Re: mth powers in GF(p^n)

In article <1127966864.451260.100020@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"Wolff" <daniel.wolff@xxxxxxxx> writes:
>Hello, any hints on showing that (p^n-1,m)=1 implies each member of
>GF(p^n) is an mth power?

Well, its true for 0, since 0^m = 0. So it remains to prove it for
nonzero elements of GF(p^n). The nonzero elements form an abelian group of
order p^n-1, so what you are trying to prove is best deduced from:

Lemma: If G is a finite abelian group of order k, and (k,m) = 1, then
every element of G is an m-th power.

Proof. Let g in G, o=Order(g). The o divides Order(G) = k, so (o,m) = 1,
so there exist integers a,b with ao + bm = 1. So g = g^(ao + bm) = ...

Derek Holt.
.