# Do unique $n^{th}$ roots commute in groups?

Let $$G$$ be a group, and let $$g \in G$$. Call $$k$$ an $$n^{th}$$ root of $$g$$ if $$k \in G$$, $$k^n = g$$.

Suppose that $$g,h \in G$$ and $$g$$ has an unique $$n^{th}$$ root, which will be denoted by $$g^{\frac{1}{n}}$$.

If $$gh = hg$$, is it true that $$g^{\frac{1}{n}}h=hg^{\frac{1}{n}}$$?

For some reason, this seems quite plausible to me, and I can't find a simple counterexample. Note that if $$g=e$$ (id. element), then any unique $$n^{th}$$ root must be $$e$$, since $$e$$ will always be an $$n^{th}$$ root of itself, so this special case will not provide any counterexamples.

• Are you missing exponents on the $h$s? Jun 19, 2022 at 23:41
• @Vincent $G = D_6$, the element $r$ has the unique square-root $r^2$. Of course, the result in the question does not say much here because $r^2$ is in the span of $r$, but still, this demonstrates that even elements in finite groups can have unique $n$th roots. Jul 15, 2022 at 17:11
• To your first comment, I wasn't saying that every element has an $n$th root for every $n$. I was saying that if an element has an $n$th root, say $k$, and it has finite order, say $m$, then it has (at least) $n$ many $n$th roots. These are $k^{im+1}$ for $0\le i\le n-1$. You can easily check that, since $k$ has order $nm$, we get $(k^{im+1})^n=(k^{nm})^ik^n=ek^n=g$. Jul 15, 2022 at 17:25
• @DavidSheard This is not true because the $k^{im+1}$ are not necessarily distinct for all $i$. Please see my example with $D_6$ in the reply to Vincent. Jul 15, 2022 at 17:28
• You are quite right, so you can have uniqueness in both the finite and infinite case. I was being quite cavalier and not worrying to check that incidental remark before I made. In some circumstances what I said is true. I think a sufficient condition is that there is an injection $\mathbb{Z}_{mn}=\langle a\rangle$ into your group for some $m>1$ such that $a\mapsto k$ and $a^n\mapsto g$...or something like that Jul 15, 2022 at 17:41

Notice that $$(h^{-1}g^{\frac{1}{n}}h)^n = h^{-1}gh = g,$$ so $$h^{-1}g^{\frac{1}{n}}h= g^{\frac{1}{n}}$$ by uniqueness.