这是我第一次在 StackExchange 上发帖，因此对于任何格式错误深表歉意。
我有一份教授提供的原始 .tex 文件，其中包含一组课堂笔记，我正尝试将其与我在整个课程中通过 LyX 编写的另一组笔记合并。我想使用 LyX，而不是在另一个 TeX 编辑器中合并，因为我发现它更具可读性且更易于编辑。
从“文件 -> 导入 -> LaTeX (纯文本)”导入时，只有一小部分 (3/32 页) 显示在 LyX 编辑器中。我已修改 LyX 序言以匹配我的原始 .tex 的序言，如下所示：
\documentclass[12pt]{article}
\topmargin -1mm % -10mm
\oddsidemargin 0mm % 7mm
\evensidemargin 0mm % 7mm
\textwidth 15.5cm
\textheight 22.5cm %
\parindent 0pt
\usepackage{amsmath}
\usepackage{amscd}
\usepackage{amsthm}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{positioning, calc}
\setlength{\parindent}{0pt}
\setlength{\parskip}{0pt}
\def\beginpf{\noindent {\bf Proof. }}
\newcommand{\boite}{\mbox{} \hfill $\square$}
\def\endpf{\boite\medskip}
\newcommand{\ndiv}{\hspace{-4pt}\not{\hspace{3pt}|}\hspace{0pt}}
\def\NN{{\mathbb N}}
\def\N{{\mathbb N}}
\def\QQ{{\mathbb Q}}
\def\Q{{\mathbb Q}}
\def\RR{{\mathbb R}}
\def\R{{\mathbb R}}
\def\CC{{\mathbb C}}
\def\C{{\mathbb C}}
\def\ZZ{{\mathbb Z}}
\def\Z{{\mathbb Z}}
\newcommand{\GL}{\mathrm{GL}}
\newcommand{\SL}{\mathrm{SL}}
\newcommand{\OO}{\mathrm{O}}
\newcommand{\ima}{\mathrm{im}}
\newcommand{\Ker}{\mathrm{ker}}
\newcommand{\conj}{\mathrm{conj}}
\def\a{{\bf a}}
\def\e{{\bf e}}
\def\u{{\bf u}}
\def\v{{\bf v}}
\def\w{{\bf w}}
\def\x{{\bf x}}
\def\y{{\bf y}}
\def\z{{\bf z}}
\def\0{{\bf 0}}
\def\gcd{\mathop{\rm gcd}\nolimits}
\def\hcf{\mathop{\rm hcf}\nolimits}
\def\lcm{\mathop{\rm lcm}\nolimits}
\def\mod{\mathop{\rm mod}\nolimits}
\def\ds{\displaystyle}
\def\bb{\begin{bit}}
\def\eb{\end{bit}}
\def\eps{\varepsilon}
\def\LHS{{\rm LHS}}
\def\RHS{{\rm RHS}}
\def\ds{\displaystyle}
\def\pl{\polter{1}}
\swapnumbers
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{prop}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{facts}[theorem]{Facts}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{remarks}[theorem]{Remarks}
\newtheorem{example}[theorem]{Example}
\newtheorem{examples}[theorem]{Examples}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{bit}[theorem]{}
\def\FF{{\mathbb F}}
\providecommand{\abs}[1]{\lvert#1\rvert}
\providecommand{\Norm}[1]{\lVert#1\rVert}
\def\b{{\bf b}}
\def\Span{\mathop{\rm span}\nolimits}
\def\im{\mathop{\rm im}\nolimits}
\def\tr{\mathop{\rm tr}\nolimits}
\def\rank{\mathop{\rm rank}\nolimits}
%\def\ds{\displaystyle}
\def\bp{\begin{pmatrix}}
\def\ep{\end{pmatrix}}
\newcommand{\pile}[2]{\genfrac{}{}{0pt}{}{#1}{#2}}
然后对于 .tex 文件主体，以下内容导入正常，但以下 3000 多行似乎无法正确导入：
\begin{document}
\centerline{\Large{\textbf{MATH 2022 Groups and Vector Spaces}}}
\vspace{.1in}
This outline is based on the previous lecturer's lectures notes, and is a useful resource, but not exactly the same as the course as now given (though I have tried to be consistent about notation). You
should take your own notes in lectures, and may use these notes as back-up.
\section{Definition and examples of groups; subgroups and order}
${\mathbb N} = \{0, 1, 2, 3, \ldots\}$, the set of natural numbers (note that some authors omit 0, but I do not).
$\mathbb Z$, $\Q$, $\R$, and $\C$ are the sets of integers, rational numbers, real numbers, and complex numbers respectively. All come with binary operations $+$ and $\times$.
\begin{definition} Fix an integer $n \ge 1$. We let $\Z_n = \{ 0, 1, 2, \ldots, n-1\}$, and we add and multiply members of ${\mathbb Z}_n$ `modulo' $n$. That is, we add or multiply two given members of
${\mathbb Z}_n$ as usual, and then find the remainder of the answer on division by $n$. This is called the \emph{ring of integers modulo $n$}.
\end{definition}
\medskip
\textbf{Example.}
$\Z_4 = \{0, 1, 2, 3\}$ and $+$ and $\times$ are given by the tables
\begin{center}
\begin{tabular}{ c | c c c c}
$+$ & $0$ & $1$ & $2$ & $3$ \\
\hline\\[-10pt]
$0$ & $0$ & $1$ & $2$ & $3$ \\
$1$ & $1$ & $2$ & $3$ & $0$ \\
$2$ & $2$ & $3$ & $0$ & $1$ \\
$3$ & $3$ & $0$ & $1$ & $2$
\end{tabular}
\qquad
\begin{tabular}{ c | c c c c}
$\times$ & $0$ & $1$ & $2$ & $3$ \\
\hline\\[-10pt]
$0$ & $0$ & $0$ & $0$ & $0$ \\
$1$ & $0$ & $1$ & $2$ & $3$ \\
$2$ & $0$ & $2$ & $0$ & $2$ \\
$3$ & $0$ & $3$ & $2$ & $1$
\end{tabular}
\end{center}
\begin{definition} A \emph{group} is a non-empty set $G$ on which is defined an associative binary operation $\circ$ such that there is an identity $e$ ($e \circ x = x$ and $x \circ e = x$ for all
$x \in G$), and each $x \in G$ has an inverse in $G$ (an element $y$ such that $x \circ y = e$ and $y \circ x = e$).
\end{definition}
The full notation is $(G, \circ)$, but if $\circ$ is understood, then we write $G$ for short. Thus to check that $G$ is a group, you have to check four things:
\vspace{.1in}
$G$ is closed under the operation,
the operation is associative
there is an identity
every element has an inverse
\vspace{.1in}
(note that `$G$ is closed under the operation' means that if $x, y \in G$ then $x \circ y \in G$; this is not listed as an axiom, since it is part of what is meant by `$\circ$ is a binary operation on $G$';
also notice that you should avoid saying that $G$ is `closed', as that means something else - use the {\em whole phrase} `$G$ is closed under $\circ$').
\begin{examples} (1) $(\Z_4, +)$ is a group. It is closed under the operation, since any remainder modulo 4 lies in the set $\{0, 1, 2, 3\}$.
Associative. For example $(3 + 2) + 1 = 1 + 1 = 2$, while $3 + (2 + 1) = 3 + 3 = 2$. In fact it inherits associativity from $+$ for $\Z$.
Identity element $0$.
The inverses of $0,1,2,3$ are $0,3,2,1$ respectively.
\medskip
(2) More generally $(\Z_n, +)$ is a group for any $n$.
\medskip
(3) $(\Z_4, \times)$ is not a group. It is closed under the operation as before, and the operation is associative (it inherits it from $\Z$). The identity element is $1$. But $0$ has no inverse, since there
is no $y$ with $0 \cdot y = 1$.
\medskip
(4) The subset $\{1,2,3\}$ of $\Z_4$ is not a group under $\times$, since it is not closed under the operation.
\medskip
(5) The subset $\{1, 3\}$ of $\Z_4$ {\em is} a group under $\times$. It is closed under the operation since $1 \times a = a$ and $3 \times 3 = 1$.
It inherits associativity from $\times$ on $\Z_4$.
The identity element is $1$.
The inverses of $1, 3$ are $1, 3$ respectively.
\medskip
(6) $(\R,+)$ is a group. The identity element is 0, and the inverse of $x$ is $-x$.
\medskip
(7) $(\R,\times)$ is not a group. It is closed under $\times$, $\times$ is associative, the identity is 1, but 0 has no inverse.
\medskip
(8) We define $\R^* = \{ x\in \R : x\neq 0\} = \R \setminus \{0\}$. Then $(\R^*,\times)$ is a group. The identity is 1, and the inverse of $x$ is $1/x$.
\medskip
(9) $(\R,-)$ is not a group. The operation is not associative, e.g.\ $(1-2)-3 = -4$ but $1-(2-3) = 2$.
\end{examples}
\begin{definition} We say that a group $(G,\circ)$ is \emph{abelian} if the operation $\circ$ is commutative, that is, $x \circ y = y \circ x$ for all $x, y \in G$. \end{definition}
\medskip
\textbf{Examples.}
$(\Z_4,+)$, $(\R,+)$, $(\R^*,\times)$ are abelian groups. This next example is not.
\begin{example} (The dihedral group $D_3$)
Consider an equilateral triangle $XYZ$ with centroid $O$.
\[
\begin{tikzpicture}[scale=0.8]
\coordinate[label=above left:$Y$] (Y) at (0,0);
\coordinate[label=above right:$Z$] (Z) at (4,0);
\coordinate[label=above:$X$] (X) at (2,3.464);
\coordinate[label=$O$] (O) at (2,1.1547);
\coordinate[label=below:A] (A) at (2, -0.6);
\coordinate[label=right:B] (B) at (3.8,2.1939);
\coordinate[label=left:C] (C) at (0.2,2.1939);
\draw [dashed] (-0.3, -0.1732) -- (B);
\draw [dashed] (4.3, -0.1732) -- (C);
\draw [dashed] (A) -- (2,3.5);
\draw [line width=1.5pt] (X) -- (Y) -- (Z) -- cycle;
\end{tikzpicture}
\]
The isometries of the plane preserving the triangle are
\[
\begin{array}{c|l}
I & \text{do nothing}\\
R & \text{rotate about $O$ by angle $2\pi/3$ (clockwise)}\\
S & \text{rotate about $O$ by angle $2\pi/3$ (anticlockwise)}\\
A & \text{reflect in line $XO$}\\
B & \text{reflect in line $YO$}\\
C & \text{reflect in line $ZO$.}
\end{array}
\]
The \emph{dihedral group} $D_3$ is given by the set $D_3 = \{ I, R, S, A, B, C\}$ with the operation $\circ$ given by composition. Specifically, $P \circ Q$ means `do $Q$ first, then $P$'.
For example $A \circ R = C$, $R \circ A = B$, $A \circ B = S$. Then $\circ$ gives the following group table (where we write $P \circ Q$ in row $P$, column $Q$).
\begin{center}
\begin{tabular}{ c | c c c c c c }
$\circ$ & $I$ & $R$ & $S$ & $A$ & $B$ & $C$ \\
\hline\\[-10pt]
$I$ & $I$ & $R$ & $S$ & $A$ & $B$ & $C$ \\
$R$ & $R$ & $S$ & $I$ & $B$ & $C$ & $A$ \\
$S$ & $S$ & $I$ & $R$ & $C$ & $A$ & $B$ \\
$A$ & $A$ & $C$ & $B$ & $I$ & $S$ & $R$ \\
$B$ & $B$ & $A$ & $C$ & $R$ & $I$ & $S$ \\
$C$ & $C$ & $B$ & $A$ & $S$ & $R$ & $I$
\end{tabular}
\end{center}
The identity element is $I$. The inverses of $I, R, S, A, B, C$ are $I, S, R, A, B, C$ respectively.
\end{example}
当尝试删除
\end{示例}
整个正文确实导入了，但随后大量的错误阻止了文档的呈现，您建议最简单的解决方法是什么。此外（虽然从长远来看这是一个小问题）导入时仍然存在大量 LaTeX 代码环境，最简单的清理方法是什么。提前感谢，再次对任何格式错误/不必要的部分表示歉意。