Let $G = \Bbb Z \times \Bbb Z$ with group law given by addition. Let $H$ be the subgroup generated by $(2,3)$. To which group is $G/H$ isomorphic to? 
Let $G = \Bbb Z \times \Bbb Z$ with group law given by addition. Let $H$ be the subgroup generated by $(2,3)$; so $H$ consists of all elements of the form $(2a,3a)$ for some $a \in \Bbb Z$. To which group is $G/H$ isomorphic to?

My idea was to let $\varphi:G \to \Bbb Z_2 \times \Bbb Z_3$ be the map sending $(a,b) \longmapsto (a \pmod 2, b \pmod 3)$. The kernel is now $\ker \varphi=\{(a,b) \in \Bbb Z \mid \varphi(a,b)=(a \pmod 2, b \pmod 3)=(0,0) \}$ i.e. tuples where the first coordinate is a multiple of $2$ and the second a multiple of $3$ that is $\ker \varphi = H$.
Now $$G/\ker \varphi = G/H \cong \operatorname{im} \varphi$$
and $\operatorname{im} \varphi = \{(a \pmod 2, b \pmod 3) \mid (a,b) \in \Bbb Z \times \Bbb Z\} = \Bbb Z_2 \times \Bbb Z_3$ so $$G/H \cong \Bbb Z_2 \times \Bbb Z_3$$
However there is apparently a flaw in the reasoning here, but I cannot figure out where I'm going wrong. Is $\ker \varphi = H$ an untrue statement?
 A: As we've noted, the kernel of your map is strictly larger than your $H$: your $H$ consists only of those elements in which the first entry is a multiple of $2$, and the second entry is the same multiple of $3$ (that is, of the form $(2m,3m)$ for a single $m$). However, the kernel consists of all elements whose first entry is any multiple of $2$ and second entry is any multiple of $3$, elements of the form $(2r,3s)$ with $r$ and $s$ arbitrary. So $(2,0)$ lies in the kernel, but not in $H$, for example.
Now, the group is generated by the images of $(1,0)$ and $(0,1)$. Consider the image of $(1,0)$ in $G/H$. Is this element of finite order? No: because
$$\begin{align*}
k\left( (1,0)+H\right) = (0,0)+H &\iff (k,0)+H = (0,0)+H\\
&\iff (k,0)\in H\\
&\iff \exists m\text{ such that }(k,0)=(2m,3m)\\
&\iff k=0.
\end{align*}$$
So we know for sure the quotient is not finite.
Similarly, $(0,1)$ maps to an element of infinite order.
In addition, since $(2,0)+H = (0,-3)+H$, we also know the image of $(0,1)$ has a multiple that is eventually in the image of the subgroup generated by $(1,0)$. That suggests that we might try mapping both to the same cyclic group of infinite order, in such a way that they will satisfy this relation.
So perhaps we might try the map $\mathbb{Z}\times\mathbb{Z}\to\mathbb{Z}$ given by $(a,b)\longmapsto 3a-2b$. It is clear that $H$ is contained in the kernel, since $(2,3)\mapsto 3(2)-2(3)=0$. It is also clear that the map is surjective, since $(1,1)\mapsto 3-2 = 1$. So then it's a matter of checking whether the kernel is exactly $H$, or if as you did I've inadvertedly defined a map with a larger kernel than I intended.
A: A less ad hoc way to figure out the quotient is to use the Smith Normal Form. The Smith Normal Form of an integer $n\times m$ matrix is an $n\times m$ matrix in which all off-diagonal entries are $0$, and the diagonal entries $\alpha_1,\alpha_2,\ldots,\alpha_k$ (with $k=\min(n,m)$) satisfy $\alpha_1\mid \alpha_2 \mid\cdots \mid \alpha_k$. It is obtained by doing elementary row and column operations, but where you are only allowed to add integer multiples of one row/column to another, and you may only multiply a row or column by $1$ or $-1$, and not an arbitrary nonzero scalar.
To do this, given a set of $m$ generators for a subgroup $H$ of $\mathbb{Z}^n$, write the generators as columns to get an $n\times m$ matrix. Here, our generator is $(2,3)$, so the matrix is $2\times 1$.
$$\left(\begin{array}{c}2\\3\end{array}\right).$$
Now, elementary column operations correspond to replacing the generating set of $H$ with another generating set obtained from the old one by the corresponding operations. Elementary row operations correspond to performing an invertible integral linear transformation on $\mathbb{Z}^n$ to transform the generating set of $H$ into an easier generating set for an equivalent subgroup.
First, we subtract the first row from the second row. This corresponds to the automorphism of $\mathbb{Z}^2$ that sends $(1,0)$ to $(1,-1)$, and sends $(0,1)$ to $(0,1)$. This will send the generator $(2,3)$ of $H$ to $(2,-2)+(0,3) = (2,1)$. So we get
$$\left(\begin{array}{c}2\\1\end{array}\right).$$
Then exchange the two rows, which corresponds to exchanging our generators; this is the automorphism of $\mathbb{Z}^2$ that sends $(1,0)$ to $(0,1)$ and $(0,1)$ to $(1,0)$. The generator $(2,1)$ of our subgroup is sent to $(1,2)$.
$$\left(\begin{array}{c}1\\2\end{array}\right).$$
Now subtract twice the first row form the second row. This is the automorphism of $\mathbb{Z}^2$ that sends $(1,0)$ to $(1,-2)$ and sends $(0,1)$ to $(0,1)$. It sends $(1,2)$ to $(1,0)$. So we get the matrix in Smith Normal Form,
$$\left(\begin{array}{c}1\\0\end{array}\right).$$
Now it is easy to just read off what the quotient is: what is the quotient of $\mathbb{Z}^2$ by the subgroup generated by $(1,0)$? It is $\mathbb{Z}$. So the quotient is isomorphic to $\mathbb{Z}$.
You can obtain an explicit morphism by keeping track of the operations. The elementary matrices that yield the elementary row operations I performed give
$$\left(\begin{array}{rr}
1 & 0\\
-2 & 1
\end{array}\right)\left(\begin{array}{cc}0&1\\0&1\end{array}\right)\left(\begin{array}{rr}1&0\\-1&0\end{array}\right) = \left(\begin{array}{rr}
-1 & 1\\
3 & -2\end{array}\right).$$
Thus, we want the map that sends $(1,0)$ to $(-1,3)$ and sends $(0,1)$ to $(1,-2)$; this map sends $(2,3)$ to $(-2,6)+(3,-6) = (1,0)$, and the morphism onto $\mathbb{Z}$ with that kernel is given by projection along the second component. So we want the map
$$(a,b)\to \pi_2\Bigl( a(-1,3)+b(1,-2)\Bigr) = \pi_2(-a+b,3a-2b) = 3a-2b,$$
which is the map I suggested in my first answer.
A: The matrix in Smith normal form is $$\begin{pmatrix}2\\ 3\end {pmatrix}\to\begin{pmatrix}-1\\ 3\end {pmatrix}\to\begin{pmatrix}-1\\ 0\end {pmatrix}\to\begin{pmatrix}1\\ 0\end {pmatrix}$$.
I get that the so-called invariant factor is then $1$.
Thus we get $\Bbb Z$ for the quotient.
