Let m n be integers with gcd m n p for some prime p Let G b

Let m; n be integers with gcd (m; n) = p for some prime p. Let G be a group and let a G be such that a^m = e = a^n.

Show that either a = e or ord (a) = p.

Solution

(a) Assume a G such that a n 6= e. Because |G| = n, we cannot have ord(a) > n, else {e, a, . . . , an} would all be distinct and this contradicts the order of G. Let ord(a) = m n. Then |hai| = m, the cyclic subgroup generated by a. By Lagrange’s theorem, |hai| | |G| = m | n. Therefore write n = dm. Then a n = (a m) d = e d = e. This completes the proof. (b) If G is cyclic, then G = hai. Therefore G = {1, a, . . . , an1}, where each power of a is different. Further, a n = 1 by the previous statement. Hence ord(a) = n. Conversely, suppose there exists a with order n. Then |hai| = |G| and hai G, so hai = G and hence G is cyclic. (c) Suppose |G| = p with p a prime. Let a 6= e. Then hai is a subgroup of G, so by Lagrange’s theorem |hai| divides p. Since p is prime, |hai| = 1 or p, and it is not 1 by