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\ihead{Math 331, Fall 2026, Lecture 8 and 9, version \today}
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\begin{document}
\section*{Math 331 Fall 2026, Lecture 8 and 9: Groups concluded and
summarized}

\textbf{website:}
\texttt{\href{https://www.cip.ifi.lmu.de/~grinberg/t/26fa}{\texttt{https://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/t/26fa/}}}

(GPT-6 was used to polish the writing that follows.)

\setcounter{section}{1}\setcounter{subsection}{12}\setcounter{theo}{6}

\begin{remark}
\phantomsection\label{rmk.group.center.GL} The answer to the question in
Example~\ref{exa.group.center.GL} in Lecture 6--7 is \textquotedblleft
no\textquotedblright: the only central matrices in $\operatorname*{GL}%
\nolimits_{n}\left(  \mathbb{R}\right)  $ are the (nonzero scalar) multiples
$\lambda I_{n}$ of the identity matrix $I_{n}$. Let us outline how it is
proved. Here $n$ is a positive integer. For $n=1$, every invertible $1\times
1$-matrix already has the form $\lambda I_{n}$, so we may assume that $n\geq2$.

Let $A=\left(  a_{i,j}\right)  _{1\leq i,j\leq n}\in\operatorname*{GL}%
\nolimits_{n}\left(  \mathbb{R}\right)  $ be a central matrix. Thus, $A$
commutes with any invertible $n\times n$-matrix. In particular, $A$ commutes
with the permutation matrix $T_{1,2}$ that differs from $I_{n}$ in that its
first two rows (or columns) are swapped:%
\[
T_{1,2}=\left(
\begin{array}
[c]{ccccc}
& 1 &  &  & \\
1 &  &  &  & \\
&  & 1 &  & \\
&  &  & \ddots & \\
&  &  &  & 1
\end{array}
\right)
\]
(where all invisible entries are $0$'s). But $T_{1,2}\in\operatorname*{GL}%
\nolimits_{n}\left(  \mathbb{R}\right)  $ and%
\begin{align*}
AT_{1,2}  &  =\left(  A\text{ with columns }1\text{ and }2\text{
swapped}\right)  \qquad\text{and}\\
T_{1,2}A  &  =\left(  A\text{ with rows }1\text{ and }2\text{ swapped}\right)
.
\end{align*}
Since $A$ is central, we have $AT_{1,2}=T_{1,2}A$, so these two matrices are
equal. Thus, for example, by comparing the $\left(  1,2\right)  $-th entries
of $AT_{1,2}$ and $T_{1,2}A$, we get%
\[
a_{1,1}=a_{2,2}.
\]
Similarly,%
\[
a_{i,i}=a_{j,j}\qquad\text{for all }i\neq j
\]
(since we can play the same game with $T_{i,j}$ instead of $T_{1,2}$, where
$T_{i,j}$ is the permutation matrix obtained from the identity matrix $I_{n}$
by swapping its $i$-th and $j$-th rows). In other words, all diagonal entries
of $A$ are equal.

We can also consider the $n\times n$-matrix%
\[
E_{1,2}=\left(
\begin{array}
[c]{ccccc}%
1 & 1 &  &  & \\
& 1 &  &  & \\
&  & 1 &  & \\
&  &  & \ddots & \\
&  &  &  & 1
\end{array}
\right)  ;
\]
\Needspace{7\baselineskip} this is the identity matrix $I_{n}$ with an extra
$1$ in the $\left(  1,2\right)  $-th cell. This matrix $E_{1,2}$ is
invertible: its inverse is $I_{n}$ with an extra $-1$ in the same cell. Thus,
$E_{1,2}\in\operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)  $, and%
\begin{align*}
AE_{1,2}  &  =\left(  A\text{ with column }1\text{ added to column }2\right)
\qquad\text{and}\\
E_{1,2}A  &  =\left(  A\text{ with row }2\text{ added to row }1\right)  .
\end{align*}
Since $A$ is central, these two products are equal. Comparing their $\left(
2,2\right)  $-entries, we see%
\[
a_{2,1}+a_{2,2}=a_{2,2}.
\]
Therefore,%
\[
a_{2,1}=0.
\]
Similarly,%
\[
a_{j,i}=0\qquad\text{for all }j\neq i
\]
(since we can play the same game with the matrix $E_{i,j}$ instead of
$E_{1,2}$, where $E_{i,j}$ is the identity matrix $I_{n}$ with an extra $1$ in
cell $\left(  i,j\right)  $, and we compare the $(j,j)$-th entries of
$AE_{i,j}$ and $E_{i,j}A$). So our matrix $A$ is diagonal. Since all its
diagonal entries are equal, it therefore has the form%
\[
A=\operatorname*{diag}\left(  \lambda,\lambda,\ldots,\lambda\right)  =\lambda
I_{n},
\]
where $\lambda\in\mathbb{R}$ is a scalar that is nonzero (since $A$ is
invertible). Conversely, each such scalar matrix is central, as shown in
Example~\ref{exa.group.center.GL}. So we have shown that the only central
matrices in $\operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)  $ are
the $\lambda I_{n}$-matrices. Thus,%
\[
Z\left(  \operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)  \right)
=\left\{  \lambda I_{n}\ \mid\ \lambda\in\mathbb{R}\setminus\left\{
0\right\}  \right\}  .
\]

\end{remark}

Here is a trivial fact:

\begin{proposition}
\phantomsection\label{prop.sg.center.abelian} The center of a semigroup is
always an abelian semigroup.
\end{proposition}

\begin{proof}
The center is a subsemigroup by Theorem~\ref{thm.sg.center}
\textbf{(\ref{thm.sg.center.a})}. Central elements commute with everything,
hence with each other; thus, the center is abelian.
\end{proof}

\subsection{Two subgroup criteria}

\label{sec.group.subgroup-criteria}

To check that a subset $H$ of a group $G$ is a subgroup, we must normally
verify three conditions: that it is closed under multiplication (to be
precise: under the operation of $G$), that it contains the neutral element,
and that it is closed under taking inverses. In some situations, we can get by
with less work. Here are two quicker criteria.

\begin{proposition}
[shortcut to subgroups I]\phantomsection\label{prop.group.subgroup.criterion}
Let $G$ be a group (written multiplicatively), and let $H$ be a nonempty
subset of $G$. Then, $H$ is a subgroup of $G$ if and only if all $a,b\in H$
satisfy $ab^{-1}\in H$.
\end{proposition}

\begin{proof}
$\Longrightarrow:$ Assume that $H$ is a subgroup of $G$. We must prove that
all $a,b\in H$ satisfy $ab^{-1}\in H$.

Let $a,b\in H$. Then, $b^{-1}\in H$ (since $H$ is closed under taking
inverses\footnote{because $H$ is a subgroup of $G$}), and thus $ab^{-1}\in H$
(since $H$ is closed under multiplication\footnote{because $H$ is a subgroup
of $G$} and $a\in H$). This proves the \textquotedblleft$\Longrightarrow
$\textquotedblright\ direction of the proposition.

$\Longleftarrow:$ Assume that all $a,b\in H$ satisfy%
\begin{equation}
ab^{-1}\in H. \label{eq.group.subgroup.criterion}%
\end{equation}
We must show that $H$ is a subgroup of $G$.

First, there exists some $c\in H$ (since $H$ is nonempty). Consider this $c$.
By our assumption (\ref{eq.group.subgroup.criterion}) (applied to $a=c$ and
$b=c$), we then have $cc^{-1}\in H$, that is, $1\in H$. So $H$ contains the
neutral element of $G$. Therefore, each $b\in H$ satisfies $b^{-1}=1b^{-1}\in
H$ (by (\ref{eq.group.subgroup.criterion}), applied to $a=1$). So $H$ is
closed under taking inverses. Consequently, for any $a,b\in H$, we have
$a\left(  b^{-1}\right)  ^{-1}\in H$ (by our assumption
(\ref{eq.group.subgroup.criterion}), applied to $b^{-1}$ instead of $b$). Of
course, this latter relation rewrites as $ab\in H$ (since $\left(
b^{-1}\right)  ^{-1}=b$), and thus we have shown that $H$ is closed under
multiplication. So $H$ is a subgroup of $G$ (since it satisfies all three
conditions). This proves the \textquotedblleft$\Longleftarrow$%
\textquotedblright\ direction of the proposition.
\end{proof}

\begin{proposition}
[shortcut to subgroups II]\phantomsection\label{prop.group.subgroup.finite}
Let $G$ be a \textbf{finite} group. Then, any nonempty subsemigroup of $G$ is
a subgroup of $G$.
\end{proposition}

\begin{proof}
Let $T$ be a nonempty subsemigroup of $G$. We must show that $T$ is a subgroup
of $G$.

First, $T$ is closed under multiplication (since $T$ is a subsemigroup).

Pick any $c\in T$ (we can do this since $T$ is nonempty). Then,
Theorem~\ref{thm.group.powers.periodic}~\textbf{(a)} from Lecture 5 (applied
to $a=c$) shows that there exists a positive integer $n$ such that $c^{n}=1$
(since $G$ is a group, so $c\in G$ has an inverse in $G$). Consider this $n$.
But $c^{n}=\underbrace{cc\cdots c}_{n\text{ times}}$ (a nonempty product of
$c$'s because $n$ is positive), thus $c^{n}\in T$ (since $T$ is closed under
multiplication and contains $c$). In other words, $1\in T$ (since $c^{n}=1$).
This shows that $T$ contains the neutral element of $G$. Therefore, $T$ is a
submonoid of $G$.

What about inverses? Let $a\in T$ be arbitrary. Then,
Theorem~\ref{thm.group.powers.periodic}~\textbf{(a)} from Lecture 5 shows that
there exists a positive integer $n$ such that $a^{n}=1$ (again because $G$ is
a group, so $a\in G$ has an inverse). Consider such an $n$. Then,
$a^{-1}=a^{n-1}$ (since $aa^{n-1}=a^{n}=1$ and $a^{n-1}a=a^{n}=1$). But $n>0$,
thus $n-1\in\mathbb{N}$, and therefore $a^{n-1}$ is a nonnegative power of
$a$. But $T$ is a submonoid of $G$ (as we saw above), and thus is closed under
taking nonnegative powers (since a positive power is just a product of one and
the same element, while a zeroth power is just the neutral element). Thus,
from $a\in T$, we obtain $a^{n-1}\in T$. Therefore, $a^{-1}=a^{n-1}\in T$.

So we have shown that $a^{-1}\in T$ for each $a\in T$. In other words, $T$ is
closed under taking inverses. Thus, $T$ is a subgroup of $G$.
\end{proof}

Don't forget that both of the above propositions require nonemptiness! The
empty set is never a subgroup.

In practice, these shortcuts are rarely ever needed. They save writing, not
thinking. The proof ideas are more important than the results.

\subsection{A note on notation}

\label{sec.sg.multiplicative-default}

Did you grow tired of me constantly saying \textquotedblleft written
multiplicatively\textquotedblright\ in many of the above theorems? So did I. Therefore:

\begin{convention}
\phantomsection\label{conv.group.summary.notation} We shall write all
semigroups (thus, in particular, all monoids and groups) multiplicatively
unless we either say otherwise or are forced to do otherwise (e.g., because
multiplication is already taken for something else). In particular, even an
abelian group will usually be written multiplicatively.
\end{convention}

\subsection{Subgroups generated by subsets}

\label{sec.sg.generated}

\subsubsection{The semigroup case}

\label{sec.sg.generated.semigroup}

Most subsets $B$ of a semigroup $S$ will usually not be subsemigroups, because
there is no reason to expect them to be closed under multiplication. However,
we can make them closed under multiplication by repeatedly inserting the
\textquotedblleft missing\textquotedblright\ products into them. This
construction can be achieved in one step by just adding all multi-factor
products of elements of $B$ into $B$:

\begin{definition}
\phantomsection\label{def.sg.generated} Let $S$ be a semigroup. Let $B$ be a
subset of $S$.

Then, the \textbf{subsemigroup of }$S$\textbf{ generated by }$B$ is defined to
be the subset%
\[
\left\{  b_{1}b_{2}\cdots b_{k}\ \mid\ k\in\mathbb{Z}_{>0}\text{ and }%
b_{1},b_{2},\ldots,b_{k}\in B\right\}
\]
of $S$. (So it is the set of all nonempty products of elements of $B$,
including products of $1$ factor as well as products of any positive number of
factors. Note that equal factors are allowed!) We denote this subset by
$\left\langle B\right\rangle _{\operatorname*{semi}}$.
\end{definition}

\begin{proposition}
\phantomsection\label{prop.sg.generated} \leavevmode


\begin{enumerate}
[label=\textbf{(\alph*)},ref=\alph*,leftmargin=*]

\item \label{prop.sg.generated.a} This subset $\left\langle B\right\rangle
_{\operatorname*{semi}}$ is actually a subsemigroup of $S$.

\item \label{prop.sg.generated.b} It contains $B$ as a subset.

\item \label{prop.sg.generated.c} It is the smallest subsemigroup of $S$ that
contains $B$ as a subset. More precisely: Any subsemigroup $T$ of $S$ that
contains $B$ as a subset must contain $\left\langle B\right\rangle
_{\operatorname*{semi}}$ as a subset.
\end{enumerate}
\end{proposition}

\begin{proof}
\leavevmode


\begin{enumerate}
[label=\textbf{(\alph*)},ref=\alph*,leftmargin=*,topsep=0pt,partopsep=0pt]

\item \label{pf.prop.sg.generated.a} We must show that $\left\langle
B\right\rangle _{\operatorname*{semi}}$ is closed under multiplication. But
this is clear: If we multiply two products $b_{1}b_{2}\cdots b_{k}$ and
$b_{1}^{\prime}b_{2}^{\prime}\cdots b_{\ell}^{\prime}$ of elements of $B$,
then we get $b_{1}b_{2}\cdots b_{k}b_{1}^{\prime}b_{2}^{\prime}\cdots b_{\ell
}^{\prime}$, which is still a product of elements of $B$.

\item \label{pf.prop.sg.generated.b} Each $b\in B$ is a product of elements of
$B$: namely, a $1$-factor product of itself. So it belongs to $\left\langle
B\right\rangle _{\operatorname*{semi}}$. Therefore, $B\subseteq\left\langle
B\right\rangle _{\operatorname*{semi}}$.

\item \label{pf.prop.sg.generated.c} Let $T$ be a subsemigroup of $S$ that
contains $B$ as a subset. Then, it is closed under multiplication (since it is
a subsemigroup). Since it contains all elements of $B$ (because it contains
$B$ as a subset), it must therefore also contain all nonempty products of
elements of $B$, including products of multiple factors. That is, it must
contain all elements of $\left\langle B\right\rangle _{\operatorname*{semi}}$
(since the elements of $\left\langle B\right\rangle _{\operatorname*{semi}}$
are precisely all such products). In other words, it contains $\left\langle
B\right\rangle _{\operatorname*{semi}}$ as a subset. \qedhere

\end{enumerate}
\end{proof}

\begin{example}
\phantomsection\label{exa.sg.generated.singleton} Let $S$ be a semigroup, and
let $a\in S$. If $B$ is the one-element set $\left\{  a\right\}  $, then%
\[
\left\langle B\right\rangle _{\operatorname*{semi}}=\left\{  a^{1},a^{2}%
,a^{3},\ldots\right\}  ,
\]
because the nonempty products of elements of $\left\{  a\right\}  $ are
precisely $a^{1},a^{2},a^{3},\ldots$. So in particular, we recover the fact
that $\left\{  a^{1},a^{2},a^{3},\ldots\right\}  $ is a subsemigroup of $S$
(see (\ref{eq.subsg.powers}) in Lecture 6).
\end{example}

\subsubsection{The monoid case}

\label{sec.sg.generated.monoid}

We can play the same game for monoids, but the resulting submonoid will often
differ from the subsemigroup $\left\langle B\right\rangle
_{\operatorname*{semi}}$ because we have to adjoin a neutral element as well:

\begin{definition}
\phantomsection\label{def.monoid.generated} Let $M$ be a monoid. Let $B$ be a
subset of $M$.

Then, the \textbf{submonoid of }$M$\textbf{ generated by }$B$ is defined to be
the subset%
\[
\left\{  b_{1}b_{2}\cdots b_{k}\ \mid\ k\in\mathbb{N}\text{ and }b_{1}%
,b_{2},\ldots,b_{k}\in B\right\}
\]
of $M$. (So it is the set of all products of elements of $B$, but now
including the empty product, which by definition is the neutral element of
$M$. Note that the empty product can be taken even if $B$ is empty.) We denote
it by $\left\langle B\right\rangle _{\operatorname*{monoid}}$.
\end{definition}

\begin{proposition}
\phantomsection\label{prop.monoid.generated} \leavevmode


\begin{enumerate}
[label=\textbf{(\alph*)},ref=\alph*,leftmargin=*]

\item \label{prop.monoid.generated.a} This subset $\left\langle B\right\rangle
_{\operatorname*{monoid}}$ is actually a submonoid of $M$.

\item \label{prop.monoid.generated.b} It contains $B$ as a subset.

\item \label{prop.monoid.generated.c} It is the smallest submonoid of $M$ that
contains $B$ as a subset. More precisely: Any submonoid $N$ of $M$ that
contains $B$ as a subset must contain $\left\langle B\right\rangle
_{\operatorname*{monoid}}$ as a subset.
\end{enumerate}
\end{proposition}

\begin{proof}
This is an easy adaptation of the proof of Proposition~\ref{prop.sg.generated}%
. In part \textbf{(a)}, the empty product brings the neutral element into
$\left\langle B\right\rangle _{\operatorname*{monoid}}$.
\end{proof}

\begin{example}
\phantomsection\label{exa.monoid.generated.singleton} Let $M$ be a monoid, and
let $a\in M$. If $B$ is the one-element set $\left\{  a\right\}  $, then%
\[
\left\langle B\right\rangle _{\operatorname*{monoid}}=\left\{  a^{0}%
,a^{1},a^{2},a^{3},\ldots\right\}  ,
\]
because the products of elements of $\left\{  a\right\}  $ are precisely
$a^{0},a^{1},a^{2},a^{3},\ldots$. Thus, we recover the fact that $\left\{
a^{0},a^{1},a^{2},a^{3},\ldots\right\}  $ is a submonoid of $M$ (see
(\ref{eq.submon.powers}) in Lecture 6).
\end{example}

\subsubsection{The group case}

\label{sec.sg.generated.group}

\Needspace{7\baselineskip} Finally, we define the subgroup of a group $G$
generated by a given subset $B$. The idea is the same as for semigroups and
for monoids, but the execution would get a bit tricky if we tried to blindly
follow the previous method: in addition to just taking products of elements of
$B$, we would have to take their inverses, as well as products of their
inverses (and of original elements of $B$), but also inverses of such
products, and products of such inverses, and so on, ad infinitum. In
principle, such constructions can be made, but here we can find a much easier
way. Indeed, any \textquotedblleft inverse-product
matryoshka\textquotedblright\ such as $\left(  ab\right)  ^{-1}\left(
cd^{-1}e\right)  ^{-1}f\left(  g^{-1}h\right)  ^{-1}ij^{-1}$ can be unravelled
into a single product (in this example, into $b^{-1}a^{-1}e^{-1}dc^{-1}%
fh^{-1}gij^{-1}$) using Theorem \ref{thm.monoid.gen-socks} and Theorem
\ref{thm.monoid.inverse-rules}. Thus, all we need are the elements of $B$,
their inverses, and the products of these. This suggests the following simple
definition of the subgroup we are looking for:

\Needspace{16\baselineskip}

\begin{definition}
\phantomsection\label{def.group.generated} Let $G$ be a group. Let $B$ be a
subset of $G$.

Then, we define the subset
\[
B^{-1}:=\left\{  b^{-1}\ \mid\ b\in B\right\}
\]
of $G$. Subsequently, the \textbf{subgroup} of $G$ \textbf{generated by }$B$
is defined to be the submonoid of $G$ generated by $B\cup B^{-1}$. We denote
it by $\left\langle B\right\rangle $. Thus,%
\begin{equation}
\left\langle B\right\rangle :=\left\langle B\cup B^{-1}\right\rangle
_{\operatorname*{monoid}}.\label{eq.def.group.generated.def}%
\end{equation}

\end{definition}

\begin{proposition}
\phantomsection\label{prop.group.generated} \leavevmode


\begin{enumerate}
[label=\textbf{(\alph*)},ref=\alph*,leftmargin=*]

\item \label{prop.group.generated.a} This subset $\left\langle B\right\rangle
$ is actually a subgroup of $G$.

\item \label{prop.group.generated.b} It contains $B$ as a subset.

\item \label{prop.group.generated.c} It is the smallest subgroup of $G$ that
contains $B$ as a subset. More precisely: Any subgroup $H$ of $G$ that
contains $B$ as a subset must contain $\left\langle B\right\rangle $ as a subset.
\end{enumerate}
\end{proposition}

\begin{proof}
\leavevmode


\begin{enumerate}
[label=\textbf{(\alph*)},ref=\alph*,leftmargin=*,topsep=0pt,partopsep=0pt,itemsep=4pt,parsep=0pt]

\item \label{pf.prop.group.generated.a} We defined $\left\langle
B\right\rangle $ as $\left\langle B\cup B^{-1}\right\rangle
_{\operatorname*{monoid}}$, so we know it is a submonoid of $G$ (by
Proposition~\ref{prop.monoid.generated}~\textbf{(a)}, applied to $B\cup
B^{-1}$ instead of $B$). It remains to prove that it is closed under taking inverses.

First, we observe that
\begin{equation}
a^{-1}\in B\cup B^{-1}\text{ for each }a\in B\cup B^{-1}%
.\label{pf.prop.group.generated.a.1}%
\end{equation}
(\textit{Proof:} Let $a\in B\cup B^{-1}$. Thus, $a\in B$ or $a\in B^{-1}$. If
$a\in B$, then the definition of $B^{-1}$ yields $a^{-1}\in B^{-1}\subseteq
B\cup B^{-1}$, so that (\ref{pf.prop.group.generated.a.1}) is proved in this
case. Thus, only the case $a\in B^{-1}$ remains to be handled. In this case,
we have $a=b^{-1}$ for some $b\in B$ (since $a\in B^{-1}=\left\{  b^{-1}%
\ \mid\ b\in B\right\}  $). Thus, for this $b$, we have $a^{-1}=\left(
b^{-1}\right)  ^{-1}=b$ (by Theorem \ref{thm.monoid.inverse-rules}
\textbf{(a)}) and thus $a^{-1}=b\in B\subseteq B\cup B^{-1}$. So we have
proved (\ref{pf.prop.group.generated.a.1}) in both cases.)

Now, let $b\in\left\langle B\right\rangle $. Then, $b\in\left\langle
B\right\rangle =\left\langle B\cup B^{-1}\right\rangle
_{\operatorname*{monoid}}$. By Definition \ref{def.monoid.generated}, this
shows that $b$ is a product $a_{1}a_{2}\cdots a_{k}$ of some elements
$a_{1},a_{2},\ldots,a_{k}$ of $B\cup B^{-1}$, where $k\in\mathbb{N}$. Consider
these elements $a_{1},a_{2},\ldots,a_{k}$. Then, from $b=a_{1}a_{2}\cdots
a_{k}$, we obtain
\[
b^{-1}=\left(  a_{1}a_{2}\cdots a_{k}\right)  ^{-1}=a_{k}^{-1}a_{k-1}%
^{-1}\cdots a_{1}^{-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{by Theorem
\ref{thm.monoid.gen-socks}}\right)  .
\]
But all of the factors $a_{k}^{-1},a_{k-1}^{-1},\ldots,a_{1}^{-1}$ belong to
$B\cup B^{-1}$ (by (\ref{pf.prop.group.generated.a.1}), since $a_{k}%
,a_{k-1},\ldots,a_{1}$ belong to $B\cup B^{-1}$). Hence, $a_{k}^{-1}%
a_{k-1}^{-1}\cdots a_{1}^{-1}$ is a product of elements of $B\cup B^{-1}$, and
therefore belongs to $\left\langle B\cup B^{-1}\right\rangle
_{\operatorname*{monoid}}$ (by Definition \ref{def.monoid.generated}). Thus,
\[
b^{-1}=a_{k}^{-1}a_{k-1}^{-1}\cdots a_{1}^{-1}\in\left\langle B\cup
B^{-1}\right\rangle _{\operatorname*{monoid}}=\left\langle B\right\rangle .
\]


So we have proved that $b^{-1}\in\left\langle B\right\rangle $ for each
$b\in\left\langle B\right\rangle $. In other words, $\left\langle
B\right\rangle $ is closed under taking inverses. Thus it is a subgroup of
$G$; this proves part \textbf{(a)}.

\item \label{pf.prop.group.generated.b}
Proposition~\ref{prop.monoid.generated} \textbf{(b)} shows that the submonoid
$\left\langle B\cup B^{-1}\right\rangle _{\operatorname*{monoid}}$ contains
$B\cup B^{-1}$ as a subset; hence, it also contains $B$ as a subset. In other
words, $\left\langle B\right\rangle $ contains $B$ as a subset.

\item \label{pf.prop.group.generated.c} Let $H$ be a subgroup of $G$ that
contains $B$ as a subset. Since $H$ is closed under taking inverses, it
therefore also contains $B^{-1}$, and hence contains the union $B\cup B^{-1}$.
Moreover, $H$ is a submonoid of $G$ (being a subgroup of $G$).
Proposition~\ref{prop.monoid.generated}~\textbf{(c)} (applied to $G$, $H$ and
$B\cup B^{-1}$ instead of $M$, $N$ and $B$) thus shows that $H$ must contain
$\left\langle B\cup B^{-1}\right\rangle _{\operatorname*{monoid}}$ as a
subset. In other words, $H$ contains $\left\langle B\right\rangle $ as a
subset (since $\left\langle B\right\rangle =\left\langle B\cup B^{-1}%
\right\rangle _{\operatorname*{monoid}}$). This proves part \textbf{(c)}. \qedhere
\end{enumerate}
\end{proof}

\begin{example}
\phantomsection\label{exa.group.generated.singleton}  Let $G$ be a group, and
let $a\in G$. If $B$ is the one-element set $\left\{  a\right\}  $, then%
\begin{equation}
\left\langle B\right\rangle =\left\{  a^{n}\ \mid\ n\in\mathbb{Z}\right\}
.\label{eq.exa.group.generated.singleton.eq}%
\end{equation}
One way to see this is by observing that $B\cup B^{-1}=\left\{  a,a^{-1}%
\right\}  $, and that the products of $a$'s and $a^{-1}$'s are simply the
integer powers $a^{n}$ of $a$: for example, $aaa^{-1}aa^{-1}a^{-1}%
aaa^{-1}a=a^{2}$. Thus, in particular, we recover the fact that $\left\{
a^{n}\ \mid\ n\in\mathbb{Z}\right\}  $ is a subgroup of $G$. Alternatively, if
we take this fact as a given, we can use Proposition
\ref{prop.group.generated} \textbf{(c)} to show that $\left\langle
B\right\rangle \subseteq\left\{  a^{n}\ \mid\ n\in\mathbb{Z}\right\}  $,
whereas Proposition \ref{prop.group.generated} \textbf{(a), (b)} yields the
converse inclusion $\left\{  a^{n}\ \mid\ n\in\mathbb{Z}\right\}
\subseteq\left\langle B\right\rangle $ (since $\left\langle B\right\rangle $
is a subgroup containing $a$, so it contains all integer powers of $a$); thus
(\ref{eq.exa.group.generated.singleton.eq}) follows.
\end{example}

\begin{definition}
\phantomsection\label{def.group.generated.list} Let $a_{1},a_{2},\ldots,a_{k}$
be $k$ elements of a group $G$. Then, the subgroup $\left\langle \left\{
a_{1},a_{2},\ldots,a_{k}\right\}  \right\rangle $ of $G$ will be simply
denoted by $\left\langle a_{1},a_{2},\ldots,a_{k}\right\rangle $. That is, we
drop the set-braces and simply write down the elements themselves inside the
angular brackets.

In particular, $\left\langle a\right\rangle =\left\langle \left\{  a\right\}
\right\rangle =\left\{  a^{n}\ \mid\ n\in\mathbb{Z}\right\}  $ for any $a\in
G$. (This subgroup $\left\langle a\right\rangle $ is called a \textbf{cyclic
subgroup} of $G$, though a proper definition of a cyclic group will have to
wait until the next chapter.)

Similar notations will be used for semigroups and monoids: e.g., we write
$\left\langle a\right\rangle _{\operatorname*{monoid}}$ for $\left\langle
\left\{  a\right\}  \right\rangle _{\operatorname*{monoid}}$.
\end{definition}

\Needspace{15\baselineskip}
\subsubsection{Assorted examples}

\label{sec.sg.generated.examples}

\begin{example}
\phantomsection\label{exa.group.generated.Q-halves-thirds} Recall the abelian
group $\left(  \mathbb{Q},+\right)  $. Keep in mind that it is written
additively, so all products in the general definitions above must be read as
sums when working in this group.

We have the subsemigroup
\begin{align*}
\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle _{\operatorname*{semi}} &
=\left\{  b_{1}+b_{2}+\cdots+b_{k}\ \middle|\
\begin{array}
[c]{l}%
k\in\mathbb{Z}_{>0},\\
b_{1},b_{2},\ldots,b_{k}\in\left\{  \dfrac{1}{2},\dfrac{1}{3}\right\}
\end{array}
\right\}  \\
&  =\left\{  \text{all nonempty sums of }\dfrac{1}{2}\text{'s and }\dfrac
{1}{3}\text{'s}\right\}  \\
&  =\left\{  n\cdot\dfrac{1}{2}+m\cdot\dfrac{1}{3}\ \mid\ n,m\in
\mathbb{N}\text{ with }n+m>0\right\}
\end{align*}
and the submonoid
\begin{align*}
\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle _{\operatorname*{monoid}}
&  =\left\{  b_{1}+b_{2}+\cdots+b_{k}\ \middle|\
\begin{array}
[c]{l}%
k\in\mathbb{N},\\
b_{1},b_{2},\ldots,b_{k}\in\left\{  \dfrac{1}{2},\dfrac{1}{3}\right\}
\end{array}
\right\}  \\
&  =\left\{  \text{all sums of }\dfrac{1}{2}\text{'s and }\dfrac{1}{3}\text{'s
including the empty sum}\right\}  \\
&  =\left\{  n\cdot\dfrac{1}{2}+m\cdot\dfrac{1}{3}\ \mid\ n,m\in
\mathbb{N}\right\}  \\
&  =\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle _{\operatorname*{semi}%
}\cup\left\{  0\right\}
\end{align*}
and the subgroup
\begin{align*}
\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle  &  =\left\langle
\dfrac{1}{2},\dfrac{1}{3},-\dfrac{1}{2},-\dfrac{1}{3}\right\rangle
_{\operatorname*{monoid}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the inverse of }b\\
\text{in our group is }-b
\end{array}
\right)  \\
&  =\left\{  b_{1}+b_{2}+\cdots+b_{k}\ \middle|\
\begin{array}
[c]{l}%
k\in\mathbb{N},\\
b_{1},b_{2},\ldots,b_{k}\in\left\{  \dfrac{1}{2},\dfrac{1}{3},-\dfrac{1}%
{2},-\dfrac{1}{3}\right\}
\end{array}
\right\}  \\
&  =\left\{  n\cdot\dfrac{1}{2}+m\cdot\dfrac{1}{3}\ \mid\ n,m\in
\mathbb{Z}\right\}  .
\end{align*}
In the last equality, we collect the positive and negative copies of each
fraction to obtain an integer multiple of this fraction. Note that we have
used the commutativity of $+$ to normalize arbitrarily ordered sums such as
$\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2}$ into simple
linear combinations of the form $n\cdot\dfrac{1}{2}+m\cdot\dfrac{1}{3}$; this
would not work for a non-abelian group!

We have a chain of inclusions%
\[
\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle _{\operatorname*{semi}%
}\subseteq\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle
_{\operatorname*{monoid}}\subseteq\left\langle \dfrac{1}{2},\dfrac{1}%
{3}\right\rangle \subseteq\mathbb{Q}.
\]
All these inclusions are proper, i.e., each set contains an element not in the
previous set. For example:

\begin{itemize}
\item We have $\dfrac{5}{6}\in\left\langle \dfrac{1}{2},\dfrac{1}%
{3}\right\rangle _{\operatorname*{semi}}$ because $\dfrac{5}{6}=\dfrac{1}%
{2}+\dfrac{1}{3}$.

\item We have $0\in\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle
_{\operatorname*{monoid}}\setminus\left\langle \dfrac{1}{2},\dfrac{1}%
{3}\right\rangle _{\operatorname*{semi}}$.

\item Both $-\dfrac{1}{2}$ and $\dfrac{1}{6}$ belong to $\left\langle
\dfrac{1}{2},\dfrac{1}{3}\right\rangle \setminus\left\langle \dfrac{1}%
{2},\dfrac{1}{3}\right\rangle _{\operatorname*{monoid}}$. In fact, $\dfrac
{1}{6}=\dfrac{1}{2}-\dfrac{1}{3}\in\left\langle \dfrac{1}{2},\dfrac{1}%
{3}\right\rangle $ but $\dfrac{1}{6}\notin\left\langle \dfrac{1}{2},\dfrac
{1}{3}\right\rangle _{\operatorname*{monoid}}$ (since any sum $n\cdot\dfrac
{1}{2}+m\cdot\dfrac{1}{3}$ with $n,m\in\mathbb{N}$ is either $0$ or
$\geq\dfrac{1}{3}$).

\item We have $\dfrac{1}{4},\dfrac{1}{5}\in\mathbb{Q}\setminus\left\langle
\dfrac{1}{2},\dfrac{1}{3}\right\rangle $. Indeed, any $n\cdot\dfrac{1}%
{2}+m\cdot\dfrac{1}{3}$ with $n,m\in\mathbb{Z}$ can be rewritten as
$\dfrac{3n+2m}{6}$, so it becomes an integer when multiplied by $6$. But
neither $\dfrac{1}{4}$ nor $\dfrac{1}{5}$ becomes an integer when multiplied
by $6$.
\end{itemize}

The subgroup $\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle $ can be
described in a simpler way:%
\[
\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle =\left\{  \dfrac{k}%
{6}\ \middle|\ k\in\mathbb{Z}\right\}  .
\]
Indeed, as we just saw, every element $n\cdot\dfrac{1}{2}+m\cdot\dfrac{1}{3}$
of $\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle $ has the form
$\dfrac{k}{6}$ for $k=3n+2m$; conversely, any $\dfrac{k}{6}$ can be written as
$k\cdot\dfrac{1}{2}+\left(  -k\right)  \cdot\dfrac{1}{3}$.
\end{example}

So we see that the same subset of a group generates a subsemigroup, a
submonoid and a subgroup, and these are generally not the same.

\begin{sloppypar}
However, they are the same when the group is finite and the subset is nonempty:
\end{sloppypar}

\begin{proposition}
\phantomsection\label{prop.group.generated.finite} Let $G$ be a
\textbf{finite} group. Let $B\subseteq G$ be nonempty. Then,%
\[
\left\langle B\right\rangle _{\operatorname*{semi}}=\left\langle
B\right\rangle _{\operatorname*{monoid}}=\left\langle B\right\rangle .
\]

\end{proposition}

\begin{proof}
\begin{sloppypar}
We know that $\left\langle B\right\rangle _{\operatorname*{semi}}$ is a
subsemigroup of $G$, and furthermore is nonempty (since it contains $B$ as a
subset, but $B$ is nonempty). But we also know
(Proposition~\ref{prop.group.subgroup.finite}) that every nonempty
subsemigroup of a finite group is already a subgroup. So $\left\langle
B\right\rangle _{\operatorname*{semi}}$ is a subgroup of $G$.
\end{sloppypar}

But $\left\langle B\right\rangle $ is the smallest subgroup of $G$ that
contains $B$ as a subset (by Proposition \ref{prop.group.generated}
\textbf{(c)}). Since $\left\langle B\right\rangle _{\operatorname*{semi}}$ is
such a subgroup, we thus conclude that $\left\langle B\right\rangle
\subseteq\left\langle B\right\rangle _{\operatorname*{semi}}$.

On the other hand, $\left\langle B\right\rangle _{\operatorname*{semi}%
}\subseteq\left\langle B\right\rangle $, because products of elements of $B$
clearly belong to $\left\langle B\right\rangle $.

Combining these two inclusions, we obtain $\left\langle B\right\rangle
_{\operatorname*{semi}}=\left\langle B\right\rangle $.

\begin{sloppypar}
An analogous argument (now including the empty product) shows that
$\left\langle B\right\rangle _{\operatorname*{monoid}}=\left\langle
B\right\rangle $ (since submonoids are subsemigroups).
\end{sloppypar}

Thus, Proposition \ref{prop.group.generated.finite} is proved.
\end{proof}

The nonemptiness assumption in Proposition \ref{prop.group.generated.finite}
is needed: if $B=\varnothing$, then $\left\langle B\right\rangle
_{\operatorname*{semi}}=\varnothing$, whereas $\left\langle B\right\rangle
_{\operatorname*{monoid}}=\left\langle B\right\rangle =\{1\}$.

\begin{example}
\phantomsection\label{exa.group.generated.S3} Consider the symmetric group
$S_{3}$. For any two distinct elements $i$ and $j$ of $\left\{  1,2,3\right\}
$, we let $t_{i,j}$ denote the permutation that swaps $i$ with $j$ while
keeping all the other elements unchanged. Set $s_{1}=t_{1,2}$ and
$s_{2}=t_{2,3}$ and $w_{0}=t_{1,3}$. Let $c$ be the cycle $1\mapsto
2\mapsto3\mapsto1$. As we know,
\[
S_{3}=\left\{  \operatorname*{id},s_{1},s_{2},w_{0},c,c^{2}\right\}  .
\]
Then,%
\begin{align*}
\left\langle c\right\rangle  & =\left\{  c^{n}\mid n\in\mathbb{Z}\right\}
=\left\{  \operatorname*{id},c,c^{2}\right\}  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }o\left(  c\right)  =3\right)  ,\\
\left\langle s_{1}\right\rangle  & =\left\{  \operatorname*{id},s_{1}\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }o\left(  s_{1}\right)  =2\right)  ,\\
\left\langle s_{2}\right\rangle  & =\left\{  \operatorname*{id},s_{2}\right\}
,
\end{align*}
and%
\[
\left\langle s_{1},s_{2}\right\rangle =\left\{  \operatorname*{id},s_{1}%
,s_{2},w_{0},c,c^{2}\right\}
\]
(since $w_{0}=s_{1}s_{2}s_{1}$ and $c=s_{1}s_{2}$ and $c^{2}=s_{2}s_{1}$).
Thus, every one of the six elements of $S_{3}$ belongs to $\left\langle
s_{1},s_{2}\right\rangle $; the reverse inclusion holds because this is a
subgroup of $S_{3}$. The latter equality, of course, says that $\left\langle
s_{1},s_{2}\right\rangle =S_{3}$. We can say this as follows: the elements
$s_{1}$ and $s_{2}$ \textbf{generate} the group $S_{3}$.
\end{example}

More generally:

\begin{definition}
\phantomsection\label{def.group.generators} Let $a_{1},a_{2},\ldots,a_{k}$ be
some elements of a group $G$. We say that these elements \textbf{generate} the
group $G$ if $\left\langle a_{1},a_{2},\ldots,a_{k}\right\rangle =G$.
\end{definition}

Similarly we can define what it means for some elements of a monoid to
generate the monoid, or for some elements of a semigroup to generate the
semigroup; however, it should be kept in mind that the three notions are not
equivalent in the case of a group, so a careful writer must signal which one
is meant. (That is, if there is any possibility for confusion, do not say
\textquotedblleft$a_{1},a_{2},\ldots,a_{k}$ generate $G$\textquotedblright;
always say \textquotedblleft$a_{1},a_{2},\ldots,a_{k}$ generate the group
$G$\textquotedblright\ or \textquotedblleft... the monoid $G$%
\textquotedblright\ or \textquotedblleft... the semigroup $G$%
\textquotedblright.)

\subsection{A group-focused summary}

\label{sec.group.summary}

\begin{sloppypar}
We will mostly be dealing with groups in this course, not semigroups or
monoids. However, I have been stating everything in maximum possible
generality so far. Let us summarize the most important results specialized to
the case of groups.
\end{sloppypar}

\subsubsection{The main definitions and notations}

\label{sec.group.summary.definitions}

\begin{itemize}
\item A \textbf{group} is a set $G$ equipped with a binary operation $\ast$
that is associative, such that $G$ has a neutral element and each element of
$G$ has an inverse (under $\ast$). (Definition~\ref{def.sg.structures}%
~\textbf{(g)}, Lecture 2--3.) Keep in mind that a \textquotedblleft binary
operation\textquotedblright\ on $G$ in our definition of this word has to be a
map from $G\times G$ to $G$.

\item A \textbf{subgroup} of a group $\left(  G,\ast\right)  $ is a subset $H$
of $G$ that is closed under $\ast$, contains the neutral element of $G$ and is
closed under taking inverses. We write $H\leq G$ for this.
(Definition~\ref{def.group.subgroup}, Lecture 6--7.)

\item A group $\left(  G,\ast\right)  $ is said to be \textbf{abelian} if all
its elements commute (i.e., if $a\ast b=b\ast a$ for all $a,b\in G$).
(Definition~\ref{def.sg.abelian}, Lecture 6--7.)

\item To write a group $\left(  G,\ast\right)  $ in \textbf{multiplicative
notation} means that we write $ab$ for $a\ast b$, write $1$ for the neutral
element, and write $a^{-1}$ for the inverse of $a$.
(Definition~\ref{def.sg.mulnot}, Lecture 4, and
Definition~\ref{def.monoid.inverse}, Lecture 2--3.)

We use multiplicative notation by default when nothing speaks against it.

\item To write an abelian group $\left(  G,\ast\right)  $ in \textbf{additive
notation} means that we write $a+b$ for $a\ast b$, write $0$ for the neutral
element, and write $-a$ for the inverse of $a$.
(Definition~\ref{def.sg.additive}, Lecture 6--7.)

\item The \textbf{powers} of an element $a$ in a group $G$ (written
multiplicatively) are defined by%
\[
a^{n}:=%
\begin{cases}
\underbrace{aa\cdots a}_{n\text{ times}}, & \text{if }n>0;\\
1, & \text{if }n=0;\\
\left(  a^{-1}\right)  ^{-n}, & \text{if }n<0
\end{cases}
\ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{Z}.
\]
If we write $G$ additively, then we write $na$ instead of $a^{n}$.
(Definition~\ref{def.sg.powers}, Lecture 5, and
Definition~\ref{def.sg.additive}, Lecture 6--7.)

\item The \textbf{order} $o\left(  a\right)  $ of an element $a$ in a group
$G$ is the smallest positive integer $n$ such that $a^{n}=1$ if such an $n$
exists; otherwise it is $\infty$. (Definition~\ref{def.monoid.order}, Lecture 6--7.)

\item The \textbf{subgroup generated by} a subset $B\subseteq G$, denoted by
$\left\langle B\right\rangle $, consists of all finite products of elements of
$B$ and their inverses, including the empty product. It is the smallest
subgroup of $G$ containing $B$. (Definition~\ref{def.group.generated} and
Proposition~\ref{prop.group.generated} above.)
\end{itemize}

\subsubsection{The main examples}

\label{sec.group.summary.examples}

\begin{itemize}
\item $\left(  \mathbb{Z},+\right)  $, $\left(  \mathbb{Q},+\right)  $,
$\left(  \mathbb{R},+\right)  $ and $\left(  \mathbb{C},+\right)  $ are
abelian groups (but $\left(  \mathbb{N},+\right)  $ is not). They form a chain
of subgroups:%
\[
\left(  \mathbb{Z},+\right)  \leq\left(  \mathbb{Q},+\right)  \leq\left(
\mathbb{R},+\right)  \leq\left(  \mathbb{C},+\right)  .
\]
(\S \ref{sec.gps.exas}, Lecture 2--3, and \S \ref{sec.sg.substructures},
Lecture 6--7.)

\item
\begin{sloppypar}
$\left(  \mathbb{Q}\setminus\left\{  0\right\}  ,\cdot\right)  $, $\left(
\mathbb{R}\setminus\left\{  0\right\}  ,\cdot\right)  $ and $\left(
\mathbb{C}\setminus\left\{  0\right\}  ,\cdot\right)  $ are abelian groups
(but $\left(  \mathbb{Z}\setminus\left\{  0\right\}  ,\cdot\right)  $ is not a
group). They also form a chain of subgroups. (\S \ref{sec.gps.exas}, Lecture
2--3; the subgroup conditions are checked just as for the additive groups above.)
\end{sloppypar}

\item
\begin{sloppypar}
Two very small groups are the trivial group $\left(  \left\{  1\right\}
,\cdot\right)  $ and the group $\left(  \left\{  1,-1\right\}  ,\cdot\right)
$ with two elements. (\S \ref{sec.gps.exas}, Lecture 2--3.)
\end{sloppypar}

\item For any set $A$, the permutations of $A$ form a group $S_{A}$ under
composition (\S \ref{sec.gps.exas}, Lecture 2--3). This group is called the
\textbf{symmetric group} of $A$. If $A$ is finite of size $n$, then $S_{A}$ is
finite of size $\left\vert S_{A}\right\vert =n!$ (see Math 222 or Math 221).

\item If $A$ is any set, then $\left(  \mathcal{P}\left(  A\right)
,\bigtriangleup\right)  $ is a group, in which all elements have order $1$ or
$2$. If $A$ is finite of size $n$, then $\left\vert \mathcal{P}\left(
A\right)  \right\vert =2^{n}$ (see Math 222 or Math 221). (The operation
$\bigtriangleup$ is defined by $X\bigtriangleup Y=\left(  X\cup Y\right)
\setminus\left(  X\cap Y\right)  $. See \S \ref{sec.gps.exas} in Lecture 2--3;
for the orders, see Example~\ref{exa.group.order.symdiff} in Lecture 6--7.)

\item All $n\times m$-matrices with real entries form a group $\left(
\mathbb{R}^{n\times m},+\right)  $. All invertible $n\times n$-matrices form a
group $\left(  \operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)
,\cdot\right)  $. (\S \ref{sec.gps.exas}, Lecture 2--3\footnote{The addition
example there is stated for square matrices, but the same argument applies to
rectangular ones.}.)
\end{itemize}

\Needspace{11\baselineskip}
\subsubsection{The main theorems}

\label{sec.group.summary.theorems}

Let $G$ be a group, written multiplicatively. All elements below belong to $G$.

\begin{itemize}
\item Generalized associativity (Theorem~\ref{thm.sg.genass}~\textbf{(b)},
Lecture 4): Products of the form $a_{1}a_{2}\cdots a_{n}$ are well-defined
without specifying the parenthesization; they can be parenthesized at will:%
\[
a_{1}a_{2}\cdots a_{n}=\left(  a_{1}a_{2}\cdots a_{k}\right)  \left(
a_{k+1}a_{k+2}\cdots a_{n}\right)  .
\]
Here $0\leq k\leq n$ are integers, and an empty product is $1$.

\item Inverses satisfy (Theorem~\ref{thm.monoid.inverse-rules}, Lecture 2--3)%
\[
\left(  ab\right)  ^{-1}=b^{-1}a^{-1}\qquad\text{and}\qquad\left(
a^{-1}\right)  ^{-1}=a.
\]


\item Solving equations (Theorem~\ref{thm.monoid.cancel}~\textbf{(a), (b)},
Lecture 4):%
\[
ab=c\ \Longleftrightarrow\ b=a^{-1}c
\]
and%
\[
ba=c\ \Longleftrightarrow\ b=ca^{-1}.
\]


\item Cancellation (Theorem~\ref{thm.monoid.cancel}~\textbf{(c), (d)}, Lecture
4):%
\[
ab=ac\ \Longleftrightarrow\ b=c
\]
and%
\[
ba=ca\ \Longleftrightarrow\ b=c.
\]


\item For any $a\in G$, the maps $L_{a}$ (left multiplication by $a$) and
$R_{a}$ (right multiplication by $a$) are bijective.
(Corollary~\ref{cor.monoid.sudoku}, Lecture 4.)

\item Rules of exponents: For all $a\in G$ and all $n,m\in\mathbb{Z}$, we have%
\begin{align*}
a^{n}a^{m}  &  =a^{n+m},\\
\left(  a^{n}\right)  ^{m}  &  =a^{nm},\\
a^{-n}  &  =\left(  a^{-1}\right)  ^{n}=\left(  a^{n}\right)  ^{-1}%
\end{align*}
(Theorems~\ref{thm.monoid.powers.add}, \ref{thm.monoid.powers.mul}
and~\ref{thm.monoid.powers.neg}, and Lemma~\ref{lem.monoid.powers.neg-pow},
Lecture 5). Moreover, if $a,b\in G$ commute, then $\left(  ab\right)
^{n}=a^{n}b^{n}$ for all $n\in\mathbb{Z}$
(Theorem~\ref{thm.monoid.powers.prod}, Lecture 5).

(The integer-exponent version of $(a^{n})^{m}=a^{nm}$ and the rule
$(ab)^{n}=a^{n}b^{n}$ for commuting $a$ and $b$ are on homework set \#2; don't
use them without proof yet.)

\item If $G$ is finite, then each $a\in G$ has finite order $o\left(
a\right)  $ (Theorem~\ref{thm.group.powers.periodic}~\textbf{(a)}, Lecture 5).
Denoting this order by $n$, we have%
\[
\left\{  \text{all powers of }a\right\}  =\left\{  a^{0},a^{1},\ldots
,a^{n-1}\right\}  ,
\]


\begin{sloppypar}
and furthermore, these $n$ powers $a^{0},\allowbreak a^{1},\allowbreak
\ldots,\allowbreak a^{n-1}$ are distinct.
(Theorem~\ref{thm.monoid.order.powers}~\textbf{(a), (c)}, Lecture 6--7.)
\end{sloppypar}

The powers form a cycle under right multiplication by $a$, since
$a^{n-1}a=a^{n}=1=a^{0}$. Its shape is shown schematically in the following
picture:%
\[
\begin{tikzpicture}[>=Stealth, line width=0.65pt,
every node/.style={inner sep=2pt},
power/.style={circle,draw,minimum size=11mm,inner sep=1pt}]
\def\cycleradius{1.85}
\node[power] (a0) at (90:\cycleradius) {$a^0$};
\node[power] (a1) at (30:\cycleradius) {$a^1$};
\node[power] (a2) at (-30:\cycleradius) {$a^2$};
\node[power] (an2) at (-150:\cycleradius) {$a^{n-2}$};
\node[power] (an1) at (150:\cycleradius) {$a^{n-1}$};
\draw[->] (72:\cycleradius)
arc[start angle=72,end angle=48,radius=\cycleradius]
node[midway,above right=3pt] {$\cdot a$};
\draw[->] (12:\cycleradius)
arc[start angle=12,end angle=-12,radius=\cycleradius]
node[midway,right=3pt] {$\cdot a$};
\draw[densely dotted] (-48:\cycleradius)
arc[start angle=-48,end angle=-78,radius=\cycleradius];
\node at (-90:\cycleradius) {$\cdots$};
\draw[densely dotted] (-102:\cycleradius)
arc[start angle=-102,end angle=-132,radius=\cycleradius];
\draw[->] (-168:\cycleradius)
arc[start angle=-168,end angle=-192,radius=\cycleradius]
node[midway,left=3pt] {$\cdot a$};
\draw[->] (132:\cycleradius)
arc[start angle=132,end angle=108,radius=\cycleradius]
node[midway,above left=3pt] {$\cdot a$};
\end{tikzpicture}
\]


(Note that if $a=1$, then this cycle consists of just a single node $a^{0}$
with a single loop back to it.)

\item Any subgroup $H$ of a group $G$ becomes a group itself if we restrict
the multiplication of $G$ to $H$ (Theorem~\ref{thm.group.subgroup}, Lecture 6--7).
\end{itemize}


\end{document}