Letting F be a finite field, how would one show that the multiplicative group must be cyclic?(adsbygoogle = window.adsbygoogle || []).push({});

I know that if the order of F = n, then the multiplicative group (say, F*) has order n - 1 = m. Then g^m = 1 for all g belonging to F*.

Thanks for your time and help.

dogma

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# Multiplicative groups

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