我在我的网盘里面传了Ctex基础版：
http://yun.baidu.com/share/link?shareid=1831040314&uk=2148406042
同时还有一个简短的教程：
http://yun.baidu.com/share/link?shareid=1854151229&uk=2148406042
软件安装了之后会有Winedit这个程序的图标（一支笔和一张纸）出现，点开这个图标，新建文件之后就可以直接写论文代码。只有不到300M，因为没有中文包（自己吐槽：那为什么叫Ctex？！）和其它许多功能（比如我发现beamer功能似乎就有问题），但是我想对于数学和物理的论文排版是没有问题的。
为了方便同学们使用，我自己贴一段代码（这些内容都可以在教程中找到），内容是我上个学期的某些抽象代数习题的答案（之前闲来无事把解答写成了代码），其中定义了环境等等基本的要素。
% !Mode::"TeX:UTF-8"
\documentclass{article}
\usepackage{amssymb}
\begin{document}
\title{2013 Abstract Algebra Extra Exercises}
\maketitle
\section{Week 2}
5. The example is given by
$$G=\left\{\left(\begin{array}{cccc} 1 & a & b\\ 0 & 1 & c \\ 0 & 0 & 1\end{array}\right):a,b,c\in\mathbb{Z}_p\right\}.$$
\section{Week 4}
1. A theorem on the number of fixed points:
Suppose that a finite group $G$ acts transitively on a finite set $X$, whose cardinal number equals $n\geq2$. Let $G_0$ be the set of elements of $G$ with no fixed points. Then the ration $\frac{|G_0|}{|G|}\geq\frac{1}{n}.$ Moreover, if $n$ is not a power of prime, the inequality becomes strict.
\emph{Proof.} Let $N(g)$ be the number of fixed points of $g\in G$. Then by Burnside's Lemma, we have the following inequality:
$$\frac{1}{|G|}\sum_{g\in G} N(g)=1,$$
which is due to the transitivity of the action. We claim another inequality:
$$\frac{1}{|G|}\sum_{g\in G}N^2(g)\geq2.$$
In fact, the action of $G$ induces an action on the product set $X\times X.$ The number $N^2(g)$ is hence the number of $X\times X$ fixed by $g$(since, for example, if $i_1,i_2,...,i_k\in X$ are fixed by $g$, then the points $(i_u,i_v)\in X\times X$ are fixed by $g$, too.). Also, number of orbits of $G$ in $X\times X$ must exceed 2: by decomposing of $X\times X$ into the diagonal and its complement it will be suffice to verify this assertion(if the action is 2-transitive, then $\sum_{g\in G}N^2(g)=2|G|,$ which is a standard result). By Burnside's Lemma, the number of orbits on $X\times X$ equals
$$\frac{1}{|G|}\sum_{g\in G}N^2(g),$$
and the inequality is thus proved. We also obtain that the equality holds iff $G$ acts doubly transitively.
Next, we observe that $g\in G_0$ is equivalent to $N(g)=0$. Also, we have $1\leq N(g)\leq n$. Thus, $g\in G\setminus G_0$ iff $(N(g)-1)(N(g)-n)\leq0$, and if $g\in G$, $(N(g)-1)(N(g)-n)=n.$ Thus,
$$\sum_{g\in G\setminus G_0}(N(g)-1)(N(g)-n)\leq0,$$
and so we have
$$\frac{1}{|G|}\sum_{g\in G}(N(g)-1)(N(g)-n)\leq\frac{1}{|G|}\sum_{g\in G_0}(N(g)-1)(N(g)-n)=\frac{n|G_0|}{|G|}.$$
By the preceding inequality, we assert that $\frac{1}{|G|}\sum_{g\in G}N^2(g)\geq2$, and hence
$$\frac{1}{|G|}\sum_{g\in G}(N(g)-1)(N(g)-n)=\frac{1}{|G|}\sum_{g\in G}N^2(g)-(n+1)N(g)+n\geq2-(n+1)+n=1.$$
Thus, $1\leq\frac{n|G_0|}{|G|}.$
Now let $G$ act on the collection of all the cosets $G/H=\{g_iH:\bigcup_{i\in I}\},$ where the rules are given by $g\circ g_iH=gg_iH.$ Obviously, $G$ acts on $G/H$ transitively, and the cardinal number of the set $G/H$ equals $\frac{|G|}{|H|}$. Thus, we can apply the proved theorem. We see that $g_iH$ is a fixed point of $g$ iff $gg_iH=g_iH,$ i.e., $g\in g_iHg_i^{-1}$. Conversely, if $g=g_ihg_i^{-1},$ then $g$ must fix $g_iH.$ Thus, The set of elements with at least one fixed points coincide with the union of conjugate class $\bigcup_{g\in G}gHg^{-1}$; the set with no fixed points is hence $G-\bigcup_{g\in G}gHg^{-1}$. Apply the preceding theorem, we obtain
$$\frac{\left|G-\bigcup_{g\in G}gHg^{-1}\right|}{|G|}\geq\frac{1}{|G/H|}=\frac{|H|}{|G|}.$$
Hence,
$$\left|G-\bigcup_{g\in G}gHg^{-1}\right|\geq|H|.$$
2.
\emph{Proof.} We consider under what cases the set $I$ becomes a subgroup of $G$. Obviously, for any $g\in I$, we have $e=\varphi(e)=\varphi(gg^{-1})=g^{-1}\phi(g^{-1}).$ Thus, $\varphi(g^{-1})=g,$ and the set $I$ is hence closed under the operation of taking inverse. If we hope $I$ to be a subgroup, the for any $x,y\in I$, we must have $xy\in I$, i.e., $\varphi(xy)=\varphi(x)\varphi(y)=x^{-1}y^{-1}=y^{-1}x^{-1}\in I\Longleftrightarrow xy=yx$. Hence, $I$ is a subgroup iff the elements commute with each other. We claim that under the assumption $|I|>\frac{3}{4}|G|,$ the elements of $I$ must commute with each other. Otherwise, there exists $x,y$ s.t. $xy\neq yx$. Thus, the number of elements of $C(x)$ cannot exceed a half of $|G|$ since $C(x)$ is a proper subgroup of $G$. Let the number of elements of $I$ which do not commute with $x$ be $n$. Then we see that $|C(x)\bigcap I|\leq\frac{1}{2}|G|,$ and hence $n=|I-C(x)|>\frac{1}{4}|G|.$ That is, there are $t(>\frac{1}{4}|G|)$ elements $x_1,x_2,...,x_t$ s.t. $xx_i\notin I$ since neither of them commutes with $x$, contradicting $|I|>\frac{3}{4}|G|.$ Hence, $I$ must be an Abelian subgroup of $G$, and should coincide with $G$ itself.
\end{document}
http://yun.baidu.com/share/link?shareid=1831040314&uk=2148406042
同时还有一个简短的教程：
http://yun.baidu.com/share/link?shareid=1854151229&uk=2148406042
软件安装了之后会有Winedit这个程序的图标（一支笔和一张纸）出现，点开这个图标，新建文件之后就可以直接写论文代码。只有不到300M，因为没有中文包（自己吐槽：那为什么叫Ctex？！）和其它许多功能（比如我发现beamer功能似乎就有问题），但是我想对于数学和物理的论文排版是没有问题的。
为了方便同学们使用，我自己贴一段代码（这些内容都可以在教程中找到），内容是我上个学期的某些抽象代数习题的答案（之前闲来无事把解答写成了代码），其中定义了环境等等基本的要素。
% !Mode::"TeX:UTF-8"
\documentclass{article}
\usepackage{amssymb}
\begin{document}
\title{2013 Abstract Algebra Extra Exercises}
\maketitle
\section{Week 2}
5. The example is given by
$$G=\left\{\left(\begin{array}{cccc} 1 & a & b\\ 0 & 1 & c \\ 0 & 0 & 1\end{array}\right):a,b,c\in\mathbb{Z}_p\right\}.$$
\section{Week 4}
1. A theorem on the number of fixed points:
Suppose that a finite group $G$ acts transitively on a finite set $X$, whose cardinal number equals $n\geq2$. Let $G_0$ be the set of elements of $G$ with no fixed points. Then the ration $\frac{|G_0|}{|G|}\geq\frac{1}{n}.$ Moreover, if $n$ is not a power of prime, the inequality becomes strict.
\emph{Proof.} Let $N(g)$ be the number of fixed points of $g\in G$. Then by Burnside's Lemma, we have the following inequality:
$$\frac{1}{|G|}\sum_{g\in G} N(g)=1,$$
which is due to the transitivity of the action. We claim another inequality:
$$\frac{1}{|G|}\sum_{g\in G}N^2(g)\geq2.$$
In fact, the action of $G$ induces an action on the product set $X\times X.$ The number $N^2(g)$ is hence the number of $X\times X$ fixed by $g$(since, for example, if $i_1,i_2,...,i_k\in X$ are fixed by $g$, then the points $(i_u,i_v)\in X\times X$ are fixed by $g$, too.). Also, number of orbits of $G$ in $X\times X$ must exceed 2: by decomposing of $X\times X$ into the diagonal and its complement it will be suffice to verify this assertion(if the action is 2-transitive, then $\sum_{g\in G}N^2(g)=2|G|,$ which is a standard result). By Burnside's Lemma, the number of orbits on $X\times X$ equals
$$\frac{1}{|G|}\sum_{g\in G}N^2(g),$$
and the inequality is thus proved. We also obtain that the equality holds iff $G$ acts doubly transitively.
Next, we observe that $g\in G_0$ is equivalent to $N(g)=0$. Also, we have $1\leq N(g)\leq n$. Thus, $g\in G\setminus G_0$ iff $(N(g)-1)(N(g)-n)\leq0$, and if $g\in G$, $(N(g)-1)(N(g)-n)=n.$ Thus,
$$\sum_{g\in G\setminus G_0}(N(g)-1)(N(g)-n)\leq0,$$
and so we have
$$\frac{1}{|G|}\sum_{g\in G}(N(g)-1)(N(g)-n)\leq\frac{1}{|G|}\sum_{g\in G_0}(N(g)-1)(N(g)-n)=\frac{n|G_0|}{|G|}.$$
By the preceding inequality, we assert that $\frac{1}{|G|}\sum_{g\in G}N^2(g)\geq2$, and hence
$$\frac{1}{|G|}\sum_{g\in G}(N(g)-1)(N(g)-n)=\frac{1}{|G|}\sum_{g\in G}N^2(g)-(n+1)N(g)+n\geq2-(n+1)+n=1.$$
Thus, $1\leq\frac{n|G_0|}{|G|}.$
Now let $G$ act on the collection of all the cosets $G/H=\{g_iH:\bigcup_{i\in I}\},$ where the rules are given by $g\circ g_iH=gg_iH.$ Obviously, $G$ acts on $G/H$ transitively, and the cardinal number of the set $G/H$ equals $\frac{|G|}{|H|}$. Thus, we can apply the proved theorem. We see that $g_iH$ is a fixed point of $g$ iff $gg_iH=g_iH,$ i.e., $g\in g_iHg_i^{-1}$. Conversely, if $g=g_ihg_i^{-1},$ then $g$ must fix $g_iH.$ Thus, The set of elements with at least one fixed points coincide with the union of conjugate class $\bigcup_{g\in G}gHg^{-1}$; the set with no fixed points is hence $G-\bigcup_{g\in G}gHg^{-1}$. Apply the preceding theorem, we obtain
$$\frac{\left|G-\bigcup_{g\in G}gHg^{-1}\right|}{|G|}\geq\frac{1}{|G/H|}=\frac{|H|}{|G|}.$$
Hence,
$$\left|G-\bigcup_{g\in G}gHg^{-1}\right|\geq|H|.$$
2.
\emph{Proof.} We consider under what cases the set $I$ becomes a subgroup of $G$. Obviously, for any $g\in I$, we have $e=\varphi(e)=\varphi(gg^{-1})=g^{-1}\phi(g^{-1}).$ Thus, $\varphi(g^{-1})=g,$ and the set $I$ is hence closed under the operation of taking inverse. If we hope $I$ to be a subgroup, the for any $x,y\in I$, we must have $xy\in I$, i.e., $\varphi(xy)=\varphi(x)\varphi(y)=x^{-1}y^{-1}=y^{-1}x^{-1}\in I\Longleftrightarrow xy=yx$. Hence, $I$ is a subgroup iff the elements commute with each other. We claim that under the assumption $|I|>\frac{3}{4}|G|,$ the elements of $I$ must commute with each other. Otherwise, there exists $x,y$ s.t. $xy\neq yx$. Thus, the number of elements of $C(x)$ cannot exceed a half of $|G|$ since $C(x)$ is a proper subgroup of $G$. Let the number of elements of $I$ which do not commute with $x$ be $n$. Then we see that $|C(x)\bigcap I|\leq\frac{1}{2}|G|,$ and hence $n=|I-C(x)|>\frac{1}{4}|G|.$ That is, there are $t(>\frac{1}{4}|G|)$ elements $x_1,x_2,...,x_t$ s.t. $xx_i\notin I$ since neither of them commutes with $x$, contradicting $|I|>\frac{3}{4}|G|.$ Hence, $I$ must be an Abelian subgroup of $G$, and should coincide with $G$ itself.
\end{document}
