Quotient space of mod n. I'm learning abstract algebra and I'm having trouble understanding what $\Bbb {Z}_n$ actually is. My book just defines it as the quotient space modulo $n$. I know this is a set. I would assume that this is also a group under various operations but again, I don't know what the individual elements are or how they are determined. 
The other part of this is the definition of cosets. I know that it says that $g \in G, H \leq G, gH= \{h: h \in H\}$ for the left. My question is, are we literally taking an element of $G$ and multiplying it with every element of $H$ or is this a some sort of composition? How does this work for say $SL(2,R)$ in $GL(2,R)$? Are we taking an element of the set of invertible matrices and multiplying it with each element of the set of matrices with determinate $1$?
I was asked in a homework to find the cosets of $9\Bbb {Z}$ in $\Bbb {Z}$ and $9\Bbb {Z}$ in $3 \Bbb {Z}$.
Secondly, I was asked to find the cosets of $<6>$ in $\Bbb {Z}_{12}$ and then $<6>$ in $<2>$ in $\Bbb {Z}_{12}$.
I have the answers to both of these questions but I don't know how to interpret them.
If anyone could explain this to me, I would be very thankful. I apologize for the lengthiness of this question. I have tried seeking other texts books and google but everything is defined so generally, that it is difficult to understand what is actually going on.
 A: There's a lot of competing ideas involved in a quotient group, so it's hard to know which might make it "easy to understand" (you're in good company-the notion of "quotient ___" is often one of the hardest concepts in any structure for people to wrap their heads around).
Because the integers are kind of user-friendly, a lot of texts introduce quotient groups by using "the integers modulo $n$" as an example. Unfortunately, this in and of itself can be confusing, since there are two possible operations in the integers (addition, and multiplication), and "abstract" groups use the notation of one (multiplication) to describe the other (addition, the only operation for which the integers form a group).
So let's look at a "really simple example". You've probably seen this before, and not even realized it.
Let's us call two integers of equal parity if they are both even, or both odd. It's not hard to see that if $a,b$ are both even, say $a = 2m$ and $b = 2n$, that:
$a - b = 2m - 2n = 2(m-n)$, so their difference is an even number.
If $a,b$ are both odd, say $a = 2k+1$ and $b = 2t + 1$, we see that:
$a - b = 2k + 1 - (2t + 1) = 2k + 1 - 2t - 1 = 2k - 2t = 2(k-t)$, so that here, again, their difference is an even number.
We can "turn this around", I claim if $a - b$ is even, they are of equal parity:
Case 1) $a$ is odd:
We know that $a - b$ is even, say $a - b = 2k$. We can also write $a = 2m + 1$, since $a$ is odd.
Thus $b = a - (a - b) = 2m + 1 - 2k = 2(m-k) + 1$, which is odd.
Case 2) $a$ is even:
Let $a = 2n$. Then $b = a - (a-b) = 2n - 2k = 2(n-k)$, which is likewise even.
We can summarize this by saying:
$a,b$ are of equal parity if and only if $a - b$ is an even number. Now the set of even numbers is also the set of doubles $a + a$ of any integer $a$, so we can re-state this as:
$a,b$ are of equal parity $\iff a-b \in 2\Bbb Z$.
Note we can add "parities" just like we can add integers:
even + even = even.
even + odd = odd.
odd + even = odd.
odd + odd = even.
Seen this way, it doesn't matter which "particular" numbers we take parities of, any examples will do: to calculate "odd + odd" we can use $3$ and $7$, or $1$ and $-1$, or $13$ and $2015$. It is "customary", historically, to use $0$ as a "stand-in" for every even number, and $1$ as a stand-in for every odd number.
If we forget these are "stand-ins" or representatives it is easy to forget that:
$1 + 1 = 0$ isn't really a statement that $1$ and $1$ makes $0$, it merely is a short-hand for "odd and odd makes even".
So, here the thing: the set "odd integers" is the set $1 + 2\Bbb Z$ (we acknowledge this by writing, as we have $a = 2k + 1$, for example). This is a coset. The "other" coset, the set of even integers is, of course, $2\Bbb Z$. We can, if we wish, write this equally well as $0 + 2\Bbb Z$.
We have "reduced" the information about integers, to just one aspect of an integer: parity. In doing this, we regard any even integer, or any odd integer, as "equivalent" to any other of the same parity. This is a really "coarse" filter, but it respects addition.
Now, the next step, is to see that "chopping up" the integers into two classes (even and odd) is essentially all about the integer $2$. And there's really nothing special about $2$, we can use $n$, for any natural number $n$, obtaining $n$ "classes", of the form $qn + r$, where $r \in \{0,1,2,\dots, n-1\}$. These are the cosets $r + n\Bbb Z$. Note that $r +n + n\Bbb Z = r + n\Bbb Z$, for example if $n = 12$, adding $12$ to each element of:
$3 + 12\Bbb Z = \{\dots,-21, -9,3,15,27,\dots\}$ just "shifts it over" by $12$, giving us the same set back again (since it extends infinitely in both directions). Every element of $3 + 12\Bbb Z$ is "$3$ more than a multiple of $12$". It's exactly as if we have taken the infinite LINE of integers, and wrapped it around into a circle that repeats every $12$ "clicks" (like a clock). This effectively "shrinks" ALL integers into $12$ "slots": the cosets, each of which is an infinite "stack" of integers above each "slot". Essentially, we stop keeping track of "absolute" size, and just keep track of "relative size" (how far past the repeating cycle of $12$ steps past the $0$ we start off with, in either direction).

For an abstract group, we replace the $+$ we use for $\Bbb Z_n = \Bbb Z/n\Bbb Z$ with the operation $\ast$ of $G$. Often the asterisk is omitted. The difference $a - b$ becomes $ab^{-1}$ (or,if you prefer, $a \ast (b^{-1})$). Instead of writing $r + n\Bbb Z$, we write $aH$. For example, if $H = \{e,h_1,h_2\}$, then $aH = \{a,ah_1,ah_2\}$. It turns out to be important whether or not $aH = Ha$, because it is that condition that allows us to "multiply" cosets. We don't have to worry about that detail in the integers, because addition is commutative, which makes $r + n\Bbb Z = n\Bbb Z + r$ automatic. Whether or not $aH = Ha$ will depend on $a$, and also on $H$. If it is true for any $a$ (so that it depends on $H$ alone), we say $H$ is a normal subgroup. These are important, and it's not self-evident that normality is crucial to the study of groups.
Unfortunately, this post is too long as it is, and I'm afraid the rest will not fit in the margin.
A: Reminder. Let $E$ be a set and $\mathcal{R}$ be an equivalence relation on $E$, for all $x\in E$, one can define : $$[x]:=\{y\in E\textrm{ s.t. }x\mathcal{R}y\}\subset E.$$
Let define the set of equivalence classes by :
$$E/\mathcal{R}:=\{[x];x\in E\}\subset\mathcal{P}(E).$$
Let $E:=\mathbb{Z}$ and for $n\in\mathbb{N}^*$, $\mathcal{R}_n=\mod n$ (check by your own that $\mod n$ is reflexive, transitive and symmetric), in this precise case, let : $$\mathbb{Z}_n:=E/\mathcal{R}_n.$$
Then, one have : $$\mathbb{Z}_n:=\{[m];m\in\mathbb{Z}\}.$$
Let $m\in\mathbb{Z}$, there exists unique integers $(q,r)\in\mathbb{Z}\times\mathbb{N}$ such that $m=nq+r$ with $r<n$. Hence, one have $[m]=[r]$ and then $[m]\in\{[0],\ldots,[n-1]\}.$ This far, one has : $$\mathbb{Z}_n\subseteq\{[0],\ldots,[n-1]\}.$$
The reverse inclusion is clearly true and one has : $$\mathbb{Z}_n=\{[0],\ldots,[n-1]\}.$$
Using words, $\mathbb{Z}_n$ is the set formed by the class of remainders in the euclidian division by $n$. Notice that for all $m\in\mathbb{Z}$, one also has : $$\mathbb{Z}_n=\{[m],[m+1],\ldots,[m+n-1]\}.$$
Let $(G,\star)$ be a group and $H$ be a subgroup of $G$, then for all $g\in G$ let : $$gH:=\{g\star h;h\in H\}.$$
So, yes we're literally looking to all the "multiplications" of $g$ with an element of $H$. In your example, if $A\in SL(2,\mathbb{R})$, then $A.GL(2,\mathbb{R})=\{AB;B\in GL(2,\mathbb{R})\}$.
Define $\mathcal{R}$ by $\forall (x,y)\in G^2,x\mathcal{R}y\Leftrightarrow\exists h\in H\textrm{ s.t. }x=hy$, $\mathcal{R}$ is an equivalence relation on $G$ and notice that : $$\forall g\in G,[g]=gH.$$
A: Edit 
I'll include the relevant part of the discussion with the OP in the comments. The elements of the quotient space $\Bbb Z_n$ are the cosets of the subgroup $n\Bbb Z$ of $\Bbb Z$. So:


*

*We have a group $G=\Bbb Z$ which gives us context. Note that $\Bbb Z$ is a group under addition and not under multiplication (we would need to include rationals for the inverses under the latter). 

*We have a subgroup $H=n\Bbb Z$, which inherits the group operation of $\Bbb Z$ (addition).

*The cosets in your notation are the elements $gH=\{gh:h\in H\}$ for $g$ in $G$. Note that this "multiplication" is abstract: it is supposed to be universal notation for the underlying group's operation, and does not necessarily stand for the usual arithmetic multiplication. In this case, the group's operation is addition, so the cosets are $m+n\Bbb Z = \{m+h:h\in n\mathbb Z\}$ for any integer $m$. 

*Powers $a^k$ stand for $a\cdot a\cdots a$ ($k$ times) in the abstract setting. In this case, that would mean $a+a+\cdots+a$ ($k$ times) or, in standard arithmetic notation, $ka$. The cyclic group $\langle 6\rangle$ in this case would be the group containing $6, 2\cdot6, 3\cdot6, \dots$ etc. Note that it has a finite number of elements because of the clock-like behavior (explained below). Similarly, we should replace $a^{-1}$ by $-a$.



The group $\mathbb Z_n$ has $n$ elements, which correspond to the cosets of $0,\dots,n-1$. These cosets are actually the sets $$\{0+kn:k\in\mathbb Z\},\dots,\{n-1+kn: k\in\mathbb Z\}.$$
Their group operation is inherited from $\mathbb Z$ and behaves much like adding hours in the clock: adding (the coset containing) $1$ to (the coset containing) $n-1$ takes us back to (the coset containing) $0$. This is because $n$ and $0$ are in the same set defined by the equivalence relation (the difference between them is an integer multiple of $n$).
We can picture the addition of any two of those cosets as a translation of the second set by an element of the first set, that is $$\{a+kn:k\in\mathbb Z\} +\{b+kn:k\in\mathbb Z\} = \{a+b+kn:k\in\mathbb Z\}.$$ The most important thing to note here is that the set in the right hand side is the same regardless of which element of the first set we chose to make the translation of the second set (we could have used $a+n$ instead of $a$ in the right hand side). That makes everything work nicely and is not obvious, it must be proven by using the definition.
