Prove that $P =\{ aH : a \in G \}$ is a partition of G 
*

*Let $\mathbb{Z}_{12} =\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ with the operation of addition modulo 12. For each fixed element $a\in\mathbb{Z}_{12}$ define the set $a +H:=\{ a + h : h \in H\}$.
For each element $a\in\mathbb{Z}_{12}$, write down the set $a + H$. How many distinct ones are there? (I have already figure out how to solve this)


And I need help for:
What can you say about the set of sets $P =\{ a + H : a \in\mathbb{Z}_{12}\}$ ?


*Suppose that $G$ is a group and that $H$ is a subgroup of $G$. For each fixed element $a\in G$ define the set $aH$ by $aH =\{ah : h\in H\}$. Prove that $P =\{ aH : a\in G \}$ is a partition of $G$.


Thank you!
 A: The comment by @Nex really is all there is.
Given: $G$ is a group, and $H$ is a subgroup of $G$ (i.e. $H$ is closed under multiplication, and under taking inverses).
First note that $aH = H \Leftrightarrow a \in H$.
Suppose $a \in H$. Then, $\forall$ $h \in H$, $ah \in H$ (proving $aH \subseteq H$) and $\forall$ $h \in H$, $h' = a^{-1}h \in H$ satisfies $ah' \in H$ (proving $H \subseteq aH$).
On the other hand, if $aH = H$ (in particular, if $aH \subseteq H$) then, $\exists$ $h, h' \in H$ satisfying $ah = h'$ which implies $a = h'h^{-1} \in H$.
Next note that $a_{1}, a_{2} \in G \Longrightarrow$ either $a_{1}H = a_{2}H$ or $a_{1}H \cap a_{2}H = \emptyset$.
Suppose $a_{1}H \cap a_{2}H \neq \emptyset$. That is, $\exists$ $x \in a_{1}H \cap a_{2}H$. That is, $\exists$ $h_{1}, h_{2} \in H$ such that $x = a_{1}h_{1} = a_{2}h_{2}$. But then, $a_{1}h_{1} = a_{2}h_{2}$ implies $a_{2}^{-1}a_{1} = h_{2}h_{1}^{-1} \in H$. Thus, $a_{2}^{-1}a_{1}H = H$; and hence, $a_{2}H = a_{2}a_{2}^{-1}a_{1}H = a_{1}H$.
We have thus shown the elements of $P$ are disjoint. To show that $P$ forms a partition of $G$, the elements of $P$ must 'cover' $G$ i.e. we need $\forall$ $g \in G$, $\exists$ $aH \in P$ s.t. $g \in aH$. But since $H$ is a subgroup, $e \in H$, and $g = ge \in gH$ for every $g \in G$. Hence, the set of cosets $P$ forms a partition of the group.
Finally note that (if $G$ is a finite group) the cardinality of every coset $aH \in P$ is the same. That is, for every $a \in G$, $|aH| = |eH| = |H|$. This is because the set-function $H \to aH$ defined by $h \mapsto ah$ is bijective, with the inverse function $aH \to H$ defined by $h \mapsto a^{-1}h$. But finite sets in bijection have the same cardinality.
Then, it is clear that $|H| . |P| = |G|$.
$|P|$ is usually written as $[G:H]$ and is called the index of $H$ in $G$. The result, $|G|=|H|.[G:H]$ will be referred to as Lagrange's formula in any standard abstract algebra text you may be using.
Coming back to part 1 of your question, first show that $H = \left\{0, 4, 8\right\}$ forms a subgroup (by showing that it is closed under addition mod 12, and under taking inverses). Then, since $|G| = 12$, and $|H| = 3$, $|P| = [G:H]$ must be $4$.

EDIT: This is typically how most first texts on Abstract Algebra prove the so-called 'Lagrange's theorem'. It is certainly the fastest way to get to this result without introducing more 'jargon'.
But really, the reference to the set-function $H \to aH$ in the proof should act as a pointer that there are neater ways to go about this. That would involve postponing any discussions on 'index of subgroups' until after group actions on sets are discussed (in particular, the group action of $G$ on its underlying set by left-multiplication, etc). Then, one may talk about transitive actions, and orbits: cosets turn out to be a specific instance of orbits, and one can prove a more general version of Lagrange's theorem.
Aluffi's textbook, Algebra: Chapter 0, provides a most beautiful treatment of groups in Chapters II and IV - an approach that I think should be adopted by all course instructors. The approach works fairly well even if you expunge the frequent harks to categories (although I don't think there's any good reason to abstain from talking about categories when you're doing Abstract Algebra).
A: It is common to prove first that $aH = bH$ if and only if $b^{-1}a \in H$.
First, suppose $aH = bH$. This means that for any $h \in H$, we have $ah \in bH$, that is: $ah = bh'$ for some (most likely different) $h' \in H$. So:
$a = b(bh')h^{-1}$
$b^{-1}a = h'h^{-1} \in H$, by closure.
On the other hand, suppose $b^{-1}a \in H$, say $b^{-1}a = h$. Let $bh'$ be any element of $bH$. Since $H$ is a subgroup, we have $h' = eh' = h(h^{-1}h')$ with $h^{-1}h'\in H$.
Hence $bh' = (bh)(h^{-1}h') = b(b^{-1}a)(h^{-1}h') = a(h^{-1}h') \in aH$, so $bH \subseteq aH$.
Now let $ah''$ be any element of $aH$. Again, we may write $h'' = (h^{-1}h)h''$.
Since $b^{-1}a = h$, we have $(b^{-1}a)^{-1} = h^{-1}$, that is: $a^{-1}b = a^{-1}(b^{-1})^{-1} = h^{-1} \in H$.
So $ah'' = (ah^{-1})(hh'') = a(a^{-1}b)(hh'') = b(hh'') \in bH$, showing $aH \subseteq bH$, and together with the first inclusion above, we have $aH = bH$.
Now we seek to prove that if $xH \cap yH \neq \emptyset$, that $xH = yH$ (so that $G$ is the disjoint union of cosets $gH$, that is, these cosets partition $G$; the fact that every $g \in G$ lies in a coset $gH$ is pretty self-evident, since $g = ge \in gH$, because $e \in H$).
So let $z = xh = yh' \in xH \cap yH$. Then $y^{-1}x = h'h^{-1} \in H$, and we are done.

Another approach commonly used is the following: Any equivalence relation on a set $S$ partitions the set via the equivalence classes of the relation. Moreover, given a partition $P$ of a set $S$, the relation $x\sim y$ if $x,y$ lie in the same subset of the partition is an equivalence relation. I will not prove this here, but what it essentially says is that partitions and equivalence relations are basically two different ways of looking at the same thing.
In this view, all that is needed to to show that $\sim_H$ (where $H$ is a subgroup of $G$) given by $a\sim_H b$ if $aH = bH$ is an equivalence relation (on $G$). Since we have established above that:
$aH = bH \iff b^{-1}a \in H$, we will use the latter form, since it's easier to work with.
1.) $\sim_H$ is reflexive:
$a \sim_H a$ means that $a^{-1}a = e \in H$, which is certainly true.
2.) $\sim_H$ is symmetric:
If $a \sim_H b$, then $b^{-1}a \in H$; since $H$ is a subgroup, $h^{-1} = a^{-1}b \in H$, whence $b \sim_H a$.
3.) $\sim_H$ is transitive:
Suppose $a\sim_H b$ and $b \sim_H c$. Then $b^{-1}a, c^{-1}b \in H$. We have closure in $H$, so we have:
$c^{-1}a = (c^{-1}b)(b^{-1}a) \in H$, and so $a\sim_H c$.

The advantage to this latter view, is that we can ask: for which subgroups $H$ is it true that:
If $a\sim_H a'$, and $b \sim_H b'$, then $aa' \sim_H bb'$?
That is, for which partitions of $G$ by $H$ is it true that the partitioning respects the group operation? Such a partitioning is called a congruence, and the group $(\Bbb Z, +)$ along with the partition:
$a \equiv b \iff b - a \in 12\Bbb Z$ (here $12\Bbb Z = \{12k: k \in \Bbb Z\}$, the integral multiples of $12$).
is such a congruence, called congruence modulo $12$. That this indeed is a congruence is easily seen:
If $a' = a + 12k$ and $b' = b + 12m$, (here $k,m$ are unspecified integers) then:
$a' + b' = a + 12k + b + 12m = (a + b) + 12(k+m)$, and $k+m$ is surely an integer.
