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\ihead{Math 331, Fall 2026, Lecture 6 and 7, version \today}
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\begin{document}
\section*{Math 331 Fall 2026, Lecture 6 and 7: Abelian groups, subgroups}

\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}{7}

\subsection{Orders of elements}

\label{sec.monoid.orders}

\subsubsection{Definition and example}

The notion of a power gives rise to the notion of an order:

\begin{definition}
\phantomsection\label{def.monoid.order} Let $M$ be a monoid (written
multiplicatively). Let $a\in M$.

If there exists a positive integer $n$ such that $a^{n}=1$, then the
\textbf{smallest} such $n$ is called the \textbf{order} of $a$, and is denoted
by $o\left(  a\right)  $; and then we say that $a$ has \textbf{order} $n$, and
that $a$ has \textbf{finite order}.

If no such $n$ exists, then we say that $a$ has \textbf{infinite order}, and
write $o\left(  a\right)  =\infty$.
\end{definition}

Note that if the monoid $M$ is finite and the element $a\in M$ has an inverse,
then $a$ always has finite order by Theorem \ref{thm.group.powers.periodic}
\textbf{(a)} from Lecture 5. Infinite orders can happen for elements $a$ that
have no inverses, as well as for elements of infinite monoids (even for
elements of infinite groups).

\begin{example}
\phantomsection\label{exa.group.order.S3} Recall the symmetric group $S_{3}$.
It has six elements $\operatorname*{id},\allowbreak s_{1},\allowbreak
s_{2},\allowbreak w_{0},\allowbreak c,\allowbreak c^{-1}$, where $s_{1}$ and
$s_{2}$ are as in Example \ref{exa.group.powers.noncomm} in Lecture 5 (so
$s_{1}$ swaps $1$ with $2$, whereas $s_{2}$ swaps $2$ with $3$), while $w_{0}$
swaps $1$ with $3$, and finally $c$ is the $3$-cycle $1\rightarrow
2\rightarrow3\rightarrow1$ (so that $c^{-1}$ is the $3$-cycle $1\rightarrow
3\rightarrow2\rightarrow1$). The orders of these six elements are%
\begin{align*}
o\left(  \operatorname*{id}\right)   &  =1,\ \ \ \ \ \ \ \ \ \ o\left(
s_{1}\right)  =2,\ \ \ \ \ \ \ \ \ \ o\left(  s_{2}\right)  =2,\\
o\left(  w_{0}\right)   &  =2,\ \ \ \ \ \ \ \ \ \ o\left(  c\right)
=3,\ \ \ \ \ \ \ \ \ \ o\left(  c^{-1}\right)  =3.
\end{align*}
For example, $o\left(  c\right)  =3$ is because $c^{3}=\operatorname*{id}$
(the neutral element of $S_{3}$) while $c^{1}$ and $c^{2}$ are not
$\operatorname*{id}$.
\end{example}

\begin{example}
\phantomsection\label{exa.monoid.order.zero-signs} Let $M$ be the monoid
$\left(  \left\{  0,1,-1\right\}  ,\cdot\right)  $. Its elements have orders%
\[
o\left(  0\right)  =\infty,\ \ \ \ \ \ \ \ \ \ o\left(  1\right)
=1,\ \ \ \ \ \ \ \ \ \ o\left(  -1\right)  =2.
\]

\end{example}

We observe that $o\left(  g^{-1}\right)  =o\left(  g\right)  $ for any $g\in
S_{3}$. This generalizes to invertible elements of monoids; see homework set \#2.

\begin{remark}
\phantomsection\label{rmk.group.order-size} In group theory, the word
\textquotedblleft order\textquotedblright\ has a second meaning: The
\textbf{order} of a group $G$ means the size (aka cardinality) $\left\vert
G\right\vert $ of the group, that is, the number of all elements of $G$. This
name is a historical fossil, and is nowadays more confusing than it is useful
(in particular, as Example \ref{exa.group.order.S3} shows, the order of a
group doesn't have to equal the order of any of its elements!); thus, we will
avoid it, instead simply saying \textquotedblleft size\textquotedblright\ or
\textquotedblleft cardinality\textquotedblright.
\end{remark}

There is actually a connection between the two orders: For any finite group
$G$ and any $a\in G$, we have $o\left(  a\right)  \mid\left\vert G\right\vert
$ (that is, the order of $a$ divides the order of $G$); equivalently,
$a^{\left\vert G\right\vert }=1$ (if we write $G$ multiplicatively). This will
fall out of something we will do in the next weeks: the notion of a coset and
Lagrange's theorem for subgroups. Note that this is not true for monoids, as
Example \ref{exa.monoid.order.zero-signs} shows.

\subsubsection{The structure of the powers of a finite-order element}

The order of a finite-order element $a\in M$ determines what its powers look like:

\begin{theorem}
\phantomsection\label{thm.monoid.order.powers} Let $M$ be a monoid (written
multiplicatively). Let $a\in M$ be an element that has finite order. Let
$n=o\left(  a\right)  $. Then:

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

\item \label{thm.monoid.order.powers.a} The powers $a^{0},a^{1},\ldots
,a^{n-1}$ are distinct.

\item \label{thm.monoid.order.powers.b} The element $a$ has an inverse, namely
$a^{n-1}$.

\item \label{thm.monoid.order.powers.c} Each power of $a$ (including the
negative powers) is one of $a^{0},a^{1},\ldots,a^{n-1}$.

\item \label{thm.monoid.order.powers.d} If $M$ is finite, then $n\leq
\left\vert M\right\vert $.
\end{enumerate}
\end{theorem}

\begin{proof}
We have $n=o\left(  a\right)  $, so that $n$ is a positive integer such that
$a^{n}=1$. We shall prove part \textbf{(\ref{thm.monoid.order.powers.b})}
first, so that cancellation will be justified in part
\textbf{(\ref{thm.monoid.order.powers.a})}.

\begin{enumerate}
[label=\textbf{(\alph*)},ref=\alph*,leftmargin=*,start=2,topsep=4pt,partopsep=0pt,itemsep=4pt,parsep=0pt]

\item \label{pf.thm.monoid.order.powers.b} By the rules of exponents
(specifically, Theorem \ref{thm.sg.powers.add} \textbf{(b)}), we have
$aa^{n-1}=a^{n}=1$ (since $n=o\left(  a\right)  $) and $a^{n-1}a=a^{n}=1$.
This shows that $a^{n-1}$ is an inverse of $a$. Thus, part \textbf{(b)} is proved.

%Part (b) is proved; return to part (a).
\setcounter{enumi}{0}

\item \label{pf.thm.monoid.order.powers.a} Assume the contrary. Thus, two of
these powers are equal. That is, $a^{p}=a^{q}$ for some $n>p>q\geq0$. Consider
these $p$ and $q$. Then, the rules of exponents yield
\begin{equation}
a^{p-q}a^{q}=a^{\left(  p-q\right)  +q}=a^{p}=a^{q}=1a^{q}.
\label{pf.thm.monoid.order.powers.a.4}%
\end{equation}


But part \textbf{(\ref{thm.monoid.order.powers.b})} shows that $a$ has an
inverse. Hence Lemma~\ref{lem.monoid.powers.neg-pow} from Lecture 5 shows that
$a^{q}$ has an inverse as well. Thus, we can cancel the $a^{q}$ factor from
(\ref{pf.thm.monoid.order.powers.a.4}) (using Theorem \ref{thm.monoid.cancel}
\textbf{(d)}), obtaining $a^{p-q}=1$. Note also that $p-q$ is a positive
integer (since $p>q$) and satisfies $p-q\leq p<n$ (since $q\geq0$ and $n>p$).

But $n=o\left(  a\right)  $ is defined as the \textbf{smallest} positive
integer $k$ such that $a^{k}=1$. So there cannot be any smaller such $k$. That
is, if $k$ is a positive integer such that $a^{k}=1$, then $k\geq n$. Applying
this to $k=p-q$, we obtain $p-q\geq n$ (since $a^{p-q}=1$), which contradicts
$p-q<n$. This contradiction shows that our assumption was wrong, and so part
\textbf{(a)} is proved.

%Part (b) was proved first; continue with (c).
\setcounter{enumi}{2}

\item \label{pf.thm.monoid.order.powers.c} This is visually clear if you draw
the powers of $a$ as a cycle.

For a proper proof, define the set $P=\left\{  a^{0},a^{1},\ldots
,a^{n-1}\right\}  $. Then, we must show that $a^{m}\in P$ for each
$m\in\mathbb{Z}$. We do this by two-sided induction on $m$ (see the proof of
Theorem \ref{thm.monoid.powers.add} in Lecture 5 for a detailed explanation of
this strategy). The \textit{base case} $m=0$ is obvious (we have\footnote{We
are using the fact that $n=o\left(  a\right)  $ is a positive integer (by its
definition), so that $0\in\left\{  0,1,\ldots,n-1\right\}  $.} $a^{0}%
\in\left\{  a^{0},a^{1},\ldots,a^{n-1}\right\}  =P$). For the
\textit{induction step}, we fix an integer $m$; then we must prove the
equivalence%
\[
\left(  a^{m}\in P\right)  \ \Longleftrightarrow\ \left(  a^{m+1}\in P\right)
.
\]
We prove its \textquotedblleft$\Longrightarrow$\textquotedblright\ and
\textquotedblleft$\Longleftarrow$\textquotedblright\ directions separately:

\begin{enumerate}
\item[$\Longrightarrow:$] If $a^{m}\in P$, then%
\begin{align*}
a^{m+1}  &  =a^{m}a\\
&  \in\left\{  a^{0}a,a^{1}a,\ldots,a^{n-2}a,a^{n-1}a\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }a^{m}\in
P=\left\{  a^{0},a^{1},\ldots,a^{n-2},a^{n-1}\right\}  \right) \\
&  =\left\{  a^{1},a^{2},\ldots,a^{n-1},a^{n}\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a^{k}a=a^{k+1}\text{ for each }%
k\in\mathbb{Z}\right) \\
&  =\left\{  a^{1},a^{2},\ldots,a^{n-1},a^{0}\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a^{n}=1=a^{0}\right) \\
&  =\left\{  a^{0},a^{1},\ldots,a^{n-1}\right\}  =P.
\end{align*}
This proves the \textquotedblleft$\Longrightarrow$\textquotedblright\ direction.

\item[$\Longleftarrow:$] Conversely, if $a^{m+1}\in P$, then%
\begin{align*}
a^{m}  &  =a^{m+1}a^{-1}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a^{m+1}%
a^{-1}=a^{\left(  m+1\right)  +\left(  -1\right)  }=a^{m}\right) \\
&  \in\left\{  a^{0}a^{-1},a^{1}a^{-1},\ldots,a^{n-2}a^{-1},a^{n-1}%
a^{-1}\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left(  \text{since }a^{m+1}\in
P=\left\{  a^{0},a^{1},\ldots,a^{n-2},a^{n-1}\right\}  \right) \\
&  =\left\{  a^{-1},a^{0},\ldots,a^{n-3},a^{n-2}\right\}  \quad\left(
\text{since }a^{k}a^{-1}=a^{k-1}\text{ for each }k\in\mathbb{Z}\right) \\
&  =\left\{  a^{n-1},a^{0},\ldots,a^{n-3},a^{n-2}\right\}
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a^{n-1}=a^{-1}\text{ by part
\textbf{(b)}}\right) \\
&  =\left\{  a^{0},a^{1},\ldots,a^{n-2},a^{n-1}\right\}  =P.
\end{align*}
This proves the \textquotedblleft$\Longleftarrow$\textquotedblright\ direction.
\end{enumerate}

So both directions are proved, and the induction step is complete. This proves
that $a^{m}\in P$ for each $m\in\mathbb{Z}$. Thus, part \textbf{(c)} is proved.

\item \label{pf.thm.monoid.order.powers.d} By part \textbf{(a)}, the elements
$a^{0},a^{1},\ldots,a^{n-1}$ are $n$ distinct elements of the set $M$. Thus,
$M$ contains $n$ distinct elements. Therefore, $n\leq\left\vert M\right\vert $
if $M$ is finite. \qedhere

\end{enumerate}
\end{proof}

\begin{remark}
\phantomsection\label{rmk.group.order.one-two} Let $G$ be a group. Only one
element of $G$ has order $1$, namely the identity element $1$.

A non-identity element $a\neq1$ of $G$ has order $2$ if and only if $a^{-1}=a$
(because $a^{-1}=a$ is equivalent to $aa=1$, that is, $a^{2}=1$). Note that a
group $G$ can have infinitely many elements of order $2$.
\end{remark}

\begin{example}
\phantomsection\label{exa.group.order.symdiff} Let $A$ be a set. In the group
$\left(  \mathcal{P}\left(  A\right)  ,\bigtriangleup\right)  $, each nonempty
subset $S$ of $A$ has order $2$. This is because $S\bigtriangleup
S=\varnothing$. (The empty subset has order $1$.)
\end{example}

\begin{example}
\phantomsection\label{exa.group.order.Z-add} In the group $\left(
\mathbb{Z},+\right)  $, the element $0$ has order $1$. All other elements
$a\in\mathbb{Z}$ have order $\infty$. In fact, the $n$-th power
\textquotedblleft$a^{n}$\textquotedblright\ of an element $a\in\mathbb{Z}$ in
the group $\left(  \mathbb{Z},+\right)  $ is just $na$, which is nonzero when
$n$ is positive and $a$ is nonzero. (This is a good example of when not to use
multiplicative notation.)
\end{example}

\begin{example}
\phantomsection\label{exa.monoid.order.numbers} In the monoid $\left(
\mathbb{Z},\cdot\right)  $, the elements have the following orders:%
\begin{align*}
o\left(  1\right)   &  =1,\ \ \ \ \ \ \ \ \ \ o\left(  -1\right)  =2,\\
o\left(  a\right)   &  =\infty\ \ \ \ \ \ \ \ \ \ \text{for all }%
a\notin\left\{  1,-1\right\}  .
\end{align*}


The same applies to $\left(  \mathbb{Q},\cdot\right)  $ and $\left(
\mathbb{R},\cdot\right)  $.

But the monoid $\left(  \mathbb{C},\cdot\right)  $ has many more elements of
finite order. They are called
\textbf{\href{https://en.wikipedia.org/wiki/Root_of_unity}{\textbf{roots of
unity}}}, and have the form%
\[
e^{2\pi iq}=\cos\left(  2\pi q\right)  +i\sin\left(  2\pi q\right)
\ \ \ \ \ \ \ \ \ \ \text{for rational }q
\]
(that is, geometrically speaking, they lie on the unit circle, and their
arguments are $2\pi q$ for rational $q$; in other words, they are vertices of
regular $n$-gons inscribed into the unit circle with one vertex at $1$).
Still, \textquotedblleft most\textquotedblright\ complex numbers have infinite
order in $\left(  \mathbb{C},\cdot\right)  $.
\end{example}

\subsubsection{Finite groups of size \texorpdfstring{$4$}{4}}

As an application of orders, let us continue on our \textquotedblleft what are
the finite groups of size $k$\textquotedblright\ quest. Recall that

\begin{enumerate}
\item there is only one group of size $1$, namely the trivial group $\left(
\left\{  1\right\}  ,\cdot\right)  $ (because it must have a neutral element,
and then, having size $1$, it cannot have any further elements).

\item there is only one group of size $2$, namely $\left(  \left\{
1,f\right\}  ,\cdot\right)  $ with $ff=1$ (as we saw in \S \ref{sec.gps.exas}
in Lecture 2).

\item there is only one group of size $3$, namely $\left(  \left\{
1,f,f^{2}\right\}  ,\cdot\right)  $ with $f^{3}=1$ (as we saw in Example
\ref{exa.groups-size.3} in Lecture 4; note that we denoted $f^{2}$ by $g$, but
it is still $f^{2}$).
\end{enumerate}

We are now ready for the next step: How many groups are there of size $4$ ?

Let $G$ be a group of size $4$, written multiplicatively. Its neutral element
$1$ has order $o\left(  1\right)  =1$; its other three elements have orders
between $2$ and $4$ inclusively (since Theorem~\ref{thm.monoid.order.powers}
\textbf{(\ref{thm.monoid.order.powers.d})} tells us that each order is $\leq
4$, but only the neutral element has order $1$). We distinguish between three cases:

\begin{enumerate}
[label=\textit{Case \arabic*:},ref=\arabic*,leftmargin=*,topsep=4pt,itemsep=6pt]

\item \label{case.group.size4.cyclic} Some element $a\in G$ has order $4$.
Then, $a^{4}=1$. Moreover, Theorem~\ref{thm.monoid.order.powers}
\textbf{(\ref{thm.monoid.order.powers.a})} tells us that $a^{0},a^{1}%
,a^{2},a^{3}$ are distinct. Since $G$ has only $4$ elements, this means that
$G=\left\{  a^{0},a^{1},a^{2},a^{3}\right\}  $. Then the multiplication table
of $G$ is given by $a^{i}a^{j}=a^{i+j}$, where $a^{4}=1=a^{0}$ and
$a^{5}=a\underbrace{a^{4}}_{=1}=a$ and $a^{6}=a\underbrace{a^{5}}%
_{=a}=aa=a^{2}$. That is, this table is%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|}\hline
& $a^{0}$ & $a^{1}$ & $a^{2}$ & $a^{3}$\\\hline\hline
$a^{0}$ & $a^{0}$ & $a^{1}$ & $a^{2}$ & $a^{3}$\\\hline
$a^{1}$ & $a^{1}$ & $a^{2}$ & $a^{3}$ & $a^{0}$\\\hline
$a^{2}$ & $a^{2}$ & $a^{3}$ & $a^{0}$ & $a^{1}$\\\hline
$a^{3}$ & $a^{3}$ & $a^{0}$ & $a^{1}$ & $a^{2}$\\\hline
\end{tabular}
\]
(where the format of the table is the same as before: we put $xy$ into the
cell in the $x$-th row and the $y$-th column). This table has a very simple
form: Its first row is $\left(  a^{0},a^{1},a^{2},a^{3}\right)  $, and each
further row is obtained by cyclically rotating the previous row one step to
the left. We will soon see that this really does define a group, called the
\textbf{cyclic group of order }$4$. (If you are impatient, you can check this
by hand... though this, too, requires some patience.)

\item \label{case.group.size4.order3} No element of $G$ has order $4$, but
some element $a$ has order $3$. Arguing as in
Case~\ref{case.group.size4.cyclic}, we see that $a^{0},a^{1},a^{2}$ are three
distinct elements of $G$, so there must be one more element $b\in G$ that is
not a power of $a$ (since $\left\vert G\right\vert =4$). Consider this $b$.
What is $ab$ ?

\begin{enumerate}
\item If $ab=1$, then $b=a^{-1}=a^{2}$ (since $a^{3}=1$), which contradicts
$b$ not being any of $a^{0},a^{1},a^{2}$.

\item If $ab=a$, then (cancelling $a$) we obtain $b=1=a^{0}$, which leads to a
similar contradiction.

\item If $ab=a^{2}$, then (cancelling $a$) we find $b=a=a^{1}$, which leads to
a similar contradiction as well.

\item If $ab=b$, then $a=1$ (by cancelling $b$), which is also false (since
$a$ has order $3$, not $1$).
\end{enumerate}

So we get a contradiction, which means that Case~\ref{case.group.size4.order3}
is impossible.

\Needspace{11\baselineskip}

\item \label{case.group.size4.klein} All three non-identity elements of $G$
have order $2$. Call them $a,b,c$. Then we can easily reconstruct the
multiplication table of $G$: For instance, $aa=a^{2}=1$ since $a$ has order
$2$. Likewise, $bb=1$ and $cc=1$. Furthermore, $ab$ cannot be $a$ (since
$b\neq1$), nor can it be $b$ (since $a\neq1$), nor can it be $1$ (since
$ab=1=aa$ would yield $b=a$). Hence, $ab$ must be $c$. Similarly, $ba=c$ and
$bc=a$ and $cb=a$ and $ca=b$ and $ac=b$. Thus, the multiplication table of $G$
looks as follows:%
\[%
\begin{tabular}
[c]{|c||c|c|c|c|}\hline
& $1$ & $a$ & $b$ & $c$\\\hline\hline
$1$ & $1$ & $a$ & $b$ & $c$\\\hline
$a$ & $a$ & $1$ & $c$ & $b$\\\hline
$b$ & $b$ & $c$ & $1$ & $a$\\\hline
$c$ & $c$ & $b$ & $a$ & $1$\\\hline
\end{tabular}
\ \ \ .
\]
Is there such a group? Yes: it is precisely the group $\left(  \mathcal{P}%
\left(  A\right)  ,\bigtriangleup\right)  $ for $A=\left\{  1,2\right\}  $,
with its four elements $\varnothing,\ \left\{  1\right\}  ,\ \left\{
2\right\}  ,\ \left\{  1,2\right\}  $ renamed as $1,a,b,c$. This is known as
the \textbf{Klein four-group}.
\end{enumerate}

So we conclude that there are two essentially different (as we will later say:
non-isomorphic) groups of size $4$, namely the cyclic group and the Klein four-group.

What about size $5$ ? We will answer this later, once we have proved
Lagrange's theorem.

\subsection{The opposite group}

\label{sec.sg.opposite}

There are several ways to build new groups from old ones. The following is
perhaps the simplest:

\begin{proposition}
\phantomsection\label{prop.sg.opposite} Let $\left(  S,\ast\right)  $ be a
semigroup. Then, we can define a new binary operation $\opmul $ on $S$ by
setting%
\[
s \opmul t:=t\ast s\ \ \ \ \ \ \ \ \ \ \text{for all }s,t\in S.
\]
So $\opmul $ is the same operation as $\ast$ but with its two arguments
swapped. Then, $\left(  S,\opmul \right)  $ is again a semigroup.

The same holds if we replace \textquotedblleft semigroup\textquotedblright\ by
\textquotedblleft monoid\textquotedblright\ or \textquotedblleft
group\textquotedblright.
\end{proposition}

\begin{proof}
Let us first check associativity.

For any $a,b,c\in S$, we have%
\[
a \opmul \left(  b \opmul c\right)  =\left(  b \opmul
c\right)  \ast a=\left(  c\ast b\right)  \ast a
\]
and similarly%
\[
\left(  a \opmul b\right)  \opmul c=c\ast\left(  b\ast a\right)  .
\]
Since $\ast$ is associative, the right-hand sides of these two equalities are
equal; therefore so are the left-hand sides. That is, $a \opmul \left(  b
\opmul c\right)  =\left(  a \opmul b\right)  \opmul c$ for all $a,b,c\in S$.
In other words, $\opmul $ is associative. This shows that $\left(
S,\opmul \right)  $ is a semigroup.

If $(S,\ast)$ is a monoid with neutral element $e$, then each $s\in S$
satisfies%
\[
e\opmul s=s\ast e=s\qquad\text{and}\qquad s\opmul e=e\ast s=s.
\]
Thus, $e$ is also neutral for $\opmul $, and therefore $\left(
S,\opmul \right)  $ is a monoid.

If $(S,\ast)$ is a group and $t$ is the inverse of $s$ with respect to $\ast$,
then
\[
s\opmul t=t\ast s=e\qquad\text{and}\qquad t\opmul s=s\ast t=e.
\]
Thus, the inverses are the same for the two operations $\ast$ and $\opmul $,
so $(S,\opmul )$ is a group.
\end{proof}

\begin{definition}
\phantomsection\label{def.sg.opposite} If $\left(  S,\ast\right)  $ is a
semigroup, and $\opmul$ is the operation defined in Proposition
\ref{prop.sg.opposite}, then the semigroup $\left(  S,\opmul\right)  $ is
called the \textbf{opposite semigroup} of $\left(  S,\ast\right)  $, and is
denoted by $\left(  S,\ast\right)  ^{\operatorname*{op}}$.

Likewise, if $\left(  S,\ast\right)  $ is a monoid or a group, then the monoid
$\left(  S,\opmul\right)  $ is called the \textbf{opposite monoid} or
\textbf{opposite group} (respectively) of $\left(  S,\ast\right)  $.

In general, we cannot use multiplicative notation for both $\left(
S,\ast\right)  $ and $\left(  S,\opmul\right)  $ simultaneously, since this
would cause two different operations to both be denoted by the same $\cdot$ symbol.
\end{definition}

\subsection{Abelian groups}

\label{sec.sg.abelian}

As we said, commutativity is not a given in semigroups, but it happens often
and deserves further study. We recall its definition:

\begin{definition}
\phantomsection\label{def.sg.commute} Let $S$ be a semigroup (written
multiplicatively). Two elements $a$ and $b$ of $S$ are said to
\textbf{commute} if $ab=ba$.
\end{definition}

\begin{example}
\textbf{(a)} For a non-example, consider the symmetric group $S_{3}$ again.
The permutations $s_{1}$ and $s_{2}$ in $S_{3}$ (see Example
\ref{exa.group.order.S3}) do not commute (since $s_{1}s_{2}$ is the cycle
$1\rightarrow2\rightarrow3\rightarrow1$, whereas $s_{2}s_{1}$ is the cycle
$1\rightarrow3\rightarrow2\rightarrow1$).

\textbf{(b)} For a positive example, consider any semigroup $S$. Then, any two
powers of a single element $a\in S$ commute. That is, if $a\in S$ and $n,m>0$,
then $a^{n}$ and $a^{m}$ commute (since $a^{n}a^{m}=a^{n+m}=a^{m}a^{n}$). The
same holds for $n,m\in\mathbb{N}$ when $S$ is a monoid and for $n,m\in
\mathbb{Z}$ when $S$ is a group (or $a$ has an inverse). (Recall that
$\mathbb{N}=\left\{  0,1,2,\ldots\right\}  $.)
\end{example}

\begin{definition}
\phantomsection\label{def.sg.abelian} A semigroup (or monoid or group) $S$
(written multiplicatively) is said to be \textbf{abelian} (or
\textbf{commutative}) if all $a,b\in S$ commute, i.e., if we have $ab=ba$ for
all $a,b\in S$.
\end{definition}

Thus, a semigroup is abelian if and only if its multiplication table is
symmetric across its main diagonal.

\Needspace{5\baselineskip} Here are some examples and non-examples:

\begin{itemize}
\item The semigroup $\left(  \mathbb{Z}_{>0},+\right)  $, the monoid $\left(
\mathbb{N},+\right)  $, and the groups $\left(  \mathbb{Z},+\right)  $,
$\left(  \mathbb{C},+\right)  $ and $\left(  \mathbb{R}^{n\times n},+\right)
$ are abelian.

\item Any of the monoids $\left(  \mathbb{Z}_{>0},\cdot\right)  $, $\left(
\mathbb{N},\cdot\right)  $, $\left(  \mathbb{Z},\cdot\right)  $ and $\left(
\mathbb{C},\cdot\right)  $ is abelian, but the monoids $\left(  \mathbb{R}%
^{n\times n},\cdot\right)  $ and $\left(  \operatorname*{GL}\nolimits_{n}%
\left(  \mathbb{R}\right)  ,\cdot\right)  $ for $n>1$ are not (since matrices
do not commute, even invertible matrices).

\item For any set $A$, the group $\left(  \mathcal{P}\left(  A\right)
,\bigtriangleup\right)  $ and the monoids $\left(  \mathcal{P}\left(
A\right)  ,\cup\right)  $ and $\left(  \mathcal{P}\left(  A\right)
,\cap\right)  $ are abelian.

\item All groups of size $\leq4$ are abelian (as we have seen when we
described them). This still holds for groups of size $\leq5$ (as we will soon
see). But not for groups of size $6$: the symmetric group $S_{3}$ is not
abelian ($s_{1}$ and $s_{2}$ do not commute). Non-abelian finite groups exist
in sizes $6$ and $8$ but not $7$ and $9$; we will later learn why.

Generally, the symmetric group $S_{n}$ (that is, the group of all the $n!$
many permutations of $\left\{  1,2,\ldots,n\right\}  $) is not abelian
whenever $n\geq3$. But the symmetric group $S_{2}$ has size $2$, so it is abelian.

\item In contrast, there are non-abelian semigroups of any size $\geq2$.
Indeed, on any set $S$, we can define a binary operation $\ast$ by%
\[
a\ast b:=b\ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in S
\]
(that is, the operation that forgets its first argument and returns its
second); this operation is associative, thus defines a semigroup, and this
semigroup is non-abelian as soon as $S$ has two distinct elements.
\end{itemize}

We have been often using multiplicative notation for arbitrary semigroups. For
abelian semigroups, it has a rival: the (analogous) additive notation, in
which the operation is written as $+$. In more detail:

\begin{definition}
\phantomsection\label{def.sg.additive} Abelian semigroups (thus also abelian
monoids and abelian groups) are often \textbf{written additively} (aka in
\textbf{additive notation}); this means that the binary operation is denoted
by $+$. Correspondingly, the neutral element (when it exists) is denoted by
$0$ (not by $1$). The inverse of an element $a$ (when it exists) is then
denoted by $-a$ (not by $a^{-1}$). The $n$-th power of an element $a$ is then
denoted by $na$ (not by $a^{n}$).
\end{definition}

All this notation is merely generalizing the example of $\left(
\mathbb{Z},+\right)  $, in which the meanings of $a+b$, $0$, $-a$ and $na$ we
just defined are precisely the usual meanings of these expressions known from
middle school. Of course, you should not use additive notation when the symbol
$+$ already has a different meaning on your set.

Also, it is \textbf{not recommended} to use additive notation for non-abelian
semigroups (or monoids, or groups). Most mathematicians instinctively reorder
sums without thinking, relying instinctively on the commutativity of addition;
thus, writing a non-commutative operation as addition risks this kind of
reordering being used by accident when it is unjustified.

\Needspace{10\baselineskip}

\subsection{Subsemigroups, submonoids, subgroups}

\label{sec.sg.substructures}

We shall now define \textbf{subobjects} of the objects we have been studying:
subsemigroups of semigroups, submonoids of monoids, subgroups of groups. The
definitions are fairly unsurprising if you have seen your share of
mathematical terminology (compare subsets, vector subspaces and topological subspaces):

\begin{definition}
\phantomsection\label{def.sg.subsemigroup} Let $\left(  S,\ast\right)  $ be a
semigroup. A subset $T$ of $S$ is said to be a \textbf{subsemigroup} of
$\left(  S,\ast\right)  $ if it is closed under the operation $\ast$, that is,
if all $a,b\in T$ satisfy $a\ast b\in T$.

If this is the case, then we write $T\leq_{\operatorname*{semi}}\left(
S,\ast\right)  $, or just $T\leq_{\operatorname*{semi}}S$ when the operation
$\ast$ is clear from the context.
\end{definition}

\begin{definition}
\phantomsection\label{def.monoid.submonoid} Let $\left(  S,\ast\right)  $ be a
monoid. A subset $T$ of $S$ is said to be a \textbf{submonoid} of $\left(
S,\ast\right)  $ if it contains the neutral element of $\left(  S,\ast\right)
$ and is closed under the operation $\ast$.

If this is the case, then we write $T\leq_{\operatorname*{monoid}}\left(
S,\ast\right)  $, or just $T\leq_{\operatorname*{monoid}}S$ when the operation
$\ast$ is clear from the context.
\end{definition}

Thus, a submonoid of a monoid $\left(  S,\ast\right)  $ is the same as a
subsemigroup of $\left(  S,\ast\right)  $ that contains the neutral element of
$\left(  S,\ast\right)  $.

\begin{definition}
\phantomsection\label{def.group.subgroup} Let $\left(  G,\ast\right)  $ be a
group. A subset $H$ of $G$ is said to be a \textbf{subgroup} of $\left(
G,\ast\right)  $ if it satisfies the following three axioms:

\begin{enumerate}
\item \label{def.group.subgroup.product} it is closed under $\ast$, that is,
we have $a\ast b\in H$ for all $a,b\in H$;

\item \label{def.group.subgroup.identity} it contains the neutral element of
$\left(  G,\ast\right)  $;

\item \label{def.group.subgroup.inverse} it is closed under taking inverses,
that is, we have $a^{-1}\in H$ for all $a\in H$.
\end{enumerate}

If this is the case, then we write $H\leq\left(  G,\ast\right)  $, or just
$H\leq G$.
\end{definition}

So a subgroup of a group $\left(  G,\ast\right)  $ is the same as a submonoid
of $\left(  G,\ast\right)  $ that is closed under taking inverses (i.e.,
contains the inverse of each of its elements).

All three definitions we just gave follow the same pattern: A sub-X of a given
X (where X is one of \textquotedblleft semigroup\textquotedblright,
\textquotedblleft monoid\textquotedblright\ and \textquotedblleft
group\textquotedblright) is a subset of this X that contains (parts of) all
the defining features of this X. Here, we regard the neutral element as a
defining feature of monoids, and we regard the neutral element and inverses as
defining features of groups.

Here are some examples:

\begin{itemize}
\item We have $\mathbb{N}\leq_{\operatorname*{monoid}}\left(  \mathbb{Z}%
,+\right)  $, because $\mathbb{N}$ is a subset of $\mathbb{Z}$ that contains
the neutral element $0$ of $\left(  \mathbb{Z},+\right)  $ and is closed under
$+$ (any sum of two nonnegative integers is a nonnegative integer).

\item We have $\mathbb{Z}_{>0}\leq_{\operatorname*{semi}}\left(
\mathbb{Z},+\right)  $ (but not $\leq_{\operatorname*{monoid}}$ because
$\mathbb{Z}_{>0}$ does not contain the neutral element $0$ of $\left(
\mathbb{Z},+\right)  $).

\item We have $\mathbb{Z}\leq\mathbb{Q}$ and $\mathbb{Q}\leq\mathbb{R}$ and
$\mathbb{R}\leq\mathbb{C}$, where all the groups are understood to be additive
(i.e., read $\mathbb{Z}$, $\mathbb{Q}$ and $\mathbb{R}$ as $\left(
\mathbb{Z},+\right)  $, $\left(  \mathbb{Q},+\right)  $ and $\left(
\mathbb{R},+\right)  $).

\item Let $\operatorname*{GL}\nolimits_{n}^{\operatorname*{diag}}\left(
\mathbb{R}\right)  $ be the set of all diagonal $n\times n$-matrices with real
entries such that all their diagonal entries are nonzero. Note that the
product of two diagonal $n\times n$-matrices is again diagonal and is given by
the formula%
\[
\operatorname*{diag}\left(  a_{1},a_{2},\ldots,a_{n}\right)  \cdot
\operatorname*{diag}\left(  b_{1},b_{2},\ldots,b_{n}\right)
=\operatorname*{diag}\left(  a_{1}b_{1},a_{2}b_{2},\ldots,a_{n}b_{n}\right)
;
\]
thus, all such matrices are invertible, with inverses given by
\[
\left(  \operatorname{diag}(a_{1},a_{2},\ldots,a_{n})\right)  ^{-1}%
=\operatorname{diag}(a_{1}^{-1},a_{2}^{-1},\ldots,a_{n}^{-1}).
\]
Thus, the set $\operatorname*{GL}\nolimits_{n}^{\operatorname*{diag}}\left(
\mathbb{R}\right)  $ is a subset of $\operatorname*{GL}\nolimits_{n}\left(
\mathbb{R}\right)  $ that is closed under multiplication, contains the
identity matrix, and is closed under taking inverses. Hence,
$\operatorname*{GL}\nolimits_{n}^{\operatorname*{diag}}\left(  \mathbb{R}%
\right)  \leq\operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)  $
(where $\operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)  $ means
$\left(  \operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)
,\cdot\right)  $ as usual).

\item Let $S$ be a semigroup (written multiplicatively), and let $a\in S$ be
arbitrary. Then,%
\begin{equation}
\left\{  a^{1},a^{2},a^{3},\ldots\right\}  \leq_{\operatorname*{semi}}S,
\label{eq.subsg.powers}%
\end{equation}
because the set $\left\{  a^{1},a^{2},a^{3},\ldots\right\}  $ is closed under
multiplication (since $a^{n}a^{m}=a^{n+m}$ for all $n,m>0$).

\item Let $S$ be a monoid (written multiplicatively), and let $a\in S$ be
arbitrary. Then,%
\begin{equation}
\left\{  a^{0},a^{1},a^{2},a^{3},\ldots\right\}  \leq_{\operatorname*{monoid}%
}S, \label{eq.submon.powers}%
\end{equation}
because the set $\left\{  a^{0},a^{1},a^{2},a^{3},\ldots\right\}  $ is closed
under multiplication (since $a^{n}a^{m}=a^{n+m}$ for all $n,m\in\mathbb{N}$)
and because $1=a^{0}$.

\item Let $G$ be a group (written multiplicatively), and let $a\in G$ be
arbitrary. Then,%
\begin{equation}
\left\{  a^{n}\ \mid\ n\in\mathbb{Z}\right\}  \leq G,
\label{eq.subgroup.powers}%
\end{equation}
because the set $\left\{  a^{n}\ \mid\ n\in\mathbb{Z}\right\}  =\left\{
\ldots,a^{-2},a^{-1},a^{0},a^{1},a^{2},\ldots\right\}  $ is closed under
multiplication (since $a^{n}a^{m}=a^{n+m}$ for all $n,m\in\mathbb{Z}$) and
because $1=a^{0}$ and $\left(  a^{n}\right)  ^{-1}=a^{-n}$ (by Lemma
\ref{lem.monoid.powers.neg-pow} in Lecture 5).

\begin{sloppypar}
In contrast, $\left\{  a^{0},a^{1},a^{2},a^{3},\ldots\right\}  $ is only a
submonoid of $G$ in general, and $\left\{  a^{1},a^{2},a^{3},\ldots\right\}  $
is only a subsemigroup.
\end{sloppypar}

\item Let $A$ be a set. Let $\mathcal{P}_{\operatorname*{even}}\left(
A\right)  $ denote the set of all finite subsets of $A$ whose size is even.
This is a subgroup of $\left(  \mathcal{P}\left(  A\right)  ,\bigtriangleup
\right)  $. Why? It is closed under $\bigtriangleup$, because for every
$S,T\in\mathcal{P}_{\operatorname*{even}}\left(  A\right)  $, the size%
\begin{align*}
\left\vert S\bigtriangleup T\right\vert  &  =\left\vert \left(  S\cup
T\right)  \setminus\left(  S\cap T\right)  \right\vert
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }S\bigtriangleup T=\left(  S\cup
T\right)  \setminus\left(  S\cap T\right)  \right) \\
&  =\underbrace{\left\vert S\cup T\right\vert }_{\substack{=\left\vert
S\right\vert +\left\vert T\right\vert -\left\vert S\cap T\right\vert
\\\text{(by inclusion-exclusion:}\\\text{e.g., look at the Venn diagram)}%
}}-\left\vert S\cap T\right\vert \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}S\cap T\subseteq S\cup T\right) \\
&  =\left\vert S\right\vert +\left\vert T\right\vert -\left\vert S\cap
T\right\vert -\left\vert S\cap T\right\vert \\
&  =\underbrace{\left\vert S\right\vert }_{\substack{\text{even}\\\text{(since
}S\in\mathcal{P}_{\operatorname*{even}}\left(  A\right)  \text{)}%
}}+\underbrace{\left\vert T\right\vert }_{\substack{\text{even}\\\text{(since
}T\in\mathcal{P}_{\operatorname*{even}}\left(  A\right)  \text{)}%
}}-\underbrace{2\cdot\left\vert S\cap T\right\vert }_{\text{even}%
}\ \ \ \ \ \ \ \ \ \ \text{is even,}%
\end{align*}


\begin{sloppypar}
so we have $S\bigtriangleup T\in\mathcal{P}_{\operatorname*{even}}\left(
A\right)  $. It contains the neutral element of $\left(  \mathcal{P}\left(
A\right)  ,\bigtriangleup\right)  $, since this neutral element is
$\varnothing$ and has size $\left\vert \varnothing\right\vert =0$. Finally, it
is closed under taking inverses, since the inverse of any element
$S\in\mathcal{P}\left(  A\right)  $ is $S$ itself.
\end{sloppypar}

\item The monoid $\left(  \left\{  0,1,-1\right\}  ,\cdot\right)  $ has a
subsemigroup $\left\{  0\right\}  $ (since $0\cdot0=0$). But this subsemigroup
$\left\{  0\right\}  $ is not a submonoid of $\left(  \left\{  0,1,-1\right\}
,\cdot\right)  $, since it does not contain the neutral element $1$ of the
latter monoid.

\item What are the subgroups of $\left(  \mathbb{Z},+\right)  $ ? For any
integer $n$, the set%
\[
n\mathbb{Z}:=\left\{  \text{all integer multiples of }n\right\}
\]
is a subgroup of $\left(  \mathbb{Z},+\right)  $, since it is closed under
addition (because $na+nb=n\left(  a+b\right)  $ for all $a,b\in\mathbb{Z}$),
contains the neutral element (because $0=n0$) and is closed under taking
inverses (because $-na=n\left(  -a\right)  $ for all $a\in\mathbb{Z}$).

Note that $0\mathbb{Z}=\left\{  0\right\}  $, so that too is a subgroup. And
$1\mathbb{Z}=\mathbb{Z}$ is a subgroup as well.

Are there any others? We will see that the answer is \textquotedblleft
no\textquotedblright; we've found them all.
\end{itemize}

The most boring examples are ones that work in the greatest generality:

\begin{proposition}
\phantomsection\label{rmk.sg.trivial-substructures}\textbf{(a)} If $M$ is any
monoid (written multiplicatively), then $\left\{  1\right\}  $ and $M$ itself
are submonoids of $M$.

\textbf{(b)} If $G$ is any group (written multiplicatively), then $\left\{
1\right\}  $ and $G$ itself are subgroups of $G$.

\textbf{(c)} If $S$ is a semigroup, then $\varnothing$ and $S$ itself are
subsemigroups of $S$. (Recall that the empty set is a semigroup, but not a
monoid or a group.)
\end{proposition}

\begin{proof}
LTTR. (The subset $\left\{  1\right\}  $ is closed under taking inverses since
$1$ is its own inverse. Everything else is entirely trivial.)
\end{proof}

One important feature of subsemigroups is the following: Any subsemigroup $T$
of a semigroup $\left(  S,\ast\right)  $ becomes a semigroup in its own right,
simply by inheriting the operation $\ast$ from $S$ (that is, for any $a,b\in
T$, we define the \textquotedblleft product\textquotedblright\ $a\ast b$ in
$T$ to be their original \textquotedblleft product\textquotedblright\ $a\ast
b$ in $S$; the definition of a subsemigroup ensures that this
\textquotedblleft product\textquotedblright\ does belong to $T$). Let us
codify this:

\begin{theorem}
\phantomsection\label{thm.sg.subsemigroup} Let $\left(  S,\ast\right)  $ be a
semigroup, and $T$ a subsemigroup of $\left(  S,\ast\right)  $. Define a new
operation $\ast_{T}:T\times T\rightarrow T$ by setting
\[
a\ast_{T}b=a\ast b\ \ \ \ \ \ \ \ \ \ \text{for all }a,b\in T.
\]
Then, $\left(  T,\ast_{T}\right)  $ is again a semigroup.
\end{theorem}

\begin{proof}
Since $T$ is closed under $\ast$, the operation $\ast_{T}$ takes its values in
$T$. Thus, $\ast_{T}$ is a well-defined binary operation on $T$. The
associativity of $\ast_{T}$ follows directly from associativity of $\ast$,
since for any $a,b,c\in T$ we have%
\begin{align*}
a\ast_{T}\left(  b\ast_{T}c\right)   &  =a\ast\left(  b\ast c\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\\
\left(  a\ast_{T}b\right)  \ast_{T}c  &  =\left(  a\ast b\right)  \ast c.
\end{align*}
This shows that $\left(  T,\ast_{T}\right)  $ is a semigroup.
\end{proof}

\begin{sloppypar}
The same construction, applied to submonoids and subgroups, yields monoids and
groups again:
\end{sloppypar}

\begin{theorem}
\phantomsection\label{thm.monoid.submonoid} Let $\left(  M,\ast\right)  $ be a
monoid, and $N$ a submonoid of $\left(  M,\ast\right)  $. Then, the semigroup
$\left(  N,\ast_{N}\right)  $ (defined as in Theorem \ref{thm.sg.subsemigroup}
for $S=M$ and $T=N$) is a monoid.
\end{theorem}

\begin{proof}
The submonoid $N$ is a subsemigroup of $M$; thus, by
Theorem~\ref{thm.sg.subsemigroup}, we know that $\left(  N,\ast_{N}\right)  $
is a semigroup. The neutral element of $M$ belongs to $N$ by the definition of
a submonoid, and is still neutral in $\left(  N,\ast_{N}\right)  $ (because
$\ast_{N}$ is just a restriction of $\ast$). Thus, $\left(  N,\ast_{N}\right)
$ is a monoid.
\end{proof}

\begin{theorem}
\phantomsection\label{thm.group.subgroup} Let $\left(  G,\ast\right)  $ be a
group, and $H$ a subgroup of $\left(  G,\ast\right)  $. Then, the monoid
$\left(  H,\ast_{H}\right)  $ (defined as in Theorem
\ref{thm.monoid.submonoid} for $M=G$ and $N=H$) is a group. Moreover, the
inverse $a^{-1}$ of any element $a\in H$ computed in this group $\left(
H,\ast_{H}\right)  $ is precisely the inverse of $a$ computed in $\left(
G,\ast\right)  $.
\end{theorem}

\begin{proof}
The subgroup $H$ is a submonoid of $G$, so Theorem~\ref{thm.monoid.submonoid}
shows that $\left(  H,\ast_{H}\right)  $ is a monoid with the same neutral
element as $G$. For every $a\in H$, its inverse $a^{-1}$ in $G$ belongs to $H$
by the definition of a subgroup. Moreover, it still plays the role of inverse
of $a$ in $H$, because the operation $\ast_{H}$ of $H$ is just a restriction
of $\ast$ (so we have $a\ast_{H}a^{-1}=a\ast a^{-1}=e$ and likewise
$a^{-1}\ast_{H}a=e$). Thus, each $a\in H$ has an inverse in $H$. Therefore,
the monoid $\left(  H,\ast_{H}\right)  $ is a group.
\end{proof}

\begin{definition}
Theorem \ref{thm.sg.subsemigroup}, Theorem \ref{thm.monoid.submonoid} and
Theorem \ref{thm.group.subgroup} will often be used automatically, without
explicit mention. Thus, when we say that $H$ is a subgroup of a group $\left(
G,\ast\right)  $, we will take it as self-evident that this makes $\left(
H,\ast_{H}\right)  $ into a group (by Theorem \ref{thm.group.subgroup}), and
we shall simply refer to this group $\left(  H,\ast_{H}\right)  $ as $H$
without mentioning $\ast_{H}$. When we do want to mention its operation, we
will often write it as $\ast$ instead of $\ast_{H}$, since it is just a
restriction of $\ast$; this is a bit sloppy but should cause no trouble.
(After all, we are already using the same symbol $+$ for the operations of the
groups $\left(  \mathbb{Z},+\right)  $ and $\left(  \mathbb{Q},+\right)  $.)
Conversely, when we say that a group $\left(  H,\circledcirc\right)  $ is a
subgroup of a group $\left(  G,\ast\right)  $, this will always be understood
to mean that $H$ is a subgroup of $\left(  G,\ast\right)  $ and that
$\circledcirc$ is the operation $\ast_{H}$ obtained by restricting $\ast$ to
$H\times H$.

All this applies similarly to submonoids and subsemigroups.
\end{definition}

Thus, for example, given any set $A$, the subgroup $\mathcal{P}%
_{\operatorname*{even}}\left(  A\right)  $ of $\left(  \mathcal{P}\left(
A\right)  ,\bigtriangleup\right)  $ becomes itself a group $\left(
\mathcal{P}_{\operatorname*{even}}\left(  A\right)  ,\bigtriangleup\right)  $.

\begin{warning}
A subsemigroup $N$ of a monoid $\left(  M,\ast\right)  $ can be a monoid
itself (i.e., the semigroup $\left(  N,\ast\right)  $ can be a monoid) without
being a submonoid of $\left(  M,\ast\right)  $ ! For example, the monoid
$\left(  \left\{  0\right\}  ,\cdot\right)  $ is a subsemigroup of the monoid
$\left(  \left\{  0,1,-1\right\}  ,\cdot\right)  $, but not a submonoid of
$\left(  \left\{  0,1,-1\right\}  ,\cdot\right)  $, because its neutral
element is $0$, not $1$. A submonoid has to contain the neutral element of the
whole monoid, not its own different neutral element!
\end{warning}

\begin{remark}
Let $G$ be a group (written multiplicatively), and let $H$ be a subgroup of
$G$. Let $a\in H$. Then, any power $a^{n}$ of $a$ computed in the group $H$
equals the corresponding power $a^{n}$ computed in $G$. This is because
$a^{n}$ is defined entirely in terms of the defining features of a group
(multiplication, neutral element and inverses), and all these features for $H$
are restrictions of the corresponding features for $G$ (for example, the
inverses in $H$ are the same as in $G$). The same applies to other features
such as orders and commutativity. However, an element $a\in H$ can be central
in $H$ without being central in $G$, because the group $H$ may have fewer
elements to \textquotedblleft veto\textquotedblright\ its centrality than $G$ does!
\end{remark}

\Needspace{10\baselineskip}

\subsection{The center of a semigroup/monoid/group}

\label{sec.sg.center}

When a semigroup (or monoid, or group) is not abelian, one may still study the
extent to which commutativity holds for some of its elements. This leads to
the following definition:

\begin{definition}
\phantomsection\label{def.sg.central} An element $a$ of a semigroup $S$ is
said to be \textbf{central} if it commutes with every $b\in S$, that is, if
$ab=ba$ for all $b\in S$ (where we write $S$ multiplicatively).
\end{definition}

\begin{definition}
\phantomsection\label{def.sg.center} The \textbf{center} $Z\left(  S\right)  $
of a semigroup $S$ is the set of all central elements of $S$.
\end{definition}

\begin{theorem}
\phantomsection\label{thm.sg.center} \leavevmode


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

\item \label{thm.sg.center.a} The center $Z\left(  S\right)  $ of a semigroup
$S$ is a subsemigroup of $S$.

\item \label{thm.sg.center.b} The center $Z\left(  M\right)  $ of a monoid $M$
is a submonoid of $M$.

\item \label{thm.sg.center.c} The center $Z\left(  G\right)  $ of a group $G$
is a subgroup of $G$.
\end{enumerate}
\end{theorem}

\Needspace{4\baselineskip}

\begin{proof}
We shall write $S$, $M$ and $G$ multiplicatively. \leavevmode


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

\item \label{pf.thm.sg.center.a} We must show that $Z\left(  S\right)  $ is
closed under multiplication. That is, we must show that if $a_{1},a_{2}\in
Z\left(  S\right)  $, then $a_{1}a_{2}\in Z\left(  S\right)  $.

So let $a_{1},a_{2}\in Z\left(  S\right)  $. That is, $a_{1}$ and $a_{2}$ are
central. Now, for each $b\in S$, we have%
\[
a_{1}\underbrace{a_{2}b}_{\substack{=ba_{2}\\\text{(since }a_{2}\text{ is
central)}}}=\underbrace{a_{1}b}_{\substack{=ba_{1}\\\text{(since }a_{1}\text{
is central)}}}a_{2}=ba_{1}a_{2}.
\]
This shows that $a_{1}a_{2}$ is central. In other words, $a_{1}a_{2}\in
Z\left(  S\right)  $, just as we wished to prove. Thus, part \textbf{(a)} is proved.

\item \label{pf.thm.sg.center.b} Part \textbf{(a)} shows that $Z\left(
M\right)  $ is a subsemigroup of $M$. It remains to show that it also contains
the neutral element $1$ of $M$. In other words, we must show that $1$ is
central. But this is clear, since we have $1b=b=b1$ for each $b\in M$.

\item \label{pf.thm.sg.center.c} Part \textbf{(b)} shows that $Z\left(
G\right)  $ is a submonoid of $G$. It remains to prove that $Z\left(
G\right)  $ is closed under taking inverses. So we must show that each $a\in
Z\left(  G\right)  $ satisfies $a^{-1}\in Z\left(  G\right)  $.

Let $a\in Z\left(  G\right)  $. We must show that $a^{-1}\in Z\left(
G\right)  $.

Let $b\in G$. Since $a$ is central (because $a\in Z\left(  G\right)  $), we
have $ab=ba$. Multiplying this equality by $a^{-1}$ from the left, we obtain
$a^{-1}ab=a^{-1}ba$. In other words, $b=a^{-1}ba$ (since $\underbrace{a^{-1}%
a}_{=1}b=1b=b$). Multiplying this latter equality by $a^{-1}$ from the right,
we obtain $ba^{-1}=a^{-1}b\underbrace{aa^{-1}}_{=1}=a^{-1}b1=a^{-1}b$. In
other words, $a^{-1}b=ba^{-1}$.

Thus, we have shown that $a^{-1}b=ba^{-1}$ for each $b\in G$. This means
precisely that $a^{-1}$ is central. That is, $a^{-1}\in Z\left(  G\right)  $.
This completes our proof of part \textbf{(c)}. \qedhere

\end{enumerate}
\end{proof}

\begin{example}
\label{exa.group.center.Zabel}If $S$ is an abelian semigroup, then $Z\left(
S\right)  =S$ (since each $a\in S$ is central).
\end{example}

\begin{example}
\label{exa.group.center.Sn}Let $n\geq3$. Consider the symmetric group $S_{n}$
(that is, the group of all permutations of $\left\{  1,2,\ldots,n\right\}  $).
It can be shown (see homework set \#2) that $Z\left(  S_{n}\right)  =\left\{
\operatorname*{id}\right\}  $. That is, the \textbf{only} permutation $w\in
S_{n}$ that commutes with all permutations in $S_{n}$ is the identity
$\operatorname*{id}$.
\end{example}

\begin{example}
\label{exa.group.center.GL}Let $n\geq1$. Consider the group
$\operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)  $ of all invertible
$n\times n$-matrices with real entries (under multiplication). What is the
center $Z\left(  \operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)
\right)  $ ?

A moment's thought suggests that each diagonal matrix with all diagonal
entries equal (i.e., each matrix of the form $\lambda I_{n}%
=\operatorname*{diag}\left(  \lambda,\lambda,\ldots,\lambda\right)  $ for
$\lambda\in\mathbb{R}\setminus\left\{  0\right\}  $) is central in
$\operatorname*{GL}\nolimits_{n}\left(  \mathbb{R}\right)  $. And this is
indeed true, since every further $n\times n$-matrix $B$ satisfies $\left(
\lambda I_{n}\right)  B=\lambda B=B\left(  \lambda I_{n}\right)  $. Are there
any other central matrices in $\operatorname*{GL}\nolimits_{n}\left(
\mathbb{R}\right)  $ ?
\end{example}


\end{document}