93 lines
3.0 KiB
TeX
Executable File
93 lines
3.0 KiB
TeX
Executable File
\section{Groups}
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Group theory gives us a set tools for understanding complex systems. We can use groups to solve the Rubik's cube, to solve problems in physics and chemistry, and to understand complex geometric symmetries. It's also worth noting that all modern cryptography relies heavily on group theory.
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\definition{}
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A \textit{group} $(G, \ast)$ consists of a set $G$ and an operator $\ast$. \par
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Groups always 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: $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a,b,c \in G$
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\item There is an \textit{identity} $e \in G$, so that $a \ast e = a \ast e = a$ for all $a \in G$.
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\item For any $a \in G$, there exists a $b \in G$ so that $a \ast b = b \ast a = e$. $b$ is called the \textit{inverse} of $a$. \par
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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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\problem{}
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Is $(\mathbb{Z}_5, +)$ a group? \par
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Is $(\mathbb{Z}_5, -)$ a group? \par
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\hint{$+$ and $-$ refer to the usual operations in modular arithmetic.}
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\vfill
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\problem{}
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Show that $(\mathbb{R}, \times)$ is not a group, then make it one by modifying $\mathbb{R}$. \par
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\begin{solution}
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$(\mathbb{R}, \times)$ is not a group because $0$ has no inverse. \par
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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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\pagebreak
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\problem{}
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Let $(G, \ast)$ be a group with finitely many elements, and let $a \in G$. \par
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Show that $\exists n \in \mathbb{Z}^+$ so that $a^n = e$ \par
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\hint{$a^n = a \ast a \ast ... \ast a$ repeated $n$ times.}
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\vspace{2mm}
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The smallest such $n$ defines the \textit{order} of $g$.
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\vfill
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\problem{}
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What is the order of 5 in $(\mathbb{Z}_{25}, +)$? \par
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What is the order of 2 in $(\mathbb{Z}_{17}^\times, \times)$? \par
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\vfill
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\theorem{}
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Let $p$ be a prime number. \par
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In any group $(\mathbb{Z}_p^\times, \ast)$ there exists a $g \in \mathbb{Z}_p^\times$ where...
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\begin{itemize}
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\item The order of $g$ is $p - 1$
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\item $\{a^0,~ a^1,~ ...,~ a^{p - 2}\} = \mathbb{Z}_n^\times$
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\end{itemize}
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We call such a $g$ a \textit{generator}, since its powers generate every other element in the group.
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\begin{instructornote}
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$\mathbb{Z}_p^\times$ has $p-1$ elements. \par
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The set $\{a^0,~ a^1,~ ...,~ a^{p - 2}\}$ also has $p-1$ elements, since we start counting from zero.
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\vspace{2mm}
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The fact that the last power here is $p-2$ can be a bit confusing, but it's just the result of counting from zero.
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We could also write this set as $\{a^1,~ a^2,~ ...,~ a^{p - 1}\}$, since $a^0 = a^{p - 1}$.
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\end{instructornote}
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\vfill
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\pagebreak
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