We have seen that the order of every element divides the order of a finite group. When the order of the group is prime, this leaves very few possibilities. A non-identity element cannot have order $1$, so its order must be the prime number itself. This forces the whole group to be generated by that one element. Thus every group of prime order is cyclic. This result is one of the first examples where Lagrange's theorem describes the whole structure of a group. :::definition[Group of Prime Order] Let $(G,\circ)$ be a finite group. The group $G$ is called a $group of prime order$ if $$\begin{equation*} |G|=p \end{equation*}$$ for some prime number $p$. ::: A group of prime order has only two possible subgroup sizes: $1$ and the prime number $p$. In $\mathbb{Z}_p$ under addition, every nonzero element repeatedly added to itself reaches every residue class. Choose a prime and a nonzero element to see the entire group appear in one cycle. The identity element is the only element that fails to generate the group. :::scientific-preview[Prime Order Generator Explorer] ::: Prime order leaves no room for a proper nontrivial cyclic subgroup. Once a non-identity element generates more than one element, its generated subgroup must have all $p$ elements. :::theorem Let $(G,\circ)$ be a finite group. If $|G|$ is prime, then $G$ is cyclic. ::: :::proof Given that $(G,\circ)$ is a finite group and $|G|$ is prime. To prove that $G$ is cyclic. Let $$\begin{equation*} |G|=p, \end{equation*}$$ where $p$ is a prime number. Since $p>1$, the group $G$ contains at least one element different from the identity element. Let $a\in G$ such that $$\begin{equation*} a\neq e. \end{equation*}$$ Then $\langle a\rangle$ is a subgroup of $G$. By Lagrange's theorem, $$\begin{equation*} |\langle a\rangle|\mid |G|. \end{equation*}$$ Therefore $$\begin{equation*} |\langle a\rangle|\mid p. \end{equation*}$$ Since $p$ is prime, the only positive divisors of $p$ are $1$ and $p$. Therefore $$\begin{equation*} |\langle a\rangle|=1 \end{equation*}$$ or $$\begin{equation*} |\langle a\rangle|=p. \end{equation*}$$ Since $a\neq e$, the subgroup $\langle a\rangle$ contains at least $e$ and $a$. Therefore $$\begin{equation*} |\langle a\rangle|\neq 1. \end{equation*}$$ Thus $$\begin{equation*} |\langle a\rangle|=p. \end{equation*}$$ Since $\langle a\rangle\subseteq G$ and both sets have $p$ elements, we get $$\begin{equation*} \langle a\rangle=G. \end{equation*}$$ Hence, $G$ is cyclic. $\square$ ::: :::corollary Let $(G,\circ)$ be a finite group. If $|G|$ is prime, then every non-identity element of $G$ is a generator of $G$. ::: :::proof Given that $(G,\circ)$ is a finite group and $|G|$ is prime. To prove that every non-identity element of $G$ is a generator of $G$. Let $$\begin{equation*} |G|=p, \end{equation*}$$ where $p$ is prime, and let $a\in G$ such that $a\neq e$. By Lagrange's theorem, $$\begin{equation*} o(a)\mid p. \end{equation*}$$ Since $p$ is prime, we get $$\begin{equation*} o(a)=1 \end{equation*}$$ or $$\begin{equation*} o(a)=p. \end{equation*}$$ Since $a\neq e$, we have $o(a)\neq 1$. Therefore $$\begin{equation*} o(a)=p. \end{equation*}$$ Thus $$\begin{equation*} |\langle a\rangle|=p=|G|. \end{equation*}$$ Since $\langle a\rangle\subseteq G$, we get $\langle a\rangle=G$. Hence, every non-identity element of $G$ is a generator of $G$. $\square$ ::: This calculator checks how primality restricts element orders. Enter a proposed group order and choose whether the element is the identity or non-identity. When the group order is prime, the only possible element orders are $1$ and $p$. For a non-identity element, order $1$ is impossible, so the element must generate the whole group. :::calculator[Prime Order Group Calculator]