Prove In any cyclic group of order n there are elements of o

Prove: In any cyclic group of order n, there are elements of order k for every integer k which divides n.

Solution

G = , and a^n = e by definition. Let k be an integer which divides n, so we have n = kp, for some integer p. Then a^n = a^(kp) = (a^p)^k = e.

Now, the only thing we need to check in order to prove that the order of a^p is k is that k must be the least positive integer such that the equality is satisfied (this is what I\'m a bit unsure about, but I believe it\'s correct - I used this reasoning in another few problems).

Assume there is an integer q < k such that (a^p)^q = e

. But then we have a^(pq) = e, and pq is clearly less than kp = n, contradicting the fact that the order of a is n (i.e. that the order of G is n). Hence, a^p is an element of G of order k.