# Number of maximal subgroups in finitely generated amenable groups

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?

• Actually infinite groups with finitely many maximal subgroups are pretty rare. I'm not sure there exists any finitely generated group with this property.
– YCor
Commented Mar 24, 2020 at 11:36
• PS: actually, the first Grigorchuk group has only finitely many maximal subgroups (namely its subgroups of index $2$). This is due to E. Pervova: Everywhere dense subgroups of a group of tree automorphisms. (Russian. Russian summary) Tr. Mat. Inst. Steklova 231 (2000), Din. Sist., Avtom. i Beskon. Gruppy, 356–367; translation in Proc. Steklov Inst. Math. 2000, no. 4(231), 339–350.
– YCor
Commented Mar 26, 2020 at 13:06
• @YCor There even exist finitely generated such groups with exactly $3$ maximal subgroups. This is problem 17.17 in the Kourovka notebook and such a group is constructed in §7 of this paper. Commented Sep 28, 2021 at 8:01

$$\mathbf{Z}$$ has a maximal subgroup $$p\mathbf{Z}$$ for each prime $$p>0$$.

There's even a finitely generated solvable group with uncountably many maximal subgroups, see this MO question.

$$C_\infty$$ has infinitely many maximal subgroups (a subgroup of index $$p$$ for each prime number $$p$$). However, it is clearly amenable.
• @SeanEberhard it seems OP writes $C_\infty$ for a an infinite cyclic group. So it's finitely generated and your answer is a copy of this one.