The following statement is known to be true:
Any subgroup of a finitely generated group lies in a maximal subgroup
Proof:
Suppose, $G = \langle \{x_1, … , x_n\} \rangle$ is a counterexample. Then there exists an infinite ascending tower of proper subgroups $H_1, H_2, … $ such that $\bigcup_{i \in \mathbb{N}} H_i = G$. Then $\forall g \in G \exists i(g) \in \mathbb{N}$ such that $g \in H_{i_g}$. It follows, that $\bigcup_{j < n} H_{i(j)} = G$ which contradicts the assumption that all those subgroups are proper.
Q.E.D.
This fact gives the rise to the question:
Do all finitely generated groups have finitely many maximal subgroups?
The answer is obviously «NO» as there are two types of counterexamples coming to the mind: the free groups and the Tarski monster groups.
However, if we additionally require the group in question to be amenable, then both those examples become ruled out. So, my question is:
Do all finitely generated amenable groups have finitely many maximal subgroups?
Best Answer
$\mathbf{Z}$ has a maximal subgroup $p\mathbf{Z}$ for each prime $p>0$.