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Suppose $F$ is any field. Let $G$ and $H$ are two subgroups of $F^*$ of order $n$. Then is it true that $G=H$?

For finite field this is true. Is this true for arbitrary fields?

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Yes, this is true. For every $a\in G$, $a^n=1$. So every $a\in G$ is a root of the polynomial $$ f(x)=x^n-1\in F[x]. $$ The same holds for every $b\in H$. Because $|G|=|H|=n$, $G$ and $H$ consist of all the roots of $f(x)$ and so $G=H$.

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    $\begingroup$ Note that this implicitly uses the fact that $x^n-1$ has at most $n$ roots in $F$. $\endgroup$ – Servaes Sep 27 at 15:06

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