Let be a finite field of characteristic . Prove that for some .
Suppose has a nonzero element of additive order . That is, is minimal such that . Note that , since annihilates every element of (being the characteristic of ). If , then by the division algorithm in we have for some and with . If , then , violating the minimalness of . If , then . Since is prime, either (so ) or (so ), which both give a contradiction. Thus .
So every element of has additive order , and thus by Cauchy’s theorem is a power of .