What is a coset in a lattice and how many they are? This PDF gives the following definition of cosets for lattices: 

However, I tried to picture a lattice $L$ with basis $(2,0),(0,2)$ and $\mathbb{Z}^2$ in order to take the quotient $\mathbb{Z}^2/L$. A coset is a shift of $L$ by $v\in\mathbb{Z}^2$. If I take $v=(0,1), (0,2), (0,3), ...,(0,n)$, they all represent the same coset because I'm shifting to the same structure, right? There is also another coset, the one represented by $(1,0)$. There's also the coset represented by $(0,0)$ So there are just $3$ cosets? 
I don't know if this is right, because $(1,1) + L$ is also a shift, but in which coset is $(1,1)$ a representant?
My book says that the number of cosets relates to the determinant of the matrix of basis for the lattice $L$ like this:
$$|\mathbb{Z}^n/L| = |det \ B|$$
How is this relation given? It's not immediate for me. Maybe I understood the concept wrong.
 A: The four cosets of $\mathbb{Z}^2/L$ in your example are
$$L,\quad (0,1)+L,\quad (1,0)+L,\quad (1,1)+L.$$
Explicitly,
$$L = \{\lambda(2,0) + \mu(0,2)\;:\; \lambda,\mu\in\mathbb{Z}\} = \{(2\lambda, 2\mu)\;:\; \lambda,\mu\in\mathbb{Z}\},$$
$$(0,1) + L = \{(0,1) + \lambda(2,0) + \mu(0,2)\;:\; \lambda,\mu\in\mathbb{Z}\} = \{(2\lambda, 2\mu+1)\;:\; \lambda,\mu\in\mathbb{Z}\},$$
etc. You should try drawing these four cosets: for each one, draw a pair of coordinate axes and a dotted grid representing $\mathbb{Z}^2$, and then circle the dots that are included.
(However, there are a bunch of different ways of writing all of these elements. For instance, the set
$$(5,8) + L = \{(5,8) + \lambda'(2,0) + \mu'(0,2)\;:\; \lambda',\mu'\in\mathbb{Z}\} = \{(2\lambda'+5, 2\mu'+8)\;:\; \lambda',\mu'\in\mathbb{Z}\}$$
can easily be seen to be equal to $(1,0)+L$ by reparametrising: set $\lambda' = \lambda-2, \mu' = \mu-4$. Alternatively, just draw it - you'll see that the pictures end up the same.)
Finally, the determinant of a matrix $B$ can be calculated as follows: take the unit cube, apply $B$ to it, and the volume of the resulting parallelepiped is $\det(B)$. (If you've never seen this before, try to prove it explicitly when $B = \left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$. Draw a picture of the unit square, and then a picture of the parallelogram you get when you apply $B$ to it, and calculate its area.)
