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抽象代数代写| Order of an element

1. Order of an element

The study of cyclic subgroups motivates us to introduce the following notion.

2. Definition 1: Order of element

Given group $(G, *)$ and an element $a \in G$ consider the set of integer powers of $a$

\left{a^{k} | k \in \mathbb{Z}\right}=\left{\ldots, a^{-2}, a^{-1}, a^{0}, a, a^{2}, \ldots\right}

If this sequence is periodic with period $d \in \mathbb{Z}$, we say that element $a \in G$ has order $d$ :


Equivalently, order of $a$ is the smallest positive power ord $(a)$ such that

a^{\operatorname{ord}(a)}=e .

If there are no such power, we say that $a$ has infinite order.

3. Example 1

  1. For every integer $n \geqslant 2$, the group $\left(\mathbb{Z}_{n},+\right)$ is a cyclic group of order $n$, since the class 1 is always a generator. The class $-1$ is also a generator. We therefore see that the generator of a cyclic group need not be unique.
  2. The group $(\mathbb{Z},+)$ is cyclic, with generators 1 and $-1$. Moreover, for every $m \in \mathbb{Z}, m \mathbb{Z}$ is the cyclic subgroup of $\mathbb{Z}$ generated by $m .$ Therefore, all the subgroups of $\mathbb{Z}$ are cyclic.
  3. The group of units $\left(\mathbb{Z}{9}^{\times}, \cdot\right)$ is a cyclic group, with generator $2 .$ Indeed, as a set, $\left(\mathbb{Z}{9}\right)^{\times}={1,2,4,5,7,8}$, and

2^{1}=2,2^{2}=4,2^{3}=8 \
2^{4} \equiv 7 \quad(\bmod 9) \
2^{5} \equiv 5 \quad(\bmod 9) \
2^{6} \equiv 1 \quad(\bmod 9)

  1. For every $n,\left(\mathbb{C}^{\times}, \cdot\right)$ has a cyclic subgroup of order $n$, given by the $n$-th roots of unity:

U_{n}=\left{e^{\frac{2 i k \pi}{n}}, k \in{0, \ldots n-1}\right}

For example, for $n=2$ we get the subgroup ${1,-1}$, for $n=3$ we get $\left{1, e^{\frac{2 i \pi}{3}}, e^{\frac{4 i \pi}{3}}\right}$, and for $n=4$ we get ${1, i,-1,-i} .$ Note that the elements of $U_{n}$ are the vertices of a regular $n$-gon in the complex plane.

4. Proposition 1

If $a$ is of finite order $d$, then for any integer $n, a^{n}=e$ if and only if $d \in n \mathbb{Z}$, that is, if and only if $d$ divides $n$.

Proof. If $n=d k$ is a multiple of $d$, then we can write

a^{n}=\left(a^{d}\right)^{k}=e^{k}=e .

In the other direction, assume $a^{n}=e .$ Since by definition of order $a^{d}=e$, we have

a^{n-q d}=\left(a^{n}\right) \cdot\left(a^{d}\right)^{-q}=e

for any $q \in \mathbb{Z}$. In particular


where $r \in \mathbb{Z}$ is the remainder of division of $n$ by $d$. Since $r<d$, we must necessarily have $r=0$ by minimality of $d$ in the definition of the order.

Recall that a group $(G, *)$ is called cyclic, if there exists an element $a \in G$ such that

G=\left{a^{k} \mid k \in \mathbb{Z}\right}

In this case we write $G=\langle a\rangle .$ If $G$ has exactly $n$ elements, we will sometimes write

G=\langle a\rangle_{n} .

Element $a \in G$ is called a generator of G. Generators are not necessarily unique, so our first goal is to answer the following questions:

5. Question 1

When an element $b=a^{k} \in\langle a\rangle_{n}$ can be taken as a generator of $G=\langle a\rangle_{n} ?$

6. Question 2

More generally, what is the cyclic subgroup $\langle b\rangle$ generated by $b=a^{k} ?$

7. Example 2

Group $\left(\mathbb{Z}_{3},+\right)$ is cyclic generated by $[1]$ :

\mathbb{Z}{3}=\langle[1]\rangle{3}={[0],[1],[2]} .

It is easy to see that element [2] can be taken as another generator.

8. Example 3

Group $\left(\mathbb{Z}{4},+\right)$ is cyclic. One can choose either of $[1]$ and $[3]=[-1]$ as a generator. On the other hand, element [2] does not generate $\mathbb{Z}{4}$ since

\langle[2]\rangle={[0],[2]} \subsetneq \mathbb{Z}_{4}

9. Proposition 2

Element $b=a^{k} \in\langle a\rangle_{n}$ can be taken as a generator of $\langle a\rangle_{n}$ if and only if $\operatorname{gcd}(n, k)=1$

Proof. First we note that $a$ has order exactly $n$.

Element $b$ generates the whole group, if and only if $a \in\langle b\rangle$, i.e., for some $l>0$ we have

b^{l}=a^{l k}=a .

By Proposition 1 this happens if and only if $l k \equiv 1 \bmod n .$

This happens if and only if element $k \in \mathbb{Z}_{n}$ is multiplicatively invertible (is a unit), which si equivalent to $\operatorname{gcd}(n, k)=1$.

10. Example 4

We saw that $\mathbb{Z}_{9}^{\times}$is a cyclic group with generator $a=2$ and is of order $\varphi(9)=6 .$ By the above example, all other generators are $2^{k}$, where $k$ is any number coprime with 6, i.e. $k=1$ and $k=5$ :


With a slight modification to the above argument, we can answer the second question

11. Proposition 3

Consider any element $b=a^{k} \in\langle a\rangle_{n}$. Then

\langle b\rangle=\left\langle a^{\operatorname{gcd}(n, k)}\right\rangle_{n / \operatorname{gcd}(n, k)}=\left{a^{0}, a^{\operatorname{gcd}(n, k)}, a^{2 \operatorname{gcd}(n, k)}, \ldots, a^{n-\operatorname{gcd}(n, k)}\right},

i.e., $a^{k}$ generates a cyclic subgroup of order $n / \operatorname{gcd}(n, k)$ which can also be generated by $a^{\operatorname{gcd}(n, k)}$.

Problem 1: Every cyclic subgroup is abelian.

12. Proposition 4

Every subgroup of a cyclic group is cyclic.

Proof. We have already proved it for infinite cyclic group (see the statement about subgroups of $\mathbb{Z}$ ). Now let $H \subset\langle a\rangle_{n} .$ Choose an element $a^{k} \in H$ with the smallest possible $k>0 .$

It is an exercise to check that $H=\left\langle a^{k}\right\rangle$.

Problem 2: Let $S^{3}$ be a group of symmetries of an equilateral triangle. Find orders of all its elements.