\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: $a \ast b = b \ast a$
	\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{}
Can you construct a group that contains a single element?

\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
\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}

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 $(\mathbb{Z}/4, +)$ and $( (\mathbb{Z}/5)^\times, \times )$ have the same algebraic structure. We can translate statements about addition in $\mathbb{Z}/4$ into statements about multiplication in $(\mathbb{Z}/5)^\times$ \\

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

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

\vfill
\pagebreak

