
#1
Feb1711, 08:55 PM

P: 3

Denote Z_n=(0.1.2....n1)
Then could I generalize the number of homomorphism H:Z_n > Z_m as gcd(n, m)=#(H:Z_n > Z_m) ? (Don't consider the case H:Z > Z) For example #(H: Z_4 > Z_2)=2 #(H: Z_12 > Z_5)= 1 (obviously the trivial one) 



#2
Feb1711, 11:19 PM

P: 234

Yes, and in fact [tex]Hom_\mathbb{Z}(Z_n,Z_m) \cong Z_{(n,m)}[/tex], that is, the homomorphisms from Z_n to Z_m form a cyclic group of order gcd(n,m).



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