Quotient group - explanation over explanation I was studying subchapter on quotient groups, and basically the book goes on motivating the theorem that the set of cosets is a group (quotient group) and while reading it, I have a hard time understanding why the last line (in this pic) is true? I mean why taking the set of cosets is not the same as the union of them? Obviously I am missing something, but I hope that the provided pic bellow will be sufficient for someone to explain what. 

Let us now apply this to the situation considered above in which $\theta$ is a homomorphism from $G$ to $G'$, $\theta(G)$ is the image of $G$, and $K$ is the kernel of $\theta$. Theorem $12.11.7$ says that if $y\in\theta(G)$ then the set of $g$ in $G$ such that $\theta(g)=y$ is some coset $x_0K$. It follows that $\theta$ acts as a bijection, namely, $x_0K\mapsto y=\theta(x_0)$ from the set $\mathcal{C}$ of all these cosets onto the group $\theta(G)$. Note carefully that $\mathcal{C}$ is the set whose elements are the cosets $gK$: thus $\mathcal{C}=\{gK:g\in G\}$ (and this is not the same as the union $\cup_g gK$ of the cosets, for this union is $G$;)

Thanks!
 A: The simple answer is that {$1, 2, 3, 4$}$\ne${{$1,2$},{$3, 4$}}. The elements of the set $\mathcal{C}$ are not elements of the set $G$, they are subsets of elements of $G$. So the elements of $\mathcal{C}$ are subsets of the forms $xK, x\in G$: they are a way of dividing the group into a collection of subsets. The union of the subsets is infact $G$, but this only expresses that $\mathcal{C}$ divides all of $G$. The subsets are also disjoint. So $\mathcal{C}$ is just a way of cleanly dividing the group, a partition.
A: The way I like to think of it is: Lagrange's Theorem tells us that a subgroup $H$ of a finite group $G$ partitions $G$ into "$H$-sized pieces".
A homomorphism $h:G \to G'$ with kernel $K$ basically shrinks every "$K$-sized piece" (the cosets of $K$) down to a single point; one of the nice things about groups is how uniform this structure is, all the cosets $gK$ shrink down the same amount.
One of the difficulties with quotient groups that takes some getting used to, is the 2-tiered structure: we start with $G$, which is made up of elements, and we conglomerate some of these elements together to make elements of a(n obviously smaller) group, whose elements are sets of elements ("chunks" of $G$ that all move "parallel to $H$", the (normal) subgroup).
This becomes even more confusing when you consider "quotient groups of quotient groups", like for example:
$\Bbb Z_{12}/\langle \overline{4} \rangle$ which is actually:
$(\Bbb Z/12\Bbb Z)/(4\Bbb Z/12\Bbb Z)$.
One of the things often glossed over in a course on groups, is that groups are based on SETS, and sets feature this particular kind of "layering": a set of subsets of a set is a totally different animal than a set composed of elements of the original set, the sets $\{a\}$ and $\{\{a\}\}$ are two entirely different things.
So a nice feature of the First Isomorphism Theorem (which it looks like you are on the verge of proving) is that we can think of quotient groups as homomorphic images of another group, rather than the more complicated explicit construction of coset-elements.
