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](https://math.stackexchange.com/questions/4409106/is-there-a-compact-metrizable-non-abelian-torsion-free-group?noredirect=1#comment9226292_4409106), and YCor pointed out that the Heisenberg group over the $p$-adic integers is a suitable example.



Compact Lie groups as quotients of torsion-free compact metrizable groups
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.


