Created the Abstract Algebra theorems and definitions cheat sheet

part-2/chapters/chapter-11/chapter-11.tex

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+\chapter{Fundamental Theorem of Finite Abelian Groups}
+\subimport{./}{the-fundamental-theorem.tex}
+\subimport{./}{the-isomorphism-classes-of-abelian-groups.tex}
+\subimport{./}{proof-of-the-fundamental-theorem.tex}

part-2/chapters/chapter-11/proof-of-the-fundamental-theorem.tex

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+\section{Proof of the Fundamental Theorem}
+
+\begin{lemma}
+	Let $G$ be a finite Abelian group of order $p^nm$, where $p$ is a prime that does not divide $m$. Then $G = H \times K$, where $H = \{x \in G\ \vert\ x^{p^n} =e\}$ and $K =\{x \in G\ \vert\ x^m = e\}$. Moreover, $\abs{H}=p^n$.
+\end{lemma}
+
+\begin{lemma}
+	Let $G$ be an Abelian group of prime-power order and let $a$ be an element of maximum order in $G$. Then $G$ can be written in the form $\lr{a} \times K$.
+\end{lemma}
+
+\begin{lemma}
+	A finite Abelian group of prime-power order is an internal direct product of cyclic groups.
+\end{lemma}
+
+\begin{lemma}
+	Suppose that $G$ is a finite Abelian group of prime-power order. If $G=H_1 \times H_2 \times \dots \times H_m$ and $G=K_1 \times K_2 \times \dots \times K_n$, where the $H$'s and $K$'s are nontrivial cyclic subgroups with $\abs{H_1} \geq \abs{H_2} \geq \dots \geq \abs{H_m}$ and $\abs{K_1} \geq \abs{K_2} \geq \dots \geq \abs{K_n}$, then $m=n$ and $\abs{H_i} = \abs{K_i}$ for all $i$.
+\end{lemma}

part-2/chapters/chapter-11/the-fundamental-theorem.tex

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+\section{The Fundamental Theorem}
+
+\begin{theorem}[Fundamental Theorem of Finite Abelian Groups]
+	Every finite Abelian group is a direct product of cyclic groups of prime-power order. Moreover, the number of terms in the product and the orders of the cyclic groups are uniquely determined by the group.
+\end{theorem}

part-2/chapters/chapter-11/the-isomorphism-classes-of-abelian-groups.tex

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+\section{The Isomorphism Classes of Abelian Groups}
+
+\begin{remark}[Greedy Algorithm for an Abelian Group of Order $\mathbf{p^n}$]
+	The Fundamental Theorem is extremely powerful. As an application, we can use it as an algorithm for constructing all Abelian groups of any order. Let's look at Abelian groups of a certain order $n$, where $n$ has two or more distinct prime divisors.
+	\begin{enumerate}
+		\item Compute the orders of the elements of the group $G$
+		\item Select an element $a_1$ of maximum order and define $G_1 = \lr{a_1}$. Set $i = 1$.
+		\item If $\abs{G} = \abs{G_i}$, stop. Otherwise, replace $i$ by $i + 1$.
+		\item Select an element $a_i$ of maximum order $p^k$ such that $p^k \leq \abs{G}/\abs{G_{i-1}}$ and none of $a_i, a^p_i,a^{p^2}_i, \dots, a^{p^{k-1}}_i$ is in $G_{i-1}$, and define $G_i=G_{i-1} \times \lr{a_i}$.
+		\item Return to step 3.
+	\end{enumerate}
+\end{remark}
+
+\begin{corollary}[Existence of Subgroups of Abelian Groups]
+	If $m$ divides the order of a finite Abelian group $G$, then $G$ has a subgroup of order $m$.
+\end{corollary}