134 lines
3.6 KiB
TeX
Executable File
134 lines
3.6 KiB
TeX
Executable File
\section{Groups}
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\definition{}
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A \textit{group} $(G, \ast)$ consists of a set $G$ and an operator $\ast$. \\
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A group must have the following properties: \\
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\begin{enumerate}
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\item $G$ is closed under $\ast$. In other words, $a, b \in G \implies a \ast b \in G$.
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\item $\ast$ is associative: $(a \ast b) \ast c = a \ast (b \ast c)$
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\item There is an \textit{identity} $\overline{0} \in G$, so that $a \ast \overline{0} = a$ for all $a \in G$.
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\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 $a$. \\
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This element is written as $-a$ if our operator is addition and $a^{-1}$ otherwise.
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\end{enumerate}
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Any pair $(G, \ast)$ that satisfies these properties is a group.
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\definition{}
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Note that our definition of a group does \textbf{not} state that $a \ast b = b \ast a$. \\
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Many interesting groups do not have this property. \\
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Those that do are called \textit{abelian} groups.
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\problem{}
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Is $(\mathbb{Z}/5, +)$ a group? \\
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Is $(\mathbb{Z}/5, -)$ a group? \\
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\hint{$+$ and $-$ refer to our usual definition of modular arithmetic.}
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\vfill
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\problem{}
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$(\mathbb{R}, \times)$ is not a group. \\
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Make it one by modifying $\mathbb{R}$. \\
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\begin{solution}
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$(\mathbb{R}, \times)$ is not a group because $0$ has no inverse. \\
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The solution is simple: remove the problem.
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\vspace{3mm}
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$(\mathbb{R} - \{0\}, \times)$ is a group.
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\end{solution}
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\vfill
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\problem{}
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What is the smallest group we can create?
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\begin{solution}
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Let $(G, \circledcirc)$ be our group, where $G = \{\star\}$ and $\circledcirc$ is defined by the identity $\star \circledcirc \star = \star$
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Verifying that the trivial group is a group is trivial.
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\end{solution}
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\vfill
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\problem{}
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Let $G$ be the set of all bijections $A \to A$. \\
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Let $\circ$ be the usual composition operator. \\
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Is $(G, \circ)$ a group?
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\vfill
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\pagebreak
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\problem{}
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Show that a group has exactly one identity element.
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\vfill
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\problem{}
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Show that each element in a group has exactly one inverse.
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\vfill
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\problem{}
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Let $(G, \ast)$ be a group and $a, b, c \in G$. Show that...
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\begin{itemize}
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\item $a \ast b$ and $a \ast c \implies b = c$
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\item $b \ast a$ and $c \ast a \implies b = c$
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\end{itemize}
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What does this mean intuitively?
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\vfill
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\problem{}
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Let $(G, \ast)$ be a finite group (i.e, $G$ has finitely many elements), and let $g \in G$. \\
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Show that $\exists~n \in Z^+$ so that $g^n = \overline{0}$ \\
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\hint{$g^n = g \ast g \ast ... \ast g$ $n$ times.}
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\vspace{2mm}
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The smallest such $n$ defines the \textit{order} of $(G, \ast)$.
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\vfill
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\problem{}
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What is the order of 5 in $(\mathbb{Z}/25, +)$? \\
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What is the order of 2 in $((\mathbb{Z}/17)^\times, \times)$? \\
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\vfill
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\problem{}
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Show that if $G$ has four elements, $(G, \ast)$ is abelian.
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\vfill
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\pagebreak
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\definition{}
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Recall your tables from \ref{modtables}: \\
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\begin{center}
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\begin{tabular}{c | c c c c}
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+ & 0 & 1 & 2 & 3 \\
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\hline
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0 & 0 & 1 & 2 & 3 \\
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1 & 1 & 2 & 3 & 0 \\
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2 & 2 & 3 & 0 & 1 \\
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3 & 3 & 0 & 1 & 2 \\
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\end{tabular}
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\hspace{1cm}
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\begin{tabular}{c | c c c c}
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\times & 1 & 2 & 3 & 4 \\
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\hline
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1 & 1 & 2 & 4 & 3 \\
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2 & 2 & 4 & 3 & 1 \\
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3 & 4 & 3 & 1 & 2 \\
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4 & 3 & 1 & 2 & 4 \\
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\end{tabular}
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\end{center}
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Look at these tables and convince yourself that $(\mathbb{Z}/4, +)$ and $( (\mathbb{Z}/5)^\times, \times )$ are the same group. \\
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We say that two such groups are \textit{isomorphic}.
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\vspace{2mm}
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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 $(\mathbb{Z}/5)^\times$ \\
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\pagebreak
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