1

I am pretty sure that it has been already discussed many times, but I couldn't find.

Well, the problem is to show that $\mathrm{Hom}(\mathbb Z/n\mathbb Z, \mathbb Z/m\mathbb Z)\simeq\mathbb Z/(n,m)\mathbb Z$.

I have proven that $Hom(Z/nZ, Z/mZ)$ is precisely $Ann(nZ) = \{x \in Z/mZ: nx = 0\}. $ Thus my problem reduces to proving $Ann(nZ) = Z/(n,m)Z$, but I haven't gotten success so far.

user26857
  • 53,190
Invincible
  • 2,778
  • 1
    When you say "module homomorphisms", do you mean modules over $\Bbb Z$, or something else? – Arthur Dec 26 '16 at 18:49

2 Answers2

2

We have $$ \operatorname{Hom}_{\mathbb{Z}}\left ( \mathbb{Z}/n\mathbb{Z},\mathbb{Z}/m\mathbb{Z} \right )\cong \mathbb{Z}/n\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}/m\mathbb{Z}\cong \mathbb{Z}/\left ( n,m \right )\mathbb{Z}, $$ see here, and here.

Community
  • 1
Dietrich Burde
  • 140,055
1

Let $d = (n,m)$. Then $m/d$ is relatively prime to $n/d$. Thus, $m$ divides $nx$ if and only if $m/d$ divides $x$. Hence $x = km/d$ for some integer $k$. It follows that $\operatorname{Ann}(n\Bbb Z)$ is the cyclic group generated by $m/d$. The order of $m/d$ in $\Bbb Z/m\Bbb Z$ is $d$.

Community
  • 1
kobe
  • 43,217