Prove that G has exactly gcd(m,n) elements a such that a^m = e.?
Let G be a cyclic group of finite order n, and let m be a positive integer. Prove that G has exactly gcd(m,n) elements a such that a^m = e.
Given Hint: In any group G, for any g in G and m in the positive integers, g^m = e if, and only if, the order of g divides m. My teacher proved this was true for us as a hint to use for the above problem.
My trouble is I don't see how the hint helps exactly because I've played around with this problem for a few hours now and I see several ways of going about doing it, and I'm not sure which if any are right. Can anyone tell me the proper way to go about doing this problem?