Help understanding how to determine a quotient group I would like some help in how to determine a quotient group, given two groups.
An example: The group $(\mathbb{E},+) $ in $(\mathbb{Z} , +)$.
I know the set of the quotient group is all left cosets of $( \mathbb{Z} , + )$ right? So the operation in the quotient group should be by definition: $a \mathbb{E} * b \mathbb{E} = ab \mathbb{E}$.
I'm just not sure how to piece this stuff together to easily determine the quotient group. Aren't there endless left cosets of $( \mathbb{Z} , + )$?...Or none? This is confusing me greatly..
 A: Let's look at a slightly more "in-depth" example: Consider the subgroup $7\Bbb Z$ of $\Bbb Z$ which consists of all (positive, negative and $0$) multiples of $7$. Let's "expand" this a bit:
$7\Bbb Z = \{\dots,-21,-14,-7,0,7,14,21,28,\dots\}$
What set do we get if we add $1$ to everything in this set?
$1 + 7\Bbb Z = \{\dots,-20,-13,-6,1,8,15,22,29,\dots\}$
If we shift "over $2$" we get:
$2 + 7\Bbb Z = \{\dots,-20,-12,-5,2,9,16,23,30,\dots\}$
Now here is a curious thing: If we take something in, say $3 + 7\Bbb Z$, and add it to something in $2 + 7\Bbb Z$, we'll get something in $5 + 7\Bbb Z $. For example:
$10 + 9 = 19$ and $19$ is $5$ more than $14$.
This process "chops up" $\Bbb Z$ into $7$ pieces (each of which is "infinite") and in every piece, the numbers are all "$7$ apart".
So the "$k$" in $k + 7\Bbb Z$ measures how far to the right (right = counting up) we are from a multiple of $7$.
If you imagine the integers as an infinite string of beads, we are "wrapping the beads into a circle" so that the $7$th bead winds up in the same place as the "$0$-bead". In fact, any two beads that are a multiple of $7$ apart on our original string, wind up in the same place on our "$7$ position circle".
You've seen this before: on a clock (there it's "modulo $12$" instead of $7$, or "modulo $24$" if you use military time).
This has the effect of effectively "setting all multiples of $7$ equal to $0$". Since we know that $14$ (for example) isn't actually $0$, we don't say:
$14 = 0$, but rather, $14$ is equivalent to $0$ since it is a multiple of $7$ away from $0$ (if $6$ was "as high as we could count before starting over", $14$ would actually BE  $\ 0$).
It's somewhat "magical" that all this actually WORKS, in that we don't get any contradictions or confusion this way. Part of this has to do with the fact that addition is commutative: that is, that $k + m = m+k$ for any two integers $k,m$. With a group in general, we need a more special condition on the subgroup to make this all work out (it has to be normal).
It turns out that with a general group $G$ and a subgroup $H$, that the set:
$(aH)(bH) = \{(ah)(bh'): h,h' \in H\}$ doesn't always equal some other coset $kH$, so subgroups for which this DOES happen are "special".
If $G$ is abelian, of course, then $(ah)(bh') = a(hb)h' = a(bh)h' = (ab)(hh')$ and this DOES happen. So quotient groups of abelian groups are "easier to understand".
In this special case of $G = \Bbb Z$ and $H = ${even integers}, this process is called "the arithmetic of parity" (reasoning by evens and odds). This "trick" (reducing mod $2$) turns an "infinite number of cases" to just TWO cases, and often that is all we need.
A: If you mean the quotient of the group $(\mathbb{Z},+)$ inside $(\mathbb{R},+)$: indeed there are infinitely many cosets: for each $x \in [0,1)$, the set $x + \mathbb{Z} = \{x+z : z \in \mathbb{Z}\}$ is a coset. These cosets are all different, and these are all the cosets.
The resulting quotient group is the group $([0,1),+)$ where addition works modulo $1$ (we identify each cosets with its generator in $[0,1)$). This is the circle group $\mathbb{R}/\mathbb{Z}$.
If $\mathbb{E}$ is the set of even integers: there are two cosets: the set of odd integers and the set of even integers. The resulting quotient group is the group $\mathbb{Z}/2\mathbb{Z}$ with two elements.
Note that you're working in additive groups; the operation on cosets is $(a+\mathbb{Z}) + (b + \mathbb{Z}) = a+b+\mathbb{Z}$. When $a$ is odd, $a+\mathbb{Z}$ is the set of odd integers; when $a$ is even, $a + \mathbb{Z}$ is the set of even integers.
Denote the cosets by $X$ (even integers) and $Y$ (odd integers). Then the group operation yields $X+X=X$, $X+Y=Y$ and $Y+Y=X$.
