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# 数学代写|现代代数代考Modern Algebra代写|ISOMORPHISM THEOREMS AND SOLVABLE GROUPS

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## 数学代写|现代代数代考Modern Algebra代写|ISOMORPHISM THEOREMS AND SOLVABLE GROUPS

The following two isomorphism theorems are fundamental in the study of groups.
Theorem 54.1 (First Isomorphism Theorem). Assume that $H$ and $K$ are subgroups of $a$ group $G$, with $K \triangleleft G$. Then $H K$ is a subgroup of $G, K \triangleleft H K$, and
$$\frac{H}{H \cap K} \approx \frac{H K}{K}$$
PROOF. We shall use Theorem 7.1 to verify that $H K$ is a subgroup. [If neither $H$ nor $K$ is normal in $G$, then $H K$ might not be a subgroup. For example, $\left\langle\left(\begin{array}{ll}1 & 2\end{array}\right)\right\rangle\left\langle\left(\begin{array}{ll}1 & 3\end{array}\right)\right\rangle$ is not a subgroup of $S_3$.]

Assume $h, h_1, h_2 \in H$ and $k, k_1, k_2 \in K$. Clearly $e \in H K$. And $h_2^{-1} k_1 h_2 \in K$ because $K \triangleleft G$, so
$$h_1 k_1 h_2 k_2=h_1 h_2\left(h_2^{-1} k_1 h_2\right) k_2 \in H K$$
Finally, if $h k \in H K$, then
$$(h k)^{-1}=k^{-1} h^{-1}=h^{-1} h k^{-1} h^{-1}=h^{-1}\left(h k^{-1} h^{-1}\right) \in H K$$
because $K \triangleleft G$. Thus $H K$ is a subgroup.
Obviously, $K \triangleleft H K$ because $K \triangleleft G$, so $H K / K$ is a group. Define $\theta: H \rightarrow H K / K$ by $\theta(h)=h K$ for each $h \in H$. Problem 54.1 asks you to verify that $\theta$ is a homomorphism of $H$ onto $H K / K$, and that $\operatorname{Ker} \theta=H \cap K$. The theorem now follows from the Fundamental Homomorphism Theorem.

## 数学代写|现代代数代考Modern Algebra代写|ALTERNATING GROUPS

In Section 7, an element of $S_n$ was defined to be even or odd depending on whether it can be written as a product of an even or an odd number of transpositions, respectively. However, it was not proved that the terms even and odd are well defined for permutations. Following is a proof. All permutations are assumed to be elements of $S_n$. Familiarity with Section 6 is assumed.

Each element of $S_n$ can be written as a product of disjoint cycles. And each $k$-cycle can be written as a product of $k-1$ transpositions:
$$\text { for } k \geq 2,\left(a_1 a_2 \ldots a_k\right)=\left(a_1 a_k\right) \ldots\left(a_1 a_3\right)\left(a_1 a_2\right) \text {. }$$
Let $N$ denote the mapping from $S_n$ to the set of non-negative integers defined by
\begin{aligned} & N((1))=0, \ & N\left(\left(a_1 a_2 \ldots a_k\right)\right)=k-1, \text { for } k \geq 2, \end{aligned}
and, if $\alpha$ is a product of $m$ mutually disjoint cycles $\alpha_1, \alpha_2, \ldots, \alpha_m$ of lengths $k_1, k_2, \ldots, k_m$, respectively, then
$$N(\alpha)=\sum_{i=1}^m N\left(\alpha_i\right)=\sum_{i=1}^m\left(k_i-1\right) .$$
We shall prove that in any factorization of $\alpha$ as a product of transpositions, the number of factors will be even iff $N(\alpha)$ is even; therefore, it also will be odd iff $N(\alpha)$ is odd.

We need the following two observations. (Remember that in this book permutations are composed from right to left.)
\begin{aligned} & \left(a c_1 \ldots c_r b d_1 \ldots d_s\right)(a b)=\left(\begin{array}{llll} a d_1 & \ldots & \left.d_s\right)\left(b c_1 \ldots c_r\right. \end{array}\right) \ & \left(a c_1 \ldots c_r\right)\left(b d_1 \ldots d_s\right)(a b)=\left(a d_1 \ldots d_s b c_1 \ldots c_r\right) \end{aligned}
Therefore, if $a$ and $b$ belong to the same cycle in the cyclic decompositions of $\alpha$, then $N(\alpha(a b))=N(\alpha)-1$, and if $a$ and $b$ belong to different cycles in the cyclic decomposition of $\alpha$, then $N(\alpha(a b))=N(\alpha)+1$. In either case,
$$N(\alpha(a b)) \equiv N(\alpha)+1(\bmod 2) .$$

# 现代代数代写

## 数学代写|现代代数代考Modern Algebra代写|ISOMORPHISM THEOREMS AND SOLVABLE GROUPS

$$\frac{H}{H \cap K} \approx \frac{H K}{K}$$

$$h_1 k_1 h_2 k_2=h_1 h_2\left(h_2^{-1} k_1 h_2\right) k_2 \in H K$$

$$(h k)^{-1}=k^{-1} h^{-1}=h^{-1} h k^{-1} h^{-1}=h^{-1}\left(h k^{-1} h^{-1}\right) \in H K$$

## 数学代写|现代代数代考Modern Algebra代写|ALTERNATING GROUPS

$S_n$的每个元素都可以写成不相交循环的乘积。每个$k$ -循环可以写成$k-1$换位的乘积:
$$\text { for } k \geq 2,\left(a_1 a_2 \ldots a_k\right)=\left(a_1 a_k\right) \ldots\left(a_1 a_3\right)\left(a_1 a_2\right) \text {. }$$

\begin{aligned} & N((1))=0, \ & N\left(\left(a_1 a_2 \ldots a_k\right)\right)=k-1, \text { for } k \geq 2, \end{aligned}

$$N(\alpha)=\sum_{i=1}^m N\left(\alpha_i\right)=\sum_{i=1}^m\left(k_i-1\right) .$$

\begin{aligned} & \left(a c_1 \ldots c_r b d_1 \ldots d_s\right)(a b)=\left(\begin{array}{llll} a d_1 & \ldots & \left.d_s\right)\left(b c_1 \ldots c_r\right. \end{array}\right) \ & \left(a c_1 \ldots c_r\right)\left(b d_1 \ldots d_s\right)(a b)=\left(a d_1 \ldots d_s b c_1 \ldots c_r\right) \end{aligned}

$$N(\alpha(a b)) \equiv N(\alpha)+1(\bmod 2) .$$

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。