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Let $G$ be a connected topological group and let $g$ be any element of $G$ (or a compact abelian topological group). If $k\in \mathbb{N}$ is it true that the equation

$$x^k=g$$

always has a solution in $G$?

Dietrich Burde
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User
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    What have you tried? $SL_2(\Bbb R)$ has elements which have no square root. – Dietrich Burde Jul 29 '22 at 18:50
  • I don't know what to try at all. I am working on a problem in ergodic theory and I have a measure space $(G, \mathcal{G}, m)$, where $G$ is a connected compact abelian group and $m$ is the Haar measure. I would like to have a transformation $R$ acting $G$ that would act as a multiplication by the solution of the equation $x^k=g$ for some specific element $g$ in $G$ and so I wonder if this more general result is true, which would allow me to consider such a transformation. – User Jul 29 '22 at 18:55
  • Yes, by no means does it have to be as general as my question. Connected, compact and abelian would work for me. – User Jul 29 '22 at 18:58
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    Every compact connected Lie group is divisible, so has $k$-th roots, because its exponential map is surjective. – Dietrich Burde Jul 29 '22 at 19:02

1 Answers1

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Let $A$ be a connected compact abelian group. Its Pontryagin dual $A^{\vee}$ is then a torsion-free discrete abelian group (see e.g. this math.SE answer) which determines $A$ since $A$ must be the Pontryagin dual of it.

Your question is whether the multiplication-by-$k$ map $A \xrightarrow{k} A$ map is always surjective, or equivalently has trivial cokernel; the Pontryagin dual question is whether the multiplication-by-$k$ map $A^{\vee} \xrightarrow{k} A^{\vee}$ is always injective, or equivalently has trivial kernel. But this is precisely the condition that $A^{\vee}$ is torsion-free, so it holds.

Qiaochu Yuan
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