# Number of group homomorphisms from Ｚ

1. Feb 23, 2008

### hmw

1. The problem statement, all variables and given/known data
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.

Last edited: Feb 23, 2008
2. Feb 23, 2008

### StatusX

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.