Number of group homomorphisms from Z

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SUMMARY

The number of group homomorphisms from the cyclic group Zn to Zm is equal to gcd(n, m). Any homomorphism can be expressed in the form f([x]) = [kx], where k is an integer. The well-defined nature of these maps requires that m divides kn, leading to the conclusion that nontrivial homomorphisms exist only when gcd(n, m) > 1. For instance, with n = 4 and m = 6, k can take the value of 3, demonstrating the existence of nontrivial homomorphisms.

PREREQUISITES
  • Understanding of group theory concepts, specifically cyclic groups.
  • Familiarity with the properties of homomorphisms in algebra.
  • Knowledge of the greatest common divisor (gcd) and its implications in number theory.
  • Basic modular arithmetic and its applications in group mappings.
NEXT STEPS
  • Study the properties of cyclic groups and their homomorphisms in detail.
  • Learn about the implications of gcd in algebraic structures.
  • Explore examples of homomorphisms between different cyclic groups.
  • Investigate the relationship between homomorphisms and modular arithmetic.
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Mathematics students, particularly those studying abstract algebra, group theory enthusiasts, and educators seeking to deepen their understanding of group homomorphisms and their properties.

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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.
 

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