Groups (not necessarily finite) with a given number of maximal subgroups

Question

It is somewhat easy to see that a group $G$ with exactly one maximal subgroup $M$ must be cyclic: any element in $G\setminus M$ generates $G$.

EDIT: @YCor pointed out in the comments that this argument only works for groups in which every proper subgroup is contained in a maximal subgroup. See bottom of post for elaboration.

It is somewhat less easy to see that a group $G$ with exactly two maximal subgroups must also be cyclic. I proved this:

Theorem: Let $G$ be a group (we are not given that it is finite, though we will see that it must be) that has exactly two maximal subgroups. Then $G$ must be finite and cyclic.

Proof:

Let $M_1,M_2$ be the maximal subgroups of $G$. Suppose $(M_1\cup M_2)\subsetneq G$. Then $G$ must be cyclic, because any element in $G\setminus (M_1\cup M_2)$ generates the whole group. So either $G\cong\mathbb{Z}$, or $G$ is finite. But $\mathbb{Z}$ has infinitely many maximal subgroups, so $G$ must be finite and cyclic.

Suppose instead that $G=M_1\cup M_2$. This is not possible, because no group can be a union of two proper subgroups. To see this, consider an element $a\in M_1\setminus M_2$ and $b\in M_2\setminus M_1$. Then $ab\in G\setminus (M_1\cup M_2)$, but the latter set is empty. $\blacksquare$

EDIT: As @YCor pointed out in the comments below, the two proofs above only work for groups in which every proper subgroup is contained in a maximal subgroup.

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If $G$ has exactly three maximal subgroups $M_1,M_2,M_3$, then again we can split cases:

Case 1: $(M_1\cup M_2\cup M_3)\subsetneq G$.

In this case, any element in $G\setminus (M_1\cup M_2\cup M_3)$ generates $G$, so $G$ is cyclic.

Case 2: $G=M_1\cup M_2\cup M_3$

In this case, Scorza's theorem tells us that $|G:M_1|=|G:M_2|=|G:M_3|=2$. This is possible if, for example, $G\cong V_4$, the Klein 4-group. $V_4$ has exactly three maximal subgroups, but it is not cyclic.

But it is possible for a group to be a union of three maximal subgroups, even though those three maximal subgroups are not the only maximal subgroups. For example, the elementary abelian $2$-group of order $8$ is a union of three maximal subgroups, but it has more than three maximal subgroups.

Question: Is there a "nice" (I admit I have no precise definition, I'm referring to the know-it-when-you-see-it sort of "nice") classification of groups with exactly $3$ maximal subgroups?

Further question: I don't expect there is a detailed classification of groups with exactly $n$ maximal subgroups for every $n$, because that would yield a classification of all finite groups, and if such a thing had been achieved, I think I would have heard of it.

But what are some interesting things that can be said about the collection of groups with exactly $n$ maximal subgroups, for $n\geq 3$?

Answer 1

A finite group with exactly three maximal subgroups must be cyclic (of order divisible by exactly 3 primes), or a 2-group. It's known that a noncyclic finite group must have more than one conjugacy class of maximal subgroups. So if $G$ has three maximal subgroups, one of them is normal and the other two are in conjugacy classes of size at most two. But if $\{ M_{2}, M_{3} \}$ is a conjugacy class of maximal subgroups, then $M_{2} = N_{G}(M_{2})$ and $|G : N_{G}(M_{2})| = 2$, which forces $M_{2} \unlhd G$. This implies that $G$ is nilpotent. From this, it follows that either $G$ is a $p$-group with exactly three maximal subgroups, or $G$ is a product of three cyclic groups of prime power order. But a non-cyclic $p$-group has at least $p+1$ maximal subgroups, and so $p = 2$ is the only option.