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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.

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  • $\begingroup$ Are you missing exponents on the $h$s? $\endgroup$
    – Randall
    Commented Jun 19, 2022 at 23:41
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    $\begingroup$ @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. $\endgroup$ Commented Jul 15, 2022 at 17:11
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    $\begingroup$ 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 $(k^{im+1})^n=(k^{nm})^ik^n=ek^n=g$. $\endgroup$ Commented Jul 15, 2022 at 17:25
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    $\begingroup$ @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. $\endgroup$ Commented Jul 15, 2022 at 17:28
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    $\begingroup$ 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 $\endgroup$ Commented Jul 15, 2022 at 17:41

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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.

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  • $\begingroup$ I was conjugating by the wrong element. Thanks! $\endgroup$ Commented Jun 19, 2022 at 23:46
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    $\begingroup$ You are very welcome! $\endgroup$
    – Ruy
    Commented Jun 19, 2022 at 23:46

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