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# 数学代写|现代代数代考Modern Algebra代写|MATH611

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## 数学代写|现代代数代考Modern Algebra代写|Kernel of a Homomorphism

For any homomorphism $\phi$ from the group $G$ to the group $G^{\prime}, \operatorname{ker} \phi$ is a normal subgroup of $G$.

Proof The identity $e$ is in $\operatorname{ker} \phi$ since $\phi(e)=e^{\prime}$, so $\operatorname{ker} \phi$ is always nonempty. If $a \in \operatorname{ker} \phi$ and $b \in \operatorname{ker} \phi$, then $\phi(a)=e^{\prime}$ and $\phi(b)=e^{\prime}$. Also, by Theorem 3.32, $\phi\left(b^{-1}\right)=[\phi(b)]^{-1}$, so
\begin{aligned} \phi\left(a b^{-1}\right) & =\phi(a) \phi\left(b^{-1}\right) \ & =\phi(a)[\phi(b)]^{-1} \ & =e^{\prime} \cdot\left(e^{\prime}\right)^{-1} \ & =e^{\prime}, \end{aligned}
and therefore $a b^{-1} \in \operatorname{ker} \phi$. Thus, by Theorem 3.12, $\operatorname{ker} \phi$ is a subgroup of $G$.
To show that $\operatorname{ker} \phi$ is normal, let $x \in G$ and $a \in \operatorname{ker} \phi$. Then
\begin{aligned} \phi\left(x a x^{-1}\right) & =\phi(x) \phi(a) \phi\left(x^{-1}\right) & & \text { since } \phi \text { is a homomorphism } \ & =\phi(x) \cdot e^{\prime} \cdot \phi\left(x^{-1}\right) & & \text { since } a \in \operatorname{ker} \phi \ & =\phi(x) \cdot \phi\left(x^{-1}\right) & & \ & =e^{\prime} & & \text { by part b of Theorem 3.32. } \end{aligned}
by part b of Theorem 3.32.
Thus $x a x^{-1}$ is in ker $\phi$, and ker $\phi$ is a normal subgroup by Theorem 4.20.
The mapping $\phi$ in Theorem 4.25 has $H$ as its kernel, and this shows that every normal subgroup of $G$ is the kernel of a homomorphism. Combining this fact with Theorem 4.26, we see that the normal subgroups of $G$ and the kernels of the homomorphisms from $G$ to another group are the same subgroups of $G$.

We can now prove that every homomorphic image of $G$ is isomorphic to a quotient group of $G$.

## 数学代写|现代代数代考Modern Algebra代写|Homomorphic Image $\Rightarrow$ Quotient Group

Let $G$ and $G^{\prime}$ be groups with $G^{\prime}$ a homomorphic image of $G$. Then $G^{\prime}$ is isomorphic to a quotient group of $G$.

Proof Let $\phi$ be an epimorphism from $G$ to $G^{\prime}$, and let $K=\operatorname{ker} \phi$. For each $a K$ in $G / K$, define $\theta(a K)$ by
$$\theta(a K)=\phi(a)$$
First, we need to prove that this rule defines a mapping. For any $a K$ and $b K$ in $G / K$,
\begin{aligned} a K=b K & \Leftrightarrow b^{-1} a K=K \ & \Leftrightarrow b^{-1} a \in K \ & \Leftrightarrow \phi\left(b^{-1} a\right)=e^{\prime} \ & \Leftrightarrow \phi\left(b^{-1}\right) \phi(a)=e^{\prime} \ & \Leftrightarrow[\phi(b)]^{-1} \phi(a)=e^{\prime} \ & \Leftrightarrow \phi(a)=\phi(b) \ & \Leftrightarrow \theta(a K)=\theta(b K) . \end{aligned}

Thus $\theta$ is a well-defined mapping from $G / K$ to $G^{\prime}$, and the $\Leftarrow$ parts of the $\Leftrightarrow$ statements show that $\theta$ is one-to-one as well.
We shall show that $\theta$ is an isomorphism from $G / K$ to $G^{\prime}$. Since
\begin{aligned} \theta(a K \cdot b K) & =\theta(a b K) \ & =\phi(a b) \ & =\phi(a) \cdot \phi(b) \ & =\theta(a K) \cdot \theta(b K), \end{aligned}
$\theta$ is a homomorphism. To show that $\theta$ is onto, let $a^{\prime}$ be arbitrary in $G^{\prime}$. Since $\phi$ is an epimorphism, there exists an element $a$ in $G$ such that $\phi(a)=a^{\prime}$. Then $a K$ is in $G / K$, and
$$\theta(a K)=\phi(a)=a^{\prime} .$$

# 现代代数代写

## 数学代写|现代代数代考Modern Algebra代写|Kernel of a Homomorphism

\begin{aligned} \phi\left(a b^{-1}\right) & =\phi(a) \phi\left(b^{-1}\right) \ & =\phi(a)[\phi(b)]^{-1} \ & =e^{\prime} \cdot\left(e^{\prime}\right)^{-1} \ & =e^{\prime}, \end{aligned}

\begin{aligned} \phi\left(x a x^{-1}\right) & =\phi(x) \phi(a) \phi\left(x^{-1}\right) & & \text { since } \phi \text { is a homomorphism } \ & =\phi(x) \cdot e^{\prime} \cdot \phi\left(x^{-1}\right) & & \text { since } a \in \operatorname{ker} \phi \ & =\phi(x) \cdot \phi\left(x^{-1}\right) & & \ & =e^{\prime} & & \text { by part b of Theorem 3.32. } \end{aligned}

## 数学代写|现代代数代考Modern Algebra代写|Homomorphic Image $\Rightarrow$ Quotient Group

$$\theta(a K)=\phi(a)$$

\begin{aligned} a K=b K & \Leftrightarrow b^{-1} a K=K \ & \Leftrightarrow b^{-1} a \in K \ & \Leftrightarrow \phi\left(b^{-1} a\right)=e^{\prime} \ & \Leftrightarrow \phi\left(b^{-1}\right) \phi(a)=e^{\prime} \ & \Leftrightarrow[\phi(b)]^{-1} \phi(a)=e^{\prime} \ & \Leftrightarrow \phi(a)=\phi(b) \ & \Leftrightarrow \theta(a K)=\theta(b K) . \end{aligned}

\begin{aligned} \theta(a K \cdot b K) & =\theta(a b K) \ & =\phi(a b) \ & =\phi(a) \cdot \phi(b) \ & =\theta(a K) \cdot \theta(b K), \end{aligned}
$\theta$是同态。为了表明$\theta$是对的，让$a^{\prime}$在$G^{\prime}$中是任意的。因为$\phi$是一个外胚，所以在$G$中存在一个元素$a$，使得$\phi(a)=a^{\prime}$。然后$a K$在$G / K$中，和
$$\theta(a K)=\phi(a)=a^{\prime} .$$

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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。