$\mathbb{Z^2}/\langle(x, y)\rangle$ as a direct sum of cyclic group

Question

For each $(x,y)\in \mathbb{Z}^2$, let $\langle(x, y)\rangle$denote the subgroup of $\mathbb{Z}^2$ generated by $(x,y)$. Express $\mathbb{Z}^2/\langle(x, y)\rangle$ as a direct sum of cyclic group.

Note $\mathbb{Z}^2/\langle(x, y)\rangle$ has at most two generators, so $\mathbb{Z}^2/\langle(x, y)\rangle\cong \mathbb{Z}^2$, $\mathbb{Z}\oplus \mathbb{Z}_n$, $\mathbb{Z}_m\oplus \mathbb{Z}_n$, $ \mathbb{Z}_n$ or $\mathbb{Z}$. I think it should be the second case. But I don't know how to figure it out. Any suggestion? Thanks

Well, to get you started, it must have rank $1$. So, it looks like $\mathbb{Z}\times\text{torsion}$. Unless, of course, $(x,y)=(0,0)$ :) — Alex Youcis, Aug 13 '13 at 23:47
Adding to Thomas Andrews hint, try to find an automorphism of $\mathbb{Z}^2$ that makes your situation look a little nicer. — Alex Youcis, Aug 14 '13 at 00:22

score 1 · Answer 1 · answered Aug 14 '13 at 00:00

The product of cyclic groups depends on the values of $x$ and $y$ (from the way you pose the question, it might appear you expect the answer to be independent of $x$ and $y$).

In general, when you want to understand a quotient one good line of attack is to use the isomorphism theorem (aka the First Iso Theorem). So, try to find an epimorphism (that is, a surjective homomorphism) $\psi : \mathbb Z^2 \to G$ such that the kernel of $\psi$ is precisely $\langle (x,y)\rangle$. Then you'd know that the quotient is isomorphic go $G$. Now, depending on the values of $x$ and $y$ choose the group $G$ and the homomorphism $\psi$.

Try to first fix some values for $x$ and $y$ to see what makes sense. Then do the general case.

$\mathbb{Z^2}/\langle(x, y)\rangle$ as a direct sum of cyclic group

1 Answers1