Why does this tensor product has exactly 2 elements? I am trying to prove that $(\mathbb{Z}/10\mathbb{Z})\otimes (\mathbb{Z}/12\mathbb{Z}) \cong (\mathbb{Z}/2\mathbb{Z})$.
I understand now that every element in $(\mathbb{Z}/10\mathbb{Z})\otimes (\mathbb{Z}/12\mathbb{Z})$ can be written as $[m]_{10} \otimes [1]_{12}$. If $m$ is even, then $[m]_{10} \otimes [1]_{12} = [0]_{10} \otimes [0]_{12}$; if $m$ is odd, then $[m]_{10} \otimes [1]_{12} = [1]_{10} \otimes [1]_{12}$. Also it is clear that $[0]_{10} \otimes [0]_{12}$ is the zero element in the $\mathbb{Z}$-module $(\mathbb{Z}/10\mathbb{Z})\otimes (\mathbb{Z}/12\mathbb{Z})$ and $2 \cdot ([1]_{10} \otimes [1]_{12}) = [0]_{10} \otimes [0]_{12}$.
So now to show $(\mathbb{Z}/10\mathbb{Z})\otimes (\mathbb{Z}/12\mathbb{Z}) \cong (\mathbb{Z}/2\mathbb{Z})$ I need to show only one more thing:
$$[1]_{10} \otimes [1]_{12} \neq [0]_{10} \otimes [0]_{12}.$$
So my question is: how can I distinguish these two elements?
 A: The easiest way I can think of is to create a non-trivial bilinear map from $\mathbb Z_{10}\times \mathbb Z_{12}\to \mathbb Z_{2}$. Which one? The one we learn at the mother's teat: Multiplication itself!
Let $$\phi([a]_{10},[b]_{12})=[ab]_2$$
Easy exercise: Check that this is well-defined and bilinear and non-zero.
Remember that every bilinear map from $\mathbb Z_{10}\times \mathbb Z_{12}$ to another $\mathbb Z$ module filters through $\mathbb Z_{10}\otimes \mathbb Z_{12}$. Now if the tensor product $\mathbb Z_{10}\otimes \mathbb Z_{12}$ was zero, then how could the non-trivial bilinear map we just found above filter through it? So it must be that there is at-least ONE non-zero element in $\mathbb Z_{10}\otimes \mathbb Z_{12}$. But you showed that there are at-most two elements in $\mathbb Z_{10}\otimes \mathbb Z_{12}$. 
So there are exactly two elements in $\mathbb Z_{10}\otimes \mathbb Z_{12}$ and consequently it is $\mathbb Z_2$.
A: This is really a special case of a more general statement. Namely, 
$$\mathbb{Z}/n\mathbb{Z}\otimes \mathbb{Z}/m\mathbb{Z}\simeq \mathbb{Z}/gcd(n,m)\mathbb{Z}\;.$$
To see this, you can simply use the bilinear properties of the tensor product. More precisely, pick a pure tensor of the form $a\otimes b$. Then $n(a\otimes b)=(na)\otimes b=0\otimes b=0$, so $a\otimes b$ is $n$-torsion. Similarily, $m(a\otimes b)=(a\otimes mb)=a\otimes 0=0$ and $(a\otimes b)$ is $m$-torsion. Now the group on the left is certainly cyclic, generated by $1\otimes 1$ (you can check this by writing out a general element of the tensor product). Define a homomorphism
$$\phi:\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z}\otimes \mathbb{Z}/m\mathbb{Z}$$
by sending the generator $1$ to $1\otimes 1$. Can you show that the map is surjective and calculate its kernel using the torsion properties of the elements? Then you can use 1st iso to get the result.
