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 in Word Problem Answers by anonymous

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:

To avoid this verification in future, please log in or register.

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
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
78,524 questions
82,362 answers
1,902 comments
63,398 users