# Finding a Quotient of $\mathbb{Z}^2$

abstract-algebragroup-theory

Let $$r, p,q \in \mathbb{Z}$$ such that $$\gcd(p,q)=1$$ and $$r\geq 1$$. What is the quotient $$\mathbb{Z}^2/\mathbb{Z}(rp,-rq)$$? Is it isomorphic to $$\mathbb{Z}\times \mathbb{Z}_r$$?

I tried to define a map
$$\mathbb{Z}\times \mathbb{Z}_r \to \mathbb{Z}^2/\mathbb{Z}(rp,-rq)$$ by $$(1,0) \mapsto \overline{(1,0)}$$ and $$(0,\overline{1}) \mapsto \overline{(p,-q)}$$. This map is injective. But I am not sure whether $$\{(1,0),(p,-q)\}$$ generates $$\mathbb{Z}^2$$? If it generates then the above map is surjective also.

Instead of $$\{(1, 0), (p, -q)\}$$, you need to make use of Bezout's identity.
Since $$p, q$$ are prime to each other, there exist $$u, v \in \Bbb Z$$ such that $$pu + qv = 1$$. This can be restated as: the matrix $$A = \begin{pmatrix}p & v\\-q & u\end{pmatrix}$$ is in $$\operatorname{GL}_2(\Bbb Z)$$.
This allows us to view $$A$$ as an isomorphism $$f$$ from $$\Bbb Z^2$$ to itself, namely we define $$f(v)=Av$$ for any (column) vector $$v \in \Bbb Z^2$$. The inverse homomorphism is given by $$v \mapsto A^{-1}v$$.
Under this isomorphism, the image of the vector $$(r, 0)$$ is just $$(rp, -rq)$$. Therefore we have $$\Bbb Z^2 / \Bbb Z(rp, -rq) \simeq \Bbb Z^2 / \Bbb Z(r, 0)$$, and it should be quite clear that the latter is isomorphic to $$\Bbb Z \times \Bbb Z/r\Bbb Z$$.