 Since the order of "F""q" is "q" − 1, its primitive roots have order "q" − 1, which implies that "x"("q" − 1)/"p""n" or "x""m" and all its powers have an order which is a power of "p".So, "P" is a subgroup where all its elements have orders which are powers of "p".There are "pn" choices for both "a" and "b", making |"P"| = "p"2"n".