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

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

In this chapter, we will prove the single most important theorem in finite group theory-Lagrange’s Theorem. In his book on abstract algebra, I. N. Herstein likened it to the ABC’s for finite groups. But first we introduce a new and powerful tool for analyzing a group – the notion of a coset. This notion was invented by Galois in 1830, although the term was coined by G. A. Miller in 1910.
Definitions Coset of $\boldsymbol{H}$ in $G$
Let $G$ be a group and let $H$ be a nonempty subset of $G$. For any $a \in G$, the set ${a h \mid h \in H}$ is denoted by $a H$. Analogously, $\mathrm{Ha}={h a \mid h \in H}$ and $a H a^{-1}=\left{a h a^{-1} \mid h \in\right.$ $H}$. When $H$ is a subgroup of $G$, the set $a H$ is called the left coset of $H$ in $G$ containing a, whereas $H a$ is called the right coset of $H$ in $G$ containing $a$. In this case, the element $a$ is to denote the number of elements in the set $a H$, and $|H a|$ to denote the number of elements in $\mathrm{Ha}$.

• EXAMPLE 1 Let $G=S_3$ and $H={(1),(13)}$. Then the left cosets of $H$ in $G$ are:
(1) $H=H$,
(12) $H={(12),(12)(13)}={(12),(132)}=(132) H$
(13) $H={(13),(1)}=H$
(23) $H={(23),(23)(13)}={(23),(123)}=(123) H$
• EXAMPLE 2 Let $\mathcal{K}=\left{R_0, R_{180}\right}$, in $D_4$, the dihedral group of order 8. Then,
\begin{aligned} R_0 \mathcal{K} &=\mathcal{K}, \ R_{90} \mathcal{K} &=\left{R_{90}, R_{270}\right}=R_{270} \mathcal{K}, \ R_{180} \mathcal{K} &=\left{R_{180}, R_0\right}=\mathcal{K}, \ V \mathcal{K} &={V, H}=H \mathcal{K}, \ D \mathcal{K} &=\left{D, D^{\prime}\right}=D^{\prime} \mathcal{K} . \end{aligned}
• EXAMPLE 3 Let $H={0,3,6}$ in $Z_9$ under addition. In the case that the group operation is addition, we use the notation $a+H$ instead of $a H$. Then the cosets of $H$ in $Z_9$ are
\begin{aligned} &0+H={0,3,6}=3+H=6+H \ &1+H={1,4,7}=4+H=7+H \ &2+H={2,5,8}=5+H=8+H \end{aligned}

The three preceding examples illustrate a few facts about cosets that are worthy of our attention. First, cosets are usually not subgroups. Second, $a H$ may be the same as $b H$, even though $a$ is not the same as $b$. Third, since in Example 1 (12) $H=$ ${(12),(132)}$ whereas $H(12)={(12),(123)}, a H$ need not be the same as $H a$.

These examples and observations raise many questions. When does $a H=b H$ ? Do $a H$ and $b H$ have any elements in common? When does $a H=H a$ ? Which cosets are subgroups? Why are cosets important? The next lemma and theorem answer these questions. (Analogous results hold for right cosets.)

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

Also, by property 1 of the lemma, $a \in a H$. Thus, each member of $G$ belongs to one of the cosets $a_i H$. In symbols,
$$G=a_1 H \cup \cdots \cup a_r H .$$
Now, property 5 of the lemma shows that this union is disjoint, so that
$$|G|=\left|a_1 H\right|+\left|a_2 H\right|+\cdots+\left|a_r H\right| .$$
Finally, since $\left|a_i H\right|=|H|$ for each $i$, we have $|G|=r|H|$.
We pause to emphasize that Lagrange’s Theorem is a subgroup candidate criterion; that is, it provides a list of candidates for the orders of the subgroups of a group. Thus, a group of order 12 may have subgroups of order $12,6,4,3,2,1$, but no others. Warning! The converse of Lagrange’s Theorem is false. For example, a group of order 12 need not have a subgroup of order 6 . We prove this in Example 5.

As we shall see in later chapters there are several theorems that guarantee the existence of subgroups of certain particular orders in finite groups.

A special name and notation have been adopted for the number of left (or right) cosets of a subgroup in a group. The index of a subgroup $H$ in $G$ is the number of distinct left cosets of $H$ in $G$. This number is denoted by $|G: H|$. As an immediate consequence of the proof of Lagrange’s Theorem, we have the following useful formula for the number of distinct left (or right) cosets of $H$ in $G$.

# 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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