After proving that cyclic groups are abelian, we now examine how the size of a cyclic group is determined by the order of its generator. If a cyclic group is generated by an element $a$, then all elements are powers of $a$. Therefore the group is finite exactly when the powers of $a$ eventually return to the identity. This connects cyclic groups with the earlier section on order of an element. In this lesson, students will prove the finite and infinite cases and learn how the order of a generator controls the whole group. :::theorem Let $(G,\circ)$ be a finite group and let $a\in G$ such that $$\begin{equation*} G=\langle a\rangle. \end{equation*}$$ If $|G|=n$, then $$\begin{equation*} o(a)=n. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a finite group, $a\in G$, $G=\langle a\rangle$, and $|G|=n$. To prove that $$\begin{equation*} o(a)=n. \end{equation*}$$ Since $G=\langle a\rangle$, every element of $G$ is an integral power of $a$. Since $|G|=n$, the distinct powers are $$\begin{equation*} e,a,a^2,\ldots,a^{n-1}. \end{equation*}$$ Therefore $$\begin{equation*} a^n=e. \end{equation*}$$ Also, if $1\leq r<n$, then $$\begin{equation*} a^r\neq e, \end{equation*}$$ because otherwise the list of powers would repeat before producing $n$ distinct elements. Thus the least positive integer $r$ such that $a^r=e$ is $n$. Hence, $$\begin{equation*} o(a)=n. \end{equation*}$$ $\square$ ::: :::theorem Let $(G,\circ)$ be a group and let $a\in G$ such that $$\begin{equation*} G=\langle a\rangle. \end{equation*}$$ If $$\begin{equation*} o(a)=n, \end{equation*}$$ then $$\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\} \end{equation*}$$ and $$\begin{equation*} |G|=n. \end{equation*}$$ ::: :::proof Given that $(G,\circ)$ is a group, $a\in G$, $G=\langle a\rangle$, and $o(a)=n$. To prove that $$\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\} \end{equation*}$$ and $|G|=n$. Since $o(a)=n$, we have $$\begin{equation*} a^n=e. \end{equation*}$$ Also, the elements $$\begin{equation*} e,a,a^2,\ldots,a^{n-1} \end{equation*}$$ are distinct. Let $x\in G$. Since $G=\langle a\rangle$, there exists $k\in\mathbb{Z}$ such that $$\begin{equation*} x=a^k. \end{equation*}$$ By the division algorithm, there exist $q,r\in\mathbb{Z}$ such that $$\begin{equation*} k=nq+r,\quad 0\leq r<n. \end{equation*}$$ Now $$\begin{align*} x &=a^k\\ &=a^{nq+r}\\ &=(a^n)^q\circ a^r\\ &=e^q\circ a^r\\ &=a^r. \end{align*}$$ Thus every element of $G$ belongs to $\{e,a,a^2,\ldots,a^{n-1}\}$. Therefore $$\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\}. \end{equation*}$$ Since these $n$ elements are distinct, we get $$\begin{equation*} |G|=n. \end{equation*}$$ Hence, $$\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\} \end{equation*}$$ and $|G|=n$. $\square$ ::: If $o(a)=n$, the powers of $a$ move through $n$ distinct elements and then return to the identity. Choose $n$ and the preview displays the cycle $e,a,a^2,\ldots,a^{n-1},a^n=e$. This makes the reason for $|\langle a\rangle|=o(a)$ visible as a finite loop of powers. :::scientific-preview[Finite Cyclic Cycle Explorer] ::: :::corollary Let $(G,\circ)$ be a group and let $a\in G$ such that $G=\langle a\rangle$. Then $G$ is infinite if and only if $a$ has infinite order. ::: :::proof Given that $(G,\circ)$ is a group, $a\in G$, and $G=\langle a\rangle$. To prove that $G$ is infinite if and only if $a$ has infinite order. [1] To prove that if $G$ is infinite, then $a$ has infinite order. If possible let $a$ have finite order. Let $$\begin{equation*} o(a)=n. \end{equation*}$$ Then by the previous theorem, $$\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\}. \end{equation*}$$ Therefore $G$ is finite. A contradiction. Hence, if $G$ is infinite, then $a$ has infinite order. [2] To prove that if $a$ has infinite order, then $G$ is infinite. If possible let $G$ be finite. Since $G=\langle a\rangle$, two positive powers of $a$ must be equal. Thus there exist $r,s\in\mathbb{N}$ such that $$\begin{equation*} r<s \end{equation*}$$ and $$\begin{equation*} a^r=a^s. \end{equation*}$$ Then $$\begin{align*} a^r=a^s &\implies a^{s-r}=e. \end{align*}$$ Since $s-r\in\mathbb{N}$, this contradicts that $a$ has infinite order. A contradiction. Hence, if $a$ has infinite order, then $G$ is infinite. Hence, $G$ is infinite if and only if $a$ has infinite order. $\square$ ::: :::example The group $(\mathbb{Z},+)$ is cyclic with generator $1$. The element $1$ has infinite order under addition because no positive multiple of $1$ is $0$. Therefore $(\mathbb{Z},+)$ is an infinite cyclic group. ::: :::example The group $(\mathbb{Z}_{10},+_{10})$ is cyclic with generator $\overline{1}$. Since $$\begin{equation*} 10\overline{1}=\overline{0} \end{equation*}$$ and no smaller positive multiple gives $\overline{0}$, we get $$\begin{equation*} o(\overline{1})=10. \end{equation*}$$ Therefore $|\mathbb{Z}_{10}|=10$. ::: :::solved-problem Let $(G,\circ)$ be a cyclic group generated by $a$. If $o(a)=7$, list all elements of $G$. ::: :::solution Let $(G,\circ)$ be a cyclic group generated by $a$. Given that $$\begin{equation*} o(a)=7. \end{equation*}$$ To list all elements of $G$. Since $G=\langle a\rangle$ and $o(a)=7$, we get $$\begin{equation*} G=\{e,a,a^2,a^3,a^4,a^5,a^6\}. \end{equation*}$$ Hence, the elements of $G$ are $$\begin{equation*} e,a,a^2,a^3,a^4,a^5,a^6. \end{equation*}$$ $\square$ ::: Use the calculator below to list the elements of a finite cyclic group generated by $a$ when $o(a)=n$. Enter $n$. The calculator displays the formal list $e,a,a^2,\ldots,a^{n-1}$. This reinforces the relation between the order of a generator and the size of a finite cyclic group. :::calculator[Finite Cyclic Group List Calculator] ::: :::exercise [1] If $G=\langle a\rangle$ and $o(a)=5$, list the elements of $G$. [2] If $G=\langle a\rangle$ and $|G|=12$, find $o(a)$. [3] Give an example of an infinite cyclic group. [4] Give an example of a finite cyclic group of order $8$. [5] Prove that if $G=\langle a\rangle$ and $a$ has infinite order, then all powers $a^k$ are distinct. ::: :::answer [1] $G=\{e,a,a^2,a^3,a^4\}$. [2] $o(a)=12$. [3] $(\mathbb{Z},+)$ is an infinite cyclic group. [4] $(\mathbb{Z}_8,+_8)$ is a finite cyclic group of order $8$. [5] If $a^r=a^s$ with $r<s$, then $a^{s-r}=e$, contradicting infinite order. ::: :::faq Q: What is the size of $\langle a\rangle$ when $o(a)=n$? A: It has $n$ elements. Q: When is a cyclic group infinite? A: It is infinite exactly when its generator has infinite order. Q: What are the elements of a finite cyclic group generated by $a$? A: They are $e,a,a^2,\ldots,a^{n-1}$ when $o(a)=n$. Q: Can a finite cyclic group have an infinite-order generator? A: No. A finite cyclic group has a generator of finite order equal to the group size. ::: :::call-to-action[Generators of Finite Cyclic Groups] subtitle: Continue to the greatest common divisor test for powers that generate a finite cyclic group. button: Next Lesson | /abstract-algebra/introduction-to-groups/cyclic-groups/generators-of-finite-cyclic-groups button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/cyclic-groups/cyclic-groups-are-abelian :::