For a Lie Group $\mathfrak{G}$ and any neighbourhood $\mathcal{V}\subset\mathfrak{G}$ of the identity $\mathrm{id}\in\mathfrak{G}$, $\exists$ neighbourhood $\mathcal{U}\subset\mathcal{V}$ of $\mathrm{id}$ such that every element of $\mathcal{U}$ has a unique square root in that neighbourhood, i.e. $\forall z\in\mathcal{U}\, \exists x\in\mathcal{U}\text{ s.t. }x^2=z$ as well as $x,\,x^2,y,\,\,y^2\in\mathcal{U},\,x^2 = y^2\Rightarrow\,x=y$.
I am almost certain that one can prove the same thing for a general topological group. It's easy to prove the first bit, i.e. since $x\mapsto x^2$ is cts and $\mathrm{id}^2=\mathrm{id}$, choose any neigbourhood $\mathcal{V}$ of $\mathrm{id}$ then by continuity $\exists$ a neighbourhood $\mathcal{U}$ of $\mathrm{id}$ such that $\mathcal{U}\,\mathcal{U}\subseteq\mathcal{V}$, then $\mathcal{W}=\mathcal{U\,U\cap U}$ is a neighbourhood of $\mathrm{id}$ wherein every element has a square root. But I can't seem to show that this square root is unique.
Question 1: Can anyone give a proof that the square root in $\mathcal{W}$ above is unique?
I would be also interested in any proof that needs to use the venerable Prof Gleason's "No Small Subgroups" (NSS) condition, i.e. for a topological group for which one can say there is a neighbourhood of the identity $\mathcal{G}$ wherein there are no subgroups other than the trivial one $\{\mathrm{id}\}$. Otherwise put, the powers of any $\gamma\in\mathcal{G}$ to wit $\mathrm{id},\,\gamma^{\pm1},\,\gamma^{\pm2},\,\gamma^{\pm3},\,\cdots$ must eventually "escape" from $\mathcal{G}$ i.e $\exists n\in\mathcal{N} \ni \gamma^n\not\in \mathcal{G}$. On reflexion, I think that my "intuition" that there should be a proof for a general topological group may be flawed and may be instead grounded on this kind of NSS behaviour: for Lie groups, uniqueness works because squaring puts a member a distance $d$ (measured, say by the Euclidean norm in the Lie algebra) from the identity at least a distance $(2 - \epsilon) d$, where you can bound $\epsilon$ by taking a small enough neighbourhood.
Question 2: As for question 1, but also assuming the no small subgroups condition.
Possible Question 3: If you can prove uniqueness with NSS condition but not otherwise, can you give a proof, or even any reasonably thought out hunch or opinion as to whether the NSS condition is essential.
In the Cantor cube, $\:\operatorname{id}\:$ has square roots other than itself that are arbitrarily close to itself.
The $\:tri$adic rationals$\:$ do not have small subgroups but do have
elements arbitrarily close to $\:\operatorname{id}\:$ that don't have square roots.
In abelian groups, $\;\;\;\; x+x \: = \: y+y \;\; \implies \;\; \{\operatorname{id}\hspace{.02 in},x\hspace{-0.05 in}+\hspace{-0.04 in}(\hspace{-0.03 in}-y)\hspace{-0.02 in}\} \:$ is a subgroup $\:\:\:$.
In abelian groups, if there are elements with more than one
small square root then there are small subgroups of order 2.
The only part this answer leaves open is whether or not there is a non-abelian
group that satisfies the no small subgroups condition but has arbitrarily
small non-$\operatorname{id}$ elements with multiple correspondingly small square roots.