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Question

在问题 4 中，你以开始，matrix但以结束bmatrix。

你必须写\end{document}而不是你的\end[document]。

为了使您的代码可编译，我必须添加标题、作者和软件包。我将您的代码更改amsmath为。bmatrix\newpage\clearpage

最后，我在问题 1 中更改了您的代码，以向您展示如何使用更好的代码来排版数学。例如，完整的方程式（您的行 Z(G) = ...）最好在环境equation（编号和标记）或equation*（未编号，未标记）下编写。我建议您在互联网上搜索排版数学的简介。或者查看“LaTeX2e（不那么）简短的介绍”的第 3 章（texdoc lshort在您的终端/控制台上）。

因此使用代码

\documentclass{article} % <===================================================
\usepackage{amsmath} % <======================================================
\title{test title}   % <======================================================
\author{Joe Doe}     % <======================================================
\begin{document}
\maketitle
%\clearpage % <===============================================================
\section*{Question 1}
A set $S$ along with a binary operator $\star$ to be a group means that 
given any two elements $x$ and $y$ that belongs to set $S$ "combines" 
to create a new element called $x \star y \in S$.  

An abelian group is a set $S$ with a binary operator $\star$ which 
satisfies the following conditions:
\begin{enumerate}
  \item $\forall x, y, z \in S$, we have 
    $x \star (y \star z) = (x \star y) \star z$.
  \item For any $x \in S$, there exists $y \in S$ such that 
    $(x \star y) = e$.
  \item There is an element $e \in S$, such that 
    $(x \star e) = x, \forall x \in S$.
  \item $\forall x, y \in S$ we have $(x \star y) = (y \star x)$.
\end{enumerate}

The center of a group $\mathcal{G}$ is the sub group consisting of those 
elements that 'commute' with every other elements.

\begin{equation*}
\mathcal{Z}(\mathcal{G}) = \{ x \in \mathcal{G}: x \star g = g \star x, 
  \forall g \in \mathcal{G} \}
\end{equation*}

Therefore the center of any abelian group is the entire group.

\section*{Question 2}

Matrix $A =
\begin{bmatrix}
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0
\end{bmatrix}$
and Matrix $B = \begin{bmatrix} 1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1
\end{bmatrix}$
\begin{enumerate}
\item $A^2 =
\begin{bmatrix}
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0
\end {bmatrix}$
\item $A^3$ =
$\begin{bmatrix}
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0
\end {bmatrix}$
\item  $A^4$ = 
$\begin {bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}$
\item $B^2$ =
$\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}$
\item{AB =}
$\begin{bmatrix}
0&0&1&0\\
0&1&0&0\\
0&0&0&1\\
1&0&0&0
\end{bmatrix}$
\item{$A^2B$=}
$\begin{bmatrix}
0&1&0&0\\
0&0&0&1\\
1&0&0&0\\
0&0&1&0
\end{bmatrix}$
\item{$A^3B$=}
$\begin {bmatrix}
0 & 0 & 0 & 1\\
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0
\end{bmatrix}$
\end{enumerate}
\text Matrix $A^4$ and $B^2$ are equal. And since $A^4$ and $B^2$
are identity matrix, combination with either of them will generate the multiple and matrix. Therefore there will be no other matrices that can be formed from multiplying combinations of A and B.

\section*{Question 3}
A group is nothing but set of elements and these matrices form are the elements that form a group. Ant matrix combination of A and B will give the resulting matrixfrom group. These 8 elements makes it a finite group. This group is definitely abelian because elements inside this group satisfies all the conditions required to be abelion. Example:
\begin{enumerate}
\item A$\star$B = B$\star$A
\item{A$\star$(B$\star$A) = (A$\star$B)$\star$A}
\item{B$\star$B = I}
\item{and A$\star$I = A}
\end{enumerate}
Since this group is an abelion, all of their elements form a center. Example
\begin{enumerate}
\item AB
\item{$B^2$}
\end{enumerate}

\section*{Question 4}
The square ABCD $\begin{bmatrix} A&B\\C&D \end{bmatrix}$ is represented by q =
$\begin{bmatrix}% <============================== bmatrix instead of matrix !!!!!!!!
A\\B\\C\\D
\end{bmatrix}$
\begin{enumerate}
\item Iq = $\begin{bmatrix} A\\B\\C\\D \end{bmatrix}$ There is no change in square. It is same.
\item Aq = $\begin{bmatrix} A\\C\\B\\D \end{bmatrix}$ Aq is the square ABCD rotated $90^{\circ}$ counter clockwise.
\item Bq = $\begin{bmatrix} B\\C\\D\\A \end{bmatrix}$ Bq is the square reflected at y = -x axis or line AD.
\item $A^2q$ = $\begin{bmatrix} C\\D\\A\\B \end{bmatrix}$ $A^2q$ is the square ABCD reflected at y = 0 axis or the line between AB and CD.
\item $A^3q$ = $\begin{bmatrix} D\\A\\B\\C \end{bmatrix}$ $A^3q$ is the square ABCD rotated $90^{\circ}$ clockwise.
\item ABq = $\begin{bmatrix} C\\B\\D\\A \end{bmatrix}$  ABq is the square ABCD reflected at line AD or y = -x axis and then rotated $90^{\circ}$ counter clockwise.
\item BAq = $\begin{bmatrix} B\\D\\A\\C \end{bmatrix}$ is the square ABCD rotated $90^{\circ}$ counter clockwise and then reflected at y = -x axis or line BA.
\end{enumerate}
%\clearpage

\section*{Question 5}
Yes there are 2 more symmetries of a square which are not included. For example:
\begin{enumerate}
\item $A^2Bq$ = $\begin{bmatrix} B\\D\\A\\C \end{bmatrix}$ $A^2Bq$ is the square ABCD reflected at y = -x axis (line AD) and then reflected at y = 0 axis (line between AC and BD)
\item $A^3Bq$ = $\begin{bmatrix} D\\B\\A\\C \end{bmatrix}$ $A^3Bq$ is the square ABCD reflected at y = -x axis (line AD) and then rotated $90{\circ}$ clockwise.
\end{enumerate}
In question~2, 8 matrices form a group. However, in a question 4, the particular group consist of 9 matrices that are form a group. The matrices in both the groups are solution for all the possible combination of the three matrices. There are 9 different solutions instead of 8 because in question~2 we found two matrices having the same solution. For example, matrices used in question~2 $A^4$ and $B^2$ and matrix used in question~4 I is an identity matrix.

\end{document} % <======================================================

编译上述代码（0 个错误，0 个警告）将得到结果（参见问题 1 中更好的数学集合）：

Answer 1