Why is $(\mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z})/(a_1+a_2+a_3=0)\cong \mathbb{Z}$?

Question

Why is $(\mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z})/(a_1+a_2+a_3=0)\cong \mathbb{Z}$?

That is: What would be the isomorphism to see this? And, in general, is there a way to find an isomorphism $G/N\cong H$? The isomorphm theorems don't seem to be applicable?

Answer 1

First define a group homomorphism $\varphi \colon \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$ by $\varphi(a_1,a_2,a_3) = a_1 + a_2 + a_3$. Note that this is surjective and $\text{ker} (\varphi) = \{ (a_1,a_2,a_3) \in \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z} \mid a_1 + a_2 + a_3 =0 \}$. Now you can conclude using the first isomorphism theorem, which gives you an explicit isomorphism.

This is the usual strategy to prove that $G/N \cong H$: find a surjective group homomorphism $G \to H$ with kernel $N$.