How many homomorphisms are? How many Homomorphisms are be between $\mathbb{Z} \times (\mathbb{Z} / 6\mathbb{Z}) $  to  $ (\mathbb{Z} /2\mathbb{Z} ) \times (\mathbb{Z} /60\mathbb{Z}) $
I know the answer is 1440 , But have no idea how to show it.
 A: You can check that the following equality holds:
$\hom(\mathbb Z\times \mathbb Z/6\mathbb Z,\mathbb Z/2\mathbb Z\times \mathbb Z/60\mathbb Z)\simeq \hom(\mathbb Z,\mathbb Z/2\mathbb Z)\times\hom(\mathbb Z,\mathbb Z/60\mathbb Z)\times\hom(\mathbb Z/6\mathbb Z,\mathbb Z/2\mathbb Z)\times\hom(\mathbb Z/6\mathbb Z,\mathbb Z/60\mathbb Z)$
(or if you want, this follows from the universal properties of product and coproduct, which coincide for abelian groups). Thus we can count elements of the RHS separately and then multiply them. Now, for every abelian group $G$, homomorphisms $\mathbb Z\to G$ are uniquely determined by the image of $1$, and every $g\in G$ determines an homomorphism via $1\mapsto g$. Therefore $|\hom(\mathbb Z,\mathbb Z/2\mathbb Z)|=2$ and $|\hom(\mathbb Z,\mathbb Z/60\mathbb Z)|=60$.
The group $\mathbb Z/6\mathbb Z$ is cyclic as well, so homomorphisms $\mathbb Z/6\mathbb Z\to G$ are again uniquely determined by the image of $[1]$. But now not all $g\in G$ determine homomorphisms $\mathbb Z/6\mathbb Z\to G$. When $G=\mathbb Z/2\mathbb Z$ there is not much choice for the image of $[1]$: either it's $[0]$ or $[1]$. The first choice yields the trivial homomorphism and the second corresponds to the quotient of $\mathbb Z/6\mathbb Z$ over its unique cyclic subgroup of order $3$, so it yields an homomorphism as well. Therefore $|\hom(\mathbb Z/6\mathbb Z,\mathbb Z/2\mathbb Z)|=2$. For the last term, notice that the image of $\mathbb Z/6\mathbb Z$ inside $\mathbb Z/60\mathbb Z$ must lie inside the unique subgroup of $\mathbb Z/60\mathbb Z$ of order $6$, call it $H$. Also, any homomorphism $\mathbb Z/6\mathbb Z\to H$ can be seen as a homomorphism $\mathbb Z/6\mathbb Z\to \mathbb Z/60\mathbb Z$ by composing with inclusion. Therefore counting homomorphisms $\mathbb Z/6\mathbb Z\to \mathbb Z/60\mathbb Z$ is the same as counting homomorphisms $\mathbb Z/6\mathbb Z\to H\simeq \mathbb Z/6\mathbb Z$. Now you can check that these are exactly $6$, so that after all we get $2\cdot 60\cdot 2\cdot 6=1440$ homomorphisms
