After studying standard examples of groups, the next natural step is to ask whether a whole group can be produced from one element. This leads to the idea of a cyclic group. Many groups already seen in this chapter, such as $(\mathbb{Z},+)$, $(\mathbb{Z}_n,+_n)$, and the group of roots of unity, are cyclic. Cyclic groups are important because their structure is controlled by powers of a single element, so questions about the group become questions about exponents. In this lesson, students will learn the definition of a cyclic group, the meaning of a generator, and the first examples of cyclic groups. :::definition[Cyclic Group] Let $(G,\circ)$ be a group. Then $(G,\circ)$ is called a $cyclic group$ if there exists an element $a\in G$ such that $$\begin{equation*} G=\langle a\rangle=\{a^n:n\in\mathbb{Z}\}. \end{equation*}$$ The element $a$ is called a $generator$ of $G$. ::: The notation $\langle a\rangle$ means the set of all integral powers of $a$. Thus it includes positive powers, the identity element $a^0=e$, and negative powers such as $a^{-1}$ and $a^{-2}$. If every element of $G$ can be written as $a^n$ for some integer $n$, then $G$ is cyclic. A common mistake is to think that a cyclic group must have only one generator. A cyclic group must have at least one generator, but it may have several. We can see cyclic generation concretely in the additive group $\mathbb{Z}_n$. Choose a modulus $n$ and an element $a$. The preview repeatedly adds $a$ modulo $n$ until the pattern returns to $0$. If the generated set contains every residue class, then $a$ is a generator of $\mathbb{Z}_n$. :::scientific-preview[Cyclic Generation Explorer] ::: :::example The additive group $(\mathbb{Z},+)$ is cyclic. In additive notation, the powers of $1$ are integer multiples of $1$: $$\begin{equation*} n\cdot1=n \quad \forall\, n\in\mathbb{Z}. \end{equation*}$$ Therefore $$\begin{equation*} \mathbb{Z}=\langle 1\rangle. \end{equation*}$$ Also, $$\begin{equation*} \mathbb{Z}=\langle -1\rangle. \end{equation*}$$ Hence, $(\mathbb{Z},+)$ is cyclic, and both $1$ and $-1$ are generators. ::: :::example The additive group $(\mathbb{Z}_6,+_6)$ is cyclic. Since $$\begin{equation*} \langle \overline{1}\rangle=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5}\}, \end{equation*}$$ we get $$\begin{equation*} \mathbb{Z}_6=\langle \overline{1}\rangle. \end{equation*}$$ The element $\overline{5}$ is also a generator because repeated addition of $\overline{5}$ produces all residue classes modulo $6$. ::: :::theorem Let $(G,\circ)$ be a group and let $a\in G$. If $$\begin{equation*} G=\langle a\rangle, \end{equation*}$$ then $$\begin{equation*} G=\langle a^{-1}\rangle. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $a\in G$, and $G=\langle a\rangle$. To prove that $$\begin{equation*} G=\langle a^{-1}\rangle. \end{equation*}$$ Let $x\in G$. Since $G=\langle a\rangle$, there exists $n\in\mathbb{Z}$ such that $$\begin{equation*} x=a^n. \end{equation*}$$ Now $$\begin{align*} x &=a^n\\ &=(a^{-1})^{-n}. \end{align*}$$ Since $-n\in\mathbb{Z}$, we get $x\in\langle a^{-1}\rangle$. Therefore $$\begin{equation*} G\subseteq\langle a^{-1}\rangle. \end{equation*}$$ Let $y\in\langle a^{-1}\rangle$. Then there exists $m\in\mathbb{Z}$ such that $$\begin{equation*} y=(a^{-1})^m. \end{equation*}$$ Now $$\begin{align*} y &=(a^{-1})^m\\ &=a^{-m}. \end{align*}$$ Since $-m\in\mathbb{Z}$, we get $y\in\langle a\rangle=G$. Therefore $$\begin{equation*} \langle a^{-1}\rangle\subseteq G. \end{equation*}$$ Hence, $$\begin{equation*} G=\langle a^{-1}\rangle. \end{equation*}$$ $\square$ ::: The theorem says that the inverse of a generator is again a generator. This is expected from the definition because allowing all integer powers already includes negative exponents. If $a$ moves forward through the powers, then $a^{-1}$ moves through the same elements in the reverse direction. In a finite cyclic group, this reversal is especially visible in the cycle of powers. :::solved-problem Let $$\begin{equation*} G=\{6x+9y:x,y\in\mathbb{Z}\}. \end{equation*}$$ Prove that $(G,+)$ is cyclic and find a generator. ::: :::solution Let $$\begin{equation*} G=\{6x+9y:x,y\in\mathbb{Z}\}. \end{equation*}$$ To prove that $(G,+)$ is cyclic and find a generator. Let $$\begin{equation*} d=\operatorname{gcd}(6,9)=3. \end{equation*}$$ First, let $g\in G$. Then there exist $x,y\in\mathbb{Z}$ such that $$\begin{equation*} g=6x+9y. \end{equation*}$$ Now $$\begin{align*} g &=6x+9y\\ &=3(2x+3y). \end{align*}$$ Since $2x+3y\in\mathbb{Z}$, we get $$\begin{equation*} g\in3\mathbb{Z}. \end{equation*}$$ Therefore $G\subseteq3\mathbb{Z}$. Conversely, since $$\begin{equation*} 3=6(-1)+9(1), \end{equation*}$$ for every $k\in\mathbb{Z}$, $$\begin{align*} 3k &=6(-k)+9k. \end{align*}$$ Thus $3k\in G$ for every $k\in\mathbb{Z}$. Therefore $3\mathbb{Z}\subseteq G$. Hence, $$\begin{equation*} G=3\mathbb{Z}=\langle 3\rangle. \end{equation*}$$ So $(G,+)$ is cyclic with generator $3$. $\square$ ::: :::remark A cyclic group is not defined by the shape of its operation table or by being finite. It is defined by the existence of one element whose integer powers give the whole group. Thus $(\mathbb{Z},+)$ is cyclic and infinite, while $(\mathbb{Z}_n,+_n)$ is cyclic and finite. ::: Use the calculator below to test whether an element generates the additive group $\mathbb{Z}_n$. Enter $n$ and $a$. The calculator lists the elements obtained by repeated addition of $\overline{a}$ and reports whether the result is all of $\mathbb{Z}_n$. Try $n=8$ with $a=1$, $a=2$, and $a=3$ to see the difference between a generator and a non-generator. :::calculator[Generator Test for Zn] ::: :::exercise [1] Define a cyclic group. [2] What does it mean for $a$ to be a generator of $G$? [3] Prove that $(\mathbb{Z},+)$ is cyclic. [4] Find two generators of $(\mathbb{Z},+)$. [5] Determine whether $\overline{2}$ generates $(\mathbb{Z}_6,+_6)$. ::: :::answer [1] A group $G$ is cyclic if there exists $a\in G$ such that $G=\langle a\rangle=\{a^n:n\in\mathbb{Z}\}$. [2] It means every element of $G$ can be written as $a^n$ for some $n\in\mathbb{Z}$. [3] Since every integer is an integer multiple of $1$, $\mathbb{Z}=\langle 1\rangle$ under addition. [4] The generators are $1$ and $-1$. [5] No. The element $\overline{2}$ generates $\{\overline{0},\overline{2},\overline{4}\}$, not all of $\mathbb{Z}_6$. ::: :::faq Q: Is every group cyclic? A: No. A group is cyclic only when all its elements are powers of one element. Q: Can a cyclic group have more than one generator? A: Yes. For example, $(\mathbb{Z},+)$ has generators $1$ and $-1$. Q: Is every cyclic group abelian? A: Yes. This is proved in the next lesson. Q: Does $\langle a\rangle$ always contain the identity element? A: Yes. It contains $a^0=e$. ::: :::call-to-action[Cyclic Groups are Abelian] subtitle: Continue to the first structural theorem for cyclic groups: every cyclic group is abelian. button: Next Lesson | /abstract-algebra/introduction-to-groups/cyclic-groups/cyclic-groups-are-abelian button-ghost: Previous Section | /abstract-algebra/introduction-to-groups/standard-examples-of-groups/klein-four-group :::