Order of Homomorphisms and Finite Elements

Locoism
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Homework Statement


Let ψ: G→H be a homomorphism and let g ε G have finite order.
a) Show that the order of ψ(g) divides the order of g

The Attempt at a Solution


I'm really lost here, but I'm guessing we can use the fact |ψ(g)| = {e,g...,g|g|-1}
and ψ(g|g|-1) = ψ(g)ψ(g)ψ(g)ψ(g)ψ(g)... (|g|-1 times)
I still have no idea where to start.
 
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First show that

\psi(g)^{|g|}=e

Then use the general fact that if a^n=e, then |a| divides n.
 
Ah ok but I would use eH?
Also, if it is an isomorphism, could I show |ψ(g)| = |g|?
 
Locoism said:
Ah ok but I would use eH?
Also, if it is an isomorphism, could I show |ψ(g)| = |g|?

Yes to both.
 
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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