let be a finite field of odd order and we know that is a cyclic group. there exist unique non-empty sets which satisfy the following conditions:
Proof: first note that 1) & 3) implies that and thus by 4) we have for all thus by 2), is a subgroup of now fix see that and
thus which proves the uniqueness of (and therefore the uniqueness of ), because we know that a cyclic group cannot have two subgroups of the same order. Q.E.D.
Remark: the above proof also gives the set (and therefore ): suppose then and