I encountered the following exercise in Isaacs' Algebra:
"Suppose a group $G$ has only one maximal subgroup. Prove that the order of $G$ must be a power of a prime".
I think I've proven this for the case when $G$ is cyclic, based on the observation that in a cyclic group $G$ with a subgroup $H$, $H$ is maximal iff
is prime. However if $G$ is not cyclic, I cannot use the property that there always exists a subgroup of order some divisor of $|G|$. I have run out of ideas in solving this problem, how can I proceed from here? Please do not post complete solutions.