# Prove that $(\mathbb Z× \mathbb Z)/〈(0, 1)〉$ is an infinite cyclic group.

I know I'm supposed to be proving that $(\mathbb Z× \mathbb Z)/〈(0, 1)〉 \cong \mathbb Z$ by using the First Isomorphism Theorem. I'm having trouble defining a homomorphism $φ:\mathbb Z× \mathbb Z → \mathbb Z$ whose kernel is $〈(0, 1)〉$. We did a similar example in class where we proved $(\mathbb Z× \mathbb Z)/〈(1, 1)〉 \cong \mathbb Z$ and $φ:\mathbb Z× \mathbb Z → \mathbb Z$ defined by $φ((a,b))= a-b$, but I cant determine the homomorphism for this one.

• $\varphi(a,b) \mapsto a$. – Morgan Rodgers Jan 31 '17 at 21:10
• I actually had a similar suggestion, but then when I have to prove that it's a homomorphism, I get stuck. We need to prove φ((a,b)+(c,d)) = φ(a,b) + φ(c,d). So, φ((a,b)+(c,d)) = φ((a+c), (b+d)) = a+c. Then I get stuck. – user21 Jan 31 '17 at 21:14
• You also have that $\varphi((a,b))+\varphi((c,d)) = a+c$. – Morgan Rodgers Jan 31 '17 at 21:22
• @johnie4usc Maybe what you're missing is that $(a,b) + (c,d) = (a+c,b+d)$. – Viktor Vaughn Feb 1 '17 at 0:16
• Yeah, you all are both right. I was missing it and it was right in front of me. – user21 Feb 1 '17 at 15:26

Define $f:\mathbb{Z}\times \mathbb{Z}\rightarrow \mathbb{Z}:(x,y)\mapsto x$, the kernel is exactly generated by $(0,1)$ and $f$ is surjective. Thus the result indeed follows from the first isomorphism theorem.