TITLE: Left/Right Cosets
QUESTION [11 upvotes]: I am trying to understand left/right cosets in group theory.
Here is the example in my text:
Let $G = \lbrace 1, a, b, c, d ,e \rbrace$
Lets define the group operation $.$ by the following table, where the entry at row $x$ and column $y$ gives $x.y$
ex. $d.e = b$
 1 a b c d e
1 1 a b c d e
a a b 1 d e c
b b 1 a e c d
c c e d 1 b a
d d c e a 1 b
e e d c b a 1

This is no problem I understand this. But then we get the left and right cosets.
Let $G$ be a group and let $H \leq G$. A left coset of $H$ in $G$  ($G / H$) is a set of the form $gH = \lbrace gh : h \in H \rbrace$ for some $g \in G$. A right coset of $H$ in $G$ ($H$ \ $G$) is a set of the form $Hg = \lbrace hg : h \in H \rbrace$ for some $g \in G$.
Here are some examples that I am trying to figure out how they are generated. I guess i just do not understand the theory fully. I would like a little help explaining and possible a few more examples.
$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, a, b \rbrace = \lbrace \lbrace1, a, b\rbrace , \lbrace c, d, e \rbrace\rbrace$
$\lbrace 1, a, b \rbrace$ \ $\lbrace 1, a, b, c, d, e\rbrace = \lbrace \lbrace 1, a, b \rbrace, \lbrace c, d, e \rbrace \rbrace$
$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, c \rbrace = \lbrace \lbrace 1, c \rbrace, \lbrace a, d \rbrace, \lbrace b, e \rbrace \rbrace$
A few more examples which i want to figure out are:
$\lbrace 1, a, b, c, d, e \rbrace / \lbrace 1, e \rbrace = ?$
$\lbrace 1, d \rbrace$ \ $\lbrace 1, a, b, c, d, e \rbrace = ?$
Thanks!

REPLY [2 votes]: Cosets arise when you want to model the idea that certain elements of a group are effectively equal. 
To see this, rather that looking at one coset at a time it is best to think of all possible cosets of a subgroup. Then you will find that cosets partition a group $G$ into equivalence classes such that two elements of a class differ by an element of the subgroup $H$. 
For example sometimes in number theory we do not want to distinguish two numbers that differ by a multiple of a given number $n$. Say $n=4$. Then let $G$ be the set of integers under addition and $H$ be the set of multiples of $4$. The cosets of $H$ are
$$H\cdot 0=\{\ldots,-4,0,4,8,\ldots\}$$
$$H\cdot 1=\{\ldots,-3,1,5,9,\ldots\}$$
$$H\cdot 2=\{\ldots,-2,2,6,10,\ldots\}$$
$$H\cdot 3=\{\ldots,-1,3,7,11,\ldots\}$$
$$H\cdot 4=\{\ldots,0,4,8,12,\ldots\}$$
$$H\cdot 5=\{\ldots,1,5,9,13,\ldots\}$$
Note that $H\cdot 0 =H\cdot 4$ and $H\cdot 1 =H\cdot 5$. In fact there are only $4$ distinct cosets, each corresponding to one congruence class of integers modulo $4$. If we care only about the remainder of an integer after division by $4$ then all the elements in a coset are equivalent and we can think of them as a single entity.
To take another example, let $G$ be the group possible rotations of a point on a unit circle and we care only about where a point ends up on the circle after the rotation. Then let $H$ be rotations that are a multiple of $2\pi$. The cosets of $H$ will now be rotations which differ by multiples of $2\pi$ and which therefore have the same effect on the final position of the point rotated. 
In general, two elements $g_1$ and $g_2$ of $G$ are defined to be equivalent ($g_1 \equiv g_2$) if $g_1 \cdot g_2^{-1} \in H$.
You can prove the following

For all $g$ in $G$, $g \equiv g$ since the identity element belongs to $H$ as it is a subgroup.
If $g_1 \equiv g_2$ the $g_2 \equiv g_1$ since if an element belongs to $H$ then so does its inverse.
If $g_1 \equiv g_2$ and $g_2 \equiv g_3$ then $g_1 \equiv g_3$ since $H$ is closed under multiplication.

This show that $\equiv$ is a genuine equivalence relation. We define the equivalence class of an element $x$ of $G$ as
$$[x]=\{y \in G\mid x \equiv y\}$$
You can show that

$x \in [x]$
For any $x,y \in G$, either $[x] \cap [y] =\emptyset$ or $[x]=[y]$.

So the equivalence classes are either equal or disjoint and they cover $G$.
Finally, back to cosets. The equivalence classes we have defined above are the same as the right cosets. If $y \equiv x$ then $yx^{-1}=h$ for some $h \in H$ or $y=hx$. So $[x]=Hx$. If we had instead defined our equivalence relation by the condition $x^{-1}y \in H$ then we would have got the left cosets.