Symmetric group edits

Advanced/Symmetric Group/parts/0 intro.tex

@@ -2,19 +2,37 @@
 
 
 \definition{}
-Let $\Omega$ be an arbitrary set of $n$ objects. \par
-A \textit{permutation} on $\Omega$ is a bijective map $f: \Omega \to \Omega$.
+Informally, a \textit{permutation} of a collection of $n$ objects is an ordering of these $n$ objects. \par
+For example, a few permutations of $\texttt{A}, \texttt{B}, \texttt{C}, \texttt{D}$ are $\texttt{ABCD}$,
+$\texttt{BCDA}$, and $\texttt{DACB}$. \par
 
 \vspace{2mm}
 
-For example, consider the objects 1, 2, and 3. \par
-The permutation $[312]$ is given by a map $f$ defined by the following table:
+This, however, isn't the definition we'll use today. Instead of defining permutations as \say{ordered lists,}
+(as we do above), we'll define them as functions. Our first goal today is to make sense of this definition.
+
+
+
+\definition{Permutations}
+Let $\Omega$ be an arbitrary set of $n$ objects. \par
+A \textit{permutation} on $\Omega$ is a map from $\Omega$ to itself that produces a \textit{unique} output for each input. \par
+\note{In other words, if $a$ and $b$ are different, $f(a)$ and $f(b)$ must also be different.}
+
+
+\footnotetext{The words \say{function} and \say{map} are equivalent.}
+
+\vspace{2mm}
+
+For example, consider $\{1, 2, 3\}$. \par
+One permutation on this set can be defined as follows: \par
 \begin{itemize}
 	\item $f(1) = 3$
 	\item $f(2) = 1$
 	\item $f(3) = 2$
 \end{itemize}
 
+If we take the array $123$ and apply
+
 
 \problem{}
 List all permutations on three objects. \par

Advanced/Symmetric Group/parts/1 cycle.tex

@@ -2,10 +2,8 @@
 \section{Cycle Notation}
 
 \definition{Order}
-The \textit{order} of a permutation $f$ is the smallest positive $n$ so that $f^n(x) = x$ for all $x$. \par
-In other words: if we repeat this permutation $n$ times, we get back to where we started. \par
-Note that the order is given by the \textit{smallest} positive integer $n$. There may be more than one!
-
+The \textit{order} of a permutation $f$ is the \textbf{smallest} positive $n$ so that $f^n(x) = x$ for all $x$. \par
+If we repeatedly apply a permutation with order $n$, we will get back to where we started after $n$ steps. \par
 
 \vspace{2mm}
 
@@ -38,9 +36,8 @@ For example, consider $[2134]$. This permutation has order $2$, as we clearly se
 	\line{4b}{4c}
 \end{tikzpicture}
 \end{center}
-Of course, swapping the first two elements of a list twice changes nothing. \par
-Thus, $[2134]$ is its own inverse, and has an order of two. \par
-Naturally, the identity permutation has order one.
+Swapping the first two elements of a list twice changes nothing. \par
+Thus, $[2134]$ has an order of two.
 
 
 \problem{}
@@ -51,7 +48,7 @@ How about $[4321]$? \par
 
 \vfill
 
-\problem{Bonus}
+\problem{}
 Show that all permutations (on a finite set) have a well-defined order. \par
 In other words, show that there is always an integer $n$ so that $f^n(x) = x$.
 
@@ -59,12 +56,17 @@ In other words, show that there is always an integer $n$ so that $f^n(x) = x$.
 
 \definition{Composition}<compdef>
 The \textit{composition} of two permutations $f$ and $g$ is the permutation $h(x) = f(g(x))$. \par
-The usual notation for this is $f \circ g$, but we'll simply write $fg$.
+We'll denote this as $fg$---that is, by simply writing the permutations we're composing next to each other.
+
+\problem{}
+Show that function composition is associative. \par
+That is, show that $f(gh) = (fg)h$.
+
+\vfill
 
 \problem{}
 What is $[1324][4321]$? \par
 How about $[321][213][231]$? \par
-\hint{is composition is left or right-associative? See \ref{compdef}}
 
 
 \vfill