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From Proposition 2.5 from https://wstein.org/edu/2007/spring/ent/ent-html/node28.html#prop:dsols, the maximum number of roots $\alpha\in k$ of $x^n-1$ in a field $k$ is $n$. That is, there are at most $n$ many $\alpha$ such that $\alpha^n-1=0$ in $k$.

I was wondering if it is true, and if so how to prove, that this maximum implies there are at most $n$ solutions to $x^n=1$ in the corresponding multiplicative group $(k\backslash \{0\},\cdot)$.

Logically, I would assume it does, as $0$ cannot be a root of $x^n-1$ in $k$, but I am very new to group theory, and have often found that my logic is wrong.

Thanks :)

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    $\begingroup$ Precisely what do you doubt here? Likely you will get much more helpful answers if you make that clear. $\endgroup$ – Bill Dubuque Jul 13 at 18:53
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    $\begingroup$ Assuming known that a quadratic has at most two rational roots can you deduce that it has at most two integer roots, without any such doubts? $\endgroup$ – Bill Dubuque Jul 13 at 19:02
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As sets, the multiplicative group is a subset of the field. So any solution in the multiplicative group is also a solution in the field. Because the number of solutions in the field is finite, this implies that also the number of solutions in the multiplicative group is finite.

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