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Can the positive characteristic of a field $F$ divide the order of a (finite) subgroup of the multiplicative group $F^*$?

If the field is finite, the answer is negative, because otherwise by Lagrange $p$ would divide a divisor of $p^k-1$ (which is the order of $F^*$ in such a case), and hence $p^k-1$ itself, a contradiction. I'm doubtful about positive (prime) characteristic infinite fields.

Kan't
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    If $G$ is a finite subgroup of $F^*$, then its element are all roots of $X^{\lvert G\rvert}-1$, therefore they are elements of $\overline{\Bbb F}p$, therefore they are elements of some $\Bbb F{p^u}$. – Sassatelli Giulio Dec 26 '25 at 20:32

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If $F$ is a field of characteristic $p > 0$, in $F[X]$ we have \begin{equation} X^p - 1 = (X - 1)^p. \end{equation} Thus for $\alpha \in F^{\ast}$, $\alpha^p = 1$ implies $\alpha = 1$, hence $F^{\ast}$ contains no elements of order $p$, and no finite subgroup can have order divisible by $p$.

Dermot Craddock
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