# Finite groups with few conjugacy classes of maximal subgroups

1. Let $$c$$ be a positive integer, $$G$$ a finite group with at most $$c$$ conjugacy classes of maximal subgroup. What can we say about $$G$$?
2. Same question, but this time $$G$$ is a finite group with at most $$c$$ conjugacy classes of core-free maximal subgroup.

Notes:

• Question 1 is equivalent to asking about groups with at most $$c$$ primitive permutation representations (up to permutation equivalence).
• Question 2 is equivalent to asking about groups with at most $$c$$ faithful primitive permutation representations (up to permutation equivalence).
• I'm interested in the same idea, but instead of permutation equivalence, one considers permutation isomorphism. This would change the original questions so that one considers $$\textrm{Aut}(G)$$-conjugacy classes, instead of just $$G$$-conjugacy classes.

What might an answer look like:

• If $$c=1$$, then one can prove that $$G$$ must be cyclic of prime power order. Core-free will require $$G$$ is of prime order. Can one give a full classification for $$c=2,3,4,\dots$$? Non-solvable groups enter at $$c=3$$ (e.g. $$A_5$$) -- I'm presuming that $$c=2$$ will still imply solvability, although I haven't written down a proof.
• If one considers the variant mentioned in the third bullet point of the notes above -- referring to $$\textrm{Aut}(G)$$-conjugacy classes -- then one also obtains elementary-abelian groups for $$c=1$$.
• I'm presuming that this stuff has been studied before so this question is also a reference-request. There is a conjecture about upper bounds for the number of maximal subgroups -- see my answer here... But my interest in the current question is specifically about very small values of $$c$$, so that conjecture is not so relevant...

See

Kano, Mikio. On the number of conjugate classes of maximal subgroups in finite groups. Proc. Japan Acad. Ser. A Math. Sci. 55 (1979), no. 7, 264--265. link

for proofs of the following ($$G$$ finite group):

(1) Suppose every non-normal maximal subgroup of $$G$$ has the same order. Then $$G$$ is solvable.

(2) If $$G$$ is simple, then $$G$$ has at least three conjugacy classes of maximal subgroups.

It follows from this that $$G$$ is solvable if $$c = 2$$:

Proof: Let $$G$$ be a finite group with exactly two conjugacy classes of maximal subgroups, with representatives $$M$$ and $$L$$. It follows from (2) that $$G$$ is not simple, so let $$N$$ be a proper nontrivial normal subgroup of $$G$$. Now $$G/N$$ has at most two conjugacy classes of maximal subgroups, so by induction $$G/N$$ is solvable. A nontrivial solvable group always has a maximal subgroup which is normal and of prime index, so either $$M$$ or $$L$$ is normal in $$G$$. If only one of them is normal, it follows from (1) that $$G$$ is solvable. If both are normal, then $$G$$ is nilpotent since all maximal subgroups are normal.

Furthermore if $$c = 2$$, then $$|G| = p^a q^b$$ for some primes $$p$$ and $$q$$. If $$G$$ is nilpotent, this is obvious. If $$G$$ is not nilpotent, then there exists a normal maximal subgroup $$M$$ and a non-normal maximal subgroup $$L$$. Take $$p$$ to be a prime dividing $$[G:L]$$, and let $$P$$ be a $$p$$-Sylow subgroup. Any subgroup containing $$N_G(P)$$ is self-normalizing, so $$N_G(P)$$ cannot be contained in $$M$$, but it also cannot be contained in a conjugate of $$L$$ since $$p$$ divides $$[G:L]$$. Thus $$P$$ is a normal subgroup of $$G$$. Now $$P$$ is not contained in $$L$$ so $$G/P$$ has a unique maximal subgroup, and hence $$G/P$$ is cyclic of prime power order.

• Great stuff -- thanks a lot for this.. Dec 10 '18 at 10:26
• Just to note that there are many other references for solvability when $c = 2$, as the answer by M. Farrokhi D. G. shows (the reference in this answer was first I could find). The statement (2) in my answer is also in Satz 6 in the paper by Pazderski. The other references seem to use very similar arguments. Dec 10 '18 at 14:57

This is not really a proper answer - just speculation about the case $$c=3$$.

It looks as though the simple groups with $$c=3$$ are $${\rm PSL}(2,q)$$ for $$q=5,7$$, and $$2^k$$ with $$k$$ prime. I have not checked that completely, but there is certainly enough known about maximal subgroups of simple groups to do so, and I am also confident that there are no more almost simple groups (i.e. groups with nonabelian simple socle) with $$c=3$$. Note also that, in the case of $${\rm PSL}(2,7)$$ only, two of the three classes are fused under an automorphism of $$G$$.

Assuming that is correct, a non-solvable group $$G$$ with $$c=3$$ would have to be a nonsplit extension of a nilpotent group $$N = \Phi(G)$$ by one of the above simple groups.

I see from the answer of M.Farrokhi D. G. that this result has been proved by V. A. Belonogov.

By checking databases of perfect groups, I see that there are examples with $$G/\Phi(G) = A_5$$ and $$|\Phi(G)| = 2, 2^5, 2^9, 3^4, 3^8, 5^3, 5^6$$.

In fact, there is a result in group cohomology that says that, for an finite group $$G$$ with order divisible by a prime $$p$$, and any $$n>0$$, there is an irreducible module $$M$$ over $${\mathbb F}_p$$ with $$H^n(G,M) \ne 0$$. So by applying this repeatedly with $$n=2$$, you can start with any of the above simple groups $$S$$ and a prime $$p$$ dividing $$|S|$$, and keep constructing downward extensions with finite $$p$$-groups, to produce examples $$G$$ with $$c=3$$, $$G/\Phi(G) \cong S$$, and $$\Phi(G)$$ a $$p$$-group of arbitrarily large order.

$$\mathbf{Added\ later}$$: following discussion in the comments, it seems relevant to summarize how maximal subgroups of a finite group $$G$$ are related to those of its nonabelian composition factors. This theory can be found in two papers published in the 1980s, one by Aschbacher and Scott, and one by Fletcher and Kovács, which are independent. The references are at the end. It is also closely related to the O'Nan-Scott Theorem, because maximal subgroups of $$G$$ correspond to its primitive permutation representations.

Let $$M/N$$ be a chief factor of $$G$$. If $$M/N$$ is elementary abelian, then its complements $$C/N$$ in $$G/N$$ (if any) are maximal and correspond to the affine case of O'Nan-Scott.

The remaining maximal subgroups of $$G$$ are related to nonabelian chief factors $$M/N$$. We are considering complements that contain $$N$$ but not $$M$$, so we may as well assume that $$N=1$$. Note that $$M \cap C_G(M) = 1$$. Suppose first that $$C_G(M) = 1$$. Then $$G \le {\rm Aut}(M) \cong {\rm Aut}(S) \wr S_n$$, where $$M \cong S^n$$ with $$S$$ nonabelian simple.

Let $$S_1$$ be one of the copies of $$S$$ in $$M$$, and let $$N_1 = N_G(S_1)$$, and $$C_1 = C_G(S_1)$$. So $$N_1/C_1$$ is isomorphic to a subgroup of $${\rm Aut}(S)$$ containing $$S$$, i.e. an almost simple group with socle $$S$$. It turns out that each conjugacy class $$C$$ of maximal subgroups of $$N_1/C_1$$ corresponds to a conjugacy class of maximal subgroups of $$G$$ whose intersection with $$N_1$$ projects onto the groups in $$C$$. They correspond to primitive groups of product type in O'Nan-Scott.

In particular, the number of conjugacy classes of maximal subgroups of $$G$$ is greater than or equal to the number for $$N_1/C_1$$ and it is greater than this when $$n>1$$, because there are maximal subgroups containing $$M$$. In particular, if $$G$$ has $$3$$ such classes, then we must have $$n=1$$, and $$G = N_1$$, which is how we can reduce this problem to the almost simple case when $$c=3$$.

Of course there are usually other maximal subgroups associated with $$M$$ that do not contain $$M$$. There may be some whose intersection with $$M$$ maps onto the direct factors $$S_i$$ of $$M$$, and these correspond to primitive groups of diagonal type in O'Nan-Scott. Occasionally there are some that complement $$M$$: the twisted wreath product type. Finally, when $$C_G(M) \ne 1$$, and there is a chief factor $$M'$$ of $$C$$ isomorphic to $$M$$, then there may be maximal subgroups that intersect $$M \times M'$$ in a diagonal subgroup: these correspond to the primitive groups with two distinct minimal normal subgroups in O'Nan-Scott.

The code in Magma and GAP for computing maximal subgroups of finite groups uses this theory, which is how I got involved in it. It has the disadvantage that it needs to know the maximal subgroups of all almost simple groups, which means that data libraries have to be compiled containing this information, which can never be complete, but it is difficult to imagine a better way of doing this.

$$\mathbf{References}$$

M. Aschbacher and L. Scott. Maximal subgroups of finite groups. J. Algebra, 92:44–80, 1985.

F. Gross and L.G. Kovács. On normal subgroups which are direct products. J. Algebra, 90:133–168, 1984.

• Thanks Derek, this is really interesting. Let me idly speculate on the basis of 2 minutes of thought: do you think there could be a function $f(c)$ such that if $G$ is a group with at most $c$ conjugacy classes of maximal subgroups, then all simple groups involved with $G$ are of Lie type of rank at most $f(c)$, or alternating of size at most $f(c)$ (or sporadic)? I wonder if one could also bound the number of non-abelian compostion factors by a function of $c$? (I realise this is asymptotics which is not in the spirit of the original post...) Dec 10 '18 at 10:32
• The answer to your first question is yes, essentially because the number of parabolic maximal subgroups of groups of Lie type increases with their Lie rank. For example ${\rm PSL}_n(q)$ has $n-1$ of these. And $A_n$ has about $n/2$ intransitive maximal subgroups with two orbits. But $A_5 \wr C_p$ has $5$ classes of maximal subgroups for all primes $p$, so the answer to the second question is no. Dec 10 '18 at 10:51
• I think that the answer to your second question is yes if you replace composition factor by chief factor. Dec 10 '18 at 11:52
• Derek, great comments. Thanks. One thing though: it's not completely clear to me that whenever, say, $A_n$, is a section of a finite group $G$, then its intransitive maximals will yield distinct maximals for $G$... Am I just being dense? Dec 10 '18 at 14:30
• No, you are not being dense, I am using a result about maximal subgroups of subgroups of wreath products here. I will add an explanation to my answer, but that will have to wait until this evening. Dec 10 '18 at 15:03

Here are a few more references:

• S. Adnan, On groups having exactly $$2$$ conjugacy classes of maximal subgroups, Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. (8) 66 (1979), no. 3, 175–178. link
• S. Adnan, On groups having exactly $$2$$ conjugacy classes of maximal subgroups. II, Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Natur. (8) 68 (1980), no. 3, 179. link
• W. Shi, Finite groups having at most two classes of maximal subgroups of the same order, Chinese Ann. Math. Ser. A 10 (1989), no. 5, 532–537.
• V. A. Belonogov, Finite groups with three classes of maximal subgroups, Math. Sb. 131 (1968), 225–239. link
• X. Li, Finite groups having three classes of maximal subgroups of the same order, Acta Math. Sinica 1 (1994), 108–115.
• G. Pazderski, Über maximale Untergruppen endlicher Gruppen, Math. Nachr. 26 (1963/1964), 307–319.
• J. Shi, W. Shi, and C. Zhang, The type of conjugacy classes of maximal subgroups and characterization of finite groups, Comm. Algebra 38 (2010), no. 1, 143–153. link
• J. Wang, The number of maximal subgroups and their types, Pure Appl. Math. (Xi'an) 5 (1989), 24–33.

The main theorem of Belonogov's article states that

Theorem. A finite nonsolvable group has three conjugacy classes of maximal subgroups if and only if $$G/\Phi(G)$$ is isomorphic to $$PSL(2,7)$$ or $$PSL(2,2^p)$$ for some prime $$p$$.

• Thank you so much -- these references are really helpful. Dec 10 '18 at 10:26