### 数学代写|抽象代数作业代写abstract algebra代考|MATH330

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• 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|Notation for Arbitrary Groups

In group theory, we will regularly discuss the properties of an arbitrary group. In this case, instead of writing the operation as $a * b$, where * represents some unspecified binary operation, it is common to write the generic group operation as $a b$. With this convention of notation, it is also common to indicate the identity in an arbitrary group as 1 instead of $e$. In this chapter, however, we will continue to write $e$ for the arbitrary group identity in order to avoid confusion. Finally, with arbitrary groups, we denote the inverse of an element $a$ as $a^{-1} .$

This shorthand of notation should not surprise us too much. We already developed a similar habit with vector spaces. When discussing an arbitrary vector space, we regularly say, “Let $V$ be a vector space.” So though, in a strict sense, $V$ is only the set of the vector space, we implicitly understand that part of the information of a vector space is the addition of vectors (some operation usually denoted $+$ ) and the scalar multiplication of vectors.

By a similar abuse of language, we often refer, for example, to “the dihedral group $D_{n}$,” as opposed to “the dihedral group $\left(D_{n}, \circ\right)$.” Similarly, when we talk about “the group $\mathbb{Z} / n \mathbb{Z}$,” we mean $(\mathbb{Z} / n \mathbb{Z},+)$ because $(\mathbb{Z} / n \mathbb{Z}, \times)$ is not a group. And when we refer to “the group $U(n)$,” we mean the group $(U(n), \times)$. We will explicitly list the pair of set and binary operation if there could be confusion as to which binary operation the group refers. Furthermore, as we already saw with $D_{n}$, even if a group is equipped with a natural operation, we often just write $a b$ to indicate that operation. Following the analogy with multiplication, in a group $G$, if $a \in G$ and $k$ is a positive integer, by $a^{k}$ we mean
$$a^{k} \stackrel{\text { def }}{=} \overbrace{a a \cdots a}^{k \text { times }} .$$
We extend the power notation so that $a^{0}=e$ and $a^{-k}=\left(a^{-1}\right)^{k}$, for any positive integer $k$.

Groups that involve addition give an exception to the above habit of notation. In that case, we always write $a+b$ for the operation, $-a$ for the inverse, and, if $k$ is a positive integer,
$$k \cdot a \stackrel{\text { def }}{=} \overbrace{a+a+\cdots+a} .$$
We refer to $k \cdot a$ as a multiple of $a$ instead of as a power. Again, we extend the notation to nonpositive “multiples” just as above with powers.

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

The following proposition holds for any associative binary operation and does not require the other two axioms of group theory.

Proof. Before starting the proof, we define a temporary but useful notation. Given a sequence $a_{1}, a_{2}, \ldots, a_{k}$ of elements in $S$, by analogy with the $\sum$ notation, we define
$$\star_{i=1}^{k} a_{i} \stackrel{\text { def }}{=}\left(\cdots\left(\left(a_{1} \star a_{2}\right) \star a_{3}\right) \cdots a_{k-1}\right) \star a_{k}$$
In this notation, we perform the operations in (1.4) from left to right. Note that if $k=1$, the expression is equal to the element $a_{1}$.

We prove by (strong) induction on $n$, that every operation expression in $(1.4)$ is equal to $\boldsymbol{x}{i=1}^{n} a{i}$

The basis step with $n \geq 3$ is precisely the assumption that $\star$ is associative. We now assume that the proposition is true for all integers $k$ with $3 \leq$ $k \leq n$. Consider an operation expression (1.4) involving $n+1$ terms. Suppose without loss of generality that the last operation performed occurs between the $j$ th and $(j+1)$ th term, i.e.,

Since both operation expressions involve $n$ terms or less, by the induction hypothesis
$$q=\left(\star_{i=1}^{j} a_{i}\right) \star\left(\star_{i=j+1}^{n} a_{i}\right) .$$
Furthermore,
\begin{aligned} q &=\left(\star_{i=1}^{j} a_{i}\right) \star\left(a_{j+1} \star\left(\star_{i=j+2}^{n} a_{i}\right)\right) \quad \text { by the induction hypothesis } \ &=\left(\left(\star_{i=1}^{j} a_{i}\right) \star a_{j+1}\right) \star\left(\star_{i=j+2}^{n} a_{i}\right) \quad \text { by associativity } \ &=\left(\star_{i=1}^{j+1} a_{i}\right) \star\left(\star_{i=j+2}^{n} a_{i}\right) \end{aligned}
Repeating this $n-j-2$ more times, we conclude that
$$q=\star_{i=1}^{n+1} a_{i}$$
The proposition follows.

## 数学代写|抽象代数作业代写abstract algebra代考|Notation for Arbitrary Groups

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

$$k \cdot a \stackrel{\text { def }}{=} \overbrace{a+a+\cdots+a}$$

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

$$\star_{i=1}^{k} a_{i} \stackrel{\text { def }}{=}\left(\cdots\left(\left(a_{1} \star a_{2}\right) \star a_{3}\right) \cdots a_{k-1}\right) \star a_{k}$$

$$q=\left(\star_{i=1}^{j} a_{i}\right) \star\left(\star_{i=j+1}^{n} a_{i}\right) .$$

$q=\left(\star_{i=1}^{j} a_{i}\right) \star\left(a_{j+1} \star\left(\star_{i=j+2}^{n} a_{i}\right)\right) \quad$ by the induction hypothesis $\quad=\left(\left(\star_{i=1}^{j} a_{i}\right) \star a_{j+1}\right) \star\left(\star_{i}^{n}\right.$

$$q=\star_{i=1}^{n+1} a_{i}$$

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