Why isn't the number of cosets equal to cardinalities of the groups? The left coset for subgroup $H$ of $G$ and element $x \in G$ is
$$xH=\{xh : h \in H\}$$
Now why is the number of left (or right) cosets not equal to $|G||H|$?
Since if one picks $x \in G$ for every element in $G$ and then $h \in H$ for every element in $H$, then shouldn't the number of left or right cosets be $|G||H|$?
An example where it doesn't seem to be:
http://math.stanford.edu/~jelicata/109hw5sol.pdf, p. 2.
 A: An illustrative example: Consider $G=\mathbb{Z}/12=\{0,1,2,3,4,5,6,7,8,9,10,11\}$ (the integers mod $12$) and $H=\langle 4\rangle=\mathbb{Z}/3=\{0,4,8\}$.  (Some of these equalities should be isomorphisms (if one wants to be precise), and I use a slash instead of a subscript to avoid confusion with the $p$-adics).
We compute everything as follows:
\begin{align*}
0+H&=\{0,4,8\}=H&1+H&=\{1,5,9\}&2+H&=\{2,6,10\}\\
3+H&=\{3,7,11\}&4+H&=\{4,8,0\}=H&5+H&=\{5,9,1\}=1+H\\
6+H&=\{6,10,2\}=2+H&7+H&=\{7,11,3\}=3+H&8+H&=\{8,0,4\}=H\\
9+H&=\{9,1,5\}=1+H&10+H&=\{10,2,6\}=2+H&11+H&=\{11,3,7\}=3+H\\
\end{align*} 
I have written out $|G||H|=12\cdot 3=36$ elements, but we notice that each triple of elements represents a set (there are $|G|$ sets written out), AND many of the sets are the same - there are actually only $|G|/|H|=12/3=4$ different sets.
A: Another example.  Take $G = \mathbb{Z}$ and $H = 2 \mathbb{Z}$ be groups under addition. 
If you find the cosets, $n+ H = H$ is n is even. 
$n+ H = 1+H$ if $n$ is odd.  Then, 
$$G/H = \{0+H , 1+H \} $$
So $G/H$ has two cosets while $G$ and $H$ are not finite.  
A: The number of pairs $(g,h)$ with $g\in G$ and $h\in H$ is indeed $|G||H|$. But we are not counting such pairs, we're counting the number of collections there are of elements of the form $gh$ (with $g$ fixed and as $h$ varies in $H$).
One of the easiest illustration of this is with groups of order $4$. Consider $Z_4=\{\bar{0},\bar{1},\bar{2},\bar{3}\}$, the cyclic group of order four, with cyclic subgroup $\{\bar{0},\bar{2}\}$ of order two. This has exactly two cosets: $\{\bar{0},\bar{2}\}$ itself and the other coset $\{\bar{1},\bar{3}\}$. Or consider the Klein-four group $Z_2\times Z_2$. It has a subgroup $Z_2\times\{\bar{0}\}$, and its unique other coset is $Z_2\times\{\bar{1}\}$. Again, two cosets.
Keep in mind the cosets partition the group into subsets of equal size. Say you had fifteen people in a room and split them up into teams of size three - how many teams are there? A: $15/3=5$.
