G3 - factor groups

coset multiplication

a H \cdot b H = (a b) H

where $a b$ defined by the operation of $G$

since cosets have multiple representations, ie. $a H = c H$ and $b H = d H$ , so coset multiplication is not well defined as $(a b) H$ may not equal to $(c d) H$
eg: $G = S_{3}$ and $H = {e, (12)}$

\begin{matrix} H (23) = {(23), (123)} = H (123) \\ H (13) = {(13), (132)} = H (132) \\ H (23) \cdot H (13) = H (23) (13) = H (123) \\ H (123) \cdot H (132) = H (123) (132) = H e = H \end{matrix}

factor/quotient groups

let $H ⊲ G$ , then the set $G / H$ is a group under coset multiplication, where $G / H = {a H ∣ a \in G} = {H a, H b, H c, \dots}$ is the set of all cosets of $H$ in $G$

it must first be shown that the operation is well defined, ie. the correspondence $G / H \times G / H \to G / H$ is a function
let $a, a^{'}, b, b^{'} \in G$ , such that $a H = a^{'} H$ and $b H = b^{'} H$ , so need to verify that $a H b H = a^{'} H b^{'} H ⟹ a b H = a^{'} b^{'} H$
$a^{'} = a h_{1}$ and $b^{'} = b h_{2}$ for some $h_{1}, h_{2} \in H$ , thus, using associativity and that $a H = H a$ in normal subgroups: $a^{'} b^{'} H = a h_{1} b h_{2} H = a h_{1} b H = a h_{1} H b = a H b = a b H$
$e H = H$ is the identity
$a^{- 1} H$ is the inverse of $a H$
$(a H b H) c H = (a b) H c H = (a b) c H = a (b c) H = a H (b c) H$ , so it is associative
therefore, $G / H$ is a group
the converse is also true, ie. if $a H b H = a b H$ defines a group operation on the set of left cosets of $H$ in $G$ , then $H ⊲ G$

a H b H = a b H ⟺ H ⊲ G

eg: let $4 Z = {0, \pm 4, \pm 8, \dots}$ , to construct $Z / 4 Z$ , first determine the left cosets of $4 Z$ in $Z :$

\begin{aligned} 0 + 4 Z & = 4 Z \\ 1 + 4 Z & = {1, 5, 9, \dots; - 3, - 7, - 11, \dots} \\ 2 + 4 Z & = {2, 6, 10, \dots; - 2, - 6, - 10, \dots} \\ 3 + 4 Z & = {3, 7, 11, \dots; - 1, - 5, - 9, \dots} \end{aligned}

if $k \in Z$ , then $k = 4 q + r$ , where $0 \leq r < 4$ , therefore, $k + 4 Z = r + 4 q + 4 Z = r + 4 Z$
from the cayley table, it is clear that $Z / 4 Z ≅ Z_{4}$
in general, $Z / n Z ≅ Z_{n}$
eg: let $G = Z_{18}$ and $H = ⟨ 6 ⟩ = {0, 6, 12}$ , then $G / H = {0 + H, 1 + H, 2 + H, 3 + H, 4 + H, 5 + H}$
consider $(5 + H) + (4 + H) :$ $= 9 + H = 3 + 6 + H = 3 + H$ , as $H$ absorbs all multiples of $6$
note: since $H ⊲ G$ , $| a H |$ can be interpreted as either the size of the set $a H$ , or as the order of element $a H$ in $G / H$ , but the appropriate interpretation will be clear from the context