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\begin{document}
\mingb
\title{\yuan 近\ \, 世\ \, 代\ \, 數\ \\
Modern Algebra}
\author{\fs Elbert A. Walker 艾伯特\thanks{author
作者} }
\date{6 October 1994}
\maketitle
\pskaib
\begin{abstract}
This document describes by example the use of {\tt asaetr.sty}, a \LaTeX\
style that somewhat conforms to the style used in the {\em Transactions of
the ASAE}\/. \cite{dunford} An accompanying BiBTeX style file is also used
to format the bibliography in a style similar to that used by ASAE.
\end{abstract}
\mingb
\section[導 論]{\kaib 導 論 Introduction}
群論是一門領域甚廣的學科，幾乎任何數學訓練都需要它。本章將簡單介紹一些%
古典的 非-Abelian 群論。在{\underbar{第二章}}，我們將介紹同構，正規子群，
商群，直和等基本觀念。 ...
\li
群論是一門領域甚廣的學科，幾乎任何數學訓練都需要它。本章將簡單介紹一些%
古典的 非-Abelian 群論。在{\underbar{第二章}}，我們將介紹同構，正規子群，
商群，直和等基本觀念。 ...
\mingb
\section{\kaib Jordan-Holder 定理}
Let $G$ be a finite group. If $G$ is not simple, then $G$ has a normal
subgroup $G_1\neq G$ such that $G/G_1$ is simple\textit{.} Just let $G_1$ be
a maximal normal subgroup in the finite group $G$\textit{.} Similarly, $G_1$
has a normal subgroup $G_2$ such that $G_1/G_2$ is simple. Thus we get a
descending chain
\[
G=G_0\supset G_1\supset G_2\supset \cdots \supset G_n=\{e\}
\]
such that $G_{i+1}$ is normal in $G_i$, and $G_i/G_{i+1}$ is simple. Now
suppose that
\[
G=H_0\supset H_1\supset H_2\supset \ldots \supset H_m=\{e\}
\]
is another such chain of subgroups. The Jordan-Holder Theorem asserts that $%
m=n$, and that there is a one-to-one correspondence between the factor
groups $G_i/G_{i+1}$ and $H_j/H_{j+1}$ such that corresponding factor groups
are isomorphic. We will prove this remarkable theorem, but first some
notation and terminology are needed. \textbf{The groups considered are not
necessarily finite.}
\begin{definition}
\ \, 設 $G$ 為一群。 $G$ 的一正規級數就是一如下之群的鍊鎖:
\[
G=G_0\supset G_1\supset G_2\supset \ldots \supset G_n=\{e\}
\]
其中，對於每一 $i$,\ $\,$ $G_{i+1}$ 皆為 $G_i$ 之正規子群。
The groups $G_i/G_{i+1}$ are the factor groups of the chain.
The length of the chain is
the number of strict inclusions in the chain. A normal series is a
composition series if each factor group $G_i/G_{i+1}$ is a simple group $%
\neq \{e\}$. Two normal series are equivalent if there is a one-to-one
correspondence between their factor groups such that corresponding factor
groups are isomorphic.
\end{definition}
\begin{theorem}[Jordan-Holder]
\ \, Any two composition series of a group are equivalent.
\end{theorem}
\proof\ \ If a group has a composition series of length
one, then the group is simple and any two composition series are certainly
equivalent. Now suppose that a group
$G$ has a composition series
\begin{equation} \label{one}
G=G_0\supset G_1\supset G_2\supset \ldots \supset G_n=\{e\}
\end{equation}
of length $n>1$, and that if a group has a composition series of length less
than $n$, then any two composition
series of that group are equivalent. Let
\begin{equation} \label{two}
G=H_0\supset H_1\supset H_2\supset \ldots \supset H_m=\{e\}
\end{equation}
be any composition series of $G$.
Consider the series
\begin{equation} \label{three}
G=G_0\supset G_1\supset G_1\cap H_1\supset G_2\cap H_1\supset \ldots \supset
G_n\cap H_1=\{e\}
\end{equation}
and
\begin{equation} \label{four}
G=H_0\supset H_1\supset H_1\cap G_1\supset H_2\cap G_1\supset \ldots \supset
H_m\cap G_1=\{e\}
\end{equation}
Since $G_{i+1}\cap H_1$ is a normal subgroup of $G_i\cap H_1$ and $%
G_i\supset G_{i+1}$, the Third Isomorphism Theorem (\textbf{2.3.12}) yields
\[
(G_i\cap H_1)/(G_{i+1}\cap H_1)=(G_i\cap H_1)/(G_{i+1}\cap (G_i\cap
H_1))\approx
\]
\[
G_{i+1}(G_i\cap H_1)/G_{i+1},
\]
and $G_{i+1}(G_i\cap H_1)$ is a normal subgroup of $G_i$ since it is a
product of two normal subgroups. Since $G_i/G_{i+1}$ is a simple group, $%
(G_{i+1}(G_i\cap H_1))/G_{i+1}$ is either $G_i/G_{i+1}$ or $G_{i+1}/G_{i+1}$%
. That is, $G_{i+1}(G_i\cap H_1)$ is either $G_{i+1}$ or $G_i$. Therefore,
if we remove repetitions from
\[
G_1\supset (G_1\cap H_1)\supset (G_2\cap H_1)\supset \ldots \supset (G_n\cap
H_1)=\{e\},
\]
we get a composition series for $G_1$. By our induction hypothesis, the
resulting composition series is equivalent to the composition series
\[
G_1\supset G_2\supset \ldots \supset G_n=\{e\}\text{,}
\]
and hence (\ref{one}) and (\ref{three}) (with repetitions removed) are
equivalent. $\cdots$
\section{\kaib 問 題 集 Problem
set}
\begin{enumerate}
\item 試証: 若 $G$ 有一合成級數，則任何正規子群
$G$ 都有合成級數。
\item 試証: 若 $N$ 在 $G$ 中為正規，且若
$G$ 有一合成級數，則 $G/N$ 亦然。
\item Suppose that $G$ has a composition series and that $N$ is normal in $%
G $. Prove that $G$ has a composition
series of which $N$ is a member.
\item Suppose that $G$ has a composition series. Prove that any normal
series of $G$ has a refinement that
is a composition series.
\item \label{zasslemma}(\textbf{Zassenhaus's Lemma}) Let \textit{A} and $B$
be subgroups of a group $G$, and let $M$ and $N$ be normal in \textit{A} and
$B$, respectively. Prove that
\end{enumerate}
\section{圖形與表格}
\begin{figure}[htb]
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\caption[原始之圖]{原始之圖 Primitive figure.}
\end{figure}
{
\renewcommand{\footnoterule}{}
\begin{table}[hbp]
\footnotesize
\vglue0pt
\caption{出版工具 Comparison of Publishing Tools}
\begin{center}
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\begin{center}
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\begin{tabular}{crr} \thickhline
Tool & \multicolumn{1}{c}{Learning Curve %\footnote{1.0 being easiest}
}& \multicolumn{1}{c}{Support %\footnote{10.0 being the best}
}\\ \thinhline
FrameMaker & 5.0 & 6.0 \\
Troff & 10.0 & 1.0 \\
\TeX %\footnote{\TeX\ is the winner!}\/
& 7.0 & 10.0 \\ \thickhline
\end{tabular}
\linethickness{0pt}
\end{center}
\end{center}
\end{table}
}
\begin{thebibliography}{9}
\bibitem[杜甫 1964]{dunford} N.
Dunford and J. Schwartz, 泛函分析,
\bibitem{} M. Struwe, \emph{Semilinear wave equations}, Bull. Amer. Math.
Soc. \textbf{26} (1992), 53-85.
\bibitem{} W.P. Thurston, \emph{Geometry and topology of three manifolds},
Lecture notes, Princeton Univ.,
NJ, 1979.
\end{thebibliography}
\end{document}
\begin{theindex}
\item 代數 1, 12
\item series 2, 33
\subitem convergent, 22, 55
\indexspace
\end{theindex}
\end{document}