# Proof: any element of a group of prime order generates the group

The following self-contained post clarifies the proof of theorem 4, chapter 13, p. 129, of the book “A book of abstract algebra” by Charles C. Pinter.

Let $G$ be a group of prime order $|G|=p$. It will be shown that for any $a\in G$ with $a\ne e$ it holds that $\langle a\rangle=G,$ where $\langle a\rangle$ is the set generated by $a$ and $e$ is the identity element of $G$.

Since the order of group $G$ is $|G|=p<\infty$, the order $\mbox{ord}(a)$ of element $a$ is also finite, i.e. $\mbox{ord}(a)=n<\infty$ for some positive integer $n$.

Note that the set $\langle a\rangle=\left\{a^0,a^1,\ldots,a^{n-1}\right\}$ generated by $a$ is a subgroup of $G$, since $\langle a\rangle$ is closed with respect to multiplication and with respect to inverses. Indeed, for any $a^i,a^j\in\langle a\rangle$, the division algorithm of integers ensures that there exist integers $q,r$ with $0\le r such that $a^i a^j = a^{i+j}=a^{nq+r}=(a^n)^q a^r=e a^r=a^r\in \langle a \rangle$. Moreover, the inverse of any $a^i\in \langle a \rangle$ is $(a^i)^{-1}=a^{n-i}\in \langle a \rangle$.

The order $|\langle a \rangle|$ of the cyclic subgroup $\langle a \rangle$ of $G$ is equal to the order of generator $a$, that is $|\langle a \rangle|=\mbox{ord}(a)=n$.

By Lagrange’s theorem, $n$ is a factor of $p$. Since $p$ is a prime, $n=1$ or $n=p$. It follows from $a\ne e$ that $n\ne 1$, so $n=p$, which means that $|\langle a \rangle|=|G|$.

$\langle a \rangle\subseteq G$ and $|\langle a \rangle|=|G|\Rightarrow\langle a \rangle=G$.