$
    – chj
    Commented Mar 31, 2022 at 7:05

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