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Let $F$ be a field, and consider the group $G=F^\times$ with multiplication ($F^\times=F-\{0\}$). How to show that for any $t\in\mathbb{Z}_{>0}$, $G$ has at most $t$ elements of order $t$?

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up vote 6 down vote accepted

Let $a \in F^{\times}$ be an element of order $t$. Then $a^t=1$. But the polynomial $x^t-1$ has at most $t$ roots over $F$.

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So here $F$ is a finite field. We know that the group of units in a finite field is cyclic, and thus if $|F|=q=p^n$ then the group of units is a cyclic group of order $q-1$.

Also note that by induction we can show that over any field a polynomial of degree $d$ has at most $d$ roots.

Using both of these, you should try to answer your question.

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The claim is true for any field, not necessarily finite. – DonAntonio Dec 3 '12 at 2:38

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