Quotient Group

Definition

Let $φ : G \to H$ be a Homomorphism (or Group Morphism) with Kernel $K$ . The quotient group, $\frac{G}{K}$ (read $G$ modulo $K$ or simply $G$ mod $K$ ) is the group whose elements are the Fiber (Group Theory) of $φ$ with group operation defined below:

Namely if $X$ is the Fiber (Group Theory) above $a$ and $Y$ is the fiber above $b$ , then the product of $X$ with $Y$ is defined to be the fiber above the product $a b$ .

The notation implies the idea that the Kernel $K$ is a single element in the group $\frac{G}{K}$ , and we shall see in Properties of Quotient Groups that, as in the case of $\frac{Z}{n Z}$ as seen here that the other elements of $\frac{G}{K}$ are just the "translates" of the kernel $K$ . We can think of $\frac{G}{K}$ as "dividing out" by $K$ , hence the name (and why we refer to it by mod $K$ , as the $K$ is dividing it into distinct parts).

Definition using Cosets

Definition

We can define the set $\frac{G}{H}$ where $H \leq G$ as the set of left cosets of $H$ in $G$ . We have a canonical map:

π : G \to \frac{G}{H}; g \mapsto g H

Example

Consider $G = S_{3} = {1, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}$ , and let $H = ⟨ (1 3) ⟩ = {1, (1 3)}$ .

We can do the same for the right cosets:

$H 1$ is the first one
$H (1 2) = {1 (1 2), (1 3) (1 2)} = {(1 2), (1 2 3)}$ . But notice then $H (1 2) \neq (1 2) H$ so then this is not a Normal Subgroup.
By the remaining number of elements then we have $H (2 3) = {(2 3), (1 3 2)}$ .

Consider instead $K \leq G$ with $K = ⟨ (1 2 3) ⟩ = {1, (1 2 3), (1 3 2)}$ . Then:

$1 K = {1, (1 2 3), (1 3 2)} = K 1$ as expected.
We haven't seen $(1 2) K = {(1 2), (1 3), (2 3)} = K (1 2)$ since the size of our coset is the same for each, so we have the other half of the elements.
Thus $K$ is a Normal Subgroup.