The set $(\mathbb{Z}/3\mathbb{Z}) \times (\mathbb{Z}/4\mathbb{Z})$ is a ring with the component-wise addition and multiplication, the zero element $([0], [0])$ and the one element $([1], [1])$. I have to show that there is exactly one ring homomorphism $(\mathbb{Z}/3\mathbb{Z}) \times (\mathbb{Z}/4\mathbb{Z}) \to \mathbb{Z}/12\mathbb{Z}$. Then I have to show that this ring homomorphism is also an isomorphism.
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See the Chinese Remainder theorem.
$f(x,y)=-3x+4y$ is the rule for the isomorphism.
Now check that it's a surjective homomorphism (with necessarily trivial kernel).
Thus, $(1,1)$ has order $12$, and generates $\Bbb Z_3\times\Bbb Z_4$, which is cyclic.
This means that any ring homomorphism is determined by the image of $(1,1)$.
But the image has to be$1$ (assuming you require this).