Prove ${\rm Hom}_\mathbb{Z}(\mathbb{Z}_n,\mathbb{Z}_m)\cong \mathbb{Z}_{m,n}$ Prove ${\rm Hom}_\mathbb{Z}(\mathbb{Z}_n,\mathbb{Z}_m)\cong \mathbb{Z}_{m,n}$
I think we can definine a map
$$\phi: \mathbb{Z}_n \to \mathbb{Z}_m,$$
where $\phi \in {\rm Hom}_\mathbb{Z}(\mathbb{Z}_n,\mathbb{Z}_m)$
Then $\pi :{\rm Hom}_\mathbb{Z}(\mathbb{Z}_n,\mathbb{Z}_m) \to \mathbb{Z}_{m,n}$
$\phi \mapsto t$ which is the lowest common denominator of $m,n$.
Then we show this is an bijective and a ring homomorphism.
First, I'm still working through this notation and I'm not sure if things above make sense. Second defining my map $\pi$ is weird because I'm not sure how we express an element in $\mathbb{Z}_{m,n}$
 A: One can show more generally that for a one-dimensional principal ideal domain $R$ and any elements $a$ and $b$ of $R,$ we have that $$\operatorname{Hom}_R\!{\left(\frac R {aR}, \frac R {bR}\right)} \cong \frac R {\gcd(a, b) R} \cong \frac R {aR} \otimes_R \frac R {bR}.$$ Convince yourself first that $\operatorname{Hom}_R(R/I, M) \cong (0 :_M I) = \{x \in I \,|\, x \cdot m = 0\}$ for any commutative unital ring $R,$ any ideal $I$ of $R,$ and any $R$-module $M.$ Considering that $R / I$ is cyclic, a well-defined $R$-linear map $\varphi : R / I \to M$ is uniquely determined by $m = \varphi(1_R + I).$ Consequently, for any element $x$ of $I,$ it follows that $$x \cdot m = x \cdot \varphi(1_R + I) = \varphi(x + I) = \varphi(0_R + I) = 0.$$ Can you construct an isomorphism $\psi : \operatorname{Hom}_R(R/I, M) \to (0 :_M I)$? Once you have established this, apply it to the case that $I = aR$ and $M = R / bR$ to show that $\operatorname{Hom}_R(R/aR, R/bR) \cong (0 + bR :_{R/bR} a),$ where $$(0 :_{R/bR} a) = \{r + bR \,|\, ra + bR = 0 + bR\} = \{r + bR \,|\, b \text{ divides } ra\} = (b :_R a)/bR.$$ Last, you must prove that $(b :_R a)/bR \cong R / dR,$ where we denote $d = \gcd(a, b).$ Consider the map $$\gamma : R \ni r \mapsto \frac{\operatorname{lcm}(a, b)} a r + bR \in \frac{(b :_R a)}{bR}.$$ Of course, we must convince ourselves that $\gamma$ is well-defined and surjective with $\ker \gamma = dR.$ But this can be achieved by manipulating the identity $ab = \gcd(a, b) \operatorname{lcm}(a, b).$
Ultimately, the question you have asked is a special case of this with $R = \mathbb Z.$
