TITLE: A group in which every proper subgroup is contained in a maximal subgroup
QUESTION [11 upvotes]: Let $G$ be a group in which every proper subgroup is contained in a maximal subgroup of $G$.
Can we conclude that $G$ is finitely generated? (@Max commented that converse of this statement is true.)

REPLY [2 votes]: It will work when the number of maximal subgroups is finite. Then we can assume that $M_1, \dots M_n$ are all maximal subgroups in $G$. There exists $x_i \notin M_i$ as each $M_i$ is a maximal subgroup ($M_i$ is a proper subgroup of $G$). Now the group generated by $x_i$ is not contained in any of $M_i$, thus should not be a proper subgroup because of the assumption. As a result, we have:
$$ \langle x_1, x_2, \dots,x_n \rangle = G $$
We will have such a condition; for example, when all of the maximal subgroups are finite index and conjugate to each other.