Number of group homomorphisms from Z

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Homework Statement


Show that the number of group homomorphisms from Zn to Zm is equal to gcd(n,m).

my attempt:

any hom from Zn to Zm must be f([x])=[kx] where k is a common factor of n and m. I can only get this far... any help is appreciated.
 
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As you've said, all homomorphism are of the form f([x])=[kx], with k an integer. Of course, these maps are only well-defined for certain choices of k. However, it turns out the criteria is not that k be a common factor of n and m.

Specifically, we must have:

x=y (mod n) => kx=ky (mod m)

which its pretty easy to see is equivalent to:

n|z => m|kz

This will be satisfied iff m divides kn. One obvious choice is k=0, which gives the trivial map sending all [x] to [0]. This homomorphism always exists. Another choice is k=m, but this is equivalent to the trivial map since [mx]=[0] in Z_m. In general, we only need to find all the solutions k with 0<=k<m, since solutions differing by a multiple of m are easily seen to give the same map.

Nontrivial values for k will only exist if gcd(n,m)>1. For example, if n=4 and m=6, we can take k=3. Try a few more examples, and hopefully you'll see the pattern that emerges.
 
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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