这是代码，出现一个涉及缺失的错误$，但我确信它全部正确。我猜我遗漏了一些微妙的东西。以下是具体的错误消息：
Missing $ inserted. <inserted text> $ l.70 \textbf{Theorem. (Cancellation)}\
输入：
\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}
答案1
\end{matrix} \right)您错过了第 65 行的结束\left( \begin{matrix}。
此外，&该矩阵第一行的最后一个元素和倒数第二个元素之间的分隔符似乎丢失了。
您最好在 tex 源中使用更多空格，以便更容易发现此类问题，例如：
而不是这样：
Have the \emph{general linear group} $GL(n,\mathbb{Q})$ and
$e = \left( \begin{matrix}
1 & 0 & \ldots & \ldots 0\\
0 & \ddots & 0 & \ldots & 0\\
0 & \ldots & \ldots & \ldots & 1$
格式如下：
Have the \emph{general linear group} $GL(n,\mathbb{Q})$ and
$e = \left(
\begin{matrix}
1 & 0 & \ldots & \ldots & 0 \\
0 & \ddots & 0 & \ldots & 0 \\
0 & \ldots & \ldots & \ldots & 1
\end{matrix}
\right)$