Started QG handout

Advanced/Quotient Groups/parts/1 groups.tex

@@ -0,0 +1,158 @@
+\section{Groups}
+
+\definition{}
+A \textit{group} $G = (S, \ast)$ consists of a set $S$ and a binary operator $\ast$. \par
+By definition, a group always has 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 Any $a \in G$ has an \textit{inverse} $a^{-1} \in G$ that satisfies $a \ast a^{-1} = a^{-1} \ast a = e$. \par
+\end{enumerate}
+
+\note[Note]{
+	Commutativity is \textit{not} a required property of a group! \\
+	In most cases, $a \ast b \neq b \ast a$.
+}
+
+
+
+\problem{}
+Is $(\znz{5}, +)$ a group? \par
+How about $(\znz{5}, -)$? \par
+\hint{In this problem, $+$ and $-$ work just as you'd expect.}
+
+\vfill
+
+\problem{}
+What is the smallest possible group?
+
+\begin{solution}
+	Let $(G, \ast)$ be our group, where $G = \{e\}$ and $\ast$ is defined by the identity $e \ast e = e$
+
+	Verifying that the trivial group is a group is trivial.
+\end{solution}
+
+\vfill
+
+\problem{}
+How many distinct groups have two elements? \par
+\hint{
+	Two groups are \say{the same} if the elements of one can be renamed to get the other. \\
+	A group is fully defined by its multiplication table.
+}
+
+
+\vfill
+\pagebreak
+
+
+
+%\problem{}<firstcross>
+%Is $(\znz{17}, \times)$ a group? \par
+%How should we modify $\znz{17}$ to make it one?
+
+%\problem{}<secondcross>
+%Is $(\znz{6}, \times)$ a group? \par
+%How should we modify $\znz{6}$ to make it one? \par
+%\hint{
+%	Be careful, this isn't as easy as \ref{firstcross}. \\
+%	Which elements aren't invertible?
+%}
+
+
+%\definition{}
+%Building on problems \ref{num:firstcross} and \ref{num:secondcross}, we'll define $(\znz{n})^\times$ as the multiplicative
+%group of integers mod $n$. \par
+
+%Specifically, $(\znz{n})^\times$ is the set of all integers coprime to $n$. \par
+
+%\vspace{2mm}
+
+%For example, $(\znz{6})^\times = \{1, 5\}$ \par
+%and $(\znz{15})^\times = \{1, 2, 4, 7, 8, 11, 13, 14\}$ \par
+
+%\vspace{2mm}
+
+%Note that $0$ is the identity in $\znz{n}$ and $1$ is the identity in $(\znz{n})^\times$\hspace{-1.5ex}. \par
+%\note[Note]{
+%	Also, notice that we've omitted the operations $+$ and $\times$ in the two groups above. \\
+%	These operations are implicitly \say{attached} to $\znz{n}$ and $(\znz{n})^\times$\hspace{-1.5ex}, \\
+%	and we rarely write them for the sake of cleaner notation.
+%}
+
+
+\vfill
+
+
+\definition{}
+Let $G$ be a group, $a$ an element of $G$, and $n \in \mathbb{Z}^+$. \par
+$a^n$ is the defined as $a \ast a \ast ... \ast a$, repeated $n$ times.
+
+\vspace{1mm}
+
+Note that this is \textbf{not} \say{normal} exponentiation! \par
+If our group's operator is $+$ (for example, $\znz{5}$), $a^n = a + ... + a$, \par
+which you'll recognize as multipication.
+
+\vspace{1mm}
+
+Beware of this odd notation. By convention, we use \say{multiplicative} notation
+when working with groups---so, $a \ast b$ may also be written as $ab$,
+and $a \ast a \ast a$ may be written as $a^3$.
+
+\vspace{1mm}
+
+Again, remember that $a^n$ simply means \say{$\ast$ $a$ with itself $n$ times,} \par
+regardless of the specific operator our group uses.
+
+\problem{}
+Let $a$ be an element of a finite group. \par
+Show that there is a positive integer $n$ so that $a^n = e$. \par
+
+\vspace{2mm}
+
+The smallest such $n$ defines the \textit{order} of $g$.
+
+\vfill
+
+\problem{}
+Find the order of 5 in $(\znz{25}, +)$. \par
+%Find the order of 2 in $((\znz{17})^\times, \times)$. \par
+Find the order of 2 in $(\znz{7}, +)$. \par
+
+\vfill
+\pagebreak
+
+
+\definition{}
+Let $G$ be a group. \par
+We say a $g \in G$ is a \textit{generator} of $G$
+if every element in $G$ can be written as some power of $g$.
+
+\vspace{2mm}
+
+If $G$ has a generator, we say $G$ is \textit{cyclic.}
+
+\problem{}
+Find a generator of $\znz{7}$. Then, find a generator of $(\znz{7})^\times$
+
+\vfill
+
+\definition{}
+Let $G$ be a group. \par
+The \textit{order} of $G$ is the number of elements in $G$. \par
+We'll write this as $|G|$, using the same notation we use with sets. \par
+\note[Note]{
+	Don't confuse the order of an \textbf{element}
+	with the order of a \textbf{group}!
+}
+
+\problem{}
+Let $G$ be a cyclic group, and let $g$ be any generator in $G$. \par
+Show that $\text{ord}(g) = |G|$. \par
+\hint{Contradiction.}
+
+\vfill
+\pagebreak