\LaTeXparent{full.tex}
\chapter{Quotient Groups}
The elements of a group $G$ can be partitioned, using a subgroup $H$ of $G$,
into cells called {\bf cosets}. One of the nicest features of {\it ESG} will
allow you to investigate these cosets, which sometimes form a new group
called a {\bf factor} or {\bf quotient group} of $G$.
\section{Before the Lab}
You will prove the following theorem for homework:
\begin{description}
\item[Theorem 6.1:] Let $G$ be a group and $H$ a subgroup of $G$. Define $%
\sim $ by saying $a\sim g$ if and only if $ag^{-1}\in H$. Then $\sim $ is an
equivalence relation.
\end{description}
We would obtain a similar result if we defined $\sim $ by saying $a\sim g$
if and only if $g^{-1}a\in H$.
In the the theorem, we have $a=hg$, for some $h\in H.$ In fact, as we let $h$
run through all the elements of $H$, we obtain all elements of $G$ that are
related to $g$ under the equivalence relation $\sim $. The cell of the
partition created by $\sim $ that contains $g$ is denoted by $Hg$ and is
called the {\bf right coset }of $H$ in $G$\ that contains{\bf \ }$g$. In the
second case, we create $gH$, the {\bf left coset} of $H$ in $G$ that contains%
{\bf \ }$g$. Why do both $Hg$ and $gH$ contain $g$?
For the two examples below, we'll consider the subgroups $H=\{1,r_1,r_2\}$
and $K=\{1,m_1\}$ of $S_3$. We'll compute the right and left cosets for each
subgroup. Confirm the computations as you read through the example.
\begin{description}
\item[Example 1:]
$$
\begin{array}[t]{l}
H:
\text{ right cosets} \\ \\
\{1,r_1,r_2\}1=\{1,r_1,r_2\} \\
\{1,r_1,r_2\}r_1=\{1r_1,r_1r_1,r_2r_1\}=\{r_1,r_2,1\} \\
\{1,r_1,r_2\}r_2=\{1r_2,r_1r_2,r_2r_2\}=\{r_2,1,r_1\} \\
\{1,r_1,r_2\}m_1=\{1m_1,r_1m_1,r_2m_1\}=\{m_1,m_3,m_2\} \\
\{1,r_1,r_2\}m_2=\{1m_2,r_1m_2,r_2m_2\}=\{m_2,m_1,m_3\} \\
\{1,r_1,r_2\}m_3=\{1m_3,r_1m_3,r_2m_3\}=\{m_3,m_2,m_1\}
\end{array}
$$
\end{description}
There are two distinct right cosets, $\{1,r_1,r_2\}$ and $\{m_1,m_2,m_3\}$,
which form a partition of the elements of $G$.
$$
\begin{array}[t]{l}
H:
\text{ left cosets} \\ \\
1\{1,r_1,r_2\}=\{1,r_1,r_2\} \\
r_1\{1,r_1,r_2\}=\{r_11,r_1r_1,r_1r_2\}=\{r_1,r_2,1\} \\
r_2\{1,r_1,r_2\}=\{r_21,r_2r_1,r_2r_2\}=\{r_2,1,r_1\} \\
m_1\{1,r_1,r_2\}=\{m_11,m_1r_1,m_1r_2\}=\{m_1,m_2,m_3\} \\
m_2\{1,r_1,r_2\}=\{m_21,m_2r_1,m_2r_2\}=\{m_2,m_3,m_1\} \\
m_3\{1,r_1,r_2\}=\{m_31,m_3r_1,m_3r_2\}=\{m_3,m_1,m_2\}
\end{array}
$$
There are two distinct left cosets, $\{1,r_1,r_2\}$ and $\{m_1,m_2,m_3\}$.
These agree with the right cosets: $Hg=gH$ for every $g\in G$.
\begin{description}
\item[Example 2:]
$$
\begin{array}[t]{llll}
K:\text{ right cosets} & & & K:
\text{ left cosets} \\ & & & \\
\{1,m_1\}1=\{1,m_1\} & & & 1\{1,m_1\}=\{1,m_1\} \\
\{1,m_1\}r_1=\{r_1,m_2\} & & & r_1\{1,m_1\}=\{r_1,m_3\} \\
\{1,m_1\}r_2=\{r_2,m_3\} & & & r_2\{1,m_1\}=\{r_2,m_2\} \\
\{1,m_1\}m_1=\{m_1,1\} & & & m_1\{1,m_1\}=\{m_{1,}.1\} \\
\{1,m_1\}m_2=\{m_2,r_1\} & & & m_2\{1,m_1\}=\{m_2,r_2\} \\
\{1,m_1\}m_3=\{m_3,r_2\} & & & m_3\{1,m_1\}=\{m_3,r_1\}
\end{array}
$$
\end{description}
There are three distinct right cosets, $\{1,m_1\}$, $\{r_1,m_2\}$, and $%
\{r_2,m_3\}.$ There are three distinct left cosets, $\{1,m_1\}$, $%
\{r_1,m_3\} $, and $\{r_2,m_2\}$. This time, however, not all of the left
and right cosets agree. That is, there are elements $g\in S_3$ so that $%
Kg\neq gK.$ The distinction between $H$ and $K$ is crucial and at the heart
of studying quotient groups. The collection of cosets forms a new group
precisely when the left and right cosets agree. The problem, it turns out,
is in trying to define a binary operation on the cosets.
The {\bf coset operation} is best defined through an example. We only have
the operation defined on $G$ to use. Let's start by using the left cosets of
$H$. Set $X=\{1,r_1,r_2\}$ and $Y=\{m_1,m_2,m_3\}$. Choose an element from $X
$ and one from $Y,$ for example, $r_1\in X$ and $m_2\in Y$. Since $r_1m_2=m_1
$, and $m_1\in Y$, we define the coset operation $*$ by $X*Y=Y.$ No matter
which representative of the cosets $X$ and $Y$ we choose, we will obtain the
same answer in this case. So we can construct the Cayley table using the
operation $*$:%
$$
\begin{array}{ccc}
& & \\
& &
\begin{array}[t]{lll}
\ast & X & Y \\
X & X & Y \\
Y & Y & X
\end{array}
\\
& &
\end{array}
$$
Check that this forms a group of order 2. Which coset serves as the identity
element?
If we try to do this with the left cosets of $K,$ however, we run into
problems. Set $U=\{1,m_1\}$, $V=\{r_1,m_3\}$, $W=\{r_2,m_2\}.$ If we compute
$V*W$ using the elements $m_3$ and $r_2$, we have $V*W=W.$ But if we do the
calculation with $r_1$ and $m_2$, we obtain $V*W=U$. Obviously there is a
problem here---the operation is not well-defined. In the lab, we will see
how {\it ESG} demonstrates this with color.
To summarize, we have the theorem:
\begin{description}
\item[Theorem 6.2:] Let $G$ be a group and let $H$ be a subgroup of $G$.
The cosets of $H$ in $G$ form a group if and only if $Hg=gH$ for all $g\in G.
$
\end{description}
When the condition of the theorem holds, we will say that the subgroup $H$
is a {\bf normal }subgroup of $G$. The collection of cosets (right or left,
of course) is then called the {\bf quotient }or{\bf \ factor group}, $G$ mod
$H$, denoted by $G/H$. In our examples above, the subgroup $H$ is normal but
$K$ is not.
It is especially important for you to work out one example carefully on your
own before you use the computer. Do all the computations {\bf by hand} for
question 1.
\begin{enumerate}
\item[1.] Consider the subgroups $H_1=\{1,r_2,d_1,d_2\},H_2=\{1,m_1\},$ and
$H_3=\{1,r_2\}$ of $D_4.$
\begin{enumerate}
\item Find all the left cosets of the subgroups in $D_4$.
\item Find all the right cosets of the subgroups in $D_4$.
\item For which subgroups are the left and right cosets equal?
\item For each subgroup identified in part (c), construct a group table for
the quotient group $G/H_i.$ What familiar group has the same group table?
\end{enumerate}
\end{enumerate}
Be sure to bring your subgroup lattices to the lab.
\section{In the Lab}
Check your answers to question 1 with {\it ESG} before continuing with the
problems below. Choose option 3 (Subgroups and Cosets/Quotients) from the
{\bf Group Properties Menu} for $D_4$, and generate each subgroup $H_i$.
Answer ``Y'' to the question, ``Would you like to see the left cosets of
this subgroup?'' Look at the coloring of the Cayley table of $D_4,$ grouped
by the left cosets of each subgroup. In some cases, you will be asked,
``Would you like to see the quotient table?'' Be sure you understand how the
Cayley table is transformed when you answer ``Y.''
\begin{enumerate}
\item[2.] In your own words, explain how you can determine from the table
on the computer screen that the coset operation is well-defined or not
well-defined.
\end{enumerate}
For problems 3-10, you may use the computer for your computations. Answer
the following questions for each group $G$ and {\bf all} nontrivial proper
subgroups $H$ of $G$.
\begin{enumerate}
\begin{enumerate}
\item Record the {\bf distinct} left and right cosets of the subgroup $H$
in $G$.
\item Is the subgroup $H$ normal in $G$? If so, have the computer construct
a group table for the new quotient group $G/H$. What familiar group has the
same group table? Your answer should be a known group from the {\it ESG}
library. Be sure to record on your subgroup lattices which subgroups are
normal.
\end{enumerate}
\end{enumerate}
Note that there are no general formulas which help us figure out which group
$G/H$ actually is. You have to rely on computing the order of $G/H$ and on
knowing something about the various groups of that order.
\begin{enumerate}
\item[3.] $G=D_5$
\item[4.] $G=D_6$
\item[5.] $G=A_4$
\item[6.] $G=Q_6$
\item[7.] $G=D_7$
\item[8.] $G=M$
\item[9.] $G=D_8$
\item[10.] $G=Q_8$
\end{enumerate}
\section{Further Work}
\begin{enumerate}
\item[11.] Make at least two conjectures about the kinds of subgroups which
always seem to be normal in a finite group $G$. (Hint: think about the
special subgroups that we considered in an earlier lab or the index of the
subgroup).
\item[12.] Make at least two conjectures about the factor groups $D_n/H$,
where $H$ is either the commutator subgroup or center of $D_n$.
\item[13.] Prove Theorem 6.1: Let $G$ be a group and $H$ a subgroup of $G$.
Define $\sim $ by $a\sim g$ if and only if $ag^{-1}\in H$. Then $\sim $ is
an equivalence relation.
\end{enumerate}