Added initial parts of group theory handout

Advanced/Group Theory/parts/01 groups.tex

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+\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
+