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\title
\LARGE\yuan 近\ \, 世\ \, 代\ \, 數\ \cr
      Modern Algebra
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\author
\fs Elbert A. Walker 艾伯特\footnote{\footnotesize author 作者}
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\kai
\centerline{摘\ \ 要}
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{\narrower\smallskip\noindent
本數學文本件顯示, 利用 ChiTeX 可以輕鬆地編寫數學文件。除了 ChiTeX 特有的幾個
特殊指令, 使用 ChiTeX 就像使用英文的 TeX 一樣的方便。
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\noindent
{\Large\kai 1. 導 論 Introduction}

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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\charsep{7}
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。 ...}

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\noindent
{\Large\kai 2. 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 {.} Just let $G_1$ be
a maximal normal subgroup in the finite group $G$. 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. {\bf The groups considered are not
necessarily finite.}

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\noindent
{\hei 定\ 義.}
\, 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.

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\noindent
{\hei 定\ 理} [Jordan-Holder]
\ \, Any two composition series of a group are equivalent.

\noindent
{\kai 證\ 明.}\ \ 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


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\noindent
{\Large\kai 3. 問 題 集 Problem set}
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\item{1.} 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{2.}  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{3.}  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.


\end