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

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.

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 $$n\geq 3$$?

• Experimenting with GAP suggests that any group with exactly $3$ maximal subgroups is a $2$-group (and there are many such $2$-groups). Is there some simple argument for why this is true (or a counterexample)? Dec 2, 2023 at 3:48
• The group $C_{p^\infty}\times C_6$ has exactly 2 maximal proper subgroups but is not cyclic. (In your reasoning "any element in $G-M_1\cup M_2$ generates the whole group" is not true — indeed not every proper subgroup is contained in a maximal one.)
– YCor
Dec 2, 2023 at 9:04
• Oh, I see! It's strange that Zorn's Lemma can be used to prove that "every proper ideal is contained in a maximal ideal", but it can't be used to prove that "every proper subgroup is contained a maximal subgroup". Dec 2, 2023 at 10:24
• @YCor “Not every proper subgroup is contained in a maximal one.” 😱 HOW? WHY? WHAT IS THIS MAGIC? 😵 I feel like I've just been told that Santa Claus doesn't exist. (And yes, indeed, when you think of it, it's obvious, but it makes you reconsider why maximal ideals exist.) This is one for the big list. Dec 2, 2023 at 10:58
• @Gro-Tsen I suppose it is worth mentioning that if $G$ is finitely generated then every proper subgroup is indeed contained in a maximal subgroup (proved using Zorn’s Lemma as usual). Dec 2, 2023 at 13:00

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.