Proving Isomorphism of Z4 / (2Z4) and Z2

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Homework Statement


Why does it make sense (when considering Z4)to form the factor group

Z4 / (2Z4) where kZn = {0, k mod n, 2k mod n, ..., nk mod n}?

I believe that this above factor group is isomorphic to Z2, but how can I prove this?
 
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The quotient group exists because 2Z4 is a normal subgroup of Z4 (since Z4 is abelian, all its subgroups are normal). To show that it is isomorphic to Z2, list the elements of Z4/2Z4 (which are cosets), and write down the isomorphism. Then, prove what you have is indeed an isomorphism, i.e. that it has the properties of an isomorphism.
 
How many elements are in Z4/2Z4? How many groups have that many elements?
 
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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