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 22: Cosets and quotient groups

Let $G$ be a group. A subgroup of $G$ is a subset $H\subseteq G$ such that

(a)	If $h_1,h_2\in H$ then $h_1,h_2\in H,$
(b)	$1\in H,$
(c)	If $h\in H$ then $h^{-1}\in H\text{.}$

Let $H$ be a subgroup of $G\text{.}$ A coset of $H$ in $G$ is a subset $$gH=\left\{gh\hspace{0.17em}\right|\hspace{0.17em}h\in H\}$$ with $g\in G\text{.}$

$$\begin{array}{ccc}{\displaystyle G}& =& {\displaystyle \{\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}=S_3,}\\ {\displaystyle H}& =& {\displaystyle \{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}\text{.}}\end{array}$$ Then $$\begin{array}{ccc}{\displaystyle \begin{array}{c}\end{array}·H}& =& {\displaystyle \{\begin{array}{c}\end{array},\begin{array}{c}\end{array},\},}\\ {\displaystyle \begin{array}{c}\end{array}·H}& =& {\displaystyle \{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},}\\ {\displaystyle \begin{array}{c}\end{array}·H}& =& {\displaystyle \{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},}\\ {\displaystyle \begin{array}{c}\end{array}·H}& =& {\displaystyle \{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},}\\ {\displaystyle \begin{array}{c}\end{array}·H}& =& {\displaystyle \{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},}\\ {\displaystyle \begin{array}{c}\end{array}·H}& =& {\displaystyle \{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}\text{.}}\end{array}$$ There are really only $3$ cosets here since $$\begin{array}{c}\begin{array}{c}\end{array}·H=\begin{array}{c}\end{array}·H=\{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},\\ \begin{array}{c}\end{array}·H=\begin{array}{c}\end{array}·H=\{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},\\ \begin{array}{c}\end{array}·H=\begin{array}{c}\end{array}·H=\{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}\text{.}\end{array}$$ $\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.=\left\{gH\hspace{0.17em}\right|\hspace{0.17em}g\in G\}$ is the set of cosets of $H$ in $G\text{.}$ In our example $$\begin{array}{ccc}{\displaystyle \raisebox{1ex}{$S_3$}\!\left/ \!\raisebox{-1ex}{$H$}\right.}& =& {\displaystyle \{\begin{array}{c}\end{array}·H,\begin{array}{c}\end{array}·H,\begin{array}{c}\end{array}·H\}}\\ & =& {\displaystyle \{\{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},\{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},\{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}\}\text{.}}\end{array}$$

Let $G$ be a group and let $H$ be a subgroup of $G\text{.}$ Let $g\in G\text{.}$

(a)	$\text{Card}\left(gH\right)=\text{Card}\left(H\right)\text{.}$
(b)	$\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.$ is a partition of $G\text{.}$

A partition of a set $S$ is a collection $𝒮$ of subsets of $S$ such that

(a)	The union of the sets in $𝒮$ is $S,$
(b)	If $U_1,U_2\in 𝒮$ then $U_1=U_2$ or $U_1\cap U_2=\varnothing \text{.}$

Let $G$ be a group and let $H$ be a subgroup of $G\text{.}$ Then $$\text{Card}\left(G\right)=\text{Card}\left(\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.\right)\text{Card}\left(H\right)\text{.}$$

		Proof of the proposition.
	(a) To show: $\text{Card}\left(gH\right)=\text{Card}\left(H\right)\text{.}$ To show: There exists a bijective function $f:H\to gH\text{.}$ Let $$\begin{array}{cccc}f:& H& ⟶& gH\\ & h& ⟼& gh\end{array}\text{.}$$ To show: $f$ is bijective. To show: There exists a function $\phi :gH\to H$ such that $\phi \circ f={\text{id}}_H$ and $f\circ \phi ={\text{id}}_{gH}\text{.}$ Let $$\begin{array}{cccc}\phi :& gH& ⟶& H\\ & x& ⟼& g^{-1}x\end{array}$$ so that $\phi \left(x\right)=g^{-1}x\text{.}$ To show: (aa) $\phi \circ f={\text{id}}_H\text{.}$ (ab) $f\circ \phi ={\text{id}}_{gH}\text{.}$ (aa) To show: If $h\in H$ then $(\phi \circ f)\left(h\right)={\text{id}}_H\left(h\right)\text{.}$ Assume $h\in H\text{.}$ To show: $(\phi \circ f)\left(h\right)={\text{id}}_H\left(h\right)\text{.}$ $$(\phi \circ f)\left(h\right)=\phi \left(f\left(h\right)\right)=\phi \left(gh\right)=g^{-1}gh=h={\text{id}}_H\left(h\right)\text{.}$$ (ab) To show: If $x\in gH$ then $(f\circ \phi )\left(x\right)={\text{id}}_{gH}\left(x\right)\text{.}$ Assume $x\in gH\text{.}$ To show: $(f\circ \phi )\left(x\right)={\text{id}}_{gH}\left(x\right)\text{.}$ $$(f\circ \phi )\left(x\right)=f\left(\phi \left(x\right)\right)=f\left(g^{-1}x\right)=g{g}^{-1}x=x={\text{id}}_{gH}\left(x\right)\text{.}$$ (b) To show: $\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.$ is a partition of $G\text{.}$ To show: (ba) $\bigcup _{g\in G}gH=G\text{.}$ (bb) If $g_1,g_2\in G$ then $g_{1}H=g_{2}H$ or $g_{1}H\cap g_{2}H=\varnothing \text{.}$ (ba) To show: If $x\in G$ then there exists $g\in G$ such that $x\in gH\text{.}$ Assume $x\in G\text{.}$ To show: There exists $g\in G$ such that $x\in gH\text{.}$ Let $g=x\text{.}$ To show: $x\in gH\text{.}$ $$x=·1\in xH,$$ since $1\in H\text{.}$ (bb) Assume $g_1,g_2\in G$ and $g_{1}H\cap g_{2}H\ne \varnothing \text{.}$ To show: $g_{1}H=g_{2}H\text{.}$ Since $g_{1}H\cap g_{2}H\ne \varnothing$ there exists $x\in g_{1}H\cap g_{2}H\text{.}$ Let $h_1,h_2\in H$ be such that $x=g_1{h}_1$ and $x=g_2{h}_2\text{.}$ To show: $g_{1}H=g_{2}H\text{.}$ To show: (bba) $g_{1}H\subseteq g_{2}H\text{.}$ (bbb) $g_{2}H\subseteq g_{1}H\text{.}$ (bba) To show: If $y\in g_{1}H$ then $y\in g_{2}H\text{.}$ Assume $y\in g_{1}H\text{.}$ Then there exists $h\in H$ such that $y=g_{1}H\text{.}$ To show: $y\in g_{2}H\text{.}$ $$y=g_{1}h=g_1{h}_1{h}_1^{-1}h=x{h}_1^{-1}h=g_2{h}_2{h}_1^{-1}h\in g_{2}H,$$ since $H$ is a subgroup. (bbb) To show: If $z\in g_{2}H$ then $z\in g_{1}H\text{.}$ Assume $z\in g_{2}H\text{.}$ Then there exists $h\prime \in H$ such that $z=g_{2}h\prime \text{.}$ To show: $z\in g_{1}H\text{.}$ $$z=g_{2}h\prime =g_2{h}_2{h}_2^{-1}h\prime =x{h}_2^{-1}h\prime =g_1{h}_1{h}_2^{-1}h\prime \in g_{1}H,$$ since $H$ is a subgroup. So $g_{1}H=g_{2}H\text{.}$ So $\raisebox{1ex}{$G$}\!\left/ \!\raisebox{-1ex}{$H$}\right.$ is a partition of $G\text{.}$ $\square$

Notes and References

These are a typed copy of Lecture 22 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on September 13, 2011.

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