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  {\huge\itshape\bfseries Grand Unified Study Guide}

  {\Large\itshape (under construction)}

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  {\large Copyright 2024 Benji Dial}

  {\large Licensed under \href{https://creativecommons.org/licenses/by-sa/4.0/}{CC BY-SA 4.0}}

  {\large \url{https://git.benjidial.net/gust}}

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\tableofcontents

\chapter*{Preface}

This is intended to be a study guide for a number of classes I have taken. I attempt to cover all the topics listed for these in the syllabi at the universities where I took them (University of North Carolina Charlotte and University of South Carolina), although they are not necessarily presented in the same way. For example, I present topology before analysis and then cite a lot of facts from topology, even though a first-year course in analysis usually does not require such a long detour.

There are exercises and examples interspersed throughout the text. The solutions to the exercises appear in Appendix \ref{solutions-chapter}. You are encouraged to think about the examples and the exercises, and to read the proofs of the theorems, in order to enhance your understanding of the material. Exercises are freely cited by future theorems and exercises as though they have been proven.

Throughout this text, $\mathbb{N}$ denotes the non-negative integers, and $\mathbb{Z}_+$ denotes the positive integers. $\mathbb{Q}_+$ and $\mathbb{R}_+$ are the positive rationals and reals respectively. $\mathbb{Z}_\pm$, $\mathbb{Q}_\pm$, and $\mathbb{R}_\pm$ are the non-zero integers, rationals, and reals respectively. For any set $S$, we use $\mathcal{P}(S)$ for the power set of $S$.

It is not necessary to read the chapters in order. Each section begins with a list of other sections that should be read first. Sections without prerequisites are intended to be readable by students in their first year of a graduate mathematics program.

\mainmatter
\pagestyle{headings}

\chapter{Groups}

\section{Definitions and basic properties}
\label{groups-section}

\subsection*{Groups}

\begin{definition}
  \index{group}
  \index{identity!in a group|see {group}}
  \index{inverse!in a group|see {group}}
  A \textit{group} is a set $G$ along with an operation $\diamond : G^2 \to G$ satisfying:
  \begin{itemize}
    \item Associativity: for all $x, y, z \in G$, we have $(x \diamond y) \diamond z = x \diamond (y \diamond z)$.
    \item Existence of identity: there is an $e \in G$ such that, for all $x \in G$, we have $x \diamond e = x$ and $e \diamond x = x$.
    \item Existence of inverses: for each $x \in G$, there is an $x' \in G$ such that $x \diamond x' = e$.
  \end{itemize}
\end{definition}

\begin{definition}
  \index{Abelian group}
  A group $(G, \diamond)$ is \textit{Abelian} if we have $x \diamond y = y \diamond x$ for all $x, y \in G$.
\end{definition}

\begin{example}
  Let $S$ be any set. Let $B(S)$ denote the set of bijections from $S$ to itself. Then, $S$ forms a group under composition. The identity is the identity function and the inverse of a bijection is its function inverse. This group is Abelian if and only if $S$ has no more than 2 elements.
\end{example}

\begin{example}
  $\mathbb{R}$ forms an Abelian group under addition. Here, the identity is $0$ and the inverse of a number is its negation.
\end{example}

\begin{example}
  $\mathbb{R}_\pm$ forms an Abelian group under multiplication. Here, the identity is $1$ and the inverse of a number is its reciprocal.
\end{example}

Some basic properties of groups follow.

\begin{theorem}
  \label{group-basic-props}
  % careful about renumbering in the enumerates - search for references to group-basic-props
  Let $(G, \diamond)$ be a group with identity $e$. For each $x \in G$, let $x'$ be the inverse of $x$.
  \begin{enumerate}
    \item Let $x, y, z \in G$. If $x \diamond y = z \diamond y$, then $x = z$.
    \item For all $x \in G$, we have $x' \diamond x = e$ (i.e.~$x = x''$).
    \item Let $x, y, z \in G$. If $x \diamond y = x \diamond z$, then $y = z$.
    \item Suppose $f \in G$ with $x \diamond f = x$ for all $x \in G$. Then, $f = e$.
    \item Suppose $f \in G$ with $f \diamond x = x$ for all $x \in G$. Then, $f = e$.
    \item Let $x, y \in G$ with $x \diamond y = e$. Then, $x = y'$ and $y = x'$.
  \end{enumerate}
  \begin{proof}
    \begin{enumerate}
      \item $x = x \diamond e = x \diamond y \diamond y' = z \diamond y \diamond y' = z \diamond e = z$.
      \item $x' \diamond x \diamond x' = x' \diamond e = e \diamond x'$. Applying (a), we get $x' \diamond x = e$.
      \item $y = e \diamond y = x' \diamond x \diamond y = x' \diamond x \diamond z = e \diamond e = z$.
      \item $f = e \diamond f = e$.
      \item $f = f \diamond e = e$.
      \item $x = x \diamond e = x \diamond y \diamond y' = e \diamond y' = y'$ and $y = e \diamond y = x' \diamond x \diamond y = x' \diamond e = x'$.
        \qedhere
    \end{enumerate}
  \end{proof}
\end{theorem}

Often the group operation will be represented by either $+$ or $\cdot$. In these two cases we define some special notation.

If we have a group $(G, +)$, then we will write $0$ for the group identity, and $-x$ for the group inverse of an $x \in G$. For any $x, y \in G$, we will write $x - y$ for $x + -y$. We define $n \cdot x$ or $nx$ for $n \in \mathbb{Z}$ and $x \in G$ recursively:
\begin{gather*}
  0 \cdot x = 0x = 0\\
  n \cdot x = nx = (n - 1)x + x \text{ for } n > 0\\
  n \cdot x = nx = -(|n|x) \text{ for } n < 0.
\end{gather*}
If we have a group $(G, \cdot)$, then we will write $1$ for the group identity and $x^{-1}$ for the group inverse of an $x \in G$. We define $x^n$ for $n \in \mathbb{Z}$ and $x \in G$ recursively:
\begin{gather*}
  x^0 = 1\\
  x^n = x^{n - 1} \cdot x \text{ for } n > 0\\
  x^n = (x^{|n|})^{-1} \text{ for } n < 0.
\end{gather*}

\begin{theorem}
  \label{group-power-rules}
  Let $(G, \cdot)$ be a group. Then, the following are true:
  \begin{enumerate}
    \item If $x \in G$ and $n, m \in \mathbb{Z}$, then $x^{n + m} = x^n x^m$.
    \item If $x \in G$ and $n, m \in \mathbb{Z}$, then $x^{n \cdot m} = (x^{n})^m$.
  \end{enumerate}
  \begin{proof}
    \begin{enumerate}
      \item
        If $m = 0$, then $x^{n + m} = x^n = x^n \cdot 1 = x^n x^m$. Let $m > 0$ and suppose that $x^{n + m - 1} = x^n x^{m - 1}$. Then $x^{n + m} = x^{n + m - 1} x = x^n x^{m - 1} x = x^n x^m$. By induction, $x^{n + m} = x^n x^m$ for all $m \ge 0$.

        If $m < 0$, then $x^n x^m = x^{n + m + |m|} x^m = x^{n + m} x^{|m|} x^m = x^{n + m} x^{|m|} (x^{|m|})^{-1} = x^{n + m}$.
      \item
        If $m = 0$, then $x^{n \cdot m} = x^0 = 1 = (x^n)^m$. Let $m > 0$ and suppose that $x^{n \cdot (m - 1)} = (x^n)^{m - 1}$. Then $x^{n \cdot m} = x^{n \cdot (m - 1) + n} = x^{n \cdot (m - 1)} x^n = (x^n)^{m - 1} x^n = (x^n)^m$. By induction, $x^{n \cdot m} = (x^n)^m$ for all $m \ge 0$.

        Observe that, for any $k \in \mathbb{Z}$, we have $x^k \cdot x^{-k} = x^{k + {-k}} = x^0 = 1$, so $x^{-k} = (x^k)^{-1}$.

        Then, if $m < 0$, we have $(x^n)^m = ((x^n)^{|m|})^{-1} = (x^{n \cdot |m|})^{-1} = x^{-(n \cdot |m|)} = x^{n \cdot m}$.
        \qedhere
    \end{enumerate}
  \end{proof}
\end{theorem}

For a group $(G, +)$, $x \in G$, and $n, m \in \mathbb{Z}$, those equations are:
\begin{itemize}
  \item $(n + m)x = nx + mx$.
  \item $(n \cdot m)x = m(nx)$.
\end{itemize}

\subsection*{Subgroups}

\begin{definition}
  \index{subgroup}
  Let $(G, +)$ be a group. A \textit{subgroup} of $G$ is a group on a subset $H \subseteq G$ with the operation being the restriction of $+$ to $H^2$.
\end{definition}

A potentially subtle point here: recall that the operation in a group on $H$ goes from $H^2$ to $H$. Thus, we need $h_1 + h_2 \in H$ for all $h_1, h_2 \in H$ for $H$ to be a subgroup.

\begin{example}
  $(\mathbb{Z}, +)$ is a subgroup of $(\mathbb{R}, +)$.
\end{example}

\begin{example}
  $(\mathbb{Q}_\pm, \cdot)$ is a subgroup of $(\mathbb{R}_\pm, \cdot)$.
\end{example}

Notice that if $(H, +)$ is a subgroup of $(G, +)$, then the $0$ from $G$ is the $0$ of $H$ by \cref{group-basic-props}(d) or \cref{group-basic-props}(e). Additionally, if $h \in H$, then the $-h$ from $G$ is the $-h$ of $H$ by \cref{group-basic-props}(f).

\begin{theorem}
  \label{subgroup-test}
  Let $(G, +)$ be a group, and let $H \subseteq G$. Then, $H$ is a subgroup of $G$ if and only if both of the following are satisfied:
  \begin{itemize}
    \item $H$ is nonempty.
    \item For all $x, y \in H$, we have $x - y \in H$.
  \end{itemize}
  \begin{proof}
    If $H$ is a subgroup, then $H$ is nonempty because it contains its inverse. For any $x, y \in H$, we have $-y \in H$, and $H$ is closed under $+$ so $x + -y \in H$.

    On the other hand, suppose $H \subseteq G$ satisfies the two bullet points. Since $H$ is nonempty, we can take some $h \in H$ and then $0 = h - h \in H$. For any $h \in H$, we have $-h = 0 - h \in H$. For any $x, y \in H$, we have $x + y = x - {-y} \in H$. So, $H$ is closed under $+$. We have already shown that $H$ has the identity and inverse from $G$, and these still satisfy those roles when restricted to $H$. Finally, restricting $+$ to $H^2$ does not impact its associativity. So, $H$ is a group under $+$.
  \end{proof}
\end{theorem}

\begin{example}
  Let $(G, +)$ be a group and let $x \in G$. Consider $\left<x\right> = \{nx : n \in \mathbb{Z}\}$. This is a group.
  \begin{proof}
    We have $x = 1x \in \left<x\right>$, so $\left<x\right>$ is nonempty. If we take two $y, z \in \left<x\right>$, we can write $y = nx$ and $z = mx$ for some $n, m \in \mathbb{Z}$. Then, using \cref{group-power-rules}, we get $y - z = nx - mx = nx + -1 \cdot mx = nx + (-m)x = (n - m)x \in \left<x\right>$.
  \end{proof}
\end{example}

\begin{definition}
  \index{subgroup!generated by an element}
  The group $\left<x\right>$ in the previous example is called the subgroup of $(G, +)$ \textit{generated by} $x$.
\end{definition}

\begin{exercise}
  What are $\left<1\right>$, $\left<-1\right>$, $\left<2\right>$, $\left<-2\right>$ in each of $(\mathbb{R}, +)$ and $(\mathbb{R}_\pm, \cdot)$?
  \begin{solution}
    In $(\mathbb{R}, +)$:
    \begin{itemize}
      \item $\left<1\right> = \left<-1\right> = \mathbb{Z}$.
      \item $\left<2\right> = \left<-2\right> = \{2n : n \in \mathbb{Z}\}$.
    \end{itemize}
    In $(\mathbb{R}_\pm, \cdot)$:
    \begin{itemize}
      \item $\left<1\right> = \{1\}$.
      \item $\left<-1\right> = \{1, -1\}$.
      \item $\left<2\right> = \{2^n : n \in \mathbb{Z}\}$.
      \item $\left<-2\right> = \{2^n, -2^n : n \in \mathbb{Z}\}$.
        \qedhere
    \end{itemize}
  \end{solution}
\end{exercise}

\begin{definition}
  \index{order!of a group}
  \index{order!of an element in a group}
  Let $(G, +)$ be a group. The \textit{order} of $(G, +)$ is the cardinality of $G$ if that is finite, and ``$\infty$'' otherwise. We call $(G, +)$ a \textit{finite} group if its order is finite. For any $x \in G$, the \textit{order} of $x$ is the order of $\left<x\right>$. We denote the order of $G$ by $|G|$, and the order of $x$ by $|x|$.
\end{definition}

\begin{exercise}
  What are the orders of $1$, $-1$, $2$, $-2$ in $(\mathbb{R}, +)$ and $(\mathbb{R}_\pm, \cdot)$?
  \begin{solution}
    In $(\mathbb{R}, +)$, all of these have order $\infty$.

    In $(\mathbb{R}_\pm, \cdot)$, the order of $1$ is $1$, the order of $-1$ is $2$, and the orders of $2$ and $-2$ are $\infty$.
  \end{solution}
\end{exercise}

\begin{theorem}[Lagrange's theorem]
  \label{lagrange}
  \index{Lagrange's theorem}
  Let $H$ be a subgroup of a finite group $G$. Then, the order of $H$ divides the order of $G$.
  \begin{proof}
    Let $+$ stand for the operation in $G$. For each $g \in G$, define $g + H = \{g + h : h \in H\}$. I claim that the relation $\sim$ on $G$ defined by $g_1 \sim g_2$ if and only if $g_1 \in g_2 + H$ is an equivalence relation.
    \begin{itemize}
      \item For any $g \in G$, we have $0 \in H$ and $g = g + 0 \in g + H$, so $g \sim g$.
      \item If $g_1 \sim g_2$, then there is an $h \in H$ with $g_1 = g_2 + h$. We get $g_2 = g_1 + -h$, so $g_2 \sim g_1$.
      \item If $g_1 \sim g_2$ and $g_2 \sim g_3$, then there are $h_1, h_2 \in H$ with $g_1 = g_2 + h_1$ and $g_2 = g_3 + h_2$. We get $g_1 = g_3 + h_2 + h_1$, so $g_1 \sim g_3$.
    \end{itemize}
    So, $\sim$ partitions $G$. In particular, note that, given a fixed $g \in G$, we have $\{g' \in G : g' \sim g\} = g + H$. In other words, $\{g + H : g \in G\}$ partitions $G$.

    Next, I claim that each of these sets is the same size. Given two $g_1, g_2 \in G$, let $f : g_1 + H \to g_2 + H$ be defined by $f(g) = g_2 + -g_1 + g$. We will show that $f$ is a well-defined bijection.
    \begin{itemize}
      \item If $g \in g_1 + H$, then there is an $h$ with $g = g_1 + h$. We have $f(g) = g_2 + -g_1 + g_1 + h = g_2 + h$, so $f(g) \in g_2 + H$. So, $f$ is well-defined.
      \item For any $g \in g_2 + H$, there is some $h$ with $g = g_2 + h$. Then, $g_1 + h \in g_1 + H$, and $f(g_1 + h) = g_2 + -g_1 + g_1 + h = g_2 + h = g$. So, $f$ is surjective.
      \item Given two $g, g' \in g_1 + H$ with $f(g) = f(g')$, we have $g_2 + -g_1 + g = g_2 + -g_1 + g'$, so $g = g'$. Then, $f$ is injective.
    \end{itemize}
    Next, observe that $0 + H = H$. We have partitioned $G$ into sets that each have the same size as $H$, so the order of $H$ divides the order of $G$.
  \end{proof}
\end{theorem}

In particular, this means that the order of an element in a finite group divides the order of that group.

\begin{exercise}
  Let $(G, +)$ be a group. Let $\mathcal{H}$ be a set of subgroups of $(G, +)$. Show that $\bigcap \mathcal{H}$ is a subgroup of $(G, +)$.
  \begin{solution}
    We use \cref{subgroup-test}. Each $H \in \mathcal{H}$ has $0 \in H$. Then, $0 \in \bigcap \mathcal{H}$, so $\bigcap \mathcal{H}$ is not empty. If $x, y \in \bigcap \mathcal{H}$, then $x, y \in H$ for each $H \in \mathcal{H}$. So, $x - y \in H$ for each $H \in \mathcal{H}$, which means $x - y \in \bigcap \mathcal{H}$.
  \end{solution}
\end{exercise}

\begin{definition}
  \index{subgroup!generated by a set}
  \index{generating set!for a group}
  Let $(G, +)$ be a group. Given a set $S \subseteq G$, the \textit{subgroup generated by $S$} is the intersection of all subgroups of $G$ with $S$ as a subset. We denote this subgroup by $\left<S\right>$. If $\left<S\right> = G$, then we say that $S$ \textit{generates} $(G, +)$, or that $S$ is a \textit{generating set} for $(G, +)$.
\end{definition}

\begin{exercise}
  Let $S$ be a subset of a group $(G, +)$. Let $S' = S \cup \{-s : s \in S\}$. Let $T$ be the set of finite sums of (not necessarily distinct) elements in $S'$, including the empty sum. For example, if $x, y \in S$, then $T$ has $0$, $x$, $x + y$, $x - y$, $x + x$, $-x + y + y$, etc. Show that $\left<S\right> = T$.
  \begin{solution}
    It is sufficient to show that $T$ is a subgroup of $(G, +)$ containing $S$ and that every subgroup of $(G, +)$ containing $S$ must contain $T$.

    It is clear that $S \subseteq T$, since each $x \in S$ can be viewed as the sum of just $x$. To show that $T$ is a subgroup, see that $0 \in T$ so $T$ is nonempty, and that if $x, y \in T$, then we have \begin{gather*}
      x = x_1 + x_2 + \cdots + x_n,\\
      y = y_1 + y_2 + \cdots + y_m;
    \end{gather*}
    where each $x_i, y_i \in S'$, so
    \begin{gather*}
      x - y = x_1 + x_2 + \cdots + x_n - y_m - y_{m - 1} - \cdots - y_1 \in T.
    \end{gather*}
    Now let $H$ be a subgroup of $(G, +)$ with $S \subseteq H$. Since $H$ has to contain inverses for each element, we have $S' \subseteq H$, and since $0 \in H$ and $H$ is closed under the operation of the group, we get $T \subseteq H$.
  \end{solution}
\end{exercise}

From the previous exercise we can see that, given a group $(G, +)$ and a $g \in G$, we have $\left<\{g\}\right> = \left<g\right>$.

\begin{example}
  $(\mathbb{Z}, +)$ is generated by $\{1\}$. Generating sets are not unique: $(\mathbb{Z}, +)$ is also generated by $\{2, 3\}$, since if a subgroup has both $2$ and $3$ then it must have $3 - 2 = 1$.
\end{example}

\begin{exercise}
  Let $S$ be the set of bijections from $\mathbb{Q}$ to $\mathbb{Q}$. Show that the subgroup of $(S, \circ)$ generated by $\{x \mapsto qx : q \in \mathbb{Q}_\pm\} \cup \{x \mapsto x + q : q \in \mathbb{Q}\}$ is $\{mx + b : m \in \mathbb{Q}_\pm, b \in \mathbb{Q}\}$.
  \begin{solution}
    Let $A = \{x \mapsto qx : q \in \mathbb{Q}_\pm\}$, $B = \{x \mapsto x + q : q \in \mathbb{Q}\}$, $C = \{mx + b : m \in \mathbb{Q}_\pm, b \in \mathbb{Q}\}$.

    $C$ is clearly nonempty. If we have $(x \mapsto m_1 x + b_1), (x \mapsto m_2 x + b_2) \in C$, then \begin{align*}
      (x \mapsto m_1 x + b_1) \circ (x \mapsto m_2 x + b_2)^{-1}
      & = (x \mapsto m_1 x + b_1) \circ \left(x \mapsto \frac{x - b_2}{m_2}\right)\\
      & = \left(x \mapsto m_1 \cdot \frac{x - b_2}{m_2} + b_1\right)\\
      & = \left(x \mapsto \frac{m_1}{m_2} x + b_1 - \frac{m_1 b_2}{m_2} \in C\right).
    \end{align*}
    So, $C$ is a subgroup of $(S, \circ)$.

    Given an $(x \mapsto m x + b) \in C$, we can write $(x \mapsto m x + b) = (x \mapsto x + b) \circ (x \mapsto mx)$, so any subgroup of $(S, \circ)$ containing $A$ and $B$ must contain $C$.
  \end{solution}
\end{exercise}

\subsection*{Homomorphisms}

\begin{definition}
  \index{homomorphism!between groups|see {group homomorphism}}
  \index{isomorphism!between groups|see {group isomorphism}}
  \index{group!homomorphism}
  \index{group!isomorphism}
  Given two groups $(G, +)$ and $(H, +)$, a \textit{group homomorphism} from $(G, +)$ to $(H, +)$ is a function $\theta : (G, +) \to (H, +)$ such that, for all $g_1, g_2 \in G$, we have $\theta(g_1 + g_2) = \theta(g_1) + \theta(g_2)$. A \textit{group isomorphism} is a group homomorphism that is also a bijection. Two groups are \textit{isomorphic} if there is a group isomorphism between them.
\end{definition}

We will just call these \textit{homomorphisms} and \textit{isomorphisms} when it is clear from context that we mean the group kind.

\begin{example}
  The function $\theta : (\mathbb{R}, +) \to (\mathbb{R}_\pm, \cdot)$ defined by $\theta(x) = e^x$ is a homomorphism, since $e^{x + y} = e^x \cdot e^y$ for any $x, y \in \mathbb{R}$. The same function with codomain $(\mathbb{R}_+, \cdot)$ is an isomorphism.
\end{example}

\begin{exercise}
  Let $\theta : (G, +) \to (H, +)$ be a group homomorphism. Show the following:
  \begin{enumerate}
    \item $\theta(0) = 0$.
    \item For all $g \in G$, we have $\theta(-g) = -\theta(g)$.
    \item For all $g \in G$ and $z \in \mathbb{Z}$, we have $\theta(zg) = z \theta(g)$.
  \end{enumerate}
  \begin{solution}
    \begin{enumerate}
      \item $\theta(0) = \theta(0) + \theta(0) - \theta(0) = \theta(0 + 0) - \theta(0) = \theta(0) - \theta(0) = 0$.
      \item $\theta(-g) = -\theta(g) + \theta(g) + \theta(-g) = -\theta(g) + \theta(g - g) = -\theta(g) + \theta(0) = -\theta(g)$.
      \item We have $\theta(0g) = \theta(0) = 0 = 0\theta(g)$. For any $n > 0$, if we have $\theta((n - 1)g) = (n - 1)\theta(g)$, then we have $\theta(ng) = \theta((n - 1)g + g) = \theta((n - 1)g) + \theta(g) = (n - 1)\theta(g) + \theta(g) = n\theta(g)$. By induction, we have $\theta(zg) = z\theta(g)$ for all $z \ge 0$. If $z < 0$, then $\theta(zg) = \theta(-|z|g) = -\theta(|z|g) = -|z|\theta(g) = z\theta(g)$.
        \qedhere
    \end{enumerate}
  \end{solution}
\end{exercise}

\section{Factor groups, direct products and sums}
\label{group-factors-products}

\begin{center}
  \textit{Prerequisite: \cref{groups-section}}
\end{center}

\subsection*{Factor groups}

\begin{definition}
  \index{normal!subgroup}
  A subgroup $(N, +)$ of a group $(G, +)$ is called \textit{normal} in $(G, +)$ if, for all $g \in G$ and $n \in N$, we have $g + n - g \in N$. We denote this by $N \normal G$.
\end{definition}

\begin{example}
  Any subgroup of an Abelian group is normal, since $g + n - g = n$.
\end{example}

\begin{example}
  Let $(G, +)$ be any group. Let $N = \{g \in G : g + g = 0\}$. Then, $N$ is normal in $G$, since for any $g \in G$ and $n \in N$, we have $g + n - g + g + n - g = g + n + n - g = g - g = 0$, so $g + n - g \in N$.
\end{example}

\begin{example}
  Let $S$ be the set of bijections from $\mathbb{Q}$ to $\mathbb{Q}$. Let $T = \{x \mapsto qx : q \in \mathbb{Q}_\pm\}$. See that $T \subseteq S$. In particular, $(T, \circ)$ is a subgroup of $(S, \circ)$ by \cref{subgroup-test}: we certainly have $T$ nonempty, and for any $x \mapsto qx, x \mapsto rx \in T$, we have $(x \mapsto qx) \circ (x \mapsto rx)^{-1} = x \mapsto qr^{-1}x \in T$. This subgroup is not normal since for example $(x \mapsto x + 1) \circ (x \mapsto 2x) \circ (x \mapsto x + 1)^{-1} = x \mapsto 2x - 1 \not \in T$.
\end{example}

\begin{theorem}
  Let $(N, +) \normal (G, +)$. For each $g \in G$, let $g + N$ denote $\{g + n : n \in N\}$. Then, $\{g + N : g \in G\}$ forms a group under the operation $(g_1 + N) + (g_2 + N) = (g_1 + g_2) + N$ (and that operation is well-defined).
  \begin{proof}
    First we show this new operation is well defined. Let $g_1, g_2, g_3, g_4 \in G$ with $g_1 + N = g_3 + N$ and $g_2 + N = g_4 + N$. We need to show that $(g_1 + g_2) + N = (g_3 + g_4) + N$. Let $g \in (g_1 + g_2) + N$. There is an $n \in N$ with $g = g_1 + g_2 + n = g_1 + g_2 + n - g_2 + g_2$. Since $N$ is normal in $G$, there is an $n' \in N$ with $g_2 + n - g_2 = n'$. So, $g = g_1 + n' + g_2$. Since $g_1 + N = g_3 + N$, there is an $n'' \in N$ with $g_1 + n' = g_3 + n''$. So, $g = g_3 + n'' + g_2 = g_3 + g_2 - g_2 + n'' + g_2$. Normality of $N$ in $G$ gives us an $n''' \in N$ with $-g_2 + n'' + g_2 = n'''$, so $g = g_3 + g_2 + n'''$. Finally, since $g_2 + N = g_4 + N$, there is an $n'''' \in N$ with $g_2 + n''' = g_4 + n''''$, so $g = g_3 + g_4 + n''''$, and therefore $g \in (g_3 + g_4) + N$. The same argument shows that anyone in $(g_3 + g_4) + N$ is in $(g_1 + g_2) + N$.

    Associativity is immediate. The identity is $0 + N$, and the inverse of a $g + N$ is $(-g) + N$.
  \end{proof}
\end{theorem}

\begin{definition}
  \index{factor!group}
  The group defined in the previous theorem is called the \textit{factor group} of $(G, +)$ modulo $N$, and is denoted $(G / N, +)$.
\end{definition}

Note that without normality, the operation is not well defined.

\begin{example}
  Let $S$ be the set of bijections from $\mathbb{R}$ to $\mathbb{R}$ considered as a group under composition, and let $T = \{x \mapsto rx : r \in \mathbb{R}_\pm\}$. See that $T$ is a non-normal subgroup of $S$ (similar to the example above with $\mathbb{Q}$ instead of $\mathbb{R}$).

  Let $f, g, h : \mathbb{R} \to \mathbb{R}$ with \begin{gather*}
    f(x) = \sqrt[3]{x^3 + 1},\\
    g(x) = \sqrt[3]{x^3 - 1},\\
    h(x) = 2x.
  \end{gather*}
  See that $f, g \in S$ and $h \in T$. We have \begin{gather*}
    f \circ h \circ T = f \circ T,\\
    g \circ h \circ T = g \circ T;
  \end{gather*}
  but \begin{gather*}
    (f \circ g) \circ T = (x \mapsto x) \circ T = T,\\
    (f \circ h \circ g \circ h) \circ T = (x \mapsto \sqrt[3]{64x^3 - 7}) \circ T \ne T.
  \end{gather*}
  So, an operation mapping $f \circ T$ and $g \circ T$ to $(f \circ g) \circ T$ for all $f, g \in S$ would not be well-defined.
\end{example}

\begin{definition}
  \index{kernel!of a group homomorphism}
  Given two groups $(G, +)$ and $(H, +)$ and a homomorphism $\theta : (G, +) \to (H, +)$, we define the \textit{kernel} of $\theta$, denoted $\ker \theta$, as $\{g \in G : \theta(g) = 0\}$.
\end{definition}

\begin{exercise}
  Show that the kernel of a homomorphism is a normal subgroup of the homomorphism's domain.
  \begin{solution}
    Let $\theta : (G, +) \to (H, +)$ be a homomorphism. We know $\ker \theta$ is nonempty because e.g.~$0 \in \ker \theta$. If we have $x, y \in \ker \theta$, then $\theta(x - y) = \theta(x) - \theta(y) = 0 - 0 = 0$, so $x - y \in \ker \theta$. Then, \cref{subgroup-test} tells us that $\ker \theta$ is a subgroup of $G$.

    To show normality, let $g \in G$ and $k \in \ker \theta$. Then, $\theta(g + k - g) = \theta(g) + \theta(k) - \theta(g) = \theta(g) + 0 - \theta(g) = 0$, so $g + k - g \in \ker \theta$.
  \end{solution}
\end{exercise}

\begin{exercise}
  Show that the image of a group under a homomorphism is a subgroup of the homomorphism's codomain.
  \begin{solution}
    We will use \cref{subgroup-test} again. Let $\theta : (G, +) \to (H, +)$ be a homomorphism. First, $\theta[G]$ is nonempty since e.g.~$\theta(0) = 0$ so $0 \in \theta[G]$. If $x, y \in \theta[G]$, then there are $z, w \in G$ with $\theta(z) = x$ and $\theta(w) = y$, and we get $\theta(z - w) = \theta(z) - \theta(w) = x - y$ so $x - y \in \theta[G]$.
  \end{solution}
\end{exercise}

\begin{theorem}[first isomorphism theorem for groups]
  \index{first isomorphism theorem!for groups}
  Let $\theta : (G, +) \to (H, +)$ be a group homomorphism. Then, $(G / {\ker \theta}, +)$ and $\theta[G]$ are isomorphic.
  \begin{proof}
    Let $\phi : (G / {\ker \theta}, +) \to (\theta[G], +)$ be defined by $\phi(g + \ker \theta) = \theta(g)$. To show $\phi$ is well-defined, let $g_1, g_2 \in G$ with $g_1 + \ker \theta = g_2 + \ker \theta$. Since $g_1 = g_1 + 0 \in g_1 + \ker \theta$, we also have $g_1 \in g_2 + \ker \theta$. Then, there is a $g_3 \in \ker \theta$ with $g_1 = g_2 + g_3$. Then, $\theta(g_1) = \theta(g_2 + g_3) = \theta(g_2) + \theta(g_3) = \theta(g_2)$.

    Next we will show that $\phi$ is a homomorphism. For any $x, y \in G$, we have $\phi((x + \ker \theta) + (y + \ker \theta)) = \phi((x + y) + \ker \theta) = \theta(x + y) = \theta(x) + \theta(y) = \phi(x + \ker \theta) + \phi(y + \ker \theta)$.

    Finally we need to show that $\phi$ is a bijection. For every $h \in \theta[G]$, there is a $g \in G$ with $\theta(g) = h$. Then, $\phi(g + \ker \theta) = h$, so $h$ is in the range of $\phi$. If we have $g_1, g_2$ with $\phi(g_1 + \ker \theta) = \phi(g_2 + \ker \theta)$, then $\theta(g_1) = \theta(g_2)$. In particular, that gives us $g_1 - g_2 \in \ker \theta$ and $g_2 - g_1 \in \ker \theta$, so $g_1 + \ker \theta = g_2 + \ker \theta$.
  \end{proof}
\end{theorem}

\begin{example}
  Consider the set $\mathbb{Q}^{\mathbb{Q}}$ of functions from $\mathbb{Q}$ to $\mathbb{Q}$. Define $+ : (\mathbb{Q}^{\mathbb{Q}})^2 \to \mathbb{Q}^{\mathbb{Q}}$ by $(f + g)(q) = f(q) + g(q)$. See that $(\mathbb{Q}^{\mathbb{Q}}, +)$ is a group. Let $T = \{f \in \mathbb{Q}^{\mathbb{Q}} : f(0) = 0\}$. Consider the function $\theta : (\mathbb{Q}^{\mathbb{Q}}, +) \to (\mathbb{Q}, +)$ defined by $\theta(f) = f(0)$. See that $\theta$ is a surjective homomorphism, and that $\ker \theta = T$. Then, $(T, +)$ is a normal subgroup of $(\mathbb{Q}^{\mathbb{Q}}, +)$, and the groups $(\mathbb{Q}^{\mathbb{Q}}/T, +)$ and $(\mathbb{Q}, +)$ are isomorphic by the theorem.
\end{example}

\subsection*{Direct products}

Let $(G, +)$ and $(H, +)$ be two groups. We can define a new group on the Cartesian product $G \times H$ with the operation $(g_1, h_1) + (g_2, h_2) = (g_1 + g_2, h_1 + h_2)$. This operation is associative since $+$ is associative in each of $(G, +)$ and $(H, +)$. The identity here is $(0, 0)$, and the inverse of $(g, h)$ is $(-g, -h)$. The following generalizes this to arbitrarily many groups.

\begin{definition}
  \index{direct product!of groups}
  Let $\Lambda$ be an index set. For each $\lambda \in \Lambda$, let $(G_\lambda, +)$ be a group. Then, we define the \textit{direct product} of $\{(G_\lambda, +)\}_{\lambda \in \Lambda}$, denoted $\bigotimes_{\lambda \in \Lambda} (G_\lambda, +)$, as the group on the Cartesian product $\prod_{\lambda \in \Lambda} G_\lambda$ with the operation $(x_\lambda)_{\lambda \in \Lambda} + (y_\lambda)_{\lambda \in \Lambda} = (x_\lambda + y_\lambda)_{\lambda \in \Lambda}$. In the case where $\Lambda$ is finite, we also denote $\bigotimes_{i = 1}^n (G_i, +)$ by $(G_1, +) \otimes (G_2, +) \otimes \cdots \otimes (G_n, +)$.
\end{definition}

\begin{definition}
  \index{canonical injection!into a direct product of groups}
  Let $\{(G_\lambda, +)\}_{\lambda \in \Lambda}$ be a collection of groups. For each $\mu \in \Lambda$, we define the \textit{canonical injection} of $(G_\mu, +)$ into $\bigotimes_{\lambda \in \Lambda} (G_\lambda, +)$, denoted by $\iota_\mu$:
  \begin{gather*}
    \iota_\mu : (G_\mu, +) \to \bigotimes_{\lambda \in \Lambda} (G_\lambda, +),\\
    \iota_\mu(g) = \left(\begin{cases} g & \text{if } \lambda = \mu \\ 0 & \text{if } \lambda \ne \mu \end{cases}\right)_{\lambda \in \Lambda.}
  \end{gather*}
\end{definition}

Notice that each canonical injection is an injective homomorphism.

\begin{definition}
  \index{canonical projection!from a direct product of groups}
  Let $\{(G_\lambda, +)\}_{\lambda \in \Lambda}$ be a collection of groups. For each $\mu \in \Lambda$, we define the \textit{canonical projection} of $\bigotimes_{\lambda \in \Lambda} (G_\lambda, +)$ onto $(G_\mu, +)$, denoted by $\pi_\mu$:
  \begin{gather*}
    \pi_\mu : \bigotimes_{\lambda \in \Lambda} (G_\lambda, +) \to (G_\mu, +),\\
    \pi_\mu((g_\lambda)_{\lambda \in \Lambda}) = g_\mu.
  \end{gather*}
\end{definition}

Notice that each canonical projection is a surjective homomorphism.

\begin{exercise}
  Let $\{(G_\lambda, +)\}_{\lambda \in \Lambda}$ be a collection of groups. Show that for each $\mu \in \Lambda$ we have \[
    \iota_\mu[G_\mu] \normal \bigotimes_{\lambda \in \Lambda} (G_\lambda, +).
  \]
  \begin{solution}
    For any $n \in G_\mu$ and $(g_\lambda)_{\lambda \in \Lambda} \in \prod_{\lambda \in \Lambda} G_\lambda$, we have \[
      (g_\lambda)_{\lambda \in \Lambda} + \iota_\mu(n) - (g_\lambda)_{\lambda \in \Lambda} = \left(\begin{cases} g_\mu + n - g_\mu & \text{if } \lambda = \mu \\ g_\lambda + 0 - g_\lambda & \text{if } \lambda \ne \mu \end{cases}\right)_{\lambda \in \Lambda} = \iota_\mu(g_\mu + n - g_\mu). \qedhere
    \]
  \end{solution}
\end{exercise}

\begin{exercise}
  Let $(G_1, +)$ and $(G_2, +)$ be groups and consider the direct product $(G_1, +) \otimes (G_2, +)$. Show that $((G_1, +) \otimes (G_2, +)) / \iota_1[G_1]$ and $\iota_2[G_2]$ are isomorphic.
  \begin{solution}
    Consider the homomorphism $\iota_2 \circ \pi_2 : (G_1, +) \otimes (G_2, +) \to (G_1, +) \otimes (G_2, +)$. We can write $(\iota_2 \circ \pi_2)(g_1, g_2) = \iota_2(g_2) = (0, g_2)$. The kernel of this homomorphism is $\iota_1[G_1]$, and the range is $\iota_2[G_2]$. Apply the first isomorphism theorem to get what we want.
  \end{solution}
\end{exercise}

\section{Center and centralizers, Cauchy's theorem}
\label{cauchy-section}

% in this section:
%   center of a group
%   centralizer of an element
%   conjugacy classes of an element
%   cauchy's theorem

\section{Sylow's theorems}
\label{sylow-section}

% in this section:
%   sylow's theorems

\chapter{Rings}
\label{rings-chapter}

\section{Definitions and basic properties}
\label{rings-section}

\begin{definition}
  \index{ring}
  A \textit{ring} is a set $R$ with two binary operations ${+}, {\cdot} : R^2 \to R$ satisfying the following:
  \begin{itemize}
    \item Associativity of addition: for all $x, y, z \in R$, we have $(x + y) + z = x + (y + z)$.
    \item Commutativity of addition: for all $x, y \in R$, we have $x + y = y + x$.
    \item Existence of additive identity: there is a $0 \in R$ such that, for all $x \in R$, we have $x + 0 = x$.
    \item Existence of additive inverses: for all $x \in R$, there is a $-x \in R$ such that $x + {-x} = 0$.
    \item Associativity of multiplication: for all $x, y, z \in R$, we have $(x \cdot y) \cdot z = x \cdot (y \cdot z)$.
    \item Left distributivity: for all $x, y, z \in R$, we have $x \cdot (y + z) = x \cdot y + x \cdot z$.
    \item Right distributivity: for all $x, y, z \in R$, we have $(x + y) \cdot z = x \cdot z + y \cdot z$.
  \end{itemize}
\end{definition}

Note that we do not require multiplication to be commutative, and we do not require the existence of a multiplicative identity or multiplicative inverses. Some sources (e.g.~\cite{wikipedia-ring}) do require the existence of a multiplicative identity, while others (e.g.~\cite{dummit-foote,gallian,maunder,mathworld-ring}) do not.

\begin{definition}
  \index{ring!with identity}
  \index{nontrivial ring}
  \index{trivial ring}
  A ring $R$ is said to \textit{have identity} if we have the following additional property:
  \begin{itemize}
    \item Existence of multiplicative identity: there is a $1 \in R$ such that, for all $x \in R$, we have $x \cdot 1 = x$ and $1 \cdot x = 1$.
  \end{itemize}
  A ring with identity is called \textit{trivial} if $1 = 0$, and nontrivial otherwise.
\end{definition}

\begin{example}
  The integers form a ring with identity under the usual $+$ and $\cdot$. The even integers form a ring without identity under the same operations. The odd integers do not form a ring under the usual operations, since there is no additive identity.
\end{example}

\begin{definition}
  \index{commutative ring}
  A ring $R$ is called \textit{commutative} if we have the following additional property:
  \begin{itemize}
    \item Commutativity of multiplication: for all $x, y \in R$, we have $x \cdot y = y \cdot x$.
  \end{itemize}
\end{definition}

\begin{example}
  The integers and the even integers are commutative rings, since multiplication of integers is commutative. For an example of a non-commutative ring, consider the set of $2 \times 2$ matrices with integer entries. One can verify that this forms a ring under the usual matrix addition and multiplication. This ring is not commutative: \[
    \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}
      \cdot
    \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}
      =
    \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}
      \text{ but }
    \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}
      \cdot
    \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}
      =
    \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.
  \]
\end{example}

Oftentimes, we will not write $\cdot$ between members of a ring. For example, we can write the condition for a commutative ring as $xy = yx$ for all $x, y \in R$.

Given a member $x$ of a ring $R$, we recursively define $nx$ and $x^n$ for $n \in \mathbb{Z}_+$ as follows:
\begin{gather*}
  1x = x\\
  x^1 = x\\
  nx = (n - 1)x + x \text{ for } n > 1\\
  x^n = x^{n - 1} \cdot x \text{ for } n > 1
\end{gather*}
We further define $0x = 0$ and $(-n)x = -(nx)$ for $n \in \mathbb{Z}_+$, where the first $0$ is the integer $0$ and the second $0$ is the additive identity of the ring.

For clarity, we only use these notations when the operations of the ring are denoted by $+$ and $\cdot$. There is no requirement that the operations of a ring be something we think of as addition and multiplication.

We will now see some basic properties of rings.

\begin{theorem}
  Let $R$ be a ring. Then, the following are all true:
  \begin{enumerate}
    \item For all $x \in R$, we have $0 + x = x$ and $-x + x = 0$.
    \item Let $z \in R$. If $x + z = x$ for all $x \in R$, then $z = 0$.
    \item Let $x, i \in R$. If $x + i = 0$, then $i = -x$.
    \item Suppose $R$ has identity, and let $o \in R$. If $x \cdot o = x$ for all $x$ or $o \cdot x = x$ for all $x$, then $o = 1$.
    \item Suppose $R$ has identity. Then, $R$ is trivial if and only if $R$ has exactly one element.
    \item For all $x \in R$, we have $-(-x) = x$.
  \end{enumerate}
  \begin{proof}
    \begin{enumerate}
      \item Just commute the addition and apply the definitions of $0$ and $-x$.
      \item Let $z$ have that property. Then in particular $0 + z = 0$. But $0 + z = z$, so $z = 0$.
      \item Since $x + i = 0$, we have $-x + x + i = -x$. But $-x + x + i = 0 + i = i$, so $i = -x$.
      \item Let $o$ have that property. Then in particular either $1 \cdot o = 1$ or $o \cdot 1 = 1$. In either case we get $o = 1$.
      \item If $R$ is trivial, then for any $x, y \in R$ we get $x = x \cdot 0 = 0 = y \cdot 0 = y$. If $R$ is not trivial, then $0$ and $1$ are two distinct elements.
      \item $-x + x = x + -x = 0$, so $x = -(-x)$.
        \qedhere
    \end{enumerate}
  \end{proof}
\end{theorem}

\begin{definition}
  \index{subring}
  Given a ring $R$, a \textit{subring} of $R$ is a set $S \subseteq R$ such that $S$ forms a ring under the restrictions of the operations from $R$.
\end{definition}

\begin{example}
  The even integers are a subring of the integers.
\end{example}

\begin{exercise}
  \label{subring-test}
  Let $R$ be a ring and $S \subseteq R$. Show that $S$ is a subring of $R$ if and only if all of the following conditions are satisfied:
  \begin{itemize}
    \item $0 \in S$.
    \item For all $x, y \in S$, we have $x + -y \in S$.
    \item For all $x, y \in S$, we have $x \cdot y \in S$.
  \end{itemize}
  \begin{solution}
    If $S$ is a subring of $R$, then the second and third bullets are obvious. For the first bullet, let $0_R$ denote the $0$ in $R$ and $0_S$ denote the $0$ in $S$. Then, $0_R = 0_S + -0_S \in S$ by the second bullet. In fact, $0_R = 0_S + -0_S = -0_S = 0_S$.

    Now suppose $S \subseteq R$ satisfies the three bullet points. For any $x, y \in S$, we have $-y = 0 + -y$, so $-y \in S$, and $x + y = x + -(-y)$, so $x + y \in S$. Then, $+$ and $\cdot$ have the right codomain when restricted to $S$. Associativity and commutativity of addition, associativity of multiplication, and both distributivities follow from $R$ being the corresponding property of $R$. We already know that the $0$ from $R$ is in $S$, and this acts like a $0$ in $S$. We also know that, for all $y \in S$, the $-y$ from $R$ is in $S$, and this acts like a $-y$ in $S$.
  \end{solution}
\end{exercise}

\begin{definition}
  \index{ideal}
  Given a ring $R$, an \textit{ideal} of $R$ is a set $I$ satisfying:
  \begin{itemize}
    \item $I$ is a subring of $R$.
    \item Left closure: for all $r \in R$ and $x \in I$, we have $r \cdot x \in I$.
    \item Right closure: for all $r \in R$ and $x \in I$, we have $x \cdot r \in I$.
  \end{itemize}
\end{definition}

\begin{example}
  The even integers are an ideal of the integers. For an example of a subring that is not an ideal, consider the $2 \times 2$ matrices of integers again. One can consider the subset of matrices where the right column is $0$. That subset will be a subring, but it will not be an ideal.
\end{example}

\begin{definition}
  \index{ring!homomorphism}
  \index{ring!isomorphism}
  \index{homomorphism!between rings|see {ring homomorphism}}
  \index{isomorphism!between rings|see {ring isomorphism}}
  Given two rings $R$ and $Q$, a \textit{ring homomorphism} from $R$ to $Q$ is a function $\theta : R \to Q$ such that, for all $r_1, r_2 \in R$, we have $\theta(r_1 + r_2) = \theta(r_1) + \theta(r_2)$, and $\theta(r_1 \cdot r_2) = \theta(r_1) \cdot \theta(r_2)$. A ring homomorphism that is also a bijection is called a \textit{ring isomorphism}. If there is a ring isomorphism between two rings, we say that those rings are \textit{isomorphic}.
\end{definition}

We will sometimes just say a \textit{homomorphism} or an \textit{isomorphism} when it is clear from context that we mean the ring kind.

\section{Direct products and sums, polynomial rings}

You should read \cref{rings-section} first.

% in this section:
%   direct product and sum definitions
%   polynomial rings as direct sums
%   usc algebra qual, january 2013, #1(a)

\chapter{Modules}

\section{Definitions and basic properties}

You should read \cref{rings-section} first. \cref{groups-section} is referenced in some examples.

\begin{definition}
  \index{module}
  \index{R-module@$R$-module|see {module}}
  Given a ring $R$, an \textit{$R$-module} is a set $M$ with two binary operations $+ : M^2 \to M$ and $\cdot : R \times M \to M$ satisfying the following:
  \begin{itemize}
    \item Associativity of addition: for all $x, y, z \in M$, we have $(x + y) + z = x + (y + z)$.
    \item Commutativity of addition: for all $x, y \in M$, we have $x + y = y + x$.
    \item Existence of additive identity: there is a $0 \in M$ such that, for all $x \in M$, we have $x + 0 = x$.
    \item Existence of additive inverses: for all $x \in M$, there is a $-x \in M$ such that $x + {-x} = 0$.
    \item Distributivity over scalar addition: for all $x, y \in R$ and $z \in M$, we have $(x + y) \cdot z = x \cdot z + y \cdot z$.
    \item Distributivity over module addition: for all $x \in R$ and $y, z \in M$, we have $x \cdot (y + z) = x \cdot y + x \cdot z$.
    \item Associativity of scalar multiplication: for all $x, y \in R$ and $z \in M$, we have $(x \cdot y) \cdot z = x \cdot (y \cdot z)$.
    \item Unitarity: If $R$ has identity, then for all $x \in R$ we have $1 \cdot x = x$.
  \end{itemize}
\end{definition}

This is typically (e.g.~\cite{dummit-foote}) called a \textit{left $R$-module}. One can define a similar structure with $\cdot : M \times R \to M$ and call that a \textit{right $R$-module}. Here, we never refer to right $R$-modules, so we will just call this an $R$-module.

\begin{example}
  Let $R$ be any ring and let $S$ be any set. Let $R^S$ denote the set of functions from $S$ to $R$. Define the operations as follows for any $f, g \in R^S$, $s \in S$, and $r \in R$:
  \begin{gather*}
    (f + g)(s) = f(s) + g(s)\\
    (r \cdot f)(s) = r \cdot f(s)
  \end{gather*}
  Then, $R^S$ forms an $R$-module with those operations. In particular, if $S$ is a finite set, then $R^S$ can be thought of as the length $|S|$ vectors with entries in $R$, and the operations are componentwise addition and scalar multiplication.
\end{example}

\begin{example}
  Any Abelian group (see \cref{groups-section}) is a $\mathbb{Z}$-module. The addition in the module is just the operation in the group, and the scalar multiplication is repeated addition (i.e.~$4 \cdot x = x + x + x + x$).
\end{example}

\begin{example}
  Any ideal of a ring is a module over that ring, with the operations inherited from the ring.
\end{example}

\appendix

\chapter{Solutions}
\label{solutions-chapter}
\clearpage

\solutions

\chapter{List of Qualifying Exam Problems}

\section{University of South Carolina, Algebra}

\begin{tabular}{ll}
  \multicolumn{2}{c}{\textbf{January 2013}}\\[0.5\topsep]
\end{tabular}

\backmatter

\printindex

\printbibliography[heading=bibintoc]

\end{document}
