I know that if $ G\_1, G\_2 $ are cyclic groups then $ G\_1 \times G\_2 $ is cyclic if and only if $ |G\_1| $ and $ |G\_2| $ are coprimes. But I have to reponds a similar question in the context of category theory: 

show that $ \mathbb{Z}\_n $, $ \mathbb{Z}\_m $ have a product in the category of cyclic groups if and only if $ n, m $ are coprimes.

For the implication $ \leftarrow \ $ I consider that $ n,m $ coprimes $ \rightarrow \mathbb{Z}\_n \times \mathbb{Z}\_m $ is cyclic, so this group is an element of the cateogory. Now we can take the proyection 
$$
p\_1 : \mathbb{Z}\_n \times \mathbb{Z}\_m \rightarrow \mathbb{Z}\_n 
,\ \ \ \
p\_1(x,y) = x
$$ 
and 
$$ 
p\_2 : \mathbb{Z}\_n \times \mathbb{Z}\_m \rightarrow \mathbb{Z}\_m
, \ \ \ \
p\_2(x,y) = y 
$$
that are homomorphism. Easily I can show that $ \mathbb{Z}\_n \times \mathbb{Z}\_m $ and $ p\_1, p\_2 $ form a product of $ \mathbb{Z}\_n $ and $ \mathbb{Z}\_m $ in the category of cyclic groups.

But I am stuck in the proof of the implication $ \rightarrow $.
['https://math.stackexchange.com/questions/3144190', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/114670/']
This should also follow from a naive count of the number of homomorphisms between cyclic groups: specifically, $$\lvert \operatorname{Hom}(\mathbb{Z}\_a, \mathbb{Z}\_b) \rvert = \gcd(a,b).$$

Suppose $\mathbb{Z}\_p$ is the product of $\mathbb{Z}\_m$ and $\mathbb{Z}\_n$ in the category of cyclic groups. Then it satisfies universal property $$\operatorname{Hom}(\mathbb{Z}\_t, \mathbb{Z}\_p) \cong \operatorname{Hom}(\mathbb{Z}\_t, \mathbb{Z}\_m) \times \operatorname{Hom}(\mathbb{Z}\_t, \mathbb{Z}\_n)$$ for all $t$. Counting both sides, we obtain the relation $$\gcd(t,p) = \gcd(t,m) \cdot \gcd(t,n)$$ for all $t$. In particular, if $t = m$, we get $$\gcd(m,p) = m \cdot \gcd(m,n).$$ The left hand side is at most $m$, but in case $m, n$ are not coprime, $\gcd(m,n) > 1$, and the right hand side is strictly bigger than $m$. This would give a contradiction.
I think you can use the fact that every finite abelian group can be written as $(\mathbb{Z}\_{n\_1})^{\oplus m\_1}\oplus\cdots\oplus (\mathbb{Z}\_{n\_i})^{\oplus m\_i}$. So suppose now that $A$ is a categorical product of $\mathbb{Z}\_n$ and $\mathbb{Z}\_m$ in the category of cyclic groups, and let us show that it is also a categorical product in the category of finite abelian groups :

Let $X$ be a finite abelian group, together with two maps $X \to \mathbb{Z}\_m$ and $X\to \mathbb{Z}\_n$. Writing $X = (\mathbb{Z}\_{n\_1})^{\oplus m\_1}\oplus\cdots\oplus (\mathbb{Z}\_{n\_i})^{\oplus m\_i}$, and precomposing by the various inclusions, we get a families of maps $f\_{n\_k}^j : \mathbb{Z}\_{n\_k}\to\mathbb{Z}\_n$ and $g\_{n\_k}^j : \mathbb{Z}\_{n\_k}\to\mathbb{Z}\_m$, for $j\leq m\_k$. We can now use the universal property of the product in the category of cyclic groups for each pair $(f\_{n\_k}^j,g\_{n\_k}^j)$ to get maps $h\_{n\_k}^j : \mathbb{Z}\_{n\_k}\to A$. Now let us this this family of maps for the universal property of the coproduct in the category of finite abelian groups, to get a map $h : X\to A$. You can prove that this map commutes with the initial $f,g$ modulo the projection (this is a trivial diagram, if you have understood the above construction).

So if $A$ is a categorical product in the category of cyclic group, it is also a categorical product in the category of finite abelian groups, and thus it is given by the cartesian product of groups.

This proof presuppose that you already know that $\times$ and $\oplus$ are respectively the product and coproduct in the category of finite abelian groups, but I think these are easier results.