diff --git a/Advanced/Symmetric Group/main.tex b/Advanced/Symmetric Group/main.tex
index b9ca834..a9d789e 100755
--- a/Advanced/Symmetric Group/main.tex	
+++ b/Advanced/Symmetric Group/main.tex	
@@ -5,7 +5,7 @@
 	singlenumbering
 ]{../../resources/ormc_handout}
 \usepackage{../../resources/macros}
-
+\usetikzlibrary{calc}
 
 \uptitlel{Advanced 2}
 \uptitler{Winter 2023}
@@ -14,14 +14,22 @@
 
 
 
+\def\line#1#2{
+	\draw[line width = 0.3mm, ->, ocyan]
+		(#1)
+		-- ($(#1) + (0, -1)$)
+		-- ($(#2) + (0,1)$)
+		-- (#2);
+}
 
 \begin{document}
 
 	\maketitle
 
 	\input{parts/0 intro}
+	\input{parts/1 cycle}
+
 
-	% cycle notation
 	% decomposition into transpositions
 	% few more problems?
 
diff --git a/Advanced/Symmetric Group/parts/0 intro.tex b/Advanced/Symmetric Group/parts/0 intro.tex
index 9e1c32f..c70bc04 100644
--- a/Advanced/Symmetric Group/parts/0 intro.tex	
+++ b/Advanced/Symmetric Group/parts/0 intro.tex	
@@ -44,7 +44,7 @@ Why do we define permutations as a \textit{bijective} map?
 
 
 We can visualize permutations with a diagram we'll call the \say{braid.}
-The arrows in the diagram denote the image of $f$ for each possible input.
+The arrows in this diagram denote the image of $f$ for each possible input.
 Two examples are below:
 
 \vspace{2mm}
@@ -60,10 +60,10 @@ Two examples are below:
 	\node (4b) at (2, -2) {4};
 	\node (2b) at (3, -2) {2};
 
-	\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
+	\line{1a}{1b}
+	\line{2a}{2b}
+	\line{3a}{3b}
+	\line{4a}{4b}
 \end{tikzpicture}
 \hfill
 \begin{tikzpicture}[scale=0.5]
@@ -77,12 +77,14 @@ Two examples are below:
 	\node (3b) at (2, -2) {3};
 	\node (4b) at (3, -2) {4};
 
-	\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
+	\line{1a}{1b}
+	\line{2a}{2b}
+	\line{3a}{3b}
+	\line{4a}{4b}
 \end{tikzpicture}
 \hfill\null
+\vspace{2mm}
+
 
 Note that in all our examples thus far, the objects in our set have an implicit order.
 This is only for convenience. The elements of $\Omega$ are not ordered (it is a \textit{set}, after all),
@@ -113,10 +115,10 @@ The rightmost diagram uses arbitrary, meaningless labels.
 	\node (3b) at (2, -2) {3};
 	\node (4b) at (3, -2) {4};
 
-	\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
+	\line{1a}{1b}
+	\line{2a}{2b}
+	\line{3a}{3b}
+	\line{4a}{4b}
 \end{tikzpicture}
 \hfill
 \begin{tikzpicture}[scale=0.5]
@@ -130,10 +132,10 @@ The rightmost diagram uses arbitrary, meaningless labels.
 	\node (3b) at (2, -2) {3};
 	\node (2b) at (3, -2) {2};
 
-	\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
+	\line{1a}{1b}
+	\line{2a}{2b}
+	\line{3a}{3b}
+	\line{4a}{4b}
 \end{tikzpicture}
 \hfill
 \begin{tikzpicture}[scale=0.5]
@@ -147,12 +149,14 @@ The rightmost diagram uses arbitrary, meaningless labels.
 	\node (3b) at (2, -2) {$\circledcirc$};
 	\node (4b) at (3, -2) {$\boxdot$};
 
-	\draw[line width = 0.3mm, ->, ocyan] (1a.south) -- (1a.south |- 0, -0.5) -- (1b.north |- 0,-1) -- (1b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (2a.south) -- (2a.south |- 0, -0.5) -- (2b.north |- 0,-1) -- (2b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (3a.south) -- (3a.south |- 0, -0.5) -- (3b.north |- 0,-1) -- (3b.north);
-	\draw[line width = 0.3mm, ->, ocyan] (4a.south) -- (4a.south |- 0, -0.5) -- (4b.north |- 0,-1) -- (4b.north);
+	\line{1a}{1b}
+	\line{2a}{2b}
+	\line{3a}{3b}
+	\line{4a}{4b}
 \end{tikzpicture}
 \hfill\null
+\vspace{2mm}
+
 
 It shouldn't be hard to see that despite the different \say{output} order (2134 and 1432), \par
 the same permutation is depicted in all three diagrams. This example demonstrates two things:
@@ -167,7 +171,6 @@ the same permutation is depicted in all three diagrams. This example demonstrate
 
 \vspace{2mm}
 
-\vspace{1cm}
 
 Why, then, do we order our elements when we talk about permutations? As noted before, this is for convenience.
 If we assign a natural order to the elements of $\Omega$ (say, 1234), we can identify permutations by simply listing
@@ -182,7 +185,7 @@ Draw braids for $[4123]$ and $[2341]$.
 
 
 Finally, note that permutations (as defined in \ref{permadef}) are \textit{not} \say{orderings of a certain set.} \par
-They are defined as \textit{bijective maps}, which can be \textit{thought of} as orderings. \par
+They are defined as \textit{bijective maps}, which can be written as orderings of a given array. \par
 Remember: permutations are verbs!
 
 \pagebreak
\ No newline at end of file
diff --git a/Advanced/Symmetric Group/parts/1 cycle.tex b/Advanced/Symmetric Group/parts/1 cycle.tex
new file mode 100755
index 0000000..8166257
--- /dev/null
+++ b/Advanced/Symmetric Group/parts/1 cycle.tex	
@@ -0,0 +1,387 @@
+
+\section{Cycle Notation}
+
+\definition{}
+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.
+
+\vspace{2mm}
+
+For example, consider $[2134]$. This permutation has order $2$, as we clearly see below:
+
+\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 (1b) at (1, -2) {1};
+	\node (3b) at (2, -2) {3};
+	\node (4b) at (3, -2) {4};
+
+	\node (1c) at (0, -4.5) {1};
+	\node (2c) at (1, -4.5) {2};
+	\node (3c) at (2, -4.5) {3};
+	\node (4c) at (3, -4.5) {4};
+
+	\line{1a}{1b}
+	\line{2a}{2b}
+	\line{3a}{3b}
+	\line{4a}{4b}
+	\line{1b}{1c}
+	\line{2b}{2c}
+	\line{3b}{3c}
+	\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.
+
+
+\problem{}
+What is the order of $[2314]$? \par
+How about $[4321]$? \par
+\note[Note]{You shouldn't need to draw any braids to solve this problem.}
+
+
+\vfill
+
+\problem{Bonus}
+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
+\pagebreak
+
+
+As you may have noticed, the square-bracket notation we've been using thus far is a bit unwieldy.
+Permutations are verbs---but we've been referring to them using a noun (namely, their output when
+applied to an ordered sequence of numbers). Our notation fails to capture the meaning of the
+underlying object.
+
+\vspace{2mm}
+
+Think about it: is the permutation $[1234]$ different than the permutation $[12345]$? \par
+Indeed, these permutations operate on different sets---but they are both the identity! \par
+What should we do if we want to talk about the identity on $\{1, 2, ..., 10\}$?
+
+\vspace{2mm}
+
+We need something better.
+
+
+
+
+\definition{}
+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
+\note[Note]{$3 \to 3$ and $4 \to 4$ are also cycles, but we'll ignore them. One-cycles aren't aren't interesting.}
+
+\begin{center}
+\begin{tikzpicture}[scale=0.5]
+	\node (1) at (0, 0) {1};
+	\node (2) at (1, 0) {2};
+	\node (3) at (2, 0) {3};
+	\node (4) at (3, 0) {4};
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(1)
+		-- ($(1) + (0,-1)$)
+		-- ($(2) + (0,-1)$)
+		-- (2);
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(2)
+		-- ($(2) + (0, 1)$)
+		-- ($(1) + (0, 1)$)
+		-- (1);
+\end{tikzpicture}
+\end{center}
+
+
+The permutation $[431265]$ is a bit more interesting---it contains of two cycles: \par
+($1 \to 3 \to 2 \to 4 \to 1$ and $5 \to 6 \to 5$)
+
+
+\begin{center}
+\begin{tikzpicture}[scale=0.5]
+	\node (1) at (0, 0) {1};
+	\node (2) at (1, 0) {2};
+	\node (3) at (2, 0) {3};
+	\node (4) at (3, 0) {4};
+	\node (5) at (4, 0) {5};
+	\node (6) at (5, 0) {6};
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(3)
+		-- ($(3) + (0,-1)$)
+		-- ($(2) + (0,-1)$)
+		-- (2);
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(2)
+		-- ($(2) + (0,1.5)$)
+		-- ($(4) + (0,1.5)$)
+		-- (4);
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(4)
+		-- ($(4) + (0,-1.5)$)
+		-- ($(1) + (0,-1.5)$)
+		-- (1);
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(1)
+		-- ($(1) + (0,1)$)
+		-- ($(3) + (0,1)$)
+		-- (3);
+
+	\draw[line width = 0.3mm, ->, ogreen]
+		(5)
+		-- ($(5) + (0,-1)$)
+		-- ($(6) + (0,-1)$)
+		-- (6);
+
+	\draw[line width = 0.3mm, ->, ogreen]
+		(6)
+		-- ($(6) + (0,1)$)
+		-- ($(5) + (0,1)$)
+		-- (5);
+
+\end{tikzpicture}
+\end{center}
+
+
+
+\problem{}
+Find all cycles in $[5342761]$.
+
+\begin{solution}
+	\begin{center}
+	\begin{tikzpicture}[scale=0.5]
+		\node (1) at (0, 0) {1};
+		\node (2) at (1, 0) {2};
+		\node (3) at (2, 0) {3};
+		\node (4) at (3, 0) {4};
+		\node (5) at (4, 0) {5};
+		\node (6) at (5, 0) {6};
+		\node (7) at (6, 0) {7};
+
+		\draw[line width = 0.3mm, ->, ocyan]
+			(1)
+			-- ($(1) + (0,2)$)
+			-- ($(7) + (0,2)$)
+			-- (7);
+
+		\draw[line width = 0.3mm, ->, ocyan]
+			(7)
+			-- ($(7) + (0,-1.5)$)
+			-- ($(5) + (0,-1.5)$)
+			-- (5);
+
+		\draw[line width = 0.3mm, ->, ocyan]
+			(5)
+			-- ($(5) + (0,1.5)$)
+			-- ($(1) + (0.5,1.5)$)
+			-- ($(1) + (0.5,-1)$)
+			-- ($(1) + (0,-1)$)
+			-- (1);
+
+		\draw[line width = 0.3mm, ->, ogreen]
+			(2)
+			-- ($(2) + (0,-1.5)$)
+			-- ($(4) + (0,-1.5)$)
+			-- (4);
+
+		\draw[line width = 0.3mm, ->, ogreen]
+			(4)
+			-- ($(4) + (0,1)$)
+			-- ($(3) + (0,1)$)
+			-- (3);
+
+		\draw[line width = 0.3mm, ->, ogreen]
+			(3)
+			-- ($(3) + (0,-1)$)
+			-- ($(2) + (0.5,-1)$)
+			-- ($(2) + (0.5,1)$)
+			-- ($(2) + (0,1)$)
+			-- (2);
+	\end{tikzpicture}
+	\end{center}
+\end{solution}
+
+\vfill
+
+
+\problem{}
+What permutation (on five objects) consists of the cycles $3 \to 5 \to 3$ and $1 \to 2 \to 4 \to 1$?
+
+
+\begin{solution}
+	\begin{center}
+	\begin{tikzpicture}[scale=0.5]
+		\node (1) at (0, 0) {1};
+		\node (2) at (1, 0) {2};
+		\node (3) at (2, 0) {3};
+		\node (4) at (3, 0) {4};
+		\node (5) at (4, 0) {5};
+
+		\draw[line width = 0.3mm, ->, ocyan]
+			(3)
+			-- ($(3) + (0,1)$)
+			-- ($(5) + (0,1)$)
+			-- (5);
+
+		\draw[line width = 0.3mm, ->, ocyan]
+			(5)
+			-- ($(5) + (0,-1)$)
+			-- ($(3) + (0,-1)$)
+			-- (3);
+
+		\draw[line width = 0.3mm, ->, ogreen]
+			(1)
+			-- ($(1) + (0,-1)$)
+			-- ($(2) + (0,-1)$)
+			-- (2);
+
+		\draw[line width = 0.3mm, ->, ogreen]
+			(2)
+			-- ($(2) + (0,1.5)$)
+			-- ($(4) + (0,1.5)$)
+			-- (4);
+
+		\draw[line width = 0.3mm, ->, ogreen]
+			(4)
+			-- ($(4) + (0,-1.5)$)
+			-- ($(1) + (0.5,-1.5)$)
+			-- ($(1) + (0.5,1)$)
+			-- ($(1) + (0,1)$)
+			-- (1);
+	\end{tikzpicture}
+
+	This is $[41523]$
+	\end{center}
+\end{solution}
+
+
+
+\vfill
+\pagebreak
+
+\definition{}
+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}.
+
+\vspace{2mm}
+
+For example, we'll write $[2134]$ as $(12)$, which denotes the cycle $1 \to 2 \to 1$:
+
+
+\begin{center}
+\begin{tikzpicture}[scale=0.5]
+	\node (1) at (0, 0) {1};
+	\node (2) at (1, 0) {2};
+	\node (3) at (2, 0) {3};
+	\node (4) at (3, 0) {4};
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(1)
+		-- ($(1) + (0,-1)$)
+		-- ($(2) + (0,-1)$)
+		-- (2);
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(2)
+		-- ($(2) + (0, 1)$)
+		-- ($(1) + (0, 1)$)
+		-- (1);
+\end{tikzpicture}
+\end{center}
+
+
+
+As another example, $[431265]$ is $(1324)(56)$ in cycle notation. \par
+Note that we write $[431265]$ as a \textit{composition} of two cycles: \par
+applying the permutation $[431265]$ is the same as applying $(1324)$, then applying $(56)$.
+
+\begin{center}
+\begin{tikzpicture}[scale=0.5]
+	\node (1) at (0, 0) {1};
+	\node (2) at (1, 0) {2};
+	\node (3) at (2, 0) {3};
+	\node (4) at (3, 0) {4};
+	\node (5) at (4, 0) {5};
+	\node (6) at (5, 0) {6};
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(3)
+		-- ($(3) + (0,-1)$)
+		-- ($(2) + (0,-1)$)
+		-- (2);
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(2)
+		-- ($(2) + (0,1.5)$)
+		-- ($(4) + (0,1.5)$)
+		-- (4);
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(4)
+		-- ($(4) + (0,-1.5)$)
+		-- ($(1) + (0,-1.5)$)
+		-- (1);
+
+	\draw[line width = 0.3mm, ->, ocyan]
+		(1)
+		-- ($(1) + (0,1)$)
+		-- ($(3) + (0,1)$)
+		-- (3);
+
+	\draw[line width = 0.3mm, ->, ogreen]
+		(5)
+		-- ($(5) + (0,-1)$)
+		-- ($(6) + (0,-1)$)
+		-- (6);
+
+	\draw[line width = 0.3mm, ->, ogreen]
+		(6)
+		-- ($(6) + (0,1)$)
+		-- ($(5) + (0,1)$)
+		-- (5);
+
+\end{tikzpicture}
+\end{center}
+
+
+\problem{}
+Convince yourself that disjoint cycles commute. \par
+That is, $(1324)(56) = (56)(1324)$ since $(1324)$ and $(56)$ do not overlap.
+
+\problem{}
+Write the following in square-bracket notation.
+\begin{itemize}
+	\item $(12)$       \tab~\tab on a set of 2 elements
+	\item $(12)(435)$  \tab on a set of 5 elements
+	\vspace{2mm}
+	\item $(321)$      \tab~\tab on a set of 3 elements
+	\item $(321)$      \tab~\tab on a set of 6 elements
+	\vspace{2mm}
+	\item $(1234)$      \tab on a set of 4 elements
+	\item $(3412)$      \tab on a set of 4 elements
+\end{itemize}
+
+\vfill
+
+\problem{}
+Write the following in square-bracket notation.
+Be careful.
+\begin{itemize}
+	\item $(13)(243)$ \tab on a set of 4 elements
+	\item $(243)(13)$ \tab on a set of 4 elements
+\end{itemize}
+
+\vfill
+\pagebreak
\ No newline at end of file