数学代写|抽象代数作业代写abstract algebra代考|MATH1014

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|抽象代数作业代写abstract algebra代考|Dihedral Symmetries

Let $n \geq 3$ and consider a regular $n$-sided polygon, $P_{n}$. Call $V=$ $\left{v_{1}, v_{2}, \ldots, v_{n}\right}$ the set of vertices of $P_{n}$ as a subset of the Euclidean plane $\mathbb{R}^{2}$. For simplicity, we often imagine the center of $P_{n}$ at the origin and that the vertex $v_{1}$ on the positive $x$-axis.

A symmetry of a regular $n$-gon is a bijection $\sigma: V \rightarrow V$ that is the restriction of a bijection $F: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$, that leaves the overall vertex-edge structure of $P_{n}$ in place; i.e., if the unordered pair $\left{v_{i}, v_{j}\right}$ are the end points of an edge of the regular $n$-gon, then $\left{\sigma\left(v_{i}\right), \sigma\left(v_{j}\right)\right}$ is also an edge.

Consider, for example, a regular hexagon $P_{6}$ and the bijection $\sigma: V \rightarrow V$ such that $\sigma\left(v_{1}\right)=v_{2}, \sigma\left(v_{2}\right)=v_{1}$, and $\sigma$ stays fixed on all the other vertices. Then $\sigma$ is not a symmetry of $P_{6}$ because it fails to preserve the vertex-edge structure of the hexagon. As we see in Figure $1.1$, though $\left{v_{2}, v_{3}\right}$ is an edge of the hexagon, while $\left{\sigma\left(v_{2}\right), \sigma\left(v_{3}\right)\right}$ are not the endpoints of an edge of the hexagon.

To count the number of bijections on the set $V=\left{v_{1}, v_{2}, \ldots, v_{n}\right}$, we note that a bijection $f: V \rightarrow V$ can map $f\left(v_{1}\right)$ to any element in $V$; then it can map $f\left(v_{2}\right)$ to any element in $V \backslash\left{f\left(v_{1}\right)\right}$; then it can map $f\left(v_{3}\right)$ to any element in $V \backslash\left{f\left(v_{1}\right), f\left(v_{2}\right)\right}$; and so on. Hence, there are
$$n \times(n-1) \times(n-2) \times \cdots \times 2 \times 1=n !$$
distinct bijections on $V$.
However, a symmetry $\sigma \in D_{n}$ can map $\sigma\left(v_{1}\right)$ to any element in $V(n$ options), but then $\sigma$ must map $v_{2}$ to a vertex adjacent to $\sigma\left(v_{1}\right)$ (2 options). Once $\sigma\left(v_{1}\right)$ and $\sigma\left(v_{2}\right)$ are known, all remaining $\sigma\left(v_{i}\right)$ for $3 \leq i \leq n$ are determined. In particular, $\sigma\left(v_{3}\right)$ must be the vertex adjacent to $\sigma\left(v_{2}\right)$ that is not $\sigma\left(v_{1}\right) ; \sigma\left(v_{4}\right)$ must be the vertex adjacent to $\sigma\left(v_{3}\right)$ that is not $\sigma\left(v_{2}\right)$; and so on. This reasoning leads to the following proposition.

数学代写|抽象代数作业代写abstract algebra代考|Abstract Notation

We introduce a notation that is briefer and aligns with the abstract notation that we will regularly use in group theory.

Having fixed an integer $n \geq 3$, denote by $r$ the rotation of angle $2 \pi / n$, by $s$ the reflection through the $x$-axis, and by $\iota$ the identity function. In other words,
$$r=R_{2 \pi / n}, \quad s=F_{0}, \quad \text { and } \quad \iota=R_{0} \text {. }$$
In abstract notation, similar to our habit of notation for multiplication of real variables, we write $a b$ to mean $a \circ b$ for two elements $a, b \in D_{n}$. Borrowing from a theorem in the next section (Proposition 1.2.13), since $\circ$ is associative, an expression such as $r r s r$ is well-defined, regardless of the order in which we pair terms to perform the composition. In this example, with $n=4$,
$$r r s r=R_{\pi / 2} \circ R_{\pi / 2} \circ F_{0} \circ R_{\pi / 2}=R_{\pi} \circ F_{0} \circ R_{\pi / 2}=F_{\pi / 2} \circ R_{\pi / 2}=F_{\pi / 4} .$$
To simplify notations, if $a \in D_{n}$ and $k \in \mathbb{N}^{*}$, then we write $a^{k}$ to represent
$$a^{k}=\overbrace{a a a \cdots a}^{k \text { times }} .$$
Hence, we write $r^{2} s r$ for $r r s r$. Since composition o is not commutative, $r^{3} s$ is not necessarily equal to $r^{2} s r$.
From Proposition 1.1.3, it is not hard to see that
$$r^{k}=R_{2 \pi k / n} \quad \text { and } \quad r^{k} s=F_{\pi k / n}$$
where $k$ satisfies $0 \leq k \leq n-1$. Consequently, as a set
$$D_{n}=\left{\iota, r, r^{2}, \ldots, r^{n-1}, s, r s, r^{2} s, \ldots, r^{n-1} s\right}$$
The symbols $r$ and $s$ have a few interesting properties. First, $r^{n}=\iota$ and $s^{2}=\iota$. These are obvious as long as we do not forget the geometric meaning of the functions $r$ and $s$. Less obvious is the equality in the following proposition.

数学代写|抽象代数作业代写abstract algebra代考|Dihedral Symmetries

$$n \times(n-1) \times(n-2) \times \cdots \times 2 \times 1=n !$$

数学代写|抽象代数作业代写abstract algebra代考|Abstract Notation

$$r=R_{2 \pi / n}, \quad s=F_{0}, \quad \text { and } \quad \iota=R_{0} .$$

$$r r s r=R_{\pi / 2} \circ R_{\pi / 2} \circ F_{0} \circ R_{\pi / 2}=R_{\pi} \circ F_{0} \circ R_{\pi / 2}=F_{\pi / 2} \circ R_{\pi / 2}=F_{\pi / 4} .$$

$$a^{k}=\overbrace{a a a \cdots a}^{k \text { times }} .$$

$$r^{k}=R_{2 \pi k / n} \quad \text { and } \quad r^{k} s=F_{\pi k / n}$$

D_{n}=\left{\iota, $r, r^{\wedge}{2}$, Vdots, $r^{\wedge}{n-1}, s, r s, r^{\wedge}{2} s, \backslash$ dots, $r^{\wedge}{n-1} s \backslash$ 右 $}$

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