Compute this factor group: $\mathbb Z_4\times\mathbb Z_6/\langle (0,2) \rangle$ 
So I'm going through example 15.10 in Fraleigh, which is computing $G/H$, where $G = \mathbb Z_4\times\mathbb Z_6$ and $H = \langle (0,2) \rangle$.

We have $H =\{(0,2), (0,4), (0,0)\}$, so the subgroup generated by $H$ has order $3$.
Since $G$ has $24$ elements and each coset has the same number of elements as $H$, there are $24 / 3 = 8$ cosets in $G/H$.
Fraleigh says,
The first factor $\mathbb Z_4$ of $G$ is left alone.  The $\mathbb Z_6$ factor is essentially collapsed by a subgroup of order $3$, giving a factor group in the second factor of order $2$ that must be isomorphic to $\mathbb Z_2$. So $G/H$ is isomorphic to $\mathbb Z_4\times\mathbb Z_2$.
The bolded is what confuses me. Here are the elements of our factor group (the cosets of $H$ in $G$):
$(0,0) + H = H$
$(0,1) + H = \{(0,1), (0,3), (0,5)\}$
$(1,0) + H = \{(1,0), (1,2), (1,4)\}$
$(1,1) + H = \{(1,1), (1,3), (1,5)\}$
The process continues and we have $4$ more cosets.
I don't see how Fraleigh computes this factor group so quickly / without writing out the cosets.  Even with writing out the cosets, I'm not sure why it's clear that $G / H$ is isomorphic to $\mathbb Z_4\times\mathbb Z_2$.
Any help much appreciated,
Mariogs

Fact $1$: If $\gcd(|a|,|b|)=1$ then $(a,b)\cong (a)\times (b)$. In that case, you can directly say $(\mathbb Z_m\times \mathbb Z_n)/<(a,b)>\cong (\mathbb Z_m/<(a)>)\times(\mathbb Z_n/<(b)>)$ (it follows from above argument).
Fact $2$: $|(a,b)|={|a||b|\over \gcd(|a|,|b|)}$.
Fact $3$: $\mathbb Z_{mn}\cong \mathbb Z_m\times \mathbb Z_n$ if and only if $\gcd(m,n)=1$.
Now let's try find $(\mathbb Z_4\times \mathbb Z_6)/<(2,3)>$. Notice that we can not use Fact $1$ since $2$ and $3$ have order $2$ in $\mathbb Z_4$ and $\mathbb Z_6$ respectivly.
We can say that $\mathbb Z_4\times \mathbb Z_6=<(1,0),(0,1)>$ and notice it is spanned by elements of $(1,0)$, $(0,1)$ with integer coefficients (and of course evaluated mod $(4)$ and mod $(6)$). Can we find another basis ?
You can easily see that $<(2,3),(1,1)>=\mathbb Z_4\times \mathbb Z_6$ since $\begin{bmatrix} 2 & 3 \\1 & 1 \end{bmatrix}$ can be row reduced to identity matrix with integer coefficients.
$<(2,3),(1,1)>/<(2,3)>\cong <(1,1)>$ so it is a cyclic group and by Fact $2$ $|(1,1)|={6\cdot4\over2}=12$. So, $<(1,1)>\cong \mathbb Z_{12}\cong \mathbb Z_3\times \mathbb Z_4$.
Let have one more example.
$(\mathbb Z_4\times \mathbb Z_8)/<(2,4)>$. Again we can write $(\mathbb Z_4\times \mathbb Z_8)=<(1,2),(1,1)>$ and then $<(1,2),(1,1)>/<(2,4)>\cong \mathbb Z_2\times <(1,1)>$ and $|(1,1)|={8\cdot4\over 4}=8$ so
$$(\mathbb Z_4\times \mathbb Z_8)/<(2,4)>\cong \mathbb Z_2\times \mathbb Z_8.$$
