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

Note: direct product of quotients is isomorphic to quotient of direct products.
More clearly: $(G_1\times G_2)/(H_1\times H_2) \cong G_1/H_1\times G_2/H_2$ where $H_i\leq G_i$.
From that point; $$\mathbb Z_4\times \mathbb Z_6/\langle(0,2)\rangle\cong \mathbb Z_4/\langle(0)\rangle\times \mathbb Z_6/\langle(2)\rangle\cong \mathbb Z_4 \times \mathbb Z_2.$$
