Number of $Z$ homomorphisms from $Z_{n}$ to $Z_{m}$ using tensor product It is a standard result that $\textrm{Hom}_{\mathbb{Z}}(\mathbb{Z}_{n},\mathbb{Z}_{m})=\mathbb{Z}_{(n,m)}$. Is it possible to show this using the following result:

$\mathbb{Z}_{n} \otimes_{\mathbb{Z}} \mathbb{Z}_{m} = \mathbb{Z}_{(n,m)}$
This is the answer for the "opposite direction". All tensor products and homs are over $\mathbb{Z}$. 

The basic argument is based on the adjointness of Hom and tensor ( Adjointness of Hom and Tensor). For abelian groups (aka $\mathbb{Z}$-modules) we get

$Hom(M\otimes N,Q) \cong Hom(M,Hom(N,Q))$.

This is pretty clear from the universal property of tensor product: maps from $M\otimes N$  to any group are bilinear maps from $M \times N$, and those are nothing but maps from $M$ to $Hom(N,-)$.

So $Hom(\mathbb{Z}_{n} \otimes \mathbb{Z}_{m}, \mathbb{Z}_t)=Hom(\mathbb{Z}_{n}, Hom(\mathbb{Z}_{m}, \mathbb{Z}_t))=Hom(\mathbb{Z}_{n}, \mathbb{Z}_{(m,t)})=\mathbb{Z}_{(n,m,t)}$,
which is also $Hom( \mathbb{Z}_{(n,m)}, \mathbb{Z}_t)$.

Now we want to apply Yoneda embedding (see first paragraph of the accepted answer to 
Can someone explain the Yoneda Lemma to an applied mathematician? ). 

The question is to which category. That depends on what we are willing to assume known about tensor product of two cyclic groups. If we assume it is cyclic, we apply Yoneda to the category of cyclic groups and are done. If we assume it known that it is finite abelian (which is easy to see from the construction of tensor product as a quotient of an abelian group generated by $m\otimes n$ with each such symbol of finite order), and know that every finite abelian group is a product of cyclic ones, then using the obvious fact $Hom(X, A\oplus B)=Hom(X, A)\oplus Hom(X, B)$ we apply Yoneda embedding to the category of finite abelian groups and we are done. Finally, if we only know the tensor product is finitely generated abelian and know structure theorem for finitely generated abelian groups, then we need only additional equality $Hom(\mathbb{Z}_{n} \otimes \mathbb{Z}_{m}, \mathbb{Z}) = Hom(\mathbb{Z}_{n}, Hom(\mathbb{Z}_{m}, \mathbb{Z}))=0=Hom( \mathbb{Z}_{(n,m)}, \mathbb{Z})$, to apply Yoneda to the category of finitely generated abelian groups.
I think I know a direct answer 
If you prove that R/I⊗R/J ≅ R/I+J , then u can easily reach that Zn⊗ZZm=Z(n,m) by the fact that you can write Zn as Z/nZ and Z(n,m) as Z/nZ+mZ
If you don't know how to prove R/I⊗R/J ≅ R/I+J just ask again and I'll answer... It's a little bit long and you have to use nakayama's lemma twice in it.