Let $G$ be a finite group which has exactly two maximal subgroups, say $M_1$ and $M_2$. We want to show that $G$ must be cyclic.
${\bf MY TRY:}$ Suppose, contrary that $G$ is noncyclic and let $a \in G \setminus M_1$. Thus $\langle a \rangle \leq M_2$.
Also if assume that $b \in G \setminus M_2$, then $\langle b \rangle \leq M_1$. Consider $\langle ab \rangle$ which is a subgroup of $G$. Since $G$ is non-cyclic, $\langle ab \rangle <G$. If $\langle ab \rangle \leq M_2$, then $ab \in M_2$. But since $a \in M_2$, we get $b \in M_2$, a contradiction. So $\langle ab \rangle \leq M_1$. Similarly, this imlies a contradiction.
is my try true?