Created the Abstract Algebra theorems and definitions cheat sheet

part-5/chapters/chapter-32/chapter-32.tex

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+\chapter{An Introduction to Galois Theory}
+\subimport{./}{fundamental-theorem-of-galois-theory.tex}
+\subimport{./}{solvability-of-polynomials-by-radicals.tex}

part-5/chapters/chapter-32/fundamental-theorem-of-galois-theory.tex

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+\section{Fundamental Theorem of Galois Theory}
+
+\begin{definition}[Automorphism, Galois Group, Fixed Field of $\mathbf{H}$]
+	Let $\E$ be an extension field of the field $\F$. An \textit{automorphism of $\E$} is a ring isomorphism from $\E$ onto $\E$. The \textit{Galois group} of $\E$ over $\F$, $\gal(\E/\F)$, is the set of all automorphisms of $\E$ that take every element of $\F$ to itself. If $H$ is a subgroup of $\gal(\E/\F)$, then set
+	\[ \E_H = \{x \in \E\ \vert\ \phi(x) = x,\ \forall\ \phi \in H\} \]
+	is called the \textit{fixed field of $H$}.
+\end{definition}
+
+\begin{theorem}[Fundamental Theorem of Galois Theory]
+	Let $\F$ be a field of characteristic 0 or a finite field. If $\E$ is the splitting field over $\F$ for some polynomial in $\F[x]$, then the mapping from the set of subfields of $\E$ containing $\F$ to the set of subgroups of $\gal(\E/\F)$ given by $\K \to \gal(\E/\F)$ is a one-to-one correspondence. Furthermore, for any subfield $\K$ of $\E$ containing $\F$,
+	\begin{enumerate}
+		\item $[\E:\K] = \abs{\gal(\E/\K)}$ and $[\K:\F] = \abs{\gal(\E/\F)} / \abs{\gal(\E/\K)}$. [The index of $\gal(\E/\K)$ in $\gal(\E/\F)$ equals the degree of $\K$ over $\F$.]
+		\item If $\K$ is the splitting field of some polynomial in $\F[x]$, then $\gal(\E/\K)$ is a normal subgroup of $\gal(\E/\F)$ and $\gal(\K/\F)$ is isomorphic to $\gal(\E/\F)/\gal(\E/\K)$.
+		\item $\K = \E_{\gal(\E/\K)}$. [The fixed field of $\gal(\E/\K)$ is $\K$.]
+		\item If $H$ is a subgroup of $\gal(\E/\F)$, then $H=\gal(\E/\E_H)$. [The automorphism group of $\E$ fixing $\E_H$ is $H$.]
+	\end{enumerate}
+\end{theorem}

part-5/chapters/chapter-32/solvability-of-polynomials-by-radicals.tex

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+\section{Solvability of Polynomials by Radicals}
+
+\begin{definition}[Solvable by Radicals]
+	Let $\F$ be a field, and let $f(x) \in \F[x]$. We say that $f(x)$ is \textit{solvable by radicals over $\F$} if $f(x)$ splits in some extension $\F(a_1,a_2,\dots,a_n)$ of $\F$ and there exist positive integers $k_1,\dots,k_n$ such that $a_1^{k_1} \in \F$ and $a_i^{k_i} \in \F(a_1,\dots,a_{i-1})$ for $i=2,\dots,n$.
+\end{definition}
+
+\begin{definition}[Solvable Group]
+	We say that a group $G$ is \textit{solvable} if $G$ has a series of subgroups
+	\[ \{e\} = H_0 \subset H_1 \subset H_2 \subset \dots \subset H_k = G \]
+	where, for each $0 \leq i <k$, $H_i$ is normal in $H_{i + 1}$ and $H_{i + 1}/H_i$ is Abelian.
+\end{definition}
+
+\begin{theorem}[Splitting Field of $\mathbf{x^n - a}$]
+	Let $\F$ be a field of characteristic 0 and let $a \in \F$. If $\E$ is the splitting field of $x^n-a$ over $\F$, then the Galois group $\gal(\E/\F)$ is solvable.
+\end{theorem}
+
+\begin{theorem}[Factor Group of a Solvable Group is Solvable]
+	A factor group of a solvable group is solvable.
+\end{theorem}
+
+\begin{theorem}[$\mathbf{N}$ and $\mathbf{G/N}$ Implies $\mathbf{G}$ Is Solvable]
+	Let $N$ be a normal subgroup of a group $G$. If both $N$ and $G/N$ are solvable, then $G$ is solvable.
+\end{theorem}
+
+\begin{theorem}[(Galois) Solvable by Radicals Implies Solvable Group]
+	Let $\F$ be a field of characteristic 0 and let $f(x) \in \F[x]$. Suppose the $f(x)$ splits in $\F(a_1,a_2,\dots,a_t)$, where $a_1^{n_1} \in \F$ and $a_i^{n_i} \in \F(a_1,\dots,a_{i-1})$ for $i=2,\dots,t$. Let $\E$ be the splitting field for $f(x)$ over $\F$ in $\F(a_1,a_2,\dots,a_t)$. Then the Galois group $\gal(\E/\F)$ is solvable.
+\end{theorem}