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} $e \in G$, so that $a \ast e = a \ast e = a$ for all $a \in G$.
	\item For any $a \in G$, there exists a $b \in G$ so that $a \ast b = b \ast a = e$. $b$ 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 if $G$ has four elements, $(G, \ast)$ is abelian.

\vfill

\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{}
Show that...
\begin{itemize}
	\item $e^{-1} = 1$
	\item $(a^{-1})^{-1} = a$
\end{itemize}
\vfill

\problem{}
Show that $(a^m)^{-1} = (a^{-1})^m$ for all $a \in G$ and $m \in \mathbb{Z}$.
\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
\pagebreak

\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 $g^n = e$ \\
\hint{$g^n = g \ast g \ast ... \ast g$ $n$ times.}

\vspace{2mm}

The smallest such $n$ defines the \textit{order} of $g$.

\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
\pagebreak

\problem{}
Let $e, a, b, c$ be counterclockwise rotations of a square by $0, \frac{\pi}{2}, \pi,$ and $\frac{3\pi}{2}$. \\
Create a multiplication table for this group.
\vfill

\problem{}
Let $d, f, g, h$ correspond to reflections of the square along the following axis. \\
Create a multiplication table for this group.

\begin{center}
\begin{tikzpicture}[scale=2]
	\draw (0,0) -- (1,0) -- (1,1) -- (0,1) -- (0,0);

	\draw[gray] (1.25,1.25) -- (-0.25,-0.25) node[below left]{$d$};
	\draw[gray] (1.25,-0.25) -- (-0.25,1.25) node[above left]{$f$};
	\draw[gray] (0.5,-0.25) -- (0.5,1.25) node[above]{$g$};
	\draw[gray] (-0.25, 0.5) -- (1.25,0.5) node[right]{$h$};

\end{tikzpicture}
\end{center}
\vfill

\problem{}
Create a multiplication table for all symmetries of a square.
\vfill
\pagebreak

\problem{}
Create a multiplication table for all symmetries of a rhombus.
\vfill
\pagebreak

\problem{}
Find the order of each element in...
\begin{itemize}
	\item The group of symmetries of a square
	\item The group of symmetries of a rhombus
\end{itemize}


\vfill
\pagebreak