Quotient group of $\mathbb{Z} \times \mathbb{Z}$ I'm working on some practice problems and would like to get a few solutions checked (more coming!). Let $H$ be the subgroup of $\mathbb{Z} \times \mathbb{Z}$ generated by the elements $(2, 2)$ and $(3, 4)$. First give a minimal set of elements which generate $(\mathbb{Z} \times \mathbb{Z}) ~ / ~ H$ (as their images by the natural homomorphism $g \mapsto g + H$) and list all the elements in this quotient group, and then describe this quotient group.
First we simplify the presentation of $H$ a bit by simplifying the generators: if $H$ is generated by $(2, 2)$ and $(3, 4)$, then it is generated by $(2, 2)$ and $(3, 4) - (2, 2) = (1, 2)$, and so it is generated by $(2, 2) - (1, 2) = (1, 0)$ and $(1, 2)$, and so it is generated by $(1, 0)$ and $(1, 2) - 2 (1, 0) = (0, 2)$ (this could be formalized into a theorem since it's essentially the Euclidean algorithm). So geometrically $H$ spans the set of lattice points of the form $(m, 2n)$ where $m$ and $n$ are integers. Hence it partitions $(\mathbb{Z} \times \mathbb{Z})$ into exactly two cosets $H$ and $(0, 1) + H$. Since $(0, 2) + H = H$, the quotient group is generated by a single element $(0, 1) + H$, and contains two elements $H$ and $(0, 1) + H$. As such, it is isomorphic to $C_2$.
Is this a (correct and) rigorous answer and are there things that could be improved? One thing I am a bit concerned about is that the problem asks me to find a set of generators for the quotient group before asking for its elements and its description, whereas I've done the opposite - was there a better approach to finding its generators?
 A: Applying the already mentioned method of Smith Normal form: the generators matrix is
$$A=\begin{pmatrix}2&2\\3&4\end{pmatrix}\stackrel{C_2-C_1}\longrightarrow\begin{pmatrix}2&0\\3&1\end{pmatrix}\stackrel{C_1-3C_2}\longrightarrow\begin{pmatrix}2&0\\0&1\end{pmatrix}\stackrel{\begin{align*}C_1\leftrightarrow C_2\\R_1\leftrightarrow R_2\end{align*}}\longrightarrow\begin{pmatrix}1&0\\0&2\end{pmatrix}$$
Thus, $\;H\cong \Bbb Z\oplus 2\Bbb Z\;$ , and thus $\;\left(\Bbb Z\otimes\Bbb Z\right)/K\cong\Bbb Z/2\Bbb Z=:\Bbb Z_2\;$
Note the above gave precisely the same generators you got. The advantage the method of Smith Forms offers is that you can methodically apply it doing very basic row/column operations on some given relations matrix, very much like elementary operations on matrices what we know from linear algebra.
A: Let $\alpha=(2,2)$ and $\beta=(3,4)$. In general any member of $\mathbb{Z} \times \mathbb{Z}/H$ will be of the form $(a,b)+H$. We will distinguish between two cases $b$ is even and $b$ is odd.
$$
(a,b)=
\begin{cases}
 &  2a(\alpha -\beta) + \frac{b}{2}(2\beta-3\alpha), \text{ if } b \text{ is even}\\
 & 2a(\alpha -\beta) + \frac{b-1}{2}(2\beta-3\alpha)+(0,1), \text{ if } b \text{ is odd}
\end{cases}
$$
Since $\alpha, \beta \in H$ and $H \leq \mathbb{Z} \times \mathbb{Z}$, therefore
$$
(a,b)+H=
\begin{cases}
 &  H, \text{ if } b \text{ is even}\\
 & (0,1)+H, \text{ if } b \text{ is odd}
\end{cases}
$$
This gives us the minimal set of generators as well as a complete description of the quotient group. 
