handouts/Advanced/Group Theory/parts/01 groups.tex

\section{Groups}

\definition{}
A \textit{group} $(G, \ast)$ consists of a set $G$ and an operator $\ast$. \\
A group must have the following properties: \\

\begin{enumerate}
	\item $G$ is closed under $\ast$. In other words, $a, b \in G \implies a \ast b \in G$.
	\item $\ast$ is associative: $(a \ast b) \ast c = a \ast (b \ast c)$
	\item There is an \textit{identity} $\overline{0} \in G$, so that $a \ast \overline{0} = a$ for all $a \in G$.
	\item For any $a \in G$, there exists a $b \in G$ so that $a \ast b = \overline{0}$. $b$ is called the \textit{inverse} of $a$. \\
	This element is written as $-a$ if our operator is addition and $a^{-1}$ otherwise.
\end{enumerate}

Any pair $(G, \ast)$ that satisfies these properties is a group.

\definition{}
Note that our definition of a group does \textbf{not} state that $a \ast b = b \ast a$. \\
Many interesting groups do not have this property. \\
Those that do are called \textit{abelian} groups.

\problem{}
Is $(\mathbb{Z}/5, +)$ a group? \\
Is $(\mathbb{Z}/5, -)$ a group? \\
\hint{$+$ and $-$ refer to our usual definition of modular arithmetic.}

\vfill

\problem{}
$(\mathbb{R}, \times)$ is not a group. \\
Make it one by modifying $\mathbb{R}$. \\

\begin{solution}
	$(\mathbb{R}, \times)$ is not a group because $0$ has no inverse. \\
	The solution is simple: remove the problem.

	\vspace{3mm}

	$(\mathbb{R} - \{0\}, \times)$ is a group.
\end{solution}

\vfill

\problem{}
What is the smallest group we can create?

\begin{solution}
	Let $(G, \circledcirc)$ be our group, where $G = \{\star\}$ and $\circledcirc$ is defined by the identity $\star \circledcirc \star = \star$

	Verifying that the trivial group is a group is trivial.
\end{solution}
\vfill

\problem{}
Let $G$ be the set of all bijections $A \to A$. \\
Let $\circ$ be the usual composition operator. \\
Is $(G, \circ)$ a group?

\vfill
\pagebreak

\problem{}
Show that a group has exactly one identity element.
\vfill

\problem{}
Show that each element in a group has exactly one inverse.
\vfill

\problem{}
Let $(G, \ast)$ be a group and $a, b, c \in G$. Show that...
\begin{itemize}
	\item $a \ast b$ and $a \ast c \implies b = c$
	\item $b \ast a$ and $c \ast a \implies b = c$
\end{itemize}
What does this mean intuitively?
\vfill

\problem{}
Let $(G, \ast)$ be a finite group (i.e, $G$ has finitely many elements), and let $g \in G$. \\
Show that $\exists~n \in Z^+$ so that $g^n = \overline{0}$ \\
\hint{$g^n = g \ast g \ast ... \ast g$ $n$ times.}

\vspace{2mm}

The smallest such $n$ defines the \textit{order} of $(G, \ast)$.

\vfill

\problem{}
What is the order of 5 in $(\mathbb{Z}/25, +)$? \\
What is the order of 2 in $((\mathbb{Z}/17)^\times, \times)$? \\

\vfill

\problem{}
Show that if $G$ has four elements, $(G, \ast)$ is abelian.

\vfill
\pagebreak

\definition{}
Recall your tables from \ref{modtables}: \\

\begin{center}
\begin{tabular}{c | c c c c}
	+ & 0 & 1 & 2 & 3 \\
	\hline
	0 & 0 & 1 & 2 & 3 \\
	1 & 1 & 2 & 3 & 0 \\
	2 & 2 & 3 & 0 & 1 \\
	3 & 3 & 0 & 1 & 2 \\
\end{tabular}
\hspace{1cm}
\begin{tabular}{c | c c c c}
	\times & 1 & 2 & 3 & 4 \\
	\hline
	1 & 1 & 2 & 4 & 3 \\
	2 & 2 & 4 & 3 & 1 \\
	3 & 4 & 3 & 1 & 2 \\
	4 & 3 & 1 & 2 & 4 \\
\end{tabular}
\end{center}

Look at these tables and convince yourself that $(\mathbb{Z}/4, +)$ and $( (\mathbb{Z}/5)^\times, \times )$ are the same group. \\
We say that two such groups are \textit{isomorphic}.

\vspace{2mm}

Intuitively, this means that these two groups have the same algebraic structure. We can translate statements about addition in $\mathbb{Z}/4$ into statements about multiplication in $(\mathbb{Z}/5)^\times$ \\

\pagebreak