Created the Abstract Algebra theorems and definitions cheat sheet

part-5/chapters/chapter-24/applications-of-sylow-theorems.tex

@@ -0,0 +1,5 @@
+\section{Applications of Sylow Theorems}
+
+\begin{theorem}[Cyclic Groups of Order $\mathbf{pq}$]
+	If $G$ is a group of order $pq$, where $p$ and $q$ are primes, $p < q$, and $p$ does not divide $q - 1$, then $G$ is cyclic. In particular, $G$ is isomorphic to $\Z_{pq}$.
+\end{theorem}

part-5/chapters/chapter-24/chapter-24.tex

@@ -0,0 +1,5 @@
+\chapter{Sylow Theorems}
+\subimport{./}{conjugacy-classes.tex}
+\subimport{./}{the-class-equation.tex}
+\subimport{./}{the-sylow-theorems.tex}
+\subimport{./}{applications-of-sylow-theorems.tex}

part-5/chapters/chapter-24/conjugacy-classes.tex

@@ -0,0 +1,13 @@
+\section{Conjugacy Classes}
+
+\begin{definition}[Conjugacy Class of $\mathbf{a}$]
+	Let $a$ and $b$ be elements of a group $G$. We say that $a$ and $b$ are \textit{conjugate} in $G$ (and call $b$ the \textit{conjugate} of $a$) if $xax^{-1}=b$ for some $x$ in $G$. The \textit{conjugacy class of $a$} is the set $\cl(a) = \{xax^{-1}\ \vert\ x \in G\}$.
+\end{definition}
+
+\begin{theorem}[Number of Conjugates of $\mathbf{a}$]
+	Let $G$ be a finite group and let $a$ be an element of $G$. Then, $\abs{\cl(a)} = \abs{G:C(a)}$.
+\end{theorem}
+
+\begin{corollary}[$\mathbf{\abs{\cl(a)}}$ Divides $\mathbf{\abs{G}}$]
+	In a finite group, $\abs{\cl(a)}$ divides $\abs{G}$.
+\end{corollary}

part-5/chapters/chapter-24/the-class-equation.tex

@@ -0,0 +1,15 @@
+\section{The Class Equation}
+
+\begin{corollary}[Class Equation]
+	For any finite group $G$,
+	\[ \abs{G} = \sum \abs{G:C(a)} \]
+	where the sum runs over one element of $a$ from each conjugacy class of $G$.
+\end{corollary}
+
+\begin{theorem}[$\mathbf{p}$-Groups Have Nontrivial Centers]
+	Let $G$ be a nontrivial finite group whose order is a power of a prime $p$. Then $\Z(G)$ has more than one element.
+\end{theorem}
+
+\begin{corollary}[Groups of Order $\mathbf{p^2}$ Are Abelian]
+	If $\abs{G}=p^2$, where $p$ is prime, then $G$ is Abelian.
+\end{corollary}

part-5/chapters/chapter-24/the-sylow-theorems.tex

@@ -0,0 +1,29 @@
+\section{The Sylow Theorems}
+
+\begin{theorem}[Existence of Subgroups of Prime-Power Order (Sylow's First Theorem, 1872)]
+	Let $G$ be a finite group and let $p$ be a prime. If $p^k$ divides $\abs{G}$, then $G$ has at least one subgroup of order $p^k$.
+\end{theorem}
+
+\begin{definition}[Sylow $\mathbf{p}$-Subgroup]
+	Let $G$ be a finite group and let $p$ be a prime. If $p^k$ divides $\abs{G}$ and $p^{k+1}$ does not divide $\abs{G}$, then any subgroup of $G$ of order $p^k$ is called a \textit{Sylow $p$-subgroup of $G$}.
+\end{definition}
+
+\begin{corollary}[Cauchy's Theorem]
+	Let $G$ be a finite group and let $p$ be a prime that divides the order of $G$. Then $G$ has an element of order $p$.
+\end{corollary}
+
+\begin{definition}[Conjugate Subgroups]
+	Let $H$ and $K$ be subgroups of a group $G$. We say that $H$ and $K$ are \textit{conjugate} in $G$ if there is an element in $G$ such that $H = gKg^{-1}$.
+\end{definition}
+
+\begin{theorem}[Sylow's Second Theorem]
+	If $H$ is a subgroup of a finite group $G$ and $\abs{H}$ is a power of a prime $p$, then $H$ is contained in some Sylow $p$-subgroup of $G$.
+\end{theorem}
+
+\begin{theorem}[Sylow's Third Theorem]
+	Let $p$ be a prime and let $G$ be a group of order $p^km$, where $p$ does not divide $m$. Then the number $n$ of Sylow $p$-subgroups of $G$ is equal to 1 modulo $p$ and divides $m$. Furthermore, any two Sylow $p$-subgroups of $G$ are conjugate.
+\end{theorem}
+
+\begin{corollary}[A Unique Sylow $\mathbf{p}$-Subgroup Is Normal]
+	A Sylow $p$-subgroup of a finite group $G$ is a normal subgroup of $G$ if and only if it is the only Sylow $p$-subgroup of $G$.
+\end{corollary}