Is this true that for a finite group $G$, if there exists an element $a\in G$, $|G|=\text{ord}(a)$, then it is isomorphic to the cyclic group $C_{|G|}$?
2026-03-31 10:36:06.1774953366
A finite group is isomorphic to cyclic group
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It is true, since $\operatorname{ord}(a)=|G|$ we have that the subgroup generated by $a$ (which is cyclic) is a subgroup of $G$ with order $|G|$, hence equal to $G$.