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A field $F$ has at most a finite number of elements of order $\leq n$ for any $n$ in integers.

How can I prove this? I thought it's related to the fact that a polynomial of degree $n$ would have at most $n$ roots over $F$. But I am not sure. Would appreciate hints.

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    An element of order $n$ will be a root of $x^n-1$. So, straight away, there can be only finitely many elements of order dividing $n$. But I don't see immediately how that tells you something about elements of orders $\leq n$ not dividng $n$.2012-02-18
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    If $x\in F$ has order $k\geq 1$ then $x^k=1$ so either $x=1$ or $1+x+\ldots+x^{k-1}=0$.2012-02-18
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    @hoyland But you could also look at $x^k - 1$ for $k < n$, and there are only finitely many of these polynomials.2012-02-18
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    @Dylan: simple + brutal= perfect!2012-02-18

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