"Missing $ inserted" error; can't figure out what is causing it

Question

Here is the code, getting an error involving a missing $ but I'm certain it's all correct. My guess is I am missing some subtle thing. Here is the specific error message:

Missing $ inserted. <inserted text> $ l.70 \textbf{Theorem. (Cancellation)}\

And the input:

\documentclass{article}
\usepackage{amssymb, amsmath, amstext, amscd, verbatim}

\newenvironment{parts}{\vspace{1ex}\begin{enumerate}\renewcommand{\itemsep}{1ex}

\renewcommand{\labelenumi}{(\alph{enumi})}}{\end{enumerate}}

\DeclareMathOperator*{\inter}{int}
\DeclareMathOperator*{\clos}{cl}
\DeclareMathOperator*{\bound}{bd}

%\everymath{\displaystyle}

\begin{document}

\title{Notes}
\author{}
\date{}
\maketitle

\tableofcontents
\pagebreak

\noindent
\textbf{Definition.} A \emph{group} $G$ is a pair $(S,*)$. Let $S \neq \varnothing$ be a set and $*$ a binary operation satisfying
\begin{parts}
\item $*$ is associative
\item $\exists e \in S$ such that $e*a=a*e=a \forall a \in S*$
\item $\forall a \in S, \exists b \in S$ such that $a*b=b*a=e$.
\end{parts}

\noindent
\textbf{Definition.} A group $G = (S,*)$ is \emph{abelian} if $*$ is commutative.\
\noindent
\textbf{Examples.}\\
(1) $(\mathbb{Z},+)$ with $e=0, b=-a$ is an abelian group.
(2) $(\mathbb{Z},\cdot)$ is \textbf{not} a group.
(3) $(\mathbb{R}\setminus \{0\}, \cdot)$ with $e=1, b=a^{-1}$ is an abelian group.
(4) Let $n \in \mathbb{P}$ be a positive integer, $T = \{1,2,\ldots,n\}$, and $S =$ set of permutations of $T =$ set of bijections on $T$.
S is function composition, hence associative.
(5) $S_n$ is the symmetric group on $n$ letters.
Let $n = 3$.
$ \left( \begin{matrix}
1 & 2 & 3 \\
3 & 1 & 2 \end{matrix} \right) * 
\left( \begin{matrix}
1 & 2 & 3 \\
3 & 2 & 1 \end{matrix} \right) = 
\left( \begin{matrix}
1 & 2 & 3 \\
2 & 1 & 3 \end{matrix} \right)
\left( \begin{matrix}
1 & 2 & 3 \\ 
3 & 2 & 1\end{matrix} \right) *
\left( \begin{matrix}
1 & 2 & 3 \\
3 & 1 & 2 \end{matrix} \right) =
\left( \begin{matrix}
1 & 2 & 3 \\
1 & 3 & 2 \end{matrix} \right) $.
\textbf{Note.} $\left| S_n \right| = n!$
(6) $n \in \mathbb{P},\quad M_n(\mathbb{Q}),\quad S =$ the set of non-singular $n \times n$ matrices,\quad $*=$ matrix multiplication.\\
Have the \emph{general linear group} $GL(n,\mathbb{Q})$ and
$e = \left( \begin{matrix}
1 & 0 & \ldots & \ldots 0\\
0 & \ddots & 0 & \ldots & 0\\
0 & \ldots & \ldots & \ldots & 1$
$GL(n,\mathbb{Z}_p), p$ is a prime. $\mathbb{Z}_p = \left\{ [0],[1],\ldots,[p-1]\right\}$.
\textbf{Theorem. (Cancellation)}\\
If $G = (S,*)$ is a group and $a*b = a*c$, then $b=c$ (and $b*a = c*a \Rightarrow b=c$).\\
\textbf{Proof.}\\
Let $d \in S$ satisfy $d*a = a*d = e$ ($e$ is the identity element). Then $d*(a*b) = d*(a*c)$.
 By associativity, we get $(d*a)*b = (d*a)*c \Rightarrow b = e*b = e*c = c$.\\

\noindent
\textbf{Corollary.}\\
If $a,b \in G$ and $a*b = a$, then $b=e$.\\
\textbf{Proof.} $a*b = a = a*e$. So $b=e$.

\noindent
\textbf{Corollary.}\\
Let $a,b,c \in G$. If $a*b=e=a*c$, then $b=c$.\\

\noindent
Therefore,
\begin{parts}
\item identity element is unique
\item for each $a \in G$, the inverse of $a$ is unique.
\end{parts}

\textbf{Notation.}\\
For multiplication, we write $a\cdot b$ and the inverse of $a$ as $a^{-1}$. Also, we have that $e=1$.\\
For addition in abelian groups, we write $a+b$ and the inverse of $a$ as $-a$. Also, we have that $e=0$.\\

\end{document}

Answer 1

You're missing the closing \end{matrix} \right) for the \left( \begin{matrix} at line 65.

Also the & separator between the last and second last element of the first row of that matrix seems to be missing.

You'd do well to use more whitespaces in your tex sources to make these kind of problems easier to spot, for example:

Instead of this:

Have the \emph{general linear group} $GL(n,\mathbb{Q})$ and
$e = \left( \begin{matrix}
1 & 0 & \ldots & \ldots 0\\
0 & \ddots & 0 & \ldots & 0\\
0 & \ldots & \ldots & \ldots & 1$

Format it like this:

Have the \emph{general linear group} $GL(n,\mathbb{Q})$ and
$e = \left(
     \begin{matrix}
      1 & 0      & \ldots & \ldots & 0 \\
      0 & \ddots & 0      & \ldots & 0 \\
      0 & \ldots & \ldots & \ldots & 1
     \end{matrix}
\right)$