I am having a really hard time understanding the concept of quotient group. I write the ideas and the parts that need explanations in parenthesis.
Let $B \le A$ an abelian subgroup, then we define a relation $ \sim $ in $A \times A$ to be the $x \sim y $ if $x-y \in B$ (at this point I think that every element in $A$ is related with some other element in $A$ because it always happens that $x-y \in B$ since $B$ is a subgroup).
It is easily to proof that is an equivalence relation. (Yes I did it)
Now we know that is an equivalence relation we will look for the equivalence class of one element. (I don't why it was necessary that the relation was an equivalence relation). $C(x) = \{y\in A: y \sim x \}= \{ x\in A: \exists b_y \in B \text{ such that } y-x = b_y \} = \{y\in A: y= x+b_y \text{ for some } b_y \in B \} =: x + B$
Then we define the coset of $x$ with respect $B$. Since it is an equivalence relation we can define the quotient application ( why?) $\Pi : A \rightarrow A/\sim \\x \mapsto x+B$
(Here as usual the function maps element in in the first set to the second, but what is the second set? by the actual mapping in the second line, it is clear to me because the above definition that elements are mapped to a set, so for each $x \in A$ we have a set in $A/\sim$ so the second set is indeed a collection of sets, but how do you visualize that set and its relation with $A$)
So $\Pi$ is a quotient group.
There is a better or easy to understand the quotient?
Best Answer
Perhaps an example:
$\mathbb Z^+$ is the integers with the group operation of addition.
$6\mathbb Z$ is the integers that are multiple of 6. i.e. $\{\cdots, -12,-6,0,6,12,\cdots\}$
This is a subgroup of $\mathbb Z^+$
And since $\mathbb Z^+$ is an abelian group, $6\mathbb Z$ is a normal subgroup.
If we choose an element from $\mathbb Z$, say $1,$ and add it to every element in $6\mathbb Z$ we get the set $\{\cdots, -5, 1, 7, \cdots\}$ This is a coset of $6\mathbb Z.$ If we chose 7, we would get the same coset.
If we did this with an element of $6\mathbb Z$ we would get $6\mathbb Z.$ If we did this with other integers, we would get other cosets. However, there are only finitely many cosets.
$\{\cdots, 0, 6, \cdots\},\{\cdots, 1, 7, \cdots\},\{\cdots, 2, 8, \cdots\},\{\cdots, 3, \cdots\},\{\cdots, 4, \cdots\},\{\cdots, 5, \cdots\}$
As you note above, this is an equivalence relation.
We can choose one integer from each equivalence class, to represent the class. This gives us a map (homomorphism) from the integers to the set $\{0,1,2,3,4,5\}.$
This is the factor group $\mathbb Z/6\mathbb Z.$
Generalizing, if we have a subgroup, we can find some set of cosets. If the group is non-abelian, the left cosets may not equal the right cosets. However, if the subgroup is normal, they will. The set of cosets is a quotient group.