Let $G$ be a group such that every maximal subgroup is of finite index and any two maximal subgroups are conjugate and any proper subgroup is contained in a maximal subgroup . Then is $G$ cyclic ? I know that if $G$ is a finite group such that any two maximal subgroups are conjugate then $G$ is cyclic . But I cannot handle the infinite case . Please help . Thanks in advance
Here is a reduction to the finite case. Suppose $G$ is a group satisfying your requirements, and $M$ a maximal subgroup. Then all maximal subgroups are conjugate to $M$. This implies that the Frattini subgroup $\Phi(G)$ of $G$ (which is defined to be the intersection of all maximal subgroups of $G$) is the intersection of the finitely many conjugates of $M$, all of which have finite index, so $\Phi(G)$ has finite index. Therefore $G/\Phi(G)$ is finite.
Now it is easy to see that $G/\Phi(G)$ still has all maximal subgroups conjugate, so by the finite case, $G/\Phi(G)$ is cyclic. Let $\overline a$ be a generator of this group, where $a \in G$. Then we get that $G = \langle a, \Phi(G)\rangle$.
We now claim that $\langle a\rangle = G$. If not, then there is some maximal subgroup $N$ of $G$ containing $\langle a\rangle$. But by definition of the Frattini subgroup, $N$ also contains $\Phi(G)$, so then $N$ would contain $\langle a, \Phi(G)\rangle = G$. Contradiction, so there is no maximal subgroup containing $\langle a\rangle$, and therefore we conclude that $G = \langle a \rangle$.