Normal Subgroups: Why Every Kernel is a Homomorphism

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Homework Statement


Expain why every normal subgroup is the kernel of some homomorphism.


The Attempt at a Solution


Every kernel is a normal subgroup but the reverse I can't show rigorously. It seems possible how to show?
 
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What's the obvious map from G to G/N? What's its kernel?
 
I would have thought that would be easy- it's the direction emphasised in Algebra texts! Of course, this says "some" homomorphism- you have to pick the homomorphism carefully.

If H is a normal subgroup G, then we can define the "quotient group", G/H. There is an obvious homomorphism from G to H. What is the kernel of that homomorphism?

Darn, I had to stop in the middle to take a telephone call and morphism got in in front of me!
 
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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