Why is $\mathbb{Z}_4\times\mathbb{Z}_6 / \langle (0,2)\rangle$ isomorphic to $\mathbb{Z}_4\times\mathbb{Z}_2$? The subgroup generated by $\langle(0,2)\rangle$ has order $3$ and is given by $H=\{(0,2), (0,4), (0,6)\}$, so there must be $8$ elements in factor group. 
In my book, it says that "$\mathbb{Z}_6$ factor is collapsed by a subgroup of order $3$, giving a factor group in the second factor of order $2$ isomorphic to $\mathbb{Z}_2$."
I don't understand this explanation. Can anyone help me?
 A: Note that $\Bbb Z_4 \times \Bbb Z_6$ has $24$ elements, and $H = \{(0,0),(0,2),(0,4)\}$ has $3$, so $(\Bbb Z_4 \times \Bbb Z_6)/H$ has $8$ elements.
Note as well that $H = \{0\} \times \langle 2\rangle$, so it seems plausible that:
$(\Bbb Z_4 \times \Bbb Z_6)/H \cong (\Bbb Z_4/\{0\}) \times (\Bbb Z_6/\langle 2\rangle)$
But rather than prove the general theorem this is a special case of, let's just exhibit a surjective abelian group homomorphism:
$\phi: \Bbb Z_4 \times \Bbb Z_6 \to \Bbb Z_4 \times \Bbb Z_2$
with kernel $H$.
Specifically, let $\phi(a,b) = (a,b\text{ (mod }2))$, This is clearly onto, and we see at once that $H \subseteq \text{ker }\phi$.
On the other hand, if $\phi(a,b) = (0,0)$, we must have $a = 0$ (since $\phi$ is just the identity map on the first coordinate), and $b$ must be even, that is $b = 0,2,4$. This shows that $\text{ker }\phi \subseteq H$, and thus the two sets are equal.
So by the Fundamental Isomorphism Theorem, $(\Bbb Z_4 \times \Bbb Z_6)/H \cong \Bbb Z_4 \times \Bbb Z_2$.

A word about the explanation you were given: a homomorphism essentially "shrinks" its kernel to an identity. Given that all cosets of a subgroup are "the same size", the size of the kernel is "the shrinkage factor" (if the kernel had two elements, the size of the quotient group would be half the size of the original group). Note how $\phi$ in what I wrote above acts on each factor group of our direct product: it does nothing to $\Bbb Z_4$, and it identifies all the "even" elements of $\Bbb Z_6$-there are $3$ of these, so we get a cyclic subgroup of order $2$ (because $6/3 = 2$) in the quotient.
