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Describe explicitly $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) := \{\varphi:\mathbb{Z}_n \rightarrow \mathbb{Z}_m \mid \mathbb{Z}\text{-linear homomorphism}\}$

There is an answer to this question that says $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) \cong \mathbb{Z}_{(n,m)}$ where $(n,m)=\gcd(n,m)$. But I am having some trouble seeing this.

For the sake of an argument, suppose the above answer is true, that $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) \cong \mathbb{Z}_{(n,m)}$. Suppose $n=4$ and $m=8$. Then $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_4, \mathbb{Z}_8) \cong \mathbb{Z}_4$.

Now consider each of the following maps in $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_4, \mathbb{Z}_8)$:

$\varphi_1(1)=1$

$\varphi_2(1)=2$

$\varphi_3(1)=3$

$\varphi_4(1)=4$

$\varphi_5(1)=5$

$\varphi_6(1)=6$

$\varphi_7(1)=7$

Are each of these maps not unique? If there are two that are identical, could you please explain which two and why? Thank you!

My alternate answer is that $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) \cong \mathbb{Z}_m$, since I think each $\varphi \in \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m)$ is uniquely determined by where it maps $\varphi(1) \in \mathbb{Z}_m$.

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  • $\begingroup$ Note proper use of \mid, \gcd, \text, and \operatorname. $\endgroup$ – Michael Hardy Oct 2 '14 at 16:38
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Hint:

For all $k \in \Bbb Z$, let us put: $\overline{k}=k+n\mathbb Z$ and $\mathop k^\bullet=k +m\mathbb Z$

Since $\varphi$ is a morphism and $n.\overline{1} =\overline{0}$ we have necessarely : $n.\varphi(\overline 1)= \mathop0^{\bullet}$

That means that the order $\varphi(\overline 1)$ must divide $n$.

We know yet that the ordre of $\varphi(\overline 1)$ divides $m$, that means that the ordre of $\varphi(\overline 1)$ divides $\gcd(n,m)$

In your example above here is the elements order table in $\Bbb Z_8$: $$\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text{element }&0&1&2&3&4&5&6&7\\ \hline \text{order} &1&8&4&8&2&8&4&8 \\ \hline\end{array}$$

The possible images of $\overline 1$ are : $\mathop 0^{\bullet},\mathop 2^{\bullet},\mathop 4^{\bullet},\mathop 6^{\bullet}, $ elements wich order divides $4$

Details:

Let $\varphi$ a morphisme from ${\mathbb Z}_n$ to ${\mathbb Z}_m$ and $G$ the image of $\varphi$. We have $$G=\langle \varphi(\overline 1) \rangle$$ and it's order is $\delta$ a divisor of $(m,n)$. It's known that there is a unique subgroup $G$ of $\mathbb Z_m$ having $\delta$ as order so this group is $G$. It's known too that $G$ has $\phi(\delta)$ generators (where $\phi$ is the Euler indicator) . All theses generators have $\delta$ as order and are differents so they determine $\phi(\delta)$ morphisms. We conclude that the number of all morphisms is : $$\sum_{\delta|(m,n)} \phi(\delta)=(m,n)$$

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    $\begingroup$ I need another hint. I am having trouble showing that $Hom_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) \cong \mathbb{Z}_{(n,m)}$. $\endgroup$ – Sam Oct 2 '14 at 22:15
  • $\begingroup$ I just added details above $\endgroup$ – Mohamed Oct 2 '14 at 22:36
  • $\begingroup$ If $G$ has order $\delta$, then what are we taking the sum over? $\endgroup$ – Sam Oct 2 '14 at 23:07
  • $\begingroup$ for each divisor $\delta$ of $(m,n)$ we compute the number of possible $\overline 1$ images by $\varphi$ having $\delta$ as order. Their number is $\phi(\delta)$. When $\delta$ browses all possible values we get the sum. $\endgroup$ – Mohamed Oct 2 '14 at 23:12
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    $\begingroup$ Hi, back again. I understand everything you have said. Now how does one identify a morphism $\phi \in Hom_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m)$ with an element of $\mathbb{Z}_{(n,m)}$? $\endgroup$ – Sam Jan 27 '15 at 19:58
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The problem is not uniqueness, but existence. Think about your $\phi_1$: you have $\phi_1(i)=i$. But in $\mathbb{Z}_4$, $4=0$ while this doesn't hold in $\mathbb{Z}_8$. So where can $0$ go under $\phi_1$?

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  • $\begingroup$ Doesn't it map to zero? $\endgroup$ – Sam Oct 2 '14 at 16:36
  • $\begingroup$ Yes-but by the homomorphism property, it should map to $4$ as well! $\endgroup$ – Kevin Carlson Oct 2 '14 at 17:37
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it seems that the problem is : $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}_n, \mathbb{Z}_m) \cong \mathbb{Z}_{(n,m)}$. this is a proof:

the seq. $Z\to Z\to Z_m \to 0$ is exact, when the 1st homomorphism, $f$ is defined as $f(a)=ma$ and the 2nd, $g$ as $g(x)=\bar x$.
so we have the exact seq.: $$0\to Hom_Z(Z_m,Z_n)\to Hom_Z(Z,Z_n)\to Hom_Z(Z,Z_n)$$ here the 1st homomorphism, is $hom(g,1)$ (which is injective map), and the 2nd is $hom(f,1)$.

so $ Hom_Z(Z_m,Z_n)/ker (hom(g,1)) \equiv ker(hom(f,1))$.
now can you prove that $ker(hom(f,1))\equiv Z/\left ( n,m \right )Z$ ?

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