# 数学代写|抽象代数作业代写abstract algebra代考|MATH355

## 数学代写|抽象代数作业代写abstract algebra代考|The Rotation Group of a Cube and a Soccer Ball

It cannot be overemphasized that Theorem $7.4$ and Lagrange’s Theorem (Theorem 7.1) are counting theorems. ${ }^3$ They enable us to determine the numbers of elements in various sets. To see how Theorem $7.4$ works, we will determine the order of the rotation group of a cube and a soccer ball. That is, we wish to find the number of essentially different ways in which we can take a cube or a soccer ball in a certain location in space, physically rotate it, and then have it still occupy its original location.

• EXAMPLE 10 Let $G$ be the rotation group of a cube. Label the six faces of the cube 1 through 6 . Since any rotation of the cube must carry each face of the cube to exactly one other face of the cube and different rotations induce different permutations of the faces, $G$ can be viewed as a group of permutations on the set ${1,2,3,4,5,6}$. Clearly, there is some rotation about a central horizontal or vertical axis that carries face number 1 to any other face, so that $\left|\operatorname{orb}_G(1)\right|=6$. Next, we consider $\operatorname{stab}_G(1)$. Here, we are asking for all rotations of a cube that leave face number 1 where it is. Surely, there are only four such motions – rotations of $0^{\circ}, 90^{\circ}, 180^{\circ}$ and $270^{\circ}$ – about the line perpendicular to the face $0^{\circ}, 90^{\circ}, 180^{\circ}$, and $270^{\circ}$ – about the line perpendicular to the face and passing through its center (see Figure $7.2$ ). Thus, by Theorem and passing through its center (see Figure $7.2$ ). Thus, by Theorem $7.4,|G|=\left|\operatorname{orb}_G(1)\right|\left|\operatorname{stab}_G(1)\right|=6 \cdot 4=24$.
Now that we know how many rotations a cube has, it is simple to determine the actual structure of the rotation group of a cube. Recall that $S_4$ is the symmetric group of degree 4 .

## 数学代写|抽象代数作业代写abstract algebra代考|An Application of Cosets to the Rubik’s Cube

Recall from Chapter 5 that in 2010 it was proved via a computer computation, which took 35 CPU-years to complete, that every Rubik’s cube could be solved in at most 20 moves. To carry out this effort, the research team of Morley Davidson John Dethridge, Herbert Kociemba, and Tomas Rokicki applied a program of Rokicki, which built on early work of Kociemba, that checked the elements of the cosets of a subgroup $H$ of order $(8 ! \cdot 8 ! \cdot 4 !) / 2=$ $19,508,428,800$ to see if each cube in a position corresponding to the elements in a coset could be solved within 20 moves. In the rare cases where Rokicki’s program did not work, an alternate method was employed. Using symmetry considerations, they were able to reduce the approximately 2 billion cosets of $H$ to about 56 million cosets for testing. Cosets played a role in this effort because Rokicki’s program could handle the $19.5+$ billion elements in the same coset in about 20 seconds.

In this chapter, we show how to piece together groups to make larger groups. In Chapter 9, we will show that we can often start with one large group and decompose it into a product of smaller groups in much the same way as a composite positive integer can be broken down into a product of primes. These methods will later be used to give us a simple way to construct all finite Abelian groups.

# 抽象代数代考

## 数学代写|抽象代数作业代写abstract algebra代考|Properties of Cosets

• $H$ 在 $G$
让 $G$ 成为一个群体，让 $H$ 是的非空子集 $G$. 对于任何 $a \in G$ ，集合 $a h \mid h \in H$ 表示为 $a H$. 类似地， 称为左陪集 $H$ 在 $G$ 包含一个，而 $H a$ 被称为右陪集 $H$ 在 $G$ 包含 $a$. 在这种情况下，元素 $a$ 是表示集合中的元 嗉个数 $a H ，$ 和 $|H a|$ 来表示元㨞的数量 Ha.
• 示例 1 让 $G=S_3$ 和 $H=(1),(13)$. 那么左陪集 $H$ 在 $G$ 是:
(1) $H=H_{\text {, }}$
(12) $H=(12),(12)(13)=(12),(132)=(132) H$
(13) $H=(13),(1)=H$
(23) $H=(23),(23)(13)=(23),(123)=(123) H$
$\backslash$ begin ${$ aligned $} R_{-} 0 \backslash$ mathcal ${K} \&=\backslash$ mathcal ${K}$, $\backslash R_{-}{90} \backslash$ mathcal ${K} \& L \backslash \backslash$ left $\left{R_{-}{90}, R_{-}{270} \backslash\right.$ right $}=R_{-}{270} \backslash$ mathcal ${K}$, $\backslash R_{-}{18$
• 例 3 让 $H=0,3,6$ 在 $Z_9$ 补充下。在组运算是加法的情况下，我们使用符号 $a+H$ 代替 $a H$. 然后的陪 集 $H$ 在 $Z_9$ 是
$$0+H=0,3,6=3+H=6+H \quad 1+H=1,4,7=4+H=7+H 2+H=2,5,8=5+H=8$$
前面的三个例子说明了一些值得我们关注的关于陪集的事实。首先，陪集通常不是子群。第二， $a H$ 可能 与 $b H$ ，虽然 $a$ 不一样 $b$. 三、由于在例一 (12) $H=(12),(132)$ 然而 $H(12)=(12),(123), a H$ 不必相 同 $H a$.
这些例子和观察提出了许㝖问题。什么时候 $a H=b H ?$ ? 做 $a H$ 和 $b H$ 有什么共同点吗? 什么时候 $a H=H a$ ? 哪些陪集是子群? 为什么陪护很重要? 下一个引理和定理回答了这些问题。（类似的结果适 用于右陪集。)

## 数学代写|抽象代数作业代写abstract algebra代考|Lagrange’s Theorem and Consequences

$$G=a_1 H \cup \cdots \cup a_r H .$$

$$|G|=\left|a_1 H\right|+\left|a_2 H\right|+\cdots+\left|a_r H\right| .$$

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