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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:

$\mathbb{Z}_{n} \otimes_{\mathbb{Z}} \mathbb{Z}_{m} = \mathbb{Z}_{(n,m)}$

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In general, it's not true for abelian groups $G, H$ that $\text{Hom}_\mathbb{Z}(G,H) \simeq G \otimes_\mathbb{Z} H$. So any proof of the proposition you state could not use only the properties of $\text{Hom}$ and of the tensor product; it would also have to rely on the properties of cyclic groups. That being said, if there is a proof of the proposition using only those properties, it is probably not a very "natural" proof. – Bruno Joyal Jul 18 '11 at 19:12
@Bruno: Furthermore, this is not even true for cyclic groups in general. It fails for $G=\mathbb{Z}_n$ and $H=\mathbb{Z}$, so the proof would need finiteness as well. – Noah Stein Jul 18 '11 at 21:48
@Bruno Joyal: thanks! just being curious here, usually one finds unexpected connections in mathematics. – user6495 Jul 19 '11 at 1:44
Very true. Good observation anyways! – Bruno Joyal Jul 19 '11 at 15:24
You can do the opposite - knowing $\textrm{Hom}_{\mathbb{Z}}(\mathbb{Z}_{n},\mathbb{Z}_{m})=\mathbb{Z}_{(n,m)}$ for all m, n you can show $\mathbb{Z}_{n} \otimes_{\mathbb{Z}} \mathbb{Z}_{m} = \mathbb{Z}_{(n,m)}$ with a bunch of abstract nonsense. I can write the argument if you are interested. – Max Jul 21 '11 at 9:47
up vote 3 down vote accepted

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.

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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.

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