Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 24 September 2014

Lecture 23: Quotient groups

Let $G$ be a group and let $H$ be a subgroup of $G .$ The set of cosets of $H$ in $G$ is $\frac{G}{H} = {g H | g \in G},$ where $g H = {g h | h \in H}$ for $g \in G .$

A normal subgroup of $G$ is a subgroup $K$ of $G$ such that if $g \in G$ and $k \in K$ then $g k g^{- 1} \in K .$

Let $G$ be a group and let $H$ be a subgroup of $G .$ Then $\frac{G}{H}$ with $\begin{matrix} \frac{G}{H} \times \frac{G}{H} & ⟶ & \frac{G}{H} \\ (g_{1} H, g_{2} H) & ⟼ & g_{1} g_{2} H \end{matrix}$ is a well defined group if and only if $H$ is a normal subgroup of $G .$

If $\begin{matrix} G & = & {\begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}} = S_{3} and \\ H & = & {\begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}} \end{matrix}$ the cosets in $\frac{G}{H}$ are $\begin{matrix} \begin{matrix} \end{matrix} \cdot H = \begin{matrix} \end{matrix} \cdot H = {\begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}}, \\ \begin{matrix} \end{matrix} \cdot H = \begin{matrix} \end{matrix} \cdot H = {\begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}}, \\ \begin{matrix} \end{matrix} \cdot H = \begin{matrix} \end{matrix} \cdot H = {\begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}} \end{matrix}$

If we try to define $\begin{matrix} \begin{matrix} \frac{G}{H} \times \frac{G}{H} & ⟶ & \frac{G}{H} \\ (g_{1} H, g_{2} H) & ⟼ & g_{1} g_{2} H \end{matrix} & (*) \end{matrix}$ then (·H) (·H)= ·H= { , } = (·H) (·H)= ·H= { , } gives ${\begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}}^{2} = {\begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}} and {\begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}}^{2} = {\begin{matrix} \end{matrix}, \begin{matrix} \end{matrix}} .$ In other words, for this $H,$ $(*)$ is not a function! The theorem tells us that this is "because" $H$ is not normal: $\begin{matrix} \end{matrix} \in H,$ but $\begin{matrix} \end{matrix} \cdot \begin{matrix} \end{matrix} \cdot {(\begin{matrix} \end{matrix})}^{- 1} = \begin{matrix} \end{matrix} \cdot \begin{matrix} \end{matrix} \cdot \begin{matrix} \end{matrix} = {\begin{matrix} \end{matrix}}_{\begin{matrix} \end{matrix}}^{\begin{matrix} \end{matrix}} = \begin{matrix} \end{matrix}$ is not in $H .$

Proof of the theorem.

\Leftarrow :

Assume

H

is a normal subgroup.

To show:

(a)

\begin{matrix} \frac{G}{H} \times \frac{G}{H} & ⟶ & \frac{G}{H} \\ (g_{1} H, g_{2} H) & ⟼ & g_{1} g_{2} H \end{matrix}

is a function.

(b)

Using

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

as product:

(ba)	If $g_{1} H, g_{2} H, g_{3} H \in \frac{G}{H}$ then $(g_{1} H g_{2} H) g_{3} H = g_{1} H (g_{2} H g_{3} H) .$
(bb)	There exists $e H \in \frac{G}{H}$ such that if $g H \in \frac{G}{H}$ then $(e H) (g H) = g H$ and $(g H) (e H) = g H .$
(bc)	If $g H \in \frac{G}{H}$ then there exists $x H \in \frac{G}{H}$ such that $(x H) (g H) = e H$ and $(g H) (x H) = e H .$

(a)

To show: If

(g_{1} H, g_{2} H) = (g_{1}^{'} H, g_{2}^{'} H)

then

g_{1} g_{2} H = g_{1}^{'} g_{2}^{'} H .

Assume

(g_{1} H, g_{2} H) = (g_{1}^{'}, g_{2}^{'} H) .

Then

g_{1} H = g_{1}^{'} H

and

g_{2} H = g_{2}^{'} .

To show:

g_{1} g_{2} H = g_{1}^{'} g_{2}^{'} H .

To show:

g_{1} g_{2} \in g_{1}^{'} g_{2}^{'} H,

since the cosets partition

G .

We know

g_{1} \in g_{1}^{'} H

and

g_{2} \in g_{2}^{'} H .

So there exist

h_{1}, h_{2} \in H

such that

g_{1} = g_{1}^{'} h_{1}

and

g_{2} = g_{2}^{'} h_{2} .

To show:

g_{1} g_{2} \in g_{1}^{'} g_{2}^{'} H .

\begin{matrix} g_{1} g_{2} & = & g_{1}^{'} h_{1} g_{2}^{'} h_{2} \\ = & g_{1}^{'} g_{2}^{'} {(g_{2}^{'})}^{- 1} h_{1} g_{2}^{'} h_{2} \\ = & g_{1}^{'} g_{2}^{'} ({(g_{2}^{'})}^{- 1} h_{1} g_{2}^{'}) h_{2} \in g_{1}^{'} g_{2}^{'} H \end{matrix}

since

H

is a normal subgroup of

G .

(b)

Use

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

as product in

\frac{G}{H} .

(ba)

To show: If

g_{1} H, g_{2} H, g_{3} H \in \frac{G}{H}

then

(g_{1} H g_{2} H) g_{3} H = g_{1} H (g_{2} H g_{3} H) .

Assume

g_{1} H, g_{2} H, g_{3} H \in \frac{G}{H} .

To show:

(g_{1} H g_{2} H) g_{3} H = g_{1} H (g_{2} H g_{3} H) .

Then

(g_{1} H g_{2} H) g_{3} H = g_{1} g_{2} H \cdot g_{3} H = (g_{1} g_{2}) g_{3} H

and

g_{1} H (g_{2} H g_{3} H) = g_{1} H \cdot g_{2} g_{3} H = g_{1} (g_{2} g_{3}) H

and

(g_{1} g_{2}) g_{3} = g_{1} (g_{2} g_{3})

since

G

is associative.
So

(g_{1} H g_{2} H) g_{3} H = g_{1} H (g_{2} H g_{3} H) .

(bb)

To show: There exist

e H \in \frac{G}{H}

such that if

g H \in \frac{G}{H}

then

(e H) (g H) = g H

and

(g H) (e H) = g H .

Let

e H = H .

To show: if

g H \in \frac{G}{H}

then

(e H) (g H) = g H

and

(g H) (e H) = g H .

Assume

g H \in \frac{G}{H} .

To show:

(bba)	$e H \cdot g H = g H .$
(bbb)	$g H \cdot e H = g H .$

(bba)	$e H \cdot g H = 1 \cdot H \cdot g \cdot H = 1 (g) H = g H,$ since $1$ is an identity for $G .$
(bbb)	$g H e H = g H 1 \cdot H = (g 1) \cdot H = g H .$

(bc)

To show: If

g H \in \frac{G}{H}

then there exists

x H \in \frac{G}{H}

such that

(g H) (x H) = e H

and

x H g H = H .

Assume

g H \in \frac{G}{H} .

To show: There exists

x H \in \frac{G}{H}

such that

g H x H = H and x H g H = H .

Let

x H = g^{- 1} H .

To show:

(bca)	$g H x H = H .$
(bcb)	$x H g H = H .$

(bca)	$g H x H = g H g^{- 1} H = g g^{- 1} H = 1 \cdot H = H .$
(bcb)	$x H g H = g^{- 1} H g H = (g g^{- 1}) H = 1 \cdot H = H .$

This completed the proof that if

H

is a normal subgroup of

G

then

\frac{G}{H}

with

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

is a group.

$□$

Notes and References

These are a typed copy of Lecture 23 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on September 14, 2011.

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