Answer to Question #144684 in Abstract Algebra for J

Question

Assume that (G,*) is a group, and that |G|=p, where p is prime. Prove that G is cyclic.

Answer 1

Let that $(G,*)$ be a group of order$|G|=p$ , where $p$ is prime. Let $a\in G$ and $a\ne e$ where $e$ is the identity element of $(G,*)$. According to Lagrange's theorem, the order of cyclic subgroup $\langle a\rangle$ divides $|G|=p$. Therefore, $|\langle a\rangle|\in\{1,p\}$. Taking into account that the identity $e$ is a unique element of order 1, we conclude that $|\langle a\rangle|>1$, and therefore, $|\langle a\rangle|=p$. Since $\langle a\rangle\subseteq G$ and $|\langle a\rangle|=p=|G|$, we conclude that $\langle a\rangle=G.$ Consequently, G is a cyclic group generated by $a$.