# Uniquely divisible neighborhoods of identity in topological groups

Let $G$ be a (finite dimensional real) Lie group, and let $A\subset G$ be an open neighborhood of identity. If $A=\operatorname{Exp}(\mathcal{A})$ is the injective range of the exponential map from a star-shaped neighborhood $\mathcal{A}\subset\mathcal{G}$ of the origin in the Lie algebra $\mathcal{G}$ of $G$, then $A$ is a uniquely divisible neighborhood of identity, i.e., $\exists!\sqrt[m]{x}\in A$ for all $x\in A$ and $m\in\mathbb{N}$.

Question: Given a locally compact Hausdorff topological group $G$, is it always possible to find an open uniquely divisible neighborhood of identity in $G$? If no, then what are the necessary/sufficient conditions?

If $G$ is discrete the question seems to trivialize, so let's assume $G$ is non-discrete. If $G$ is not Lie then I think it needs to have non-trivial subgroups in every neighborhood of identity, but I am not sure this is related.

• Your question is very false for groups like the $p$-adic numbers, the profinite completion of the integers, etc etc. – znt Jan 3 '17 at 19:52
• Heh, I guess for the $p$-adic numbers one could use the $p$-adic numbers themselves :-) so maybe the $p$-adic integers are a better example. Yes, I just mean that in my mind your question is ludicrous because if you asked me to think of a locally compact Hausdorff topological group then because of my background the first 5 examples I'd think of would not satisfy your question. – znt Jan 3 '17 at 20:10
• @YCor: As mentioned by znt, the additive group of the $p$-adic field $\mathbb{Q}_p$ is uniquely divisible, but is not a Lie group. Does it help to add the conditions that the neighborhood can be taken arbitrarily small, and that the neighborhood is closed under taking $m$th roots? – Dave Witte Morris Jan 4 '17 at 7:42