Wielandts program on the study of X -maximal subgroups of finite - - PowerPoint PPT Presentation

Wielandt’s program on the study of X -maximal subgroups of finite groups

Danila Revin

1Sobolev Institute of Mathematics, Novosibirsk, Russia

revin@math.nsc.ru

Yichang, G2D2, August 18, 2019

The report is based on some old ideas expressed by Helmut Wielandt in his talk to the famous Santa-Cruz conference on finite groups in 1979 and in his lectures given in T¨ ubingen in 1963-1964.

• H. Wielandt, Zusammengesetzte Gruppen: H¨
• lder Programm

heute, The Santa Cruz conf. on finite groups, Santa Cruz,

• 1979. Proc. Sympos. Pure Math., 37, Providence RI: Amer.
• Math. Soc., 1980, 161–173.
• H. Wielandt, Zusammengesetzte Gruppen endlicher Ordnung,

Vorlesung an der Universit¨ at T¨ ubingen im Wintersemester 1963/64. Helmut Wielandt: Mathematical Works, Vol. 1, Group theory (ed. B. Huppert and H. Schneider, de Gruyter, Berlin, 1994), 607–516.

Notation and terminology

◮ In the talk, ’a group’ means ’a finite group’ and G is a group. ◮ Always a group G acts on a set Ω on the right:

Ω × G → Ω (ω, g) → ωg

◮ Recall, a series

G = G0 ≥ G1 ≥ · · · ≥ Gn = 1 is said to be normal (resp., subnormal, composition), if, for all i = 1, . . . , n,

• Gi G;
• Gi Gi−1 (such Gi is called subnormal in G, Gi G); and
• Gi ⊳ Gi−1 and Gi−1/Gi is simple. In this case,

the sections Gi−1/Gi are called composition factors of G.

◮ Every finite group has a composition series.

Up to isomorphism, the multiset of composition factors of G does not depend on the choose of a composition series and is an invariant of G.

◮ We fix a complete class X of finite groups.

According to Wielandt, a non-empty class X is said to be complete if X is closed under taking

• subgroups (G ∈ X and H ≤ G ⇒ H ∈ X ),
• homomorphic images

(G ∈ X and φ : G → G ∗ ⇒ G φ ∈ X ) and

• extensions (A G and A, G/A ∈ X ⇒ G ∈ X ).

◮ If G ∈ X then we say that G is an X -group. ◮ Examples of complete classes:

• The class G of finite groups is complete.
• The class S of finite solvable groups is complete.
• Take a subset π of the set P of all primes.

A group G is a π-group, if p | |G| ⇒ p ∈ π for all p ∈ P. The classes Gπ and Sπ of finite π-groups and finite solvable π-groups, respectively, are complete.

• Given n > 0, the class Gn of all groups G such that |S| < n for

every nonabelian composition factor S of G is complete.

◮ For every complete X , we have

Sπ ⊆ X ⊆ Gπ, where π = π(X ) is the set of primes p such that p divides the

• rder of some G ∈ X .

• I. X -Maximality vs maximality

A group is an active object and is acting on various sets. A source

• f actions of G is G itself: G acts
• on the set of its elements via (right) multiplication,

conjugation etc.,

• on the set of cosets of a subgroup via multiplication,
• on the set of subgroups of G via conjugation etc.

If G acts on a set then G acts transitively on every orbit. The group itself is the source of all its transitive actions: every transitive action of G is equivalent to the action of G on the cosets of a point stabilizer. Often the study of transitive actions can be reduced to the study of so-called primitive actions.

In group-theoretical terms, a transitive action of G is primitive if and only if a point stabilizer is a maximal subgroup of G. A subgroup H of G is said to be maximal if H < G and H ≤ K ≤ G implies either K = H or K = G for all subgroups K

• f G. In other words, for subgroups, the maximality means the

maximality among the proper subgroups w.r.t. inclusion. The O’Nan–Scott theorem provides some classification of the faithful primitive actions. An important case of a primitive permutation group is a so-called almost simple group. Recall, a group is said to be almost simple, if it is isomorphic to a group G such that S ∼ = Inn(S) ≤ G ≤ Aut(S) for a nonabelian simple group S which is the socle of G. Every such G can be considered as a primitive permutation group in which a point stabiliser is a maximal subgroup of G not containing S. There are no hope to obtain the complete list of maximal subgroups of the simple groups.

◮ A subgroup H of G is said to be maximal

if H < G and H ≤ K < G ⇒ K = H for all K. In other words, for subgroups, the maximality means the maximality among the proper subgroups w.r.t. inclusion.

◮ For a complete class X , a subgroup H of G is called an

X -maximal subgroup or a maximal X -subgroup if H ∈ X and H ≤ K ∈ X ⇒ K = H for all K ≤ G. In other words, for subgroups, the X -maximality means the maximality among the X -subgroups w.r.t. inclusion. Denote by mX (G) the set of X -maximal subgroups of G.

◮ If G ∈ X then mX (G) = {G} while H < G for any

maximal H.

◮ Let p be a prime. Then mGp(G) = Sylp(G). ◮ If G is a p-group and H = 1 is a maximal subgroup of G then

there are no complete classes X such that H ∈ mX (G). The definitions are similar, but the maximality and the X -maximality are absolutely distinct concepts.

Let G be an almost simple group such that S = Inn(S) ≤ G ≤ Aut(S) for nonabelian simple S. Denote X = G|S|, the (complete) class of groups whose n.-a. composition factors are of order less than |S|. Then H is a maximal subgroup of G not including S (i. e. H is a point stabilizer in G considered as a primitive permutation group) iff H ∈ mX (G). Thus, in the almost simple groups, the maximality of a subgroup means an X -maximality for some complete X .

We are interested in the following

Main Problem

Given a group G and a complete class X , determine mX (G). As well as the study of maximal subgroup is the crucial point in studying the subgroup structure of a group, studying the X -subgroups of a group (for example, a classical problem dating back to works by Galois and Jordan: determine solvable subgroups

• f the symmetric groups) can be reduced to the above problem.

• II. Reduction to factors of a subnormal series:

X -submaximal subgroups

Main Problem

Given a group G and a complete class X , determine mX (G). Following to an approach practiced in the group theory, it is natural to try to reduced this problem to the factors of a (sub)normal (for example, composition) series of G by replacing X -maximal subgroups with their projections.

Let a series of subgroups G = G0 ≥ G1 ≥ . . . ≥ Gn−1 ≥ Gn = 1 (1)

• f a group G be subnormal, i. e. Gi Gi−1 for every i = 1, . . . , n.

For a subgroup H of G we denote by Hi = (H ∩ Gi−1)Gi/Gi the projection of H on the factor G i = Gi−1/Gi. Example: G = G0 G1 G2 = 1 (this series is necessarily normal) H1 = HG1/G1 is the image of H in G/G1 under G → G/G1 and H2 = H ∩ G1 is the intersection of H and G1. Note, that in general, it may happen that:

◮ for given subgroups Hi of the factor G i of (1),

there is no subgroups of G with projections Hi,

◮ Hi = K i for subgroups H, K ≤ G and all i = 1, . . . , n,

but H and K are nonisomorphic.

Take a subnormal series G = G0 ≥ G1 ≥ . . . ≥ Gn−1 ≥ Gn = 1

• f a group G (i. e. Gi Gi−1 for every i = 1, . . . , n)

and a subgroup H ≤ G. If Hi are X -maximal in G i = Gi−1/Gi for every i = 1, . . . , n, then H ∈ mX (G). What about the converse statement?

Take a subnormal series G = G0 ≥ G1 ≥ . . . ≥ Gn−1 ≥ Gn = 1

• f a group G (i. e. Gi Gi−1 for every i = 1, . . . , n)

and a subgroup H ≤ G. If Hi are X -maximal in G i = Gi−1/Gi for every i = 1, . . . , n, then H ∈ mX (G). What about the converse statement?

Example 1 Let p be a prime and X = Gp the class of p-group. Let H ∈ mX (G), i.e., H is a Sylow p-subgroup of G, and A G. We have a normal series G A 1, HA/A is a Sylow p-subgroup of G/A, i.e. H1 = HA/A ∈ mX (G/A) = mX

• G 1

, H ∩ A is a Sylow p-subgroup of A, i.e. H2 = H ∩ A ∈ mX (A) = mX

• G 2

.

Example 2 Let X = G168 (or X = S , or X = G{2,3}), let G = PGL2(7), and let A = PSL2(7) ∼ = PGL3(2). We have a normal series G A 1, where G/A is of order 2 and A = PSL2(7) is simple. |A| = 168 = 23 · 3 · 7, |G| = 24 · 3 · 7. A Sylow 2-subgroup H of G is maximal in G. In particular, H ∈ mX (G). But H2 = H ∩ A / ∈ mX (A).

Wielandt’s idea was to find a generalization of the concept of an X -maximal subgroup which would be close to this one, informative (in particular, differ from the concept of an X -subgroup) and would well behave in the case of intersections with normal (equivalently, with subnormal) subgroups.

Definition (Wielandt, 1979)

A subgroup H of a group G is called a submaximal X -subgroup (or X -submaximal subgroup, notation H ∈ smX (G)) if there is a monomorphism φ : G → G ∗ into a finite group G ∗ such that G φ G ∗ and Hφ = K ∩ G φ for some K ∈ mX (G ∗). Briefly, H ∈ smX (G) if there are G ∗ and K ∈ mX (G ∗) such that G G ∗ and H = K ∩ G. If H ∈ smX (G) and A G then H ∩ A ∈ smX (A). mX (G) ⊆ smX (G). Example of an X -submaximal but not X -maximal subgroup: X := G{2,3} is the class of {2, 3}-groups; G := PSL2(7) = PGL3(2); a Sylow 2-subgroup of G is X -submaximal but not X -maximal.

H ∈ smX (G) if there are G ∗ and K ∈ mX (G ∗) such that G G ∗ and H = K ∩ G. mX (G) ⊆ smX (G). If H ∈ smX (G) and A G then H ∩ A ∈ smX (A). (1) every X -maximal subgroup is X -submaximal ; (2) if A G and H ≤ G is X -submaximal then H ∩ A is X -submaximal in A ; (3) not every X -subgroup is X -submaximal ? In general, arbitrary X -subgroup is not X -submaximal. An obstruction here is the Wielandt–Hartley theorem.

H ∈ smX (G) if there are G ∗ and K ∈ mX (G ∗) such that G G ∗ and H = K ∩ G. mX (G) ⊆ smX (G). If H ∈ smX (G) and A G then H ∩ A ∈ smX (A). (1) every X -maximal subgroup is X -submaximal ; (2) if A G and H ≤ G is X -submaximal then H ∩ A is X -submaximal in A ; (3) not every X -subgroup is X -submaximal. In general, arbitrary X -subgroup is not X -submaximal. An obstruction here is the Wielandt–Hartley theorem.

H ∈ smX (G) if there are G ∗ and K ∈ mX (G ∗) such that G G ∗ and H = K ∩ G. mX (G) ⊆ smX (G). If H ∈ smX (G) and A G then H ∩ A ∈ smX (A). (1) every X -maximal subgroup is X -submaximal ; (2) if A G and H ≤ G is X -submaximal then H ∩ A is X -submaximal in A ; (3) not every X -subgroup is X -submaximal. In general, arbitrary X -subgroup is not X -submaximal. An obstruction here is the Wielandt–Hartley theorem.

• III. Wielandt–Hartley theorem

Recall, π(X ) is the set of primes p such that p divides the order of some X ∈ X . Equivalently, π(X ) = {p ∈ P | Zp ∈ X }. Notice, G does not contain nontrivial X -subgroups iff |G| is divisible by no primes in π(X ). In this case we say that G is an X ′-group or G ∈ X ′. It is easy to see that the class X ′ is complete and if π(X ) = P then X ′ = {1}.

Theorem (H.Wielandt (1963), B.Hartley (1971))

Let A G and H ∈ mX (G). Then NA(H ∩ A)/(H ∩ A) ∈ X ′. In particular, assume A G and H ∈ mX (G). Then H ∩ A = 1 iff A ∈ X ′. All known proofs of the Wielandt–Hartley theorem use the Schreier conjecture on the solvability of the group of outer automorphisms

• f every simple group. The theorem is proven by Wielandt in his

lectures 1963/64. Hartley published his proof in 1971. This theorem can be found in the classical book

• M. Suzuki, Group Theory II, Springer-Verlag,

New York–Berlin–Heidelberg–Tokyo, 1986.

G ∈ X ′ if G does not contain nontrivial X -subgroups. Theorem (Wielandt–Hartley) Let A G and H ∈ mX (G). Then NA(H ∩ A)/(H ∩ A) ∈ X ′.

Corollary below is a particular case of the theorem. It is fundamental in the analysis of maximal subgroups of almost simple

• groups. It is used to find so-called novelties in such groups. It was
• btained independently (and without references to Wielandt) by

R.Wilson (1985), by M.Liebeck, C.Praeger and J.Saxl (1988) and by others and can be found in

• J. N. Bray, D. F. Holt, C. M. Roney-Dougal, The Maximal

Subgroups of the Low-Dimensional Finite Classical Groups. Cambridge, 2013. If G is almost simple with the socle S and X = G|S|, then π(X ) = P and the Wielandt–Hartley theorem implies

Corollary

Let G be almost simple with a simple socle S, and let H be a maximal subgroup of G. Then NS(H ∩ S) = H ∩ S.

G ∈ X ′ if G does not contain nontrivial X -subgroups. Theorem (Wielandt–Hartley) Let A G and H ∈ mX (G). Then NA(H ∩ A)/(H ∩ A) ∈ X ′. H ∈ smX (G) if there are G ∗ and K ∈ mX (G ∗) such that G G ∗ and H = K ∩ G.

In his talk in 1979, Wielandt announced an analog of this theorem for the case where AG.

Theorem (Wielandt–Hartley, a strong version)

Let AG and H ∈ mX (G). Then NA(H ∩ A)/(H ∩ A) ∈ X ′. Originally, Wielandt formulated the above theorem in the following equivalent form:

Theorem (Wielandt–Hartley, a strong version)

If H ∈ smX (G) then NG(H)/H ∈ X ′. Wielandt’s proof of the above statements has not been published to this day. Recently S. Skresanov found a proof of this theorem.

Theorem (Wielandt–Hartley, a strong version)

• Let AG and H ∈ mX (G). Then NA(H ∩ A)/(H ∩ A) ∈ X ′.
• If H ∈ smX (G) then NG(H)/H ∈ X ′.

Corollary 1

Let a group G have a subnormal series G = G0 ≥ G1 ≥ . . . ≥ Gn−1 ≥ Gn = 1. If H, K ∈ smX (G) satisfy Hi = K i for all i = 1, . . . , n, then H and K are conjugated in H, K.

Corollary 1.1

Let a group G have a subnormal series G = G0 ≥ G1 ≥ . . . ≥ Gn−1 ≥ Gn = 1. If H, K ∈ mX (G) satisfy Hi = K i for all i = 1, . . . , n, then H and K are conjugated in H, K.

G ∈ X ′ if G does not contain nontrivial X -subgroups. H ∈ smX (G) if there are G ∗ and K ∈ mX (G ∗) such that G G ∗ and H = K ∩ G.

Theorem (Wielandt–Hartley, a strong version)

If H ∈ smX (G) then NG(H)/H ∈ X ′. The following statement will be important for further discussion.

Corollary 2

Let G = G1 × G2 × . . . × Gn. Then smX (G) = {H1, H2, . . . , Hn | Hi ∈ smX (Gi), i = 1, 2, . . . , n}, where H1, H2, . . . , Hn ∼ = H1 × · · · × Hn. Every minimal normal subgroup of a group is a direct product of simple groups.

Proposition

For a subgroup H of a nonabelian simple group S, the following conditions are equivalent. (1) H ∈ smX (S). (2) H = S ∩ K for some K ∈ mX (AutS) where S = InnS.

• IV. How can we find X -maximal subgroups?

Main Problem

Given a group G and a complete class X , determine mX (G).

Problem I

Given simple S and a complete class X , determine smX (S).

Problem I∗

Given simple S and a complete class X , determine mX (Aut(S)). Assume, we have solved Problem I for some X and all simple S. Then have a fishing net to catch (up to conjugation) every X -maximal subgroup M of arbitrary group G.

Main Problem

Given a group G and a complete class X , determine mX (G).

Problem I

Given simple S and a complete class X , determine smX (S).

Problem I∗

Given simple S and a complete class X , determine mX (Aut(S)). Assume, we have solved Problem I for some X and all simple S. Then have a fishing net to catch (up to conjugation) every X -maximal subgroup M of arbitrary group G.

Main Problem

Given a group G and a complete class X , determine mX (G).

Problem I

Given simple S and a complete class X , determine smX (S).

Problem I∗

Given simple S and a complete class X , determine mX (Aut(S)). Assume, we have solved Problem I for some X and all simple S. Then have a fishing net to catch (up to conjugation) every X -maximal subgroup M of arbitrary group G.

Main Problem

Given a group G and a complete class X , determine mX (G).

Problem I

Given simple S and a complete class X , determine smX (S).

Problem I∗

Given simple S and a complete class X , determine mX (Aut(S)). Assume, we have solved Problem I for some X and all simple S. Then have a fishing net to catch (up to conjugation) every X -maximal subgroup M of arbitrary group G.

Homomorphic images of X -maximal subgroups exhibit an even more irregular behavior than one for intersections with (sub)normal subgroups.

Proposition (Wielandt).

In general, if A G and H ∈ mX (G) then HA/A / ∈ mX (G/A): Assume, A and B are groups and X -maximal subgroups of A are not conjugate. Let G = A ≀ B be the regular wreath product and let φ : G → B be the natural epimorphism. Then, for any X -subgroup K ≤ B (not only for K ∈ mX (B)!), there is a maximal X -subgroup H of G such that Hφ = K.

A group G is said to be X -separable, if G have a (sub)normal series G = G0 ≥ G1 ≥ . . . ≥ Gn−1 ≥ Gn = 1 such that, for every i = 1, . . . , n, the section G i = Gi−1/Gi is either an X -group or an X ′-group (i. e. does not contain non-trivial X -subgroups).

X -Reduktionssatz for X -separable groups (Chunikhin, Wielandt).

Let A be X -separable. Then, for G A, we have M ∈ mX (G) ⇒ MA/A ∈ mX (G/A) and and the map M → MA/A, mX (G) → mX (G/A) induces a bijection between the sets of conjugacy classes of X -maximal subgroups in G and G/A.

Suppose, smX (S) is known for every nonabelian simple S. Suppose, G is a group and mX (G0) is known if |G0| < |G|. We need to find our M ∈ mX (G).

◮ If G is simple then mX (G) ⊆ smX (G) is known. ◮ Let A G be minimal normal in G and let

: G → G/A be the canonical epimorphism. If A is a X - or X ′-group then H → H = HA/A is a bijection between the conjugacy classes which form mX (G) and mX (G); and mX (G) is known (|G| < |G|). Assume, A is not a X - or X ′-group. In particular, A is nonabelian and A = S1 × · · · × Sn for nonabelian simple S1, . . . , Sn conjugate in G.

◮ M ∩ A ∈ smX (A) = {H1, . . . , Hn | Hi ∈ smX (Si)} is known. ◮ M ≤ NG(M ∩ A) and |NG(M ∩ A)| < |G|,

since A is minimal normal and is not a X -group.

◮ G acts on the set of conjugacy classes of X -submaximal

subgroups of A, and NG(M ∩ A) is the stabilizer of the class containing M ∩ A. We claim that, we can reduce our problem to the same problem for the stabilizers of such classes in G.

◮ The Wielandt–Hartley Theorem implies that

NA(M ∩ A)/(M ∩ A) ∈ X ′, while M ∩ A ∈ X . So, NA(M ∩ A) is X -separable. In view of the X -Reduktionssatz there is a bijection between the sets of conjugacy classes of X -maximal subgroups in NG(M ∩ A) and NG(M ∩ A)/NA(M ∩ A).

◮ Since

NG(M ∩ A)/NA(M ∩ A) ∼ = ANG(M ∩ A)/A = NG(M ∩ A), there is a bijection between the conjugacy classes which form mX (NG(M ∩ A)) and mX (NG(M ∩ A)).

◮ Since |NG(M ∩ A)| < |G|, if we can find NG(M ∩ A), then we

can find mX (NG(M ∩ A)) and mX (NG(M ∩ A)) and M is found.

• V. When can we really find

the X -maximal subgroups?

• Problem. Given X , describe X -submaximal subgroups of the

finite simple groups. Study properties of these subgroups: conjugacy classes, the pronormality, the conjugacy in the automorphism group etc. If this problem were solved for all X , we would know the maximal subgroups of the simple and almost simple groups. But in order to describe maximal subgroups of PSLn(q) we need to know all absolutely irreducible modular representation of all almost simple

• groups. Now the latter task does not seem to be solvable.

Let us try to highlight some cases where one can achieve success in solving the problem. There are two possibilities for X : X either contains the group of order 2 or not.

• Problem. Given X , describe X -submaximal subgroups of the

finite simple groups. Study properties of these subgroups: conjugacy classes, the pronormality, the conjugacy in the automorphism group etc. If this problem were solved for all X , we would know the maximal subgroups of the simple and almost simple groups. But in order to describe maximal subgroups of PSLn(q) we need to know all absolutely irreducible modular representation of all almost simple

• groups. Now the latter task does not seem to be solvable.

Let us try to highlight some cases where one can achieve success in solving the problem. There are two possibilities for X : X either contains the group of order 2 or not.

Consider the case Z2 / ∈ X . Since X is closed under subgroups, the Cauchy theorem implies that X consists of groups of odd

• rder. The Feit–Thompson theorem together with the inclusion

Sπ ⊆ X ⊆ Gπ, where π = π(X ) and 2 / ∈ π, imply Sπ = X = Gπ.

Conjecture If Z2 /

∈ X then smX (G) = mX (G) for every G.

Proposition

If X is complete, then the following statements are equivalent. (1) smX (G) = mX (G) for every G; (2) smX (S) = mX (S) for every simple S; (3) S ∩ K ∈ mX (S) for every simple S and K ∈ mX (AutS). (4) A ∩ K ∈ mX (A) for every G, A G and K ∈ mX (G). Why we can describe X -(sub?)maximal subgroups of simple classical groups, if Z2 / ∈ X ? An X -maximal subgroup is solvable and is contained in a member if an Aschbacher class. We can apply to this member our inductive argument, as above.

Consider the case Z2 ∈ X . What can we do? Nothing, in general case. But... Notice, a Sylow 2-subgroup of G is an X -group and is contained in an X -maximal subgroup of odd index. Let us try to determine the members of mX (G) of odd index. It is easy to see that if H ≤ G is a subgroup of odd index, then every projection Hi of H on a section G i of a subnormal series is of odd index in G i. Assume, we have determined all X -submaximal subgroups of odd indices of all simple groups. Then we can apply our above arguments to arbitrary X -maximal subgroup of every finite group.

• Problem. Given X such that Z2 ∈ X , describe X -submaximal

subgroups of odd indices of the finite simple groups. An X -maximal subgroup of odd index is included in a maximal subgroup of odd index. The maximal subgroups of odd indices of simple groups are known due to W.Kantor, M.Liebeck, J.Saxl and are studied in details in works by N.Maslova. We can apply our arguments to these subgroups. In the symmetric groups, π-maximal (C.D.H.Cooper, 1971) and maximal solvable (K.Korotitskii, 2019) OI-subgroups are described.

Consider the case Z2 ∈ X . What can we do? Nothing, in general case. But... Notice, a Sylow 2-subgroup of G is an X -group and is contained in an X -maximal subgroup of odd index. Let us try to determine the members of mX (G) of odd index. It is easy to see that if H ≤ G is a subgroup of odd index, then every projection Hi of H on a section G i of a subnormal series is of odd index in G i. Assume, we have determined all X -submaximal subgroups of odd indices of all simple groups. Then we can apply our above arguments to arbitrary X -maximal subgroup of every finite group.

• Problem. Given X such that Z2 ∈ X , describe X -submaximal

subgroups of odd indices of the finite simple groups. An X -maximal subgroup of odd index is included in a maximal subgroup of odd index. The maximal subgroups of odd indices of simple groups are known due to W.Kantor, M.Liebeck, J.Saxl and are studied in details in works by N.Maslova. We can apply our arguments to these subgroups. In the symmetric groups, π-maximal (C.D.H.Cooper, 1971) and maximal solvable (K.Korotitskii, 2019) OI-subgroups are described.

Consider the case Z2 ∈ X . What can we do? Nothing, in general case. But... Notice, a Sylow 2-subgroup of G is an X -group and is contained in an X -maximal subgroup of odd index. Let us try to determine the members of mX (G) of odd index. It is easy to see that if H ≤ G is a subgroup of odd index, then every projection Hi of H on a section G i of a subnormal series is of odd index in G i. Assume, we have determined all X -submaximal subgroups of odd indices of all simple groups. Then we can apply our above arguments to arbitrary X -maximal subgroup of every finite group.

• Problem. Given X such that Z2 ∈ X , describe X -submaximal

subgroups of odd indices of the finite simple groups. An X -maximal subgroup of odd index is included in a maximal subgroup of odd index. The maximal subgroups of odd indices of simple groups are known due to W.Kantor, M.Liebeck, J.Saxl and are studied in details in works by N.Maslova. We can apply our arguments to these subgroups. In the symmetric groups, π-maximal (C.D.H.Cooper, 1971) and maximal solvable (K.Korotitskii, 2019) OI-subgroups are described.

Consider the case Z2 ∈ X . What can we do? Nothing, in general case. But... Notice, a Sylow 2-subgroup of G is an X -group and is contained in an X -maximal subgroup of odd index. Let us try to determine the members of mX (G) of odd index. It is easy to see that if H ≤ G is a subgroup of odd index, then every projection Hi of H on a section G i of a subnormal series is of odd index in G i. Assume, we have determined all X -submaximal subgroups of odd indices of all simple groups. Then we can apply our above arguments to arbitrary X -maximal subgroup of every finite group.

• Problem. Given X such that Z2 ∈ X , describe X -submaximal

subgroups of odd indices of the finite simple groups. An X -maximal subgroup of odd index is included in a maximal subgroup of odd index. The maximal subgroups of odd indices of simple groups are known due to W.Kantor, M.Liebeck, J.Saxl and are studied in details in works by N.Maslova. We can apply our arguments to these subgroups. In the symmetric groups, π-maximal (C.D.H.Cooper, 1971) and maximal solvable (K.Korotitskii, 2019) OI-subgroups are described.

• VI. What it is done?

X -maximal subgroups of minimal non-solvable groups

• Problem. Given X , describe X -submaximal subgroups of the

simple groups. Study properties of these subgroups: the conjugacy, the pronormality, the conjugacy in the automorphism group etc. As a staring point, in his talk, Wielandt suggested (Frage g) to consider the following particular case of this general problem: Problem (Wielandt). Given X , describe X -submaximal subgroups of the minimal non-solvable groups. Study properties of these subgroups: conjugacy classes, the pronormality, the intravariancy, the conjugacy in the automorphism group etc. In Wielandt’s terminology, a subgroup H of a group G is said to be intravariant if the conjugacy class of H in G is AutG-invariant,

• i. e., for any α ∈ AutG there is g ∈ G such that Hα = Hg.

According to P. Hall, a subgroup H of a group G is said to be pronormal if H and Hg are conjugate in H, Hg for every g ∈ G. Now Wielandt’s problem is solved:

• W. Guo, R., Classification and properties of the π-submaximal

subgroups in minimal nonsolvable group”, Bulletin of Mathematical Sciences, 8:2 (2018), 325–351.

• Problem. Given X , describe X -submaximal subgroups of the

simple groups. Study properties of these subgroups: the conjugacy, the pronormality, the conjugacy in the automorphism group etc. As a staring point, in his talk, Wielandt suggested (Frage g) to consider the following particular case of this general problem: Problem (Wielandt). Given X , describe X -submaximal subgroups of the minimal non-solvable groups. Study properties of these subgroups: conjugacy classes, the pronormality, the intravariancy, the conjugacy in the automorphism group etc. In Wielandt’s terminology, a subgroup H of a group G is said to be intravariant if the conjugacy class of H in G is AutG-invariant,

• i. e., for any α ∈ AutG there is g ∈ G such that Hα = Hg.

According to P. Hall, a subgroup H of a group G is said to be pronormal if H and Hg are conjugate in H, Hg for every g ∈ G. Now Wielandt’s problem is solved:

• W. Guo, R., Classification and properties of the π-submaximal

subgroups in minimal nonsolvable group”, Bulletin of Mathematical Sciences, 8:2 (2018), 325–351.

A group G is minimal nonsolvable if G is nonsolvable while every proper subgroup of G is solvable. It is well-known and easy to show that G is a minimal nonsolvable group if and only if G/Φ(G) is a minimal simple group (i. e. a nonabelian finite simple group S such that S is minimal nonsolvable) where Φ(G) is the Frattini subgroup of G i. e. the intersection of the maximal subgroups of G. If G is minimal nonsolvable then Φ(G) = F(G) is the greatest normal nilpotent subgroup (the Fitting subgroup). In 1968, J. Thompson proved that S is a minimal simple group if and only if S is isomorphic to a group in the following list: (1) L2(2p), where p is a prime; (2) L2(3p), where p is an odd prime; (3) L2(p), where p is a prime, p > 3 and p2 + 1 ≡ 0 (mod 5); (4) Sz(2p), where p is an odd prime; (5) L3(3).

Reduction of Wielandt’s problem to minimal simple groups:

• Proposition. Let N = Φ(G) and H ∈ smX (G). Then

(1) mX (G/N) = {KN/N | K ∈ mX (G)}. (2) H ∩ N is the unique Hall X -subgroup OX (N) of N. (3) HN/N ∈ smX (G/N). (4) (HN/N) prn (G/N) iff H prn G. (5) H is intravariant in G iff the conjugacy class of HN/N in G/N is invariant under the image AutG of the map AutG → AutG/N given by the rule φ → φ where φ : Ng → Ngφ for φ ∈ AutG.

• Theorem. (W.Guo, R.) Let π = π(X ) and S a minimal simple
• group. Suppose |π ∩ π(S)| > 1 and π(S) π.

Then representatives of the conjugacy classes of submaximal π-subgroups, the information of their structure, π-maximality, pronormality, intravariancy, and the action of AutS on the set of conjugacy classes of submaximal π-subgroups can be specify in the corresponding table below.

The π-submaximal subgroups of S = L2(q) where q = 2p, p is a prime, in the case 2 / ∈ π Notation: πε = π ∩ π(q − ε), ε ∈ {+, −}. Cond. H N Pro. Intra. 1 π+ = ∅ C(q−1)π 1

• 2

π− = ∅ C(q+1)π 1

• In any cases H is π-maximal.

The π-submaximal subgroups of S = L2(q) where q = 2p, p is a prime, in the case 2 ∈ π Notation: πε = π ∩ π(q − ε), ε ∈ {+, −} Cond. H N Pro. Intra. 1 Eq : C(q−1)π 1

• 2

π+ = ∅ D2(q−1)π 1

• 3

π− = ∅ D2(q+1)π 1

• In any cases H is π-maximal.

The π-submaximal subgroups of S = L2(q) where q = 3p, p is an odd prime, in the case 2 / ∈ π Notation: πε = π ∩ π(q − ε), ε ∈ {+, −}. Cond. H N Pro. Intra. 1 3 ∈ π Eq : C(q−1)π 1

• 2

3 / ∈ π and π+ = ∅ C(q−1)π 1

• 3

π− = ∅ C(q+1)π 1

• In any cases H is π-maximal.

The π-submaximal subgroups of S = L2(q) where q = 3p, p is an odd prime, in the case 2 ∈ π Notation: πε = π ∩ π(q − ε), ε ∈ {+, −} Cond. H N Pro. Intra. 1 3 ∈ π Eq : C 1

2 (q−1)π

1

• 2

π+ = {2} D(q−1)π 1

• 3

D(q+1)π 1

• 4

3 ∈ π A4 1

• In any cases H is π-maximal.

The π-submaximal subgroups of S = L2(q) where q > 3 is a prime, q2 ≡ −1 (mod 5), in the case 2 / ∈ π Notation: πε = π ∩ π(q − ε), ε ∈ {+, −}. Cond. H N Pro. Intra. 1 q ∈ π Cq : C(q−1)π 1

• 2

q / ∈ π and π+ = ∅ C(q−1)π 1

• 3

π− = ∅ C(q+1)π 1

• In any cases H is π-maximal.

The π-submaximal subgroups of S = L2(q) where q > 3 is a prime, q2 ≡ −1 (mod 5), in the case 2 ∈ π Notation: πε = π ∩ π(q − ε), ε ∈ {+, −}, δ ∈ Aut(S) \ Inn(S), |δ| = 2, Aut(S) = Inn(S), δ ∼ = S : δ ∼ = PGL2(q)

Cond. H N H is not π-max if Pro. Int. q ∈ π Cq :C 1

2(q−1)π

1

• π+ ={2}, or

π+ ={2}, 3∈π, and 3 / ∈ π, or D(q−1)π 1 q ≡48 41∗ or

• q ≡8 1

π+ = {2, 3} and q ≡72 7, 31∗∗ π− ={2}, or π− ={2}, 3∈π, and 3 / ∈ π, or D(q+1)π 1 q ≡48 7∗ or

• q ≡8 −1

π− = {2, 3} and q ≡72 41, 65∗∗ 3 ∈ π and A4 1

• q ≡8 ±3

3 ∈ π and S4 2† q ≡8 ±1

• ✪

∗ H ∼

= D8 and H ≤ S4 ∈ mπ(S); ∗∗ H ∼ = D6 and H ≤ S4 ∈ mπ(S).

† are interchanged by δ.

The π-submaximal subgroups of S = Sz(q) where q = 2p, p is an odd prime, in the case 2 / ∈ π Notation: r =

• (2q) = 2(p+1)/2

π0 = π ∩ π(q − 1), πε = π ∩ π(q − εr + 1), ε ∈ {+, −}. Cond. H N Pro. Intra. 1 π0 = ∅ C(q−1)π 1

• 2

π+ = ∅ C(q−r+1)π 1

• 3

π− = ∅ C(q−r+1)π 1

• In any cases H is π-maximal.

The π-submaximal subgroups of S = Sz(q) where q = 2p, p is an odd prime, in the case 2 ∈ π Notation: r =

• (2q) = 2(p+1)/2

π0 = π ∩ π(q − 1), πε = π ∩ π(q − εr + 1), ε ∈ {+, −}. Cond. H N Pro. Intra. 1 E 1+1

q

: C(q−1)π 1

• 2

π0 = ∅ C(q−1)π : C4 1

• 3

π+ = ∅ D2(q−r+1)π 1

• 4

π− = ∅ D2(q+r+1)π 1

• In any cases H is π-maximal.

The π-submaximal subgroups of S = L3(3) in the case π ∩ π(S) = {3, 13} H N Pro. Intra. 1 C13 : C3 1

• 2

32+1

+

1

• In any cases H is π-maximal.

The π-submaximal subgroups of S = L3(3) in the case π ∩ π(S) = {2, 13} H N Pro. Intra. 1 C13 1

• 2

SD16 1

• In any cases H is π-maximal.

The π-submaximal subgroups of S = L3(3) in the case π ∩ π(S) = {2, 13} γ ∈ Aut(S) \ Inn(S), |γ| = 2, Aut(S) = Inn(S), γ H N H is not π-max if Pro. Intra. 1 E32 : GL2(3) 2∗

• ✪

2 31+2

+

: C 2

2

1 always∗∗

• 3

GL2(3) 1 always∗∗

• 4

S4 1

• ∗ are interchanged by γ

∗∗ is contained in E32 : GL2(3)

Corollary. If G is minimal nonsolvable and H ∈ smX (G) then H is pronormal.

• VII. What it is done?

X -Reduktionssatz

The X -Reduktionssatz holds for a group A means that for any G, if G has a normal subgroup N isomorphic to A, then M ∈ mX (G) ⇒ MN/N ∈ mX (G/N) and the map M → MN/N, mX (G) → mX (G/N) induces a bijection between the sets of conjugacy classes of X -maximal subgroups in G and G/N.

Observation

If the X -Reductionssatz holds for A then the X -maximal subgroups of A are conjugate.

Proposition (Wielandt).

Let the X -submaximal subgroups of A be conjugate. Then the X -Reductionssatz holds for A. mX (G) ⊆ smX (G). Problem (Wielandt, 1963). Is it true that if the members of mX (G) are conjugate then ones of smX (G) are conjugate?

The X -Reduktionssatz holds for a group A means that for any G, if G has a normal subgroup N isomorphic to A, then M ∈ mX (G) ⇒ MN/N ∈ mX (G/N) and the map M → MN/N, mX (G) → mX (G/N) induces a bijection between the sets of conjugacy classes of X -maximal subgroups in G and G/N. Problem (Wielandt, 1963). Is it true that if the members of mX (G) are conjugate then ones of smX (G) are conjugate?

Theorem (W.Guo, DR, E.Vdovin)

For a group A, the following statements are equivalent: (1) The X -Reductionssatz holds for A; (2) The members of mX (G) are conjugate; (3) The members of smX (G) are conjugate; (4) For every composition factor S of G, one of Conditions I–VII holds for (S, X ).

In the Conditions I–VII we denote π = π(X ) = {p | p is a prime and p divides |X| for some X ∈ X } Condition I. (S, X ) satisfies Condition I either S ∈ X or |π ∩ π(S)| ≤ 1. Condition II. (S, X ) satisfies Condition II if one of the following cases holds. (1) S ≃ M11 and π ∩ π(S) = {5, 11}; (2) S ≃ M12 and π ∩ π(S) = {5, 11}; (3) S ≃ M22 and π ∩ π(S) = {5, 11}; (4) S ≃ M23 and π ∩ π(S) coincide with one of the following sets {5, 11} and {11, 23}; (5) S ≃ M24 and π ∩ π(S) coincide with one of the following sets {5, 11} and {11, 23}; (6) S ≃ J1 and π ∩ π(S) coincide with one of the following sets {3, 5}, {3, 7}, {3, 19}, and {5, 11}; (7) S ≃ J4 and π ∩ π(S) coincide with one of the following sets {5, 7}, {5, 11}, {5, 31}, {7, 29}, and {7, 43}; (8) S ≃ O′N and π ∩ π(S) coincide with one of the following sets {5, 11} and {5, 31};

(12) S ≃ Co2 and π ∩ π(S) = {11, 23}; (13) S ≃ Co3 and π ∩ π(S) = {11, 23}; (14) S ≃ M(23) and π ∩ π(S) = {11, 23}; (15) S ≃ M(24)′ and π ∩ π(S) = {11, 23}; (16) S ≃ B and π ∩ π(S) coincide with one of the following sets {11, 23} and {23, 47}; (17) S ≃ M and π ∩ π(S) coincide with one of the following sets {23, 47} and {29, 59}. Condition III. Let S be isomorphic to a group of Lie type over the field Fq of characteristic p ∈ π and let τ = (π ∩ π(S)) \ {p}. We say that (S, X ) satisfies Condition III if τ ⊆ π(q − 1) and every prime in π does not divide the order of the Weyl group of S. In order to formulate Conditions IV and V, we need the following

• notation. If r is an odd prime and q is an integer not divisible by r,

then e(q, r) is the smallest positive integer e with qe ≡ 1 (mod r).

Condition IV. Let S be isomorphic to a group of Lie type with the base field Fq of characteristic p but not isomorphic to 2B2(q),

2F 4(q) and 2G 2(q). Let 2, p ∈ π. Denote by r the minimum in

π ∩ π(S) and let τ = (π ∩ π(S)) \ {r} and a = e(q, r). We say that (S, π) satisfies Condition IV if there exists t ∈ τ with b = e(q, t) = a and one of the following statements holds. (1) S ≃ An−1(q), a = r − 1, b = r, (qr−1 − 1)r = r,

• n

r−1

• =

n

r

• , and e(q, s) = b for every s ∈ τ;

(2) S ≃ An−1(q), a = r − 1, b = r, (qr−1 − 1)r = r,

• n

r−1

• =

n

r

• + 1, n ≡ −1 (mod r), and e(q, s) = b for every

s ∈ τ; (3) S ≃ 2An−1(q), r ≡ 1 (mod 4), a = r − 1, b = 2r, (qr−1 − 1)r = r,

• n

r−1

• =

n

r

• and e(q, s) = b for every s ∈ τ;

(4) S ≃ 2An−1(q), r ≡ 3 (mod 4), a = r−1

2 , b = 2r,

(qr−1 − 1)r = r,

• n

r−1

• =

n

r

• and e(q, s) = b for every s ∈ τ;

(5) S ≃ 2An−1(q), r ≡ 1 (mod 4), a = r − 1, b = 2r, (qr−1 − 1)r = r,

• n

r−1

• =

n

r

• + 1, n ≡ −1 (mod r) and

e(q, s) = b for every s ∈ τ; (6) S ≃ 2An−1(q), r ≡ 3 (mod 4), a = r−1

2 , b = 2r,

(qr−1 − 1)r = r,

• n

r−1

• =

n

r

• + 1, n ≡ −1 (mod r) and

e(q, s) = b for every s ∈ τ; (7) S ≃ 2Dn(q), a ≡ 1 (mod 2), n = b = 2a and for every s ∈ τ either e(q, s) = a or e(q, s) = b; (8) S ≃ 2Dn(q), b ≡ 1 (mod 2), n = a = 2b and for every s ∈ τ either e(q, s) = a or e(q, s) = b.

Condition V. Let S be isomorphic to a group of Lie type with the base field Fq of characteristic p, and not isomorphic to 2B2(q),

2F 4(q) and 2G 2(q). Suppose, 2, p ∈ π. Let r be the minimum in

π ∩ π(S), let τ = (π ∩ π(S)) \ {r} and c = e(q, r). We say that (S, X ) satisfies Condition V if e(q, t) = c for every t ∈ τ and one

• f the following statements holds.

(1) S ≃ An−1(q) and n < cs for every s ∈ τ; (2) S ≃ 2An−1(q), c ≡ 0 (mod 4) and n < cs for every s ∈ τ; (3) S ≃ 2An−1(q), c ≡ 2 (mod 4) and 2n < cs for every s ∈ τ; (4) S ≃ 2An−1(q), c ≡ 1 (mod 2) and n < 2cs for every s ∈ τ; (5) S is isomorphic to one of the groups Bn(q), Cn(q), or 2Dn(q), c is odd and 2n < cs for every s ∈ τ; (6) S is isomorphic to one of the groups Bn(q), Cn(q), or Dn(q), c is even and n < cs for every s ∈ τ; (7) S ≃ Dn(q), c is even and 2n ≤ cs for every s ∈ τ; (8) S ≃ 2Dn(q), c is odd and n ≤ cs for every s ∈ τ; (9) S ≃ 3D4(q);

(10) S ≃ E6(q), and if r = 3 and c = 1 then 5, 13 ∈ τ; (11) S ≃ 2E 6(q), and if r = 3 and c = 2 then 5, 13 ∈ τ; (12) S ≃ E7(q), if r = 3 and c ∈ {1, 2} then 5, 7, 13 ∈ τ, and if r = 5 and c ∈ {1, 2} then 7 ∈ τ; (13) S ≃ E8(q), if r = 3 and c ∈ {1, 2} then 5, 7, 13 ∈ τ, and if r = 5 and c ∈ {1, 2} then 7, 31 ∈ τ; (14) S ≃ G2(q); (15) S ≃ F4(q), and if r = 3 and c = 1 then 13 ∈ τ.

Condition VI. We say that (S, X ) satisfies Condition VI if one of the following statements holds. (1) S is isomorphic to 2B2(22m+1) and π ∩ π(S) is contained in

• ne of the sets

π(22m+1 − 1), π(22m+1 ± 2m+1 + 1); (2) S is isomorphic to 2G 2(32m+1) and π ∩ π(S) is contained in

• ne of the sets

π(32m+1 − 1) \ {2}, π(32m+1 ± 3m+1 + 1) \ {2}; (3) S is isomorphic to 2F 4(22m+1) and π ∩ π(S) is contained in

• ne of the sets

π(22(2m+1) ± 1), π(22m+1 ± 2m+1 + 1), π(22(2m+1) ± 23m+2 ∓ 2m+1 − 1) π(22(2m+1) ± 23m+2 + 22m+1 ± 2m+1 − 1).

Condition VII. Let S be isomorphic to a group of Lie type with the base field Fq of characteristic p. Suppose that 2 ∈ π and 3, p ∈ π, and let τ = (π ∩ π(S)) \ {2} and ϕ = {t ∈ τ | t is a Fermat number}. We say that (S, X ) satisfies Condition VII if τ ⊆ π(q − ε), where the number ε = ±1 is such that 4 divides q − ε, and one of the following statements holds. (1) S is isomorphic to either An−1(q) or 2An−1(q), s > n for every s ∈ τ, and t > n + 1 for every t ∈ ϕ; (2) S ≃ Bn(q), and s > 2n + 1 for every s ∈ τ; (3) S ≃ Cn(q), s > n for every s ∈ τ, and t > 2n + 1 for every t ∈ ϕ; (4) S is isomorphic to either Dn(q) or 2Dn(q), and s > 2n for every s ∈ τ; (5) S is isomorphic to either G2(q) or 2G 2(q), and 7 ∈ τ; (6) S ≃ F4(q) and 5, 7 ∈ τ; (7) S is isomorphic to either E6(q) or 2E6(q), and 5, 7 ∈ τ; (8) S ≃ E7(q) and 5, 7, 11 ∈ τ; (9) S ≃ E8(q) and 5, 7, 11, 13 ∈ τ; (10) S ≃ 3D4(q) and 7 ∈ τ.

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