multiplicative groups

dogma

Letting F be a finite field, how would one show that the multiplicative group must be cyclic?

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

matt grime

google for multiplicative group finite field cyclic, and look at the first hit from planetmath

dogma

thank you once again...google is great.

dogma