群给我的直观印象

1.若一个元素在群内，则它的逆元也在群内.


2.若两个元素都在群内，则这两个元素以任意顺序互相作用后仍在群内.


3.若一个元素在子群外，另一个元素在子群内，则这两个元素互相作用在子群外.


4.若一个元素在子群外，则该元素的逆元也在子群外.

group of transformations

A group of transformations on set $ {S}$ is a set $ {G}$.$ {G}$ is a set of bijections from $ {S}$ to itself. And $ {G}$ is a group,which means that it satisfies three properties:

Identity:$ {\forall a\in G}$,$ { \exists e\in G}$ such that $ {a\circ e=e\circ a =a}$.($ {e}$ represents the identity mapping.)
Inverse:$ {\forall a\in G}$,$ {\exists a^{-1}\in G}$ such that $ {a^{-1}\circ a=a\circ a^{-1}=e}$.($ {a^{-1}}$ represents the inverse map of $ {a}$.)
combinative :$ {\forall a,b,c\in G}$,we have $ {(a\circ b)\circ c=a\circ (b\circ c)}$.(There is a prerequisite for this property,that is,$ {\forall a,b\in G,a\circ b\in G}$.The combinative property can be deduced from this prerequisite,so it is better to replace the combinative property by this prerequisite.But we have to be sure that if $ {a\circ b\in G}$,$ {a\circ b}$ should satisfy property 1 and 2,this is easy to verify. )

Now I want to talk about another point:

---

All the bijections from set $ {S}$ to $ {S}$ form a group of transformation.But a group of transformation is not necessarily consists of all the bijections from set $ {S}$ to $ {S}$.

---

The first statement of this point can be verified as follows:

1.It is easy to verify that if $ a$ is a bijection from $ S$ to $ S$,then $ a^{-1}$ is also a bijection from $ S$ to $ S$.

2.It is easy to verify that for all bijection $ a$ from $ S$ to $ S$,there exists an identity mapping $ e$ from $ S$ to $ S$ such that $ a\circ e=e\circ a=a$.

3.It is easy to verify that if $ a,b$ is two bijections from $ S$ to $ S$,then $ a\circ b$ is also a bijection from $ S$ to $ S$.And,$ a\circ b$ has an inverse $ b^{-1}\circ a^{-1}$,this inverse is also a bijection form $ S$ to $ S$ .And,$ (a\circ b )\circ e=e\circ (a\circ b)=a\circ b$.

As for the second statement,I show you some examples:

Example 1.G is the set of identity mapping of set $ {A}$.

Example 2.Let $ S$ be the set of all the real numbers,while the transformations considered to have the form $ f(x)=ax+b$.Consider the following cases,in some cases ,$ G$ is a group,while in the rest of the cases,$ G$ is not a group:

(1) $ G=\{f(x)|a=1,b$ is an odd number$ \}$.In this case,$ G$ is not a group,because if it is a group,it should have an identity element for every member in it.But for $ f(x)=x+1$,its indentity is $ g(x)=x+0$,but 0 is not an odd number.

(2)$ G=\{f(x)|a=1,b$is a positive integer or 0$ \}$.In this case,$ G$ is not a group,because it does't have an inverse for every member in it.For example,$ f(x)=x+3$.Its inverse is $ g(x)=x-3$.But -3 is a negative number.

(3)$ G=\{f(x)|a=1,b$is an even number.$ \}$.In this case,$ G$ is a group,because first,all the elements in $ G$ are bijections from $ S$ to $ S$.And,every element in $ G$ has an inverse which is also in $ G$,for example,the inverse of $ f(x)=x+2$ is $ g(x)=x-2$,-2 is an even number.And,every element in $ G$ has an indentity element which is also in $ G$ ,as 0 is an even number.And, it is easy to verify that the composition of any two elements in $ G$ is also in $ G$,as the sum of two even numbers is also an even number.

《几何与代数导引》例2.9

求以直线$x=y=z$为轴，过直线$2x=3y=-5z$的圆锥面方程.

解：

两条直线显然相交于原点.设圆锥面上的任意一点为$(x,y,z)$.我们知道直线
$2x=3y=-5z$的方向向量为$(15,10,-6)$.则直线$x=y=z$的方向向量为
$(1,1,1)$.我们知道
\begin{equation}
\cos \langle (1,1,1), (15,10,-6)\rangle=\frac{15+10-6}{\sqrt{3}\sqrt{15^2+10^2+6^2}}=\frac{1}{\sqrt{3}}
\end{equation}
则
\begin{equation}
  \frac{x+y+z}{\sqrt{3}\sqrt{x^2+y^2+z^2}}=\frac{1}{\sqrt{3}}
\end{equation}
即$xy+yz+zx=0$即为该圆锥面的方程.

《几何与代数导引》例2.8

一条直线$l_1$绕另一条直线$l_2$旋转所得的旋转面的分类讨论：
1.若直线$l_1$与直线$l_2$重合，则旋转面是一条直线$l_1$.
2.若直线$l_1$与直线$l_2$不重合，则$l_1$和$l_2$之间必有公垂线.设$l_1$的
方程为
\begin{equation}
\label{eq:1}
\begin{cases}
  x=y=0\\
z\in\bf{R}\\
\end{cases}
\end{equation}
$l_2$的方程我们再分类讨论如下：

2.1 若$l_2$的方程为
\begin{equation}
\label{eq:2}
\begin{cases}
  x=a\\
y=0\\
z\in\bf{R}
\end{cases}
\end{equation}
其中$a\in\bf{R}$.则易得旋转得到圆柱面
\begin{equation}
\label{eq:3}
\begin{cases}
x^2+y^2=a^2\\
z\in\bf{R}\\
\end{cases}
\end{equation}
2.2若$l_2$的方程为
\begin{equation}
\label{eq:4}
\begin{cases}
  x=a\\
z=ky\\
\end{cases}
\end{equation}
其中$k\in\bf{R}$.则对于旋转面上任意一点来说$(x,y,z)$，都存在该旋转面上
的相应的点$(x_0,y_0,z_0)$.使得
\begin{equation}
\label{eq:5}
x_0^2+y_0^2+z_0^2=x^2+y^2+z^2
\end{equation}
且
\begin{equation}
\label{eq:6}
z=z_0
\end{equation}

且
\begin{equation}
\label{eq:7}
\begin{cases}
  x_0=a\\
z_0=ky_0\\
\end{cases}
\end{equation}
于是我们得
\begin{equation}
\label{eq:8}
a^2+y_0^2=x^2+y^2
\end{equation}
2.1.1当$k=0$时，我们可得$y_0$可取任意值，此时旋转面的方程为
\begin{equation}
\label{eq:9}
x^2+y^2\geq a^2
\end{equation}
2.1.2当$k\neq 0$时，我们得
旋转面的方程为
\begin{equation}
\label{eq:10}
x^2+y^2=a^2+(\frac{z}{k})^2
\end{equation}
2.1.2.1当$a\neq 0$时，即
\begin{equation}
\label{eq:11}
\frac{x^2+y^2}{a^2}-\frac{z^2}{k^2a^2}=1
\end{equation}
可见是旋转单叶双曲面.
2.1.2.2当$a=0$时，
\begin{equation}
\label{eq:12}
x^2+y^2=(\frac{z^2}{k^2})
\end{equation}
此时，是圆锥面.

《几何与代数导引》例2.7.4

求$yz$面上二次曲线
\begin{equation}
  \begin{cases}
    \frac{y^2}{a^2}=2z\\
x=0\\
  \end{cases}
\end{equation}
绕$z$轴旋转所得的二次曲面的方程.


解：对于二次曲面上的任意点$p=(x,y,z)$.都存在相应的二次曲面上的点
$(x_0,y_0,z_0)$,使得
\begin{equation}
  (x-x_0,y-y_0,z-z_0)\cdot (0,0,1)=0
\end{equation}
且
\begin{equation}
  x^2+y^2+z^2=x_0^2+y_0^2+z_0^2
\end{equation}
且
\begin{equation}
  \begin{cases}
    \frac{y_0^2}{a^2}=2z_{0}\\
x_0=0\\
z_0\geq 0\\
  \end{cases}
\end{equation}
可得
\begin{equation}
x^2+y^2=2za^2
\end{equation}

《几何与代数导引》例2.7.3

求$yz$面上二次曲线
\begin{equation}
  \begin{cases}
    \frac{z^2}{c^2}-\frac{y^2}{a^2}=1\\
x=0\\
  \end{cases}
\end{equation}
绕$z$轴旋转所得的二次曲面的方程.


解：对于二次曲面上的任意点$p=(x,y,z)$.都存在相应的二次曲面上的点
$(x_0,y_0,z_0)$,使得
\begin{equation}
  (x-x_0,y-y_0,z-z_0)\cdot (0,0,1)=0
\end{equation}
且
\begin{equation}
  x^2+y^2+z^2=x_0^2+y_0^2+z_0^2
\end{equation}
且
\begin{equation}
  \begin{cases}
    \frac{z_0^2}{c^2}-\frac{y_0^2}{a^2}=1\\
x_0=0\\
y_0\geq 0\\
  \end{cases}
\end{equation}
可得
\begin{equation}
  \frac{x^2}{a^2}+\frac{y^2}{a^2}-\frac{z^2}{c^2}=-1
\end{equation}

《几何与代数导引》例2.7.2

求$yz$面上二次曲线
\begin{equation}
  \begin{cases}
    \frac{y^2}{a^2}-\frac{z^2}{c^2}=1\\
x=0\\
  \end{cases}
\end{equation}
绕$z$轴旋转所得的二次曲面的方程.


解：对于二次曲面上的任意点$p=(x,y,z)$.都存在相应的二次曲面上的点
$(x_0,y_0,z_0)$,使得
\begin{equation}
  (x-x_0,y-y_0,z-z_0)\cdot (0,0,1)=0
\end{equation}
且
\begin{equation}
  x^2+y^2+z^2=x_0^2+y_0^2+z_0^2
\end{equation}
且
\begin{equation}
  \begin{cases}
    \frac{y_0^2}{a^2}-\frac{z_0^2}{c^2}=1\\
x_0=0\\
y_0\geq 0\\
  \end{cases}
\end{equation}
可得
\begin{equation}
  \frac{x^2}{a^2}+\frac{y^2}{a^2}-\frac{z^2}{c^2}=1
\end{equation}