Finite Order of Elements in Quotient Groups

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Homework Statement


If G is a group and N is a normal subgroup of G, show that if a in G has finite order o(a), then Na in G/N has finite order m, where m divides o(a).


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The Attempt at a Solution


I have no idea where to start. The problem says to prove it by using the homomorphism of G onto G/N.
 
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Try taking a look at using Lagrange's Theorem.
 
I don't mean to dig this up, but isn't there a theorem (not sure if it lagrange) that states that the minimum order of an element to belong in a subgroup is a divisor of the group order. Couldn't one arrive at this by creating a bijection and then invoking the fact that |G:N|=|G|/|N|. <= That was something given to us in class that led to Lagrange's theorem.
 
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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