Posted on Categories:abstract algebra, 抽象代数, 数学代写

# 数学代写|抽象代数代写Abstract Algebra代考|MATH4200 Element-coset commutativity

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|抽象代数代写Abstract Algebra代考|Element-coset commutativity

If you understand cyclic groups, the previous section might leave you unimpressed. With addition based on division with remainders, perhaps it is obvious that quotient groups work tidily. But that only makes the general question more interesting. Quotient groups arise not only in cyclic groups but also in dihedral groups, where the elements are not numbers but symmetries. Symmetries have neither division nor remainders, at least in the usual sense-could there be an analogue of division with remainders for dihedral groups? And, to reiterate the earlier question, does a quotient group arise for every subgroup of every group, or not?
We will approach these questions by considering the dihedral group $D_4$. The table in Section 7.1 (refer to your copy) shows that if $H=\left{e, \rho, \rho^2, \rho^3\right}$is the four-element subgroup of rotations, then $D_4 / H \cong \mathbb{Z}_2$. But $D_4$ has many subgroups, as discussed in Section 6.6 and represented below. Do other subgroups give rise to meaningful cosets and a quotient group? Does $H=\left{e, \rho^2\right}$ give rise to a quotient group in $D_4$, for instance? If so, how many elements would you expect that quotient group to have?

Calculating cosets is straightforward, but we should think carefully about those calculations because for cyclic groups I implicitly relied on experience of division with remainders. For instance, I calculated the three cosets of $3 \mathbb{Z}$ shown below, then stopped. Probably three seemed like the right number. But why, exactly? What would happen if we calculated $3+3 \mathbb{Z}, 4+3 \mathbb{Z}$, and so on?
\begin{aligned} & 0+3 \mathbb{Z}={0+z \mid z \in 3 \mathbb{Z}}={\ldots,-6,-3,0,3,6,9, \ldots} \ & 1+3 \mathbb{Z}={1+z \mid z \in 3 \mathbb{Z}}={\ldots,-5,-2,1,4,7,10, \ldots} \ & 2+3 \mathbb{Z}={2+z \mid z \in 3 \mathbb{Z}}={\ldots,-4,-1,2,5,8,11, \ldots} \end{aligned}

## 数学代写|抽象代数代写Abstract Algebra代考|Left and right cosets

Perhaps, then, a quotient group arises for every subgroup of every group. But we should not conclude too hastily. Everything worked for $H=$ $\left{e, \rho^2\right}$ in $D_4$ due to an element-coset form of commutativity. But, with respect to commutativity, some elements are less ‘well behaved’ than $\rho^2$. So it is worth checking similarly for different subgroups. Calculating cosets for $H=\left{e, r_1\right}$, for instance, gives:
\begin{aligned} e H & =\left{e e, e r_1\right}=\left{e, r_1\right} ; \ \rho H & =\left{\rho e, \rho r_1\right}=\left{\rho, r_3\right} ; \ \rho^2 H & =\left{\rho^2 e, \rho^2 r_1\right}=\left{\rho^2, r_2\right} ; \ \rho^3 H & =\left{\rho^3 e, \rho^3 r_1\right}=\left{\rho^3, r_4\right} ; \ r_1 H & =\left{r_1 e, r_1 r_1\right}=\left{r_1, e\right} ; \ r_2 H & =\left{r_2 e, r_2 r_1\right}=\left{r_2, \rho^2\right} ; \ r_3 H & =\left{r_3 e, r_3 r_1\right}=\left{r_3, \rho\right} ; \ r_4 H & =\left{r_4 e, r_4 r_1\right}=\left{r_4, \rho^3\right} . \end{aligned}
Again this yields four distinct cosets, each appearing twice. And again $H=\left{e, r_1\right}$ is a subgroup of $D_4$, so it is closed under composition and forms a first table block. This time, the next listed coset, $\left{\rho, r_3\right}$, gives a self-contained block below the subgroup-check that the partial table below is filled in correctly. However, to the right of the subgroup, things ‘go wrong. The top row reflects the fact that $e\left{\rho, r_3\right}=\left{\rho, r_3\right}$. But the lower row contains $r_1 \rho=r_4$, not $r_3$, and $r_1 r_3=\rho^3$, not $\rho$. In other words, $\rho\left{e, r_1\right} \neq\left{e, r_1\right} \rho$.

## 数学代写|抽象代数代写Abstract Algebra代考|Element-coset commutativity

$\backslash$ left 缺少或无法识别的分隔符 是旋转的四元素子群，那么 $D_4 / H \cong \mathbb{Z}_2$. 但 $D_4$ 有许多子 组，如第 6.6 节中所讨论和下面所示。其他子群是否产生有意义的陪集和商群? 做
【left 缺少或无法识别的分隔符 产生一个商群 $D_4$ ，例如? 如果是这样，您莃望商组有多少 个元莍?

$$0+3 \mathbb{Z}=0+z|z \in 3 \mathbb{Z}=\ldots,-6,-3,0,3,6,9, \ldots \quad 1+3 \mathbb{Z}=1+z| z \in 3 \mathbb{Z}=\ldots,-5,-2,1,4,7,10, \ldots 2+3 \mathbb{Z}=2+z \mid z \in 3$$

## 数学代写|抽象代数代写Abstract Algebra代考|Left and right cosets

\left 缺少或无法识别的分隔符 在 $D_4$ 由于交换性的元素陪集形式。但是，就交换性而言，某 些元责的“表现良好”不如 $\rho^2$. 因此值得对不同的子组进行类似的检查。计算陪集
left 缺少或无法识别的分隔符，，例如，给出:
\left 缺少或无法识别的分隔符

\left 缺少或无法识别的分隔符，在子组下面给出一个独立的块 – 检查下面的部分表格是否 填写正确。然而，对于子组的右侧，事情“出了问题”。第一行反映了这样一个事实
\left 缺少或无法识别的分隔符 $\quad$. 但下排包含 $r_1 \rho=r_4$ ，不是 $r_3 ＼mathrm{~ ， 和 ~} r_1 r_3=\rho^3$ ，不是 $\rho$. 换 句话说，\left 缺少或无法识别的分隔符

## 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。