# Product in category of cyclic groups.

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 $$\rightarrow$$.

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.