Generalization:

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:

1) and

2)

3)

4)

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