Finished symmetric group handout

Advanced/Symmetric Group/main.tex

@@ -28,19 +28,31 @@
 
 	\input{parts/0 intro}
 	\input{parts/1 cycle}
+	\input{parts/2 groups}
+	\input{parts/3 subgroup}
 
 
-	% decomposition into transpositions
+	\section{Bonus problems}
-	% few more problems?
 
-	% inline functions
+	\problem{}
-	% symmetric group
+	Show that $x$ has a multiplicative inverse mod $n$ iff $\text{gcd}(x, n) = 1$
-	% order & generators
+
-	% subgroups
+	\vfill
+
+	\problem{}
+	Let $\sigma = (\sigma_1 \sigma_2 ... \sigma_k)$ be a $k$-cycle in $S_n$, and let $\tau$ be an arbitrary element of $S_n$. \par
+	Show that $\tau \sigma \tau^{-1}$ = $\bigl(\tau(\sigma_1), \tau(\sigma_2), ..., \tau(\sigma_k)\bigr)$ \par
+	\hint{As usual, $\sigma$ is a permutation. Thus, $\sigma(x)$ is the value at position $x$ after applying $\sigma$.}
+
+	\vfill
+
+	\problem{}
+	Show that the set $\Bigl\{ (1, 2),~ (1,2,...,n) \Bigr\}$ generates $S_n$.
+	\vfill
+
+	% TODO: (a second day?)
 	% alternating group
-
+	% type and sign and conjugation
-	% type and sign
 	% isomorphisms & automorphisms
 	% automorphism groups
-
 \end{document}

Advanced/Symmetric Group/parts/1 cycle.tex

@@ -1,7 +1,7 @@
 
 \section{Cycle Notation}
 
-\definition{}
+\definition{Order}
 The \textit{order} of a permutation $f$ is the smallest $n$ 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.
 
@@ -36,8 +36,9 @@ 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 twice results in the identity map. \par
+Of course, swapping the first two elements of a list twice changes nothing. \par
-$[2134]$ happens to be its own inverse. Also, the order of $[1234]$ is (of course) one.
+Thus, $[2134]$ is its own inverse, and has an order of two. \par
+Naturally, the identity permutation has order one.
 
 
 \problem{}
@@ -52,6 +53,18 @@ How about $[4321]$? \par
 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 $f^n(x) = x$.
 
+\vfill
+
+\definition{Composition}
+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 $fg$.
+
+\problem{}
+What is $[1324][4321]$? \par
+How about $[321][213][231]$? \par
+\hint{composition is left-associative, so we evaluate $abc$ as $(ab)c$}
+
+
 \vfill
 \pagebreak
 
@@ -74,7 +87,7 @@ We need something better.
 
 
 
-\definition{}
+\definition{Cycles}
 Any permutation is composed of a number of \textit{cycles}. \par
 
 For example, consider the permutation $[2134]$, which consists of one two-cycle: $1 \to 2 \to 1$ \par
@@ -217,7 +230,7 @@ Find all cycles in $[5342761]$.
 
 
 \problem{}
-What permutation (on five objects) consists of the cycles $3 \to 5 \to 3$ and $1 \to 2 \to 4 \to 1$?
+What permutation (on five objects) is formed by the cycles $3 \to 5 \to 3$ and $1 \to 2 \to 4 \to 1$?
 
 
 \begin{solution}
@@ -271,7 +284,7 @@ What permutation (on five objects) consists of the cycles $3 \to 5 \to 3$ and $1
 \vfill
 \pagebreak
 
-\definition{}
+\definition{Cycle Notation}
 We now have a solution to our problem of notation.
 Instead of referring to permutations using their output, we will refer to them using their \textit{cycles}.
 
@@ -355,12 +368,20 @@ applying the permutation $[431265]$ is the same as applying $(1324)$, then apply
 \end{tikzpicture}
 \end{center}
 
+Any permutation $\sigma$ may be written as a product (i.e, composition) of disjoint cycles $\sigma_1\sigma_2...\sigma_k$. \par
+Make sure you believe this fact. If you don't, ask an instructor. \par
+Also, the identity $f(x) = x$ is written as $()$ in cycle notation.
+
+
 
 \problem{}
 Convince yourself that disjoint cycles commute. \par
-That is, $(1324)(56) = (56)(1324)$ since $(1324)$ and $(56)$ do not overlap.
+That is, that $(1324)(56) = (56)(1324) = [431265]$ since $(1324)$ and $(56)$ do not overlap. \par
 
-\problem{}
+
+
+
+\problem{}<insquare>
 Write the following in square-bracket notation.
 \begin{itemize}
 	\item $(12)$       \tab~\tab on a set of 2 elements
@@ -372,9 +393,15 @@ Write the following in square-bracket notation.
 	\item $(1234)$      \tab on a set of 4 elements
 	\item $(3412)$      \tab on a set of 4 elements
 \end{itemize}
+\note{
+	Note that $(12)$ refers the \say{swap first two} permutation on a set of \textit{any} size. \\
+	We can now use the same name for the same permutation on two different sets! \\
+}
 
 \vfill
 
+
+
 \problem{}
 Write the following in square-bracket notation.
 Be careful.
@@ -383,5 +410,117 @@ Be careful.
 	\item $(243)(13)$ \tab on a set of 4 elements
 \end{itemize}
 
+\vfill
+
+
+
+\problem{}
+Look at the last two permutations in \ref{insquare}, $(1234)$ and $(3412)$. \par
+These are \textit{identical}---they are the same cycle written in two different ways. \par
+List all other ways to write this cycle. \hint{There are two more.}
+
+\pagebreak
+
+
+\problem{}
+What is the inverse of $(12)$? \par
+How about $(123)$? And $(4231)$? \par
+\note{
+	Note that again, we don't need to know how big our set is. \\
+	The inverse of $(12)$ is the same in all sets.
+}
+
+\vfill
+
+
+\problem{}
+Say $\sigma$ is a permutation composed of cycles $\sigma_1\sigma_2...\sigma_k$. \par
+Say we know the order of all $\sigma_i$. What is the order of $\sigma$?
+
+\begin{solution}
+	$\text{lcm}\Bigl(\text{ord}(\sigma_1),~ \text{ord}(\sigma_2),~ ..., ~ \text{ord}(\sigma_k)\Bigr)$
+\end{solution}
+
+
+\vfill
+
+\problem{}<cycletrans>
+Show that any cycle $(123...n)$ is equal to the product $(12)(23)...(n-1, n)$.
+
+\begin{solution}
+	TODO
+\end{solution}
+
+\vfill
+
+\problem{}
+Write $(7126453)$ as a product of transpositions. \par
+
+\vfill
+\pagebreak
+
+\problem{}<simpletrans>
+Show that any permutation is a product of transpositions.
+
+\begin{solution}
+	Use \ref{cycletrans}.
+\end{solution}
+
+\vfill
+
+
+\problem{}
+Show that any permutation is a product of transpositions of the form $(1, k)$. \par
+
+\begin{solution}
+	Use \ref{simpletrans} and rewrite each $(a, b)$ as $(1, a)(1, b)(1, a)$.
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+
+\problem{}
+Show that any transposition $(a, b)$ is equal to the product $(a, a+1)(a+1, b)(a, a+1)$.
+
+\begin{solution}
+	TODO
+\end{solution}
+
+\vfill
+
+
+
+\problem{}
+Show that any permutation is a product of adjacent transpositions. \par
+(An \textit{adjacent transposition} swaps two adjacent elements, and thus looks like $(n, n+1)$)
+
+\begin{solution}
+	As before, we will use \ref{simpletrans} and rewrite the transpositions it produces in a form that fits the problem.
+	We thus need to show that every transposition $(a, b)$ is a product of adjacent transpositions.
+
+	\vspace{8mm}
+
+	In the proof below, assume that $a < b$ and perform induction on $b - a$. \par
+
+	\textbf{Base Case:}\par
+	If $b - a = 1$, we clearly see that $(a, b)$ is a product of adjacent. \par
+	In fact, it \textit{is} an adjacent transposition.
+
+	\vspace{4mm}
+
+	\textbf{Induction:}\par
+	Now, say $b - a = n + 1$. \par
+	Assume that all $(a, b)$ where $b - a \leq n$ are products of adjacent transpositions.\par
+	Note that $(a, b) = (a, a+1)(a+1, b)(a, a+1)$.
+
+	\vspace{2mm}
+
+	$(a, a+1)$ is an adjacent transposition, and $b - (a+1) = n$. \par
+	Thus, $(a, b)$ is a product of adjacent transpositions.
+\end{solution}
+
+
 \vfill
 \pagebreak

Advanced/Symmetric Group/parts/2 groups.tex

@@ -0,0 +1,180 @@
+\section{Groups (review)}
+
+\definition{}
+Before we continue, we must introduce a bit of notation:
+\begin{itemize}
+	\item $S_n$ is the set of permutations on $n$ objects.
+	\item $\mathbb{Z}_n$ is the set of integers mod $n$.
+
+	\item $\mathbb{Z}_n^\times$ is the set of integers mod $n$ with multiplicative inverses, which is \par
+	the set of integers smaller than $n$ and coprime to $n$\footnotemark{}\hspace{-1ex}. \par
+	For example, $\mathbb{Z}_{12}^\times = \{1, 5, 7, 11\}$.
+
+	\footnotetext{We proved this in another handout, but you make take it as fact here.}
+\end{itemize}
+
+\problem{}
+What are the elements of $S_3$? \tab\hint{Use cycle notation}\par
+How about $\mathbb{Z}_{17}^\times$?
+
+\vfill
+
+
+\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
+\note[Note]{$+$ and $-$ refer to the usual operations in modular arithmetic.}
+\vfill
+
+
+\problem{}
+What is the smallest group?
+
+\begin{solution}
+	Let $(G, \star)$ be our group, where $G = \{x\}$ and $\star$ is defined by $x \star x = x$
+
+	Verifying that the trivial group is a group is trivial.
+\end{solution}
+
+\vfill
+\pagebreak
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+\problem{}
+Show that $S_n$ is a group under composition.
+
+\vfill
+
+\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 $g$.
+
+\begin{examplesolution}
+	We've already done a special case of this problem! \par
+	Look back through the handout and find it, then rewrite your proof for an arbitrary group.
+\end{examplesolution}
+
+
+\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
+\pagebreak
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+\definition{}<gendef>
+Let $G$ be a group, and let $g$ be an element of $G$. \par
+We say $g$ is a \textit{generator} if every other element of $G$ may be written as a power of $g$. \par
+
+\problem{}
+Say the size of a group $G$ is $n$. \par
+If $g$ is a generator, what is its order? \par
+Provide a proof.
+
+\begin{solution}
+	The order of a generator must equal the order of its group.
+\end{solution}
+
+\vfill
+
+\problem{}
+Find the only generator of $(\mathbb{Z}^+, +)$ \par
+Then, find all generators of $(\mathbb{Z}_5, +)$
+
+\vfill
+\pagebreak
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+\definition{}
+Let $S$ be a subset of the elements in $G$. \par
+We say that $S$ \textit{generates} $G$ if every element of $G$ may be written as a product of elements in $S$. \par
+\note{Note that this is an extension of \ref{gendef}.}
+
+\problem{}
+We've already found a few generating sets of $S_n$. What are they?
+
+\begin{solution}
+	The following sets generate $S_n$:
+	\begin{itemize}
+		\item All transpositions
+		\item All transpositions of the form $(1, k)$
+		\item All adjacent transpositions
+	\end{itemize}
+
+	\vspace{2mm}
+
+	The smallest generating set of $S_n$ consists of the transposition $(12)$ and the $n$-cycle $(1,2,...,n)$. \par
+	The proof of this is a bonus problem later in the handout.
+\end{solution}
+
+\vfill
+
+\problem{}
+Find the smallest set that generates $(\mathbb{Z}^+, +)$. \par
+\vfill
+
+\problem{}
+Find the smallest set that generates $(\mathbb{Z}, +)$. \par
+\vfill
+
+\pagebreak

Advanced/Symmetric Group/parts/3 subgroup.tex

@@ -0,0 +1,126 @@
+\section{Subgroups}
+
+\problem{}<s2s3share>
+What elements do $S_2$ and $S_3$ share?
+\vspace{2cm}
+
+
+
+Consider the sets $\{1, 2\}$ and $Omega_3 = \{1,2,3\}$. Clearly, $\{1, 2\} \subset \{1, 2, 3\}$. \par
+Can we say something similar about $S_2$ and $S_3$?
+
+\vspace{2mm}
+
+Looking at \ref{s2s3share}, we may want to say that $S_2 \subset S_3$ since every element of $S_2$ is in $S_3$. \par
+This reasoning, however, is not correct. Remember that $S_2$ and $S_3$ are \textit{groups}, not \textit{sets}: \par
+their elements come with structure.
+
+\vspace{2mm}
+
+Therefore, the \say{subset} relation isn't particularly useful when applied to groups. \par
+We instead use a similar relation: subgroups.
+
+\definition{}
+Let $G$ and $G'$ be groups. We say $G'$ is a \textit{subgroup} of $G$ (and write $G' \subset G$) if the following are true:\par
+(Note that $x, y$ are elements of $G$, and $xy$ is multiplication in $G$)
+\begin{itemize}
+	\item the set of elements in $G'$ is a subset of the set of elements in $G$.
+	\item the identity of $G$ is in $G'$
+	\item $x,y \in G' \implies xy \in G'$
+	\item $x \in G' \implies x^{-1} \in G'$
+\end{itemize}
+
+The above definition may look faily scary, but the idea behind a subgroup is simple. \par
+Consider $S_3$ and $S_4$, the groups of permutations of $3$ and $4$ elements. \par
+
+\vspace{2mm}
+
+Say we have a set of four elements and only look at the first three. \par
+$S_3$ fully describes all the ways we can arrange those three elements:
+
+\begin{center}
+\begin{tikzpicture}[scale=0.5]
+	\node (1a) at (0, 0.5) {1};
+	\node (2a) at (1, 0.5) {2};
+	\node (3a) at (2, 0.5) {3};
+	\node (4a) at (3, 0.5) {4};
+
+	\node (2b) at (0, -2) {2};
+	\node (3b) at (1, -2) {3};
+	\node (1b) at (2, -2) {1};
+	\node (4b) at (3, -2) {4};
+
+	\draw[line width = 0.3mm, ->, ogreen]
+		(4a)
+		-- ($(4a) + (0, -1)$)
+		-- ($(4b) + (0,1)$)
+		-- (4b);
+
+	\line{1a}{1b}
+	\line{2a}{2b}
+	\line{3a}{3b}
+
+	\node[fill=white,draw=oblue,line width=0.3mm] at (1, -0.75) {$S_3$};
+
+\end{tikzpicture}
+\end{center}
+
+
+\problem{}
+Show that $S_3$ is a subgroup of $S_4$.
+
+\vfill
+\pagebreak
+
+
+
+
+\problem{}<firstindex>
+How many subgroups of $S_4$ are equal to $S_3$?
+
+\begin{solution}
+	Four, since there are four ways to pick three things from $S_4$.
+\end{solution}
+
+\vfill
+
+\problem{}
+What is the order of $S_3$ and $S_4$? \par
+How is this related to \ref{firstindex}?
+
+\begin{solution}
+	$|S_4| = |S_3| \times [S_4 : S_3]$
+
+	\vspace{2mm}
+
+	This solution is written using index notation, but the class
+	doesn't yet need to know what it means.
+\end{solution}
+
+\vfill
+
+\problem{}
+$S_4$ also has $S_2$ and the trivial group as subgroups. \par
+How many instances of each does $S_4$ contain?
+
+\vfill
+
+
+\problem{}
+$(\mathbb{Z}_4, +)$ is also a subgroup of $S_4$. Find it! \par
+How many copies of $Z_4$ are in $S_4$? \par
+(You'll need to re-label elements, since we usually use different notation for $\mathbb{Z}_4$ and $S_4$).
+
+\begin{solution}
+	A good hint is \say{look at generators.}
+
+	\vspace{4mm}
+
+	There are four instances of $\mathbb{Z}_4$ in $S_4$, \par
+	each of which is generated by a 4-cycle of $S_n$. \par
+	(i.e, the group generated by $(1234)$ is isomorphic to $\mathbb{Z}_4$)
+\end{solution}
+
+
+\vfill
+\pagebreak