13 Quotient Groups (cont.)

left/right cosets

For any $N \leq G$ (see Subgroup) and any $g \in G$ let:
$g N = {g n | n \in N}; N g = {n g | n \in N}$
be called (respectively) a left coset and a right coset of $N$ in $G$ . Any element of a coset is called a representative for the coset.

We denote the set of left cosets by $\frac{G}{H}$ , as we see in Quotient Group.

Warning

In general $g H \neq H g$ for all $g \in G$ . Essentially, once the left/right cosets agree with each other, then we have a Normal Subgroup specifically for that $H$ .

Examples

Say we have $G = S_{3}$ (see Symmetric Group & Permutations) and $H = ⟨ (1 2) ⟩ {1, (1 2)}$ . Then the left-cosets of $H$ are;

$g \in G$	$g H$
1	${1 1, 1 (1 2)} = {1, (1 2)} = H$
$(1 2)$	${(1 2) 1, (1 2) (1 2)} = {(1 2), 1}$
$(1 3)$	${(1 3), (1 2 3)}$
$(2 3)$	${(2 3), (1 3 2)}$
$(1 2 3)$	${(1 2 3), (1 3)}$
$(1 3 2)$	${(1 3 2), (2 3)}$
Notice that these are partitioned. Namely name:

$1 H, (1 2) H = C_{1}$
$(1 3) H, (1 2 3) H = C_{2}$
$(2 3) H, (1 3 2) H = C_{3}$
creating a partition of 3 elements.

Note

For one right coset see the order matters:
$H (1 3) = {1 (1 3), (1 2) (1 3)} = {(1 3), (1 3 2)} \neq (1 3) H$

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.

Theorem

Suppose $G$ is a finite Group and $H \leq G$ (see Subgroup). Then $| H |$ divides $| G |$ (see Order (Groups)).

Proof

The left cosets of $H$ partition $G$ , so they all have size $| H |$ and are disjoint. Thus:
$
|G| = (# \text{ cosets of $H$ in $G$ })|H|
$
Thus $| H |$ divides $| G |$ .

☐

An alternative method of this is saying:

Theorem

The number of left cosets of $H$ in $G$ equals $\frac{| G |}{| H |}$

Proof

See Index (Groups).

☐

index

For $H \leq G$ (see Subgroup) the index of $H$ in $G$ is the number of cosets of $H$ in $G$ .

This is often denoted $[G \cdot H]$ or $| G : H |$ . Notice that this just $| \frac{G}{H} | = [G \cdot H]$ by definition. If $G$ is finite ( $| G | < \infty$ ) then $[G \cdot H] = \frac{| G |}{| H |}$ via Lagrange's Divisibility Theorem of Order of Subgroups.

dg-publish: true

Theorem

If $H \leq G$ (see Subgroup) and $[G \cdot H] = 2$ (see Index (Groups)) then $H ⊴ G$ .

Proof

Use 11 Basic Axioms and Examples (Groups)#25 and apply it to $H$ , using Equivalences of Group Actions.

☐

So all of this means that for any $g \in G$ then $| g |$ must divide $| G |$ :

Proof

${ord}_{G} (g) = | ⟨ g ⟩ |$ , but using Lagrange's Divisibility Theorem of Order of Subgroups then it must divide $| G |$ .

☐

But when does the set $\frac{G}{H}$ have a group structure? Namely one such that $π : G \to \frac{G}{H}$ to be a Homomorphism (or Group Morphism). The answer is that:

Claim

If $π : G \to \frac{G}{H}$ is a Homomorphism (or Group Morphism), then:

π (g_{1} g_{2}) = π (g_{1}) π (g_{2})

for all $g_{1}, g_{2} \in G$ . But all of this is only iff:

(g_{1} g_{2}) H = (g_{1} H) \cdot (g_{2} H)

So we must define the Coset Product by:

(g_{1} H) \cdot (g_{2} H) := (g_{1} g_{2}) H

We have to consider that this well defined.

coset product

We define the coset product for $H \leq G$ as:
$(g_{1} H) \cdot (g_{2} H) := (g_{1} g_{2}) H$
for all $g_{1}, g_{2} \in G$ . Then this operation is well defined iff $H ⊴ G$ .

We want to prove that this is well defined; namely if $g_{1} H = \tilde{g_{1}} H$ and $g_{2} H = \tilde{g_{2}} H$ do we have:
$(g_{1} g_{2}) H = (\tilde{g_{1}} \tilde{g_{2}}) H$

Example

To highlight that it is not always well-defined, consider the example $H = {1, (1 3)}$ for $G = S_{3}$ . We had:

Notice $(1 2) H \cdot (2 3) H = (1 2) (2 3) H = (1 2 3) H$ . Namely if the green coset is $C_{2}$ and the red one is $C_{3}$ then does it always equal $C_{3}$ (as it did in our calculation).

What we see is that $(1 3 2) H \cdot (2 3) H = (1 3 2) (2 3) H = (1 3) H$ is in $C_{1}$ instead, which is different. Thus we needed the other condition of being a Normal Subgroup to be actually well defined.

The Proof

Let's do the proof using the importance of Normal Subgroup:

Proof

( $\Rightarrow$ ): We want to show that $H ⊴ G$ . Take any $h \in H$ and $g \in G$ . We got to show that $g h g^{- 1} \in H$ .

First notice that $g h \in g H$ . That means that since $g \in g H$ then $(g h) H = g H$ are the same cosets. But then because $g_{1} H = g_{2} H ⟺ g_{2}^{- 1} g_{1} \in H ⟺ g_{1}^{- 1} g_{2} \in H$ , then we can compare $(g h) H$ and $g H$ . Namely:
$(g h) H \cdot g^{- 1} H = (g h g^{- 1}) H$
and:
$g H \cdot g^{- 1} H = (g g^{- 1}) H = H$
Thus $(g h g^{- 1}) H = H$ so then $g h g^{- 1} \in H$ , as we wanted.

( $\Leftarrow$ ): TBD

☐

As a consequence, for any $K ⊴ G$ , the set $\frac{G}{K}$ is a Group with the operation defined in the Coset Product. The projection map $π : G \to \frac{G}{K} : g \mapsto g K$ is a Homomorphism (or Group Morphism). The identity of $\frac{G}{K}$ is the coset $1 K$ since:

(g K) (1 K) = (g 1) K = g K = (1 g) K = (1 K) (g K)

Further notice that $(g K)^{- 1} = g^{- 1} K$ .

Also:

\begin{aligned} Ker (π) & = {g \in G | g K = 1 K} \\ = {g \in G | g \in K} \\ = K \end{aligned}

so as a consequence, every Normal Subgroup is a Kernel of a Homomorphism (or Group Morphism).

That means that whenever you have some $K ⊴ G$ then there is some corresponding $π$ map where $Ker (π) = K$ .

So for any $φ : G \to H$ Homomorphism (or Group Morphism) then $Ker (φ) ⊴ G$ .

But how does $\frac{G}{Ker (φ)}$ fit into the picture?