# 数学代写|现代代数代写Modern Algebra代考|Cyclic Groups

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## 数学代写|现代代数代写Modern Algebra代考|Cyclic Groups

In the last section, a group $G$ was defined to be cyclic if there exists an element $a \in G$ such that $G=\langle a\rangle$. It may happen that there is more than one element $a \in G$ such that $G=\langle a\rangle$. For the additive group $\mathbf{Z}$, we have $\mathbf{Z}=\langle 1\rangle$ and also $\mathbf{Z}=\langle-1\rangle$, since any $n \in \mathbf{Z}$ can be written as $(-n)(-1)$. Here, $(-n)(-1)$ does not indicate a product but rather a multiple of -1 , as described in Section 3.2.
Generator
Any element $a$ of the group $G$ such that $G=\langle a\rangle$ is a generator of $G$.
If $a$ is a generator of a multiplicative group $G$, then $a^{-1}$ is also, since any element $x \in G$ can be written as
$$x=a^n=\left(a^{-1}\right)^{-n}$$
for some integer $n$, and $G=\langle a\rangle=\left\langle a^{-1}\right\rangle$. For the additive group $G$, if $a$ is a generator of $G$, then $-a$ is also, since any $x \in G$ can be written as
$$x=n a=(-n)(-a)$$
for some integer $n$, and $G=\langle a\rangle=\langle-a\rangle$.

$$\mathbf{Z}_n={[0],[1], \ldots,[n-1]}$$
is a cyclic group with generator [1], since any $[k]$ in $\mathbf{Z}_n$ can be written as
$$[k]=k[1]$$
where $k[1]$ indicates a multiple of [1] as described in Section 3.2. Elements other than [1] may also be generators. To illustrate this, consider the particular case
$$\mathbf{Z}_6={[0],[1],[2],[3],[4],[5]} .$$
The element [5] is also a generator of $\mathbf{Z}_6$ since [5] is the additive inverse of [1]. The following list shows how $\mathbf{Z}_6$ is generated by [5]-that is, how $\mathbf{Z}_6$ consists of multiples of [5].
\begin{aligned} & 1[5]=[5] \ & 2[5]=[5]+[5]=[4] \ & 3[5]=[5]+[5]+[5]=[3] \ & 4[5]=[2] \ & 5[5]=[1] \ & 6[5]=[0] \end{aligned}
The cyclic subgroups generated by the other elements of $\mathbf{Z}_6$ under addition are
\begin{aligned} & \langle[0]\rangle={[0]} \ & \langle[2]\rangle={[2],[4],[0]} \ & \langle[3]\rangle={[3],[0]} \ & \langle[4]\rangle={[4],[2],[0]}=\langle[2]\rangle . \end{aligned}
Thus [1] and [5] are the only elements that are generators of the entire group.

## 数学代写|现代代数代写Modern Algebra代考|Generators of a Finite Cyclic Group

Let $G=\langle a\rangle$ be a cyclic group of order $n$. Then $a^m$ is a generator of $G$ if and only if $m$ and $n$ are relatively prime.
$p \Leftarrow q$ Proof On the one hand, if $m$ is such that $m$ and $n$ are relatively prime, then $d=$ $(m, n)=1$, and $a^m$ is a generator of $G$ by Theorem 3.26.
$p \Rightarrow q \quad$ On the other hand, if $a^m$ is a generator of $G$, then $a=\left(a^m\right)^p$ for some integer $p$. By part b of Theorem 3.21, this implies that $1 \equiv m p(\bmod n)$. That is,
$$1-m p=n q$$
for some integer $q$. This gives
$$1=m p+n q,$$
and it follows from Theorem 2.12 that $(m, n)=1$.

The Euler phi-function $\phi(n)$ was defined for positive integers $n$ in Exercise 23 of Section 2.8 as follows: $\phi(n)$ is the number of positive integers $m$ such that $1 \leq m \leq n$ and $(m, n)=1$. It follows, from Theorems 3.21 and 3.28, that the cyclic group $\langle a\rangle$ of order $n$ has $\phi(n)$ distinct generators.

Example 9 Let $G=\langle a\rangle$ be a cyclic group of order 10. The positive integers less than 10 and relatively prime to 10 are $1,3,7$, and 9 . Therefore, all generators of $G$ are included in the list
$a, a^3, a^7$, and $a^9$, and $G$ has $\phi(10)=4$ distinct generators.

# 现代代数代考

## 数学代写|现代代数代写Modern Algebra代考|Cyclic Groups

$$x=a^n=\left(a^{-1}\right)^{-n}$$

$$x=n a=(-n)(-a)$$

$$\mathbf{Z}_n={[0],[1], \ldots,[n-1]}$$

$$[k]=k[1]$$

$$\mathbf{Z}_6={[0],[1],[2],[3],[4],[5]} .$$

\begin{aligned} & 1[5]=[5] \ & 2[5]=[5]+[5]=[4] \ & 3[5]=[5]+[5]+[5]=[3] \ & 4[5]=[2] \ & 5[5]=[1] \ & 6[5]=[0] \end{aligned}
$\mathbf{Z}_6$的其他元素在加法作用下生成的循环子群为
\begin{aligned} & \langle[0]\rangle={[0]} \ & \langle[2]\rangle={[2],[4],[0]} \ & \langle[3]\rangle={[3],[0]} \ & \langle[4]\rangle={[4],[2],[0]}=\langle[2]\rangle . \end{aligned}

## 数学代写|现代代数代写Modern Algebra代考|Generators of a Finite Cyclic Group

$p \Leftarrow q$ 证据一方面，如果 $m$ 是这样的 $m$ 和 $n$ 是相对优质的吗 $d=$ $(m, n)=1$，和 $a^m$ 是的生成器 $G$ 根据定理3.26。
$p \Rightarrow q \quad$ 另一方面，如果 $a^m$ 是的生成器 $G$那么， $a=\left(a^m\right)^p$ 对于某个整数 $p$． 根据定理3.21的b部分，这意味着 $1 \equiv m p(\bmod n)$． 也就是说，
$$1-m p=n q$$

$$1=m p+n q,$$

$a, a^3, a^7$和$a^9$, $G$有$\phi(10)=4$个不同的生成器。

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