Abstract Algebra Homomorphism Proof

The_Iceflash
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Homework Statement



Let G and H be two groups. If f: G \rightarrow H is a homomorphism, a \in G and b = f(a). If ord(a) = n, ord(b) = m, then n is a multiple of m. (Let e_{1} be the identity of G and e_{2} be the identity of H)

I have to prove that n is a multiple of m.

Homework Equations


N/A

The Attempt at a Solution



I know an = ?
and that bn= ?
then I conclude from what I get that n and m are multiples.

Can I have some help with this proof? Thanks.
 
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a^n = e (in G)
b^m = e (in H)...
 
A homomorphism has what property?
 
Consider these ideas:

(1) Why is n >= m ?
(2) With n >= m established, you can write n = q m + r, where 0 <= r < m and q > 0.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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