After proving the greatest common divisor test for generators, the prime-order case becomes especially simple. If a cyclic group has prime order $p$, then every non-identity element is a generator. This happens because every exponent $1,2,\ldots,p-1$ is relatively prime to $p$. Prime-order cyclic groups are important because they have the simplest possible nontrivial generator structure. In this lesson, students will prove the prime-order generator theorem and apply it to concrete finite cyclic groups. :::theorem Let $(G,\circ)$ be a cyclic group and let $p$ be a prime number. If $$\begin{equation*} |G|=p, \end{equation*}$$ then every non-identity element of $G$ is a generator of $G$. ::: :::proof Given that $(G,\circ)$ is a cyclic group, $p$ is a prime number, and $|G|=p$. To prove that every non-identity element of $G$ is a generator of $G$. Since $G$ is cyclic, there exists $g\in G$ such that $$\begin{equation*} G=\langle g\rangle. \end{equation*}$$ Let $x\in G$ such that $$\begin{equation*} x\neq e. \end{equation*}$$ Since $G=\langle g\rangle$, there exists $r\in\mathbb{Z}$ such that $$\begin{equation*} x=g^r. \end{equation*}$$ Because $x\neq e$, the exponent $r$ is not divisible by $p$. Therefore $$\begin{equation*} \operatorname{gcd}(r,p)=1. \end{equation*}$$ By the generator criterion for finite cyclic groups, $$\begin{equation*} G=\langle g^r\rangle. \end{equation*}$$ Since $x=g^r$, we get $$\begin{equation*} G=\langle x\rangle. \end{equation*}$$ Hence, every non-identity element of $G$ is a generator of $G$. $\square$ ::: This proof avoids an unnecessary detour. Since the group is already cyclic, every element is some power of a fixed generator. A non-identity element cannot correspond to an exponent divisible by $p$, because such a power would be the identity. Primality then forces the exponent to be relatively prime to $p$, so the element is a generator. In a cyclic group of prime order, every non-identity element is a generator. Choose a prime $p$ and the preview marks $0$ as the identity in $\mathbb{Z}_p$ and every nonzero residue as a generator. It also shows that each exponent from $1$ to $p-1$ has greatest common divisor $1$ with $p$. :::scientific-preview[Prime Order Generator Explorer] ::: :::example Let $G=\langle a\rangle$ and $|G|=5$. Then $$\begin{equation*} G=\{e,a,a^2,a^3,a^4\}. \end{equation*}$$ Since $5$ is prime, each of $$\begin{equation*} a,\;a^2,\;a^3,\;a^4 \end{equation*}$$ is a generator of $G$. The only element that is not a generator is $e$. ::: :::example In the additive group $(\mathbb{Z}_7,+_7)$, every nonzero residue class is a generator: $$\begin{equation*} \overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6}. \end{equation*}$$ For instance, $$\begin{equation*} \langle \overline{3}\rangle=\mathbb{Z}_7 \end{equation*}$$ because $\operatorname{gcd}(3,7)=1$. ::: :::corollary Let $(G,\circ)$ be a cyclic group and let $p$ be a prime number. If $|G|=p$, then $G$ has exactly $p-1$ generators. ::: :::proof Given that $(G,\circ)$ is a cyclic group, $p$ is a prime number, and $|G|=p$. To prove that $G$ has exactly $p-1$ generators. By the previous theorem, every non-identity element of $G$ is a generator. The group $G$ has $p$ elements, and exactly one of them is the identity element. Therefore the number of non-identity elements is $$\begin{equation*} p-1. \end{equation*}$$ Hence, $G$ has exactly $p-1$ generators. $\square$ ::: :::solved-problem Let $G$ be a cyclic group of order $11$. How many generators does $G$ have? ::: :::solution Let $G$ be a cyclic group of order $11$. To find the number of generators of $G$. Since $11$ is prime, every non-identity element of $G$ is a generator. The group has $11$ elements and one identity element. Therefore the number of generators is $$\begin{align*} 11-1&=10. \end{align*}$$ Hence, $G$ has $10$ generators. $\square$ ::: :::solved-problem Determine all generators of $(\mathbb{Z}_{13},+_{13})$. ::: :::solution Let $(\mathbb{Z}_{13},+_{13})$ be the additive group modulo $13$. To determine all generators. Since $13$ is prime, every nonzero residue class is a generator. Therefore all generators are $$\begin{equation*} \overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11},\overline{12}. \end{equation*}$$ Hence, these are all generators of $\mathbb{Z}_{13}$. $\square$ ::: :::remark The theorem assumes the group is cyclic. Later, after Lagrange's theorem is available, one proves the stronger result that every group of prime order is cyclic. At this stage, we use only the cyclic hypothesis and the generator criterion already proved. ::: Use the calculator below to list all generators of $\mathbb{Z}_p$ when $p$ is prime. Enter a prime number $p$. The calculator returns all nonzero residue classes, because each of them generates the additive group modulo $p$. :::calculator[Prime Cyclic Generator Calculator] ::: :::exercise [1] State the prime-order generator theorem for cyclic groups. [2] How many generators does a cyclic group of order $17$ have? [3] List all generators of $(\mathbb{Z}_5,+_5)$. [4] Why is the identity element not a generator when $|G|>1$? [5] If $G=\langle a\rangle$ and $|G|=7$, determine whether $a^4$ is a generator. ::: :::answer [1] If $G$ is cyclic of prime order $p$, then every non-identity element of $G$ is a generator. [2] It has $16$ generators. [3] The generators are $\overline{1},\overline{2},\overline{3},\overline{4}$. [4] The identity generates only $\{e\}$, not the whole group when $|G|>1$. [5] Yes, because $\operatorname{gcd}(4,7)=1$. ::: :::faq Q: In a cyclic group of prime order, which elements are generators? A: Every non-identity element is a generator. Q: How many generators does a cyclic group of prime order $p$ have? A: It has $p-1$ generators. Q: Does the identity generate a nontrivial prime-order group? A: No. The identity generates only the trivial subgroup. Q: Why does prime order make the generator test simple? A: Every integer from $1$ to $p-1$ is relatively prime to $p$. ::: :::call-to-action[Even Cyclic Groups] subtitle: Continue to even cyclic groups and prove that they contain exactly one element of order two. button: Next Lesson | /abstract-algebra/introduction-to-groups/cyclic-groups/even-cyclic-groups button-ghost: Previous Lesson | /abstract-algebra/introduction-to-groups/cyclic-groups/generators-of-finite-cyclic-groups :::