Number of group homomorphisms from Z

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The number of group homomorphisms from Zn to Zm is equal to gcd(n, m). Any homomorphism can be expressed as f([x]) = [kx], where k is an integer that must satisfy specific divisibility conditions. The key condition is that m must divide kn, which leads to the conclusion that nontrivial homomorphisms exist only when gcd(n, m) > 1. The trivial homomorphism, where k=0, always exists, while other values of k yield nontrivial maps depending on the gcd. This establishes a clear relationship between the number of homomorphisms and the greatest common divisor of n and m.
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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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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