Let F be a finite field.

How do I prove that the order of F is always of order pn where p is prime?

**Answer**

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 q≠p be any other prime dividing |F|. Then by Cauchy’s Theorem, there is an element x∈F whose order in (F,+) is q.

Then q⋅x=0. But we also have p⋅x=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⋅(p⋅x)+b⋅(q⋅x)=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 p∈N since, by the pigeonhole principle, there must exist distinct n1<n2 both in the set {1,2,…,|F|+1} such that 1+1+⋯+1⏟n1=1+1+⋯+1⏟n2, so that 1+1+⋯+1⏟n2−n1=0. In fact, this argument also implies p≤n.

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