Order of finite fields is pnp^n

Let F be a finite field.
How do I prove that the order of F is always of order pn where p is prime?


Let p be the characteristic of a finite field F.Note 1 Then since 1 has order p in (F,+), we know that p divides |F|. Now let qp be any other prime dividing |F|. Then by Cauchy’s Theorem, there is an element xF whose order in (F,+) is q.

Then qx=0. But we also have px=0. Now since p and q are relatively prime, we can find integers a and b such that ap+bq=1.

Thus (ap+bq)x=x. But (ap+bq)x=a(px)+b(qx)=0, giving x=0, which is not possible since x has order at least 2 in (F,+).

So there is no prime other than p which divides |F|.

Note 1: Every finite field has a characteristic pN since, by the pigeonhole principle, there must exist distinct n1<n2 both in the set {1,2,,|F|+1} such that 1+1++1n1=1+1++1n2, so that 1+1++1n2n1=0. In fact, this argument also implies pn.

Source : Link , Question Author : Mohan , Answer Author : Maximilian Janisch

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