handouts/Advanced/Cryptography/parts/2 groups.tex

\section{Groups}

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.

\definition{}
A \textit{group} $(G, \ast)$ consists of a set $G$ and an operator $\ast$. \par
Groups always 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: $(a \ast b) \ast c = a \ast (b \ast c)$ for all $a,b,c \in G$
	\item There is an \textit{identity} $e \in G$, so that $a \ast e = a \ast e = a$ for all $a \in G$.
	\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
	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.

\problem{}
Is $(\mathbb{Z}_5, +)$ a group? \par
Is $(\mathbb{Z}_5, -)$ a group? \par
\hint{$+$ and $-$ refer to the usual operations in modular arithmetic.}
\vfill


\problem{}
Show that $(\mathbb{R}, \times)$ is not a group, then make it one by modifying $\mathbb{R}$. \par

\begin{solution}
	$(\mathbb{R}, \times)$ is not a group because $0$ has no inverse. \par
	The solution is simple: remove the problem.

	\vspace{3mm}

	$(\mathbb{R} - \{0\}, \times)$ is a group.
\end{solution}
\vfill


\problem{}
What is the smallest group we can create?

\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

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

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

%\problem{}
%Show that $(\mathbb{Z}_n^\times, \times)$ is a group for any $n \in \mathbb{Z}^+$.
%\vfill

%\problem{}
%Let $(G, \ast)$ be a group and $a, b, c \in G$. Show that...
%\begin{itemize}
%	\item $a \ast b = a \ast c \implies b = c$
%	\item $b \ast a = c \ast a \implies b = c$
%\end{itemize}
%This means that we can \say{cancel} operations in groups, much like we do in algebra.
%\vfill
%\pagebreak




% \problem{}
% Let $G$ be the set of all bijections $A \to A$. \par
% Let $\circ$ be the usual composition operator. \par
% Is $(G, \circ)$ a group?
%
% \vfill
%
% \definition{}
% Note that our definition of a group does \textbf{not} state that $a \ast b = b \ast a$. \par
% Many interesting groups do not have this property.
% Those that do are called \textit{abelian} groups. \par
%
% \vspace{2mm}
%
% One example of a non-abelian group is the set of invertible 2x2 matrices under matrix multiplication. In this handout, all % groups are abelian.
%
%
%
% \problem{}
% Show that if $G$ has four elements, $(G, \ast)$ is abelian.

\problem{}
Let $(G, \ast)$ be a group with finitely many elements, and let $a \in G$. \par
Show that $\exists n \in \mathbb{Z}^+$ so that $a^n = e$ \par
\hint{$a^n = a \ast a \ast ... \ast a$ repeated $n$ times.}

\vspace{2mm}

The smallest such $n$ defines the \textit{order} of $g$.

\vfill

\problem{}
What is the order of 5 in $(\mathbb{Z}_{25}, +)$? \par
What is the order of 2 in $(\mathbb{Z}_{17}^\times, \times)$? \par

\vfill


\theorem{}
Let $p$ be a prime number. \par
In any group $(\mathbb{Z}_p^\times, \ast)$ there exists a $g \in \mathbb{Z}_p^\times$ where...
\begin{itemize}
	\item The order of $g$ is $p - 1$
	\item $\{a^0,~ a^1,~ ...,~ a^{p - 2}\} = \mathbb{Z}_n^\times$
\end{itemize}
We call such a $g$ a \textit{generator}, since its powers generate every other element in the group.

\begin{instructornote}
	$\mathbb{Z}_p^\times$ has $p-1$ elements. \par
	The set $\{a^0,~ a^1,~ ...,~ a^{p - 2}\}$ also has $p-1$ elements, since we start counting from zero.

	\vspace{2mm}

	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.
	We could also write this set as $\{a^1,~ a^2,~ ...,~ a^{p - 1}\}$, since $a^0 = a^{p - 1}$.
\end{instructornote}

\vfill
\pagebreak