96 lines
2.7 KiB
TeX
Executable File
96 lines
2.7 KiB
TeX
Executable File
\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
|
|
|