diff --git a/Advanced/Symmetric Group/main.tex b/Advanced/Symmetric Group/main.tex index a9d789e..b7ab866 100755 --- a/Advanced/Symmetric Group/main.tex +++ b/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 - % few more problems? + \section{Bonus problems} - % inline functions - % symmetric group - % order & generators - % subgroups + \problem{} + Show that $x$ has a multiplicative inverse mod $n$ iff $\text{gcd}(x, n) = 1$ + + \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 + % type and sign and conjugation % isomorphisms & automorphisms % automorphism groups - \end{document} diff --git a/Advanced/Symmetric Group/parts/1 cycle.tex b/Advanced/Symmetric Group/parts/1 cycle.tex index 8166257..b80ccbe 100755 --- a/Advanced/Symmetric Group/parts/1 cycle.tex +++ b/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 -$[2134]$ happens to be its own inverse. Also, the order of $[1234]$ is (of course) one. +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. \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{} 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{} +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{} +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 \ No newline at end of file diff --git a/Advanced/Symmetric Group/parts/2 groups.tex b/Advanced/Symmetric Group/parts/2 groups.tex new file mode 100755 index 0000000..6eb4216 --- /dev/null +++ b/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{} +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 diff --git a/Advanced/Symmetric Group/parts/3 subgroup.tex b/Advanced/Symmetric Group/parts/3 subgroup.tex new file mode 100644 index 0000000..1bbfc72 --- /dev/null +++ b/Advanced/Symmetric Group/parts/3 subgroup.tex @@ -0,0 +1,126 @@ +\section{Subgroups} + +\problem{} +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{} +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