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?
asked Sep 25, 2012

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2 Answers

Could I see how your teacher proved the original problem to see the clue?
answered Feb 14, 2013 by anonymous
if f(x),g(x)  ϵ f (x) then prove that (f(x)g(x))' = f'(x)g(x)+f(x)g'(x)
answered Jun 17, 2014 by Braham Dev Yadav
if f(x),g(x) ϵ f (x) then prove that (f(x)g(x))' = f'(x)g(x)+f(x)g'(x)

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