Suppose that $G$ is a group with more than one element, $G$ had no proper, non-trivial subgroup then prove that $|G|$ is prime.
$G$ is finite
If not then for any $x\neq e$ we have $\langle x^2 \rangle$ a non-trivial subgroup of $G$. Hence $G$ is finite.
Now given $G$ is finite.Let $|G|=m$ For any $x\neq e$ we have $\langle x \rangle$ a subgroup of $G$.
Now because there exists no non trivial subgroup, we have $\langle x\rangle=G$
Hence $G=\langle x \rangle$
Hence $G$ is cyclic.
How do I show that $|G|$ is prime?
Kindly do not use Cauchy Theorem. Use Lagrange's Theorem only, or topics taught before Lagrange Theorem.