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# 数学代写|抽象代数代写Abstract Algebra代考|MATH393 Groups ℤ12 and ⟨𝑔⟩: Elements match up

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## 数学代写|抽象代数代写Abstract Algebra代考|Groups ℤ12 and ⟨𝑔⟩: Elements match up

Let’s begin by revisiting an earlier example. Recall that given a group element $g$, the cyclic subgroup $\langle g\rangle$ is defined by $\langle g\rangle=\left{g^k \mid k \in \mathbb{Z}\right}$, i.e., the set of all integer powers of g.

Example 13.16 Let $g$ be an element of a multiplicative group with ord $(g)=12$. By Theorem 13.17, the distinct elements of $\langle g\rangle$ are given by $\langle g\rangle=\left{\varepsilon, g^1, g^2, g^3, \ldots, g^{11}\right}$, where $\varepsilon=g^0$. We thus have the correspondence $k \leftrightarrow g^k$ between $\mathbb{Z}{12}$ and $\langle g\rangle$. Moreover, $\langle g\rangle$ behaves like $\mathbb{Z}{12}$; e.g.,
$$g^9 \cdot g^7=g^{9+7}=g^{16}=g^{12+4}=g^{12} \cdot g^4=\varepsilon \cdot g^4=g^4,$$
so that $g^9 \cdot g^7=g^4$, which is just like $9+7=4$ in $\mathbb{Z}{12}$. Thus, we’ve said that the groups $\mathbb{Z}{12}$ and $\langle g\rangle$ are essentially the same. To make this notion of sameness more precise, consider the function $\theta: \mathbb{Z}{12} \rightarrow\langle g\rangle$ where $\theta(k)=g^k$ for all $k \in \mathbb{Z}{12}$. For instance, $\theta(7)=g^7$.

Remark. The function $\theta$ associates to each $k \in \mathbb{Z}_{12}$ the element $g^k \in\langle g\rangle$. This is no different from the correspondence $k \leftrightarrow g^k$. But using the language of a function here provides a powerful tool that will allow us to derive and prove various features of groups that are essentially the same.

We will prove that $\theta$ is one-to-one and onto. To show that $\theta$ is one-to-one, we begin by assuming that $\theta(a)=\theta(b)$, where $a, b \in \mathbb{Z}_{12}$. Then we show that $a=b$.

## 数学代写|抽象代数代写Abstract Algebra代考|Groups ℤ12 and ⟨𝑔⟩: Operations match up

As in Section 16.1, let $g$ be a group element with ord $(g)=12$. Consider again the function $\theta: \mathbb{Z}{12} \rightarrow\langle g\rangle$ where $\theta(k)=g^k$ for all $k \in \mathbb{Z}{12}$. We saw that $\theta$ is a bijection; i.e., it’s one-to-one and onto. Therefore, $\theta$ allows the elements in $\mathbb{Z}_{12}$ and $\langle g\rangle$ to “match up” with each other.

We also recall from Example $13.16$ that the operations of $\mathbb{Z}{12}$ and $\langle g\rangle$ match up as well. For instance, since $12=0$ in $\mathbb{Z}{12}$ and $g^{12}=\varepsilon$ in $\langle g\rangle$, we have $9+7=4$ in $\mathbb{Z}{12}$, which is just like $g^9 \cdot g^7=g^{9+7}=g^4$ in $\langle g\rangle$. Addition in $\mathbb{Z}{12}$ feels like multiplication in $\langle g\rangle$, and the function $\theta$ can more precisely capture this intuition.

Using the law of exponents (i.e., $g^{a+b}=g^a \cdot g^b$ ), we have
$$\theta(9+7)=g^{9+7}=g^9 \cdot g^7=\theta(9) \cdot \theta(7),$$
so that $\theta(9+7)=\theta(9) \cdot \theta(7)$.
Let’s dig deeper into the equation $\theta(9+7)=\theta(9) \cdot \theta(7)$. As shown in the diagram below:

• $\theta(9+7)$ means first add 9 and 7 in $\mathbb{Z}_{12}$ and then apply $\theta$ to the sum.
• $\theta(9) \cdot \theta(7)$ means first apply $\theta$ to each of 9 and 7 and then multiply them in $\langle g\rangle$.

## 数学代写|抽象代数代写Abstract Algebra代考|Groups

$\mathbb{Z} 12$ and $\langle g\rangle$ : Elements match up

〈left 的分隔符缺失或无法识别 ，即g的所有整数帛的集合。

$$g^9 \cdot g^7=g^{9+7}=g^{16}=g^{12+4}=g^{12} \cdot g^4=\varepsilon \cdot g^4=g^4,$$

## 数学代写|抽象代数代写Abstract Algebra代考|Groups ℤ12 and ⟨𝑔⟩: Operations match up

$$\theta(9+7)=g^{9+7}=g^9 \cdot g^7=\theta(9) \cdot \theta(7),$$

• $\theta(9+7)$ 表示先将 9 和 7 相加 $\mathbb{Z}_{12}$ 然后申请 $\theta$ 到总和。
• $\theta(9) \cdot \theta(7)$ 表示先申请 $\theta$ 到 9 和 7 中的每一个, 然后将它们相乘 $\langle g\rangle$.

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