TITLE: $ (\mathbb{Z} \times \mathbb{Z}) / \langle (a, b ) \rangle$ is isomorphic to $ \mathbb{Z} \times \mathbb{Z} _d $, where $ d = \gcd(a,b)$
QUESTION [13 upvotes]: $  \frac{\mathbb{Z} \times  \mathbb{Z}}{\langle (a, b ) \rangle}$ is isomorphic to $ \mathbb{Z} \times  \mathbb{Z} _d $, where $ d = \gcd(a,b)$.
Here, $\langle (a, b ) \rangle$ is a group generated by $ ( a, b) \in \mathbb{Z} \times  \mathbb{Z}$ .
I tried to use the First isomorphism theorem, putting a homomorphism map $ \phi : \mathbb{Z} \times  \mathbb{Z} \rightarrow \mathbb{Z} \times  \mathbb{Z}_d $ by $ (x, y) \mapsto (bx-ay, xy) $.
Then ker $ \phi  = \langle (a, b ) \rangle $ and so $  \frac{\mathbb{Z} \times  \mathbb{Z}}{\langle(a,b)\rangle}$ is isomorphic to Im  $\phi = \mathbb{Z} \times  \mathbb{Z} _d $
Is my approach right? I'm not sure. Thanks!

REPLY [3 votes]: You are on the right track with respect to the method, but your proposed homomorphism is none.
By the extended Euklidean algorithm, there are integers $u,v\in\mathbb Z$ such that $ua+vb=d$. The homomorphism 
$$\begin{align}\phi\colon\mathbb Z\times\mathbb Z&\to\mathbb Z\times (\mathbb Z/d\mathbb Z)\\(x,y)&\mapsto(\frac bdx-\frac ady,ux+vy+d\mathbb Z)\end{align}$$
is onto because $\phi(v,-u)=(1,0+d\mathbb Z)$ and $\phi(\frac ad,\frac bd)=(0,1+d\mathbb Z)$. Note that $\phi(x,y)=(0,0+d\mathbb Z)$ implies $(x,y)=c\cdot(a,b)$ with $c\in\mathbb Q$ from the first component, i.e. $(x,y)=c'\cdot (\frac ad,\frac bd)$ with $c'\in\mathbb Z$; but then the second component is $c'+d\mathbb Z$ (because $u\frac ad+v\frac bd=1$), so that $\ker\phi\subseteq \langle(a,b)\rangle$. From $\phi(a,b)=0$ we finally obtain $\ker \phi=\langle(a,b)\rangle$ and that $\phi$ induces an isomorphism
$$ (\mathbb Z\times \mathbb Z)/\langle (a,b)\rangle\cong \mathbb Z\times (\mathbb Z/d\mathbb Z).$$