Hall Subgroups of Finite Groups

Danila Olegovitch Revin and Evgenii Petrovitch Vdovin

Abstract. In this paper, we complete the classification of Hall subgroups in finite almost simple groups. We also prove that a finite group satisfies Dπ if and only if all its composition factors satisfy Dπ.

1. Introduction Let π be a set of primes. A π-subgroup H of G is called a Hall π-subgroup if |G : H| is not divided by any prime r ∈ π. Following [25], we say that a finite group G satisfies E… (or just that G is an E…-group), if G contains a Hall π-subgroup; G is a C…-group if G is an E…-group and all Hall π-subgroups of G are conjugate; and G is a D…-group if G is a C…-group and every π-subgroup of G is contained in some Hall π-subgroup of G. The set of E…-groups is known to be closed under normal subgroups and factor groups but it is not closed under extensions. Since a normal subgroup of an E…- group satisfies E…, it is important to describe finite simple groups with E…-property. Moreover, if S is a finite nonabelian simple E…-group and Aut(S) satisfies E…, then by [21, Corollary 3.6], every extension of S by a E…-group satisfies E…. Thus we also consider almost simple E…-groups. The set of C…-groups is known to be closed under extensions but it is not closed under normal subgroups. V.A. Vedernikov in [44], using CFSG, proved the following lemma.

[44, Lemma 3] Let G satisfy E…, M ≤ H ≤ G and M E G. If H/M satisﬁes E…, then H also does.

It is immediate from this lemma that the set of C…-groups is closed under factor groups. The main result of [22] states that if 2 6∈ π, then E… and C… are equivalent. In particular, in this case a ﬁnite group G satisﬁes E… if and only if every composition

2000 Mathematics Subject Classiﬁcation. Primary 20D20; Secondary 20D06, 20G40. Key words and phrases. Finite simple groups, Hall subgroups, Groups of Lie type, Algebraic groups, Alternating groups, Normalizer of maximal torus. The work is supported by RFBR, grant N 05-01-00797, Universities of Russia, grant N UR.04.01.202. The second author is partially supported by University of Padua, Italy.

°c 0000 (copyright holder) 1 2 D.O.REVIN AND E.P.VDOVIN factor of G satisfies E…. In [43], the authors complete the classification of Hall subgroups of odd order in finite simple groups, so the classification of all finite E…-groups with 2 6∈ π is obtained. The set of D…-groups is known to be closed under factor groups. L.A. Shemet- kov and V.D. Mazurov, respectively, added the following problems to “Kourovka notebook” [45]: 3.62. Is an extension of a D…-group by a D…-group again a D…-group? 13.33. Is a normal subgroup of a D…-group again a D…-group? The result by F. Gross in [22] implies an affirmative answer to the second problem in case 2 6∈ π. Using the classification of Hall subgroups of odd order in finite simple groups, the authors in [43] obtain an affirmative answer to the first problem in case 2 6∈ π. F. Gross proved the following proposition [23, Proposition 1.1].

Proposition 1.1. Let S be either An(q) or Cn(q). Assume S has a Hall π-subgroup with 3 6∈ π. Then all Hall π-subgroups of S are conjugate in S and a Hall π-subgroup of S has a Sylow tower. Further, if S ≤ G ≤ Aut S, then G has a Hall π-subgroup, all Hall π-subgroups of G are conjugate in G and a Hall π-subgroup of G is solvable. In view of this proposition, F. Gross made the following conjectures. Conjecture 1.2. The above result holds true if S is allowed to be any of 2 2 the classical simple group, i.e., if S is one of An(q), Bn(q), Cn(q), Dn(q), An(q ), 2 2 or Dn(q ). Conjecture 1.3. Let G be a finite group and π a set of primes with 3 6∈ π. Then G has a Hall π-subgroup if and only if each composition factor of G has a Hall π-subgroup. Conjecture 1.4. If H is a Hall π-subgroup of the finite simple group G and 2 3 6∈ π, then either H has a Sylow tower or H = G = B2(q). In this paper, we complete the classification of Hall subgroups in finite almost simple groups. We prove the three conjectures posed by F.Gross and give affirmative answers to Problems 3.62 and 13.33 from “The Kourovka Notebook” mentioned above. The authors thank Professor Anatolii S. Kondratiev and Professor Viktor D. Mazurov for helpful discussions.

2. Notations Our notations are standard. For abstract groups, they agree with those in [17].For a group G by Aut G we denote the automorphism group of G and Inn G is a subgroup of Aut G generated by inner automorphisms. A finite group G is called almost simple if there is a finite simple group K with K ≤ G ≤ Aut K. If n is a rational integer, π a set of primes, then by π(n) we denote the set of all prime divisors of n, by n… we denote the maximal divisor t of n such that π(t) ⊆ π. For a finite group G, π(G) = π(|G|) by definition. We say that a Hall π-subgroup M of G is nontrivial if M 6= {1} and M is not a Sylow subgroup of G. If n is a positive rational integer, r is an odd prime and (r, n) = 1, then by e(r, n) we denote the minimal positive rational integer e with ne ≡ 1(mod r). If n is an odd rational integer, let e(2, n) = 1 if n ≡ 1(mod 4) and e(2, n) = 2 if n ≡ −1(mod 4). HALL SUBGROUPS OF FINITE GROUPS 3

For groups of Lie type and linear algebraic groups, our notation agrees with that in [7] and [26], respectively. Denote by Gsc a universal group of Lie type. Then we call every factor group Gsc/Z, where Z ≤ Z(Gsc), a group of Lie type. Almost in all cases, Gsc/Z(Gsc) is simple and we say that Gad = Gsc/Z(Gsc) is of adjoint type. If G is a universal group of Lie type, then every automorphism of G/Z(G) can be considered as an automorphism of G and vice versa. We shall use this fact below without any reference. 2 2 2 2 2 2 If G is isomorphic to An(q ), Dn(q ), or E6(q ), we say that G is defined 2 3 3 3 over GF (q ), if G ' D4(q ) we say that G is defined over GF (q ) and we say that G is defined over GF (q) for other finite groups of Lie type. The field GF (q) in all cases is called the base field of G. If G is a universal finite group of Lie type with the base field GF (q), then there is a positive integer k(= |Φ+| in most cases) and a polynomial f(t) ∈ Z[t] such that |G| = f(q) · qk and (q, f(q)) = 1 (see [7, Theorems 9.4.10 and 14.3.1]). We denote this polynomial by fG(t). If G is not universal, then there is a universal group K with G = K/Z and we define fG(t) = fK (t). Assume that G is a connected simple linear algebraic group defined over an algebraically closed field of a positive characteristic p. Let σ be an endomorphism of G such that G = CG(σ) is a finite set. Then σ is said to be a Frobenius map and p0 G = O (G) appears to be a finite group of Lie type. Moreover, all finite groups of Lie type both split and twisted can be derived in this way. By [39, Theorem 30], p0 every automorphism of G = O (G) can be extended to an automorphism of G. Let M ≤ Aut G and for every x ∈ M, fix an extension x ∈ Aut G. If x ∈ M ∩ G (clearly G ≤ Aut G), then x = x ∈ G is the conjugation by x. We say that a subgroup R of G is invariant under M ≤ Aut G (or just M-invariant), if for any x ∈ M, we have that R is invariant under its (fixed before) extension x ∈ Aut G.

In particular, if M ≤ G, we have that M ≤ NG(R). If G is simply connected, p0 then G = G = O (G) and if G is of adjoint type, then G = Gb is the group of inner-diagonal automorphisms of G. If R is a σ-stable reductive subgroup of G, then R ∩ G = R ∩ G is called a reductive subgroup of G. If R is of maximal rank, then R ∩ G is also of maximal rank. Note that if R is σ-stable of maximal rank, then R = T ·G1 ∗...∗Gk, where T is some σ-stable maximal torus£ of R¤and G1,..., Gk are subsystem subgroups of G. Furthermore, G1 ∗ ... ∗ Gk = R, R . It is known that R = T G1 ∗ ... ∗ Gm. We call subgroups G1,...,Gm subsystem subgroups of G. In general, m ≤ k and, for all fii i, the base field of Gi is equal to GF (q ), where αi ≥ 1. There is a nice algorithm due to Borel and de Siebental determining all subsystem subgroups of G (see [2]). One has to consider the extended Dynkin diagram of G and remove any number of nodes. Connected components of the remaining graph are Dynkin diagrams of subsystem subgroups of G and Dynkin diagrams of all subsystem subgroups can be derived in this way. If T is a σ-stable torus of G, then T = T ∩ G = T ∩ G is called a torus of G. If T is maximal, then T is a maximal torus of G. Note that |T | = g(q), where GF (q) is the base field of G, g(t) is a polynomial of degree n dividing fG(t) and n is the rank of G. If L ≤ H ≤ G, L = L ∩ G and H = H ∩ G for some closed (but not necessarily connected) σ-stable subgroups L, H of G, then define N(H,L) by (NH (L)) ∩ H. Clearly, N(H,L) is well-defined, since if L and H are σ-stable, then NH (L) is also σ-stable. For more details, see [10, Chapter 1]. 4 D.O.REVIN AND E.P.VDOVIN

3. Preliminary results The following lemma is known. Lemma 3.1. Let G be a ﬁnite group, A a normal subgroup of G.

(1) If G satisfies E… or D…, then G/A satisfies E… or D…, respectively. (2) If A satisfies C… and G/A satisfies E… (resp., C…), then G satisfies E… (resp., C…). (3) If A is nilpotent, then G satisfies E… (resp., C…, D…) if and only if G/A satisfies E… (resp., C…, D…). (4) If G satisfies E…, then A satisfies E…. p0 Lemma 3.2. Let G = O (G) be a finite group of Lie type and X be a subgroup of Aut G. Then one of the following holds: (1) There is a proper connected σ-stable invariant under X subgroup F of G; (2) X is almost simple; (3) X ∩G lies in the normalizer of some Jordan subgroup of Inn G ' G/Z(G); (4) G = E8(q), char GF (q) > 5, Y ≤ G ∩ X ≤ NG(Y ), where Y ' A × B, A ' Alt5, and B ' Alt6 are alternating groups of degree 5 and 6, respectively, NG(Y )/Y is a four-group, CG(B) = A, CG(A) ' Sym6 is a 0 symmetric group on six letters, and B = CG(A) . The group Y is uniquely defined to within conjugacy in G. Proof. In [4, Theorem 1], this lemma is proven for a canonical Frobenius map σ. One can easily adapt this proof for an arbitrary Frobenius map. ¤

p0 Lemma 3.3. Let G = O (G) be a finite simple group of Lie type defined over a field of odd characteristic p, π a set of primes such that 2 ∈ π, J a Jordan subgroup of G, and M a Hall π-subgroup of G contained in NG(J). Then one of the following holds: (1) M is a Sylow 2-subgroup of G. (2) G ' A1(q), π ∩ π(G) = {2, 3}, |G|… = 24 or |G|… = 12, M = NG(J) ' Sym4 or M = NG(J) ' Alt4, respectively. Proof. The information about the structure of Jordan subgroups and their normalizers in all simple algebraic groups is given in [3] and may be found in Table 1. HALL SUBGROUPS OF FINITE GROUPS 5

Table 1 Jordan subgroups of simple algebraic groups

G J NG(J) 2n 2n Arn−1, n ≥ 1 r r · Sp2n(r) 2n 2n Bn, n ≥ 2 2 2 · Sym2n+1 2n 2n − C2n−1 , n ≥ 3 2 2 · O2n(2) 2n 2n + D2n−1 , n ≥ 3 2 2 · O2n(2) 2n 2n Dn+1, n ≥ 3 2 2 · Sym2n+2 3 3 3 E6 3 3 · 3 · SL3(3) 5 5 10 E8 2 2 · 2 · SL2(5) 3 3 E8 5 5 · SL3(5) 3 3 F4 3 3 · SL3(3) 3 3 G2 2 2 · SL3(2)

2n Consider all groups from Table 1 separately. Let G = Arn−1(q), J = r , 2n NG(J) = r · Sp2n(r). There can be two cases: r = 2 and r 6= 2. Assume ﬁrst that r 6= 2 and show that |NG(J)|2 < |G|2. Indeed, |G|p0 = 1 2 3 rn (rn;q−1) (q − 1)(q − 1) ... (q − 1); therefore,

h i n n n 2 3 rn rn−1 [ r −1 ]+[ r −1 ]+::: 3(r −1) |G|2 = (q − 1)(q − 1) ... (q − 1) ≥ 2 · 2 2 22 ≥ 2 2 . 2 Now,

2 4 2n |NG(J)|2 = [(r − 1)(r − 1) ... (r − 1)]2 e n [ n ]+[ n ]+::: e n n e+1 n = (2 ) · 2 2 22 ≤ (2 ) · 2 = (2 ) ,

e 2 r2−1 where e is the maximal degree such that 2 divides r − 1. Clearly, e + 1 ≤ 2 . 3(rn−1) r2−1 2 n 2 Thus, |G|2 ≥ 2 , |NG(J)|2 ≤ 2 . Evidently, |G|2 > |NG(J)|2 if n ≥ 2. If e r e 2 r n = 1, then |NG(J)|2 = 2 and |G|2 > 2 . It is clear that 2 ≤ r − 1 ≤ 2 ; hence r |NG(J)|2 ≤ 2 < |G|2. It follows that if r is odd, then NG(J) does not contain a Sylow 2-subgroup of G, and so does not contain a Hall π-subgroup of G (we assume that 2 ∈ π). Assume now that r = 2. Then h i 1 2 3 2n |G|2 = (q − 1)(q − 1) ... (q − 1) (2n, q − 1) 2 n n 2n−2 [ 2 −1 ]+[ 2 −1 ]+::: 2n+1−4 ≥ 2 · 2 2 22 > 2 .

n n2 On the other hand, |NG(J)|2 = 2 · 2 . Thus, if n ≥ 3 we obtain that |G|2 > |NG(J)|2. The case n = 1 gives us point (2) of the lemma. In case n = 2, it is easy to see that |NG(J)|2 < |G|2. Suppose G = Bn(q) or G = Dn+1(q). If NG(J) contains a nontrivial Hall π-subgroup, then, in view of Lemma 3.1, either Sym2n+1 or Sym2n+2, respectively, contains a nontrivial Hall π-subgroup. But the order of any nontrivial Hall π- subgroup of Symn for any n is divided by 3 (see [25, Theorem A4] and [41, Theo- rem A]). One can easily verify that |NG(J)|3 < |G|3, so in these two cases NG(J) does not contain nontrivial Hall π-subgroups of G. 6 D.O.REVIN AND E.P.VDOVIN

In all other cases, |G|2 > |NG(J)|2; therefore, NG(J) can not contain a Hall π-subgroup with 2 ∈ π. If G is a twisted group, we proceed in the same way. ¤

p0 Lemma 3.4. Let Y ≤ E8(q) = O (G) be chosen as in Lemma 3.2 (4). Then

NE8(q)(Y ) does not contain nontrivial Hall subgroups of E8(q).

Proof. The lemma is evident, since NG(Y ) does not contain Sylow subgroups of E8(q). ¤

p0 Lemma 3.5. Let π be a set of primes such that p 6∈ π. Let G = O (G) be a ﬁnite simple group of Lie type and M an almost simple Hall π-subgroup of G. Assume that M does not normalize any proper connected σ-stable subgroup of G. Then G, M, and π satisfy one of the following conditions.

(1) G ' A1(q), p > 5, M ' A1(4), π ∩ π(G) = {2, 3, 5}, |G|{2;3;5} = 2 |M|{2;3;5} = 2 · 3 · 5. (2) G ' B3(q), p > 7, M ' B3(2), π ∩ π(G) = {2, 3, 5, 7}, |G|{2;3;5;7} = 9 4 |M|{2;3;5;7} = 2 · 3 · 5 · 7. (3) G ' D4(q), p > 7, M ' D4(2), π ∩ π(G) = {2, 3, 5, 7}, |G|{2;3;5;7} = 12 5 2 |M|{2;3;5;7} = 2 · 3 · 5 · 7. Proof. Since M is an almost simple group, [15] implies that 2 ∈ π, hence M has odd index. By the Main Theorem of [34], if M is an almost simple subgroup of odd index of a ﬁnite simple group of Lie type and M does not normalizes any proper connected σ-stable subgroup of G, then either M is simple or G = G2(q) and M = G2(2). All possibilities for a ﬁnite simple group G to contain a simple Hall π-subgroup, with p 6∈ π, are described in [16, Corollary B]. Clearly, G2(2) is not a Hall subgroup of G2(q) for all q. ¤

Lemma 3.6. Let G be a reductive connected linear algebraic group, S a subgroup of G and T a torus (not necessarily maximal) containing S. Then CG(S) is a reductive (but not necessary connected) subgroup of maximal rank of G.

Proof. Consider CG(T ). It is known to be a connected reductive subgroup of maximal rank of G (see [26, Theorem 22.3 and Corollary 26.2A]). In particular, there exists a maximal torus of G, containing T and, therefore, S. Now we follow the proof of [27, Theorem 2.2]. ¤

Lemma 3.7. Let R be a connected closed reductive subgroup of maximal rank of G. Denote by W the Weyl group of G, by W the Weyl group of R, and by W ⊥ R R the subgroup orthogonal to WR subgroup in W . Then: (1) N (W )/W × W ⊥ ' Aut(∆ ), where ∆ stands for the Dynkin dia- W R R R R R gram of R;

(2) NW (WR)/WR ' NG(R)/R; (3) For every maximal torus T of R, we have N (R) = N (N (T )) · R = N (R) · R. G NG(T ) R NG(T ) Proof. Point (1) is proved in [8, Proposition 4]. As to point (2), let T be a maximal torus of G contained in R so that W = NG(T )/T and WR = NR(T )/T . g−1 Now let g ∈ NG(R). Then T is a maximal torus of R. Since R is connected, HALL SUBGROUPS OF FINITE GROUPS 7

g−1 c there exists c ∈ R such that T = T . It follows that cg ∈ NG(T ) ∩ NG(R) = N (R), i.e., g ∈ N (R)R. Hence we have NG(T ) NG(T )

N (R) NN (T )(R)R N (R) ∩ N (T ) N (W ) G = G ' G G ' W R . R R R ∩ NG(T ) WR The equality N (R) ∩ N (T ) = N (N (T )) is proven in [8, Proposition 2]. G G NG(T ) R (3) Clear from above. ¤

p0 Lemma 3.8. Let G = O (G) be a ﬁnite group of Lie type, G a simple connected linear algebraic group, σ a Frobenius map. Assume that M ≤ Aut G and a connected σ-stable proper subgroup R of G such that Ru(R) 6= 1 is M-invariant. Then there exists a proper σ-stable M-invariant parabolic subgroup of G. In particular, M normalizes some proper parabolic subgroup of G.

Proof. Denote Ru(R) by U. Consider the chain of subgroups: U0 = U; N0 = NG(U0); Ui = Ru(Ni−1); Ni = NG(Ui). Clearly, N0 ≤ N1 ≤ ... Since the Sdimension of G is finite, we obtain that Nm = Nm+1 for some m. Denote the union i Ni by N. By [26, Proposition 30.3], N is a proper parabolic subgroup of G. Since U0 is σ-stable and invariant under M, we obtain that N is also σ-stable and invariant under M. ¤ Lemma 3.9. Let G, G, and σ be chosen as in Lemma 3.8. Assume that M ≤ Aut G and some proper connected σ-stable subgroup R of G such that Ru(R) = 1 and Z(R) 6≤ Z(G) is invariant under M. Then there is a proper connected σ-stable reductive M-invariant subgroup of maximal rank of G. Proof. Since R is connected, Z(R) is contained in some (moreover, in every) maximal torus of R. Since Z(R) is a characteristic subgroup of R, Z(R) is σ- 0 stable and invariant under M. By Lemma 3.6, CG(Z(R)) is a connected reductive subgroup of maximal rank of G. Moreover, it is σ-stable and invariant under M. 0 Since Z(R) 6≤ Z(G), CG(Z(R)) is a proper connected σ-stable and M-invariant subgroup of G. ¤ Lemma 3.10. Let G be a finite group of Lie type over a field of odd characteristic, G is not a Ree group, and let S be a Sylow 2-subgroup of G. Then the following hold. (1) There exists a maximal torus Q such that N(G, Q) contains S. (2) If Q1 is another maximal torus of G with |N(G, Q1)|2 = |G|2, then Q and Q1 are conjugate under G. (3) NG(Q) = N(G, Q).

Proof. Clearly, we may assume G to be universal so that G = G. (1) Follows from [19, Theorem 4.10.2]. (2) Consider all types of ﬁnite groups of Lie type separately. G = An(q). If e(2, q) = 1, let T be a Cartan subgroup of G; and if e(2, q) = 2, let T be a maximal torus of order (q+1)k(q−1)n−k, where k = [(n + 1)/2]. If T is a k n−k Cartan subgroup, then N(G, T )/T ' W (An) ' Symn+1. If |T | = (q+1) (q−1) , then, by [6, Proposition 23], N(G, T )/T ' 2 o Symk. Since n+1 n+1 [ 2 ]+[ 4 ]+::: |2 o Symk |2 = 2 = | Symn+1 |2, 8 D.O.REVIN AND E.P.VDOVIN we have that |N(G, T )/T |2 = |W (An)|2 in both cases. Assume that Q is another torus with |N(G, Q)|2 = |G|2. If |Q| = |T |, then, by [6, Proposition 23], Q and T are conjugate in G. So assume that |Q| 6= |T |. By [6, Proposition 23 and Table 3], every maximal torus of G has order qn1 − 1 qnk − 1 · ... · · (q − 1)k−1 with n + ... + n = n + 1. q − 1 q − 1 1 k

Since |Q| 6= |T |, direct calculations show that |Q|2 < |T |2. In view of Lemma 3.7, N(G, Q)/Q is isomorphic to a subgroup of the Weyl group of G, so |N(G, Q)/Q|2 ≤ |W |2 = |N(G, T )/T |2. Therefore,

|N(G, Q)|2 = |N(G, Q)/Q|2 · |Q|2 < |N(G, T )/T |2 · |T |2 = |N(G, T )|2.

Since |N(G, T )|2 ≤ |G|2, we obtain that |N(G, Q)|2 < |G|2, a contradiction. There- fore, |Q| = |T |, and Q and T are conjugate in G. Furthermore, we obtain the structure of T : if e(2, q) = 1, then T is a Cartan subgroup of G and if e(2, q) = 2, n+1 n− n+1 then |T | = (q + 1)[ 2 ](q − 1) [ 2 ]. 2 2 G = An(q ). If e(2, q) = 1, let T be a Cartan subgroup, and if e(2, q) = 2 let T be a maximal torus of order (q + 1)n. Direct calculations using [10, Proposi- tion 3.3.6] show that N(G, T )/T ' 2 o Symk with k = [(n + 1)/2], if T is a Cartan n subgroup and N(G, T )/T ' Symn+1, if |T | = (q + 1) . Therefore, |N(G, T )/T |2 = |W (An)|2 in both cases. Let Q be a maximal torus with |N(G, Q)|2 = |G|2. Since every maximal torus in G has order qn1 − (−1)n1 qnk − (−1)nk · ... · · (q + 1)k−1 with n + ... + n = n + 1, q + 1 q + 1 1 k it follows that if |Q| 6= |T |, then |N(G, Q)|2 < |N(G, T )|2. Therefore, |Q| = |T | and, by [9, Proposition 8], Q and T are conjugate under G. G = Bn(q) or G = Cn(q). We do not separate these two cases, since |Bn(q)| = |Cn(q)| and W (Bn) ' W (Cn) ' 2 o Sn. If e(2, q) = 1, let T be a Cartan subgroup of G; if e(2, q) = 2, let T be a maximal torus of order (q + 1)n. Then, by [6, Propositions 24], N(G, T )/T ' W (Bn) in both cases. Assume that Q is a maximal torus of G with |N(G, Q)|2 = |G|2. By [6, Proposition 24], qk1 − 1 qkm − 1 |Q| = · ... · · (q − 1)m−1 · (qr1 + 1) · ... · (qrl + 1), q − 1 q − 1 with k1 + ... + km + r1 + ... + rl = n + 1 (r1 + ... + rl = n if m = 0). Again it is evident that if |Q| 6= |T |, then |Q|2 < |T |2. Since N(G, Q)/Q is a subgroup of W , we obtain that if |Q| 6= |T |, then |N(G, Q)|2 < |N(G, T )|2. So, |Q| = |T |, and, by [6, Proposition 24], Q and T are conjugate in G. G = Dn(q). If e(2, q) = 1, let T be a Cartan subgroup; if e(2, q) = 2, let T be a maximal torus of order (q+1)n if n is even, and of order (q+1)n−1(q−1), if n is odd. n−1 Using [6, Proposition 25], it is easy to see that N(G, T )/T ' W ' 2 h Symn if T is a Cartan subgroup or n is even and N(G, T )/T ' 2 o Symn−1 if n is odd. Anyway |N(G, T )/T |2 = |W (Dn)|2 in both cases. Using [6, Proposition 25] and [9, Proposition 10], like in previous cases we obtain the statement. 2 2 G = D2n+1(q ). Taking T to be a Cartan subgroup if e(2, q) = 1 or to be a torus of order (q +1)2n+1 if e(2, q) = 2, and using [10, Proposition 3.3.6], we obtain that N(G, T )/T ' 2oSym2n in the ﬁrst case and N(G, T )/T ' W (G) = W (D2n+1) in the second case. Anyway, |N(G, T )/T |2 = |W (Dn)|2 in both cases; thus, using [9, Proposition 10], we obtain the statement in this case. HALL SUBGROUPS OF FINITE GROUPS 9

2 2 G = D2n(q ). Let T be a Cartan subgroup if e(2, q) = 1 and a torus of order 2n−1 (q +1) (q −1) if e(2, q) = 2. In both cases, N(G, T )/T ' 2oSym2n−1. Recalling that σ can be written in the form γϕ, where γ is a graph automorphism and ϕ is a ﬁeld automorphism, denote by ρ the symmetry of the Dynkin diagram of D2n, corresponding to γ. Let H be a maximal torus split over GF (p). Then Q = gH for some g ∈ −1 −1 G and g σ(g) ∈ NG(H). Let w be the image of g σ(g) under the natural homomorphism

NG(H) → NG(H)/H ' W (D2n).

The last equality induces a σ-action on W = W (D2n). Since H is split over GF (p), H is a Cartan subgroup of G. In view of [7, Section 13.1], the action of ¿ σ on W (D2n) is given by σ(w) = w , where τ is an orthogonal transformation of ZΦ(D2n)⊗Z R corresponding to ρ. We may choose {e1 −e2, . . . , e2n−1 −e2n, e2n−1 + e2n} to be a fundamental system of Φ(D2n), where e1, . . . , e2n is an orthonormal basis of a 2n-dimensional Euclidian space. On this basis, W (D2n) is generated by permutation matrices and diagonal matrices of form diag(±1,..., ±1) with even number of −1s; furthermore, τ has the matrix diag(1,..., 1, −1). Moreover, the set of all diagonal matrices forms a normal elementary Abelian 2-subgroup D of W . In ¿ −1 view of [10, Proposition 3.3.6], N(G, Q)/Q ' CW;(w) = {v ∈ W |v wv = w} = {v ∈ W |σ(v)wv−1 = w}.

Consider W1 = h W, τ i ' 2oSym2n. Clearly, CW;(w) = CW (τw) = CW1 (τw)∩ W . Let D1 = h D, τ i (clearly D1 is elementary Abelian), M the group of all permutation matrices so that M ' Sym2n. Then W1 = D1 h M, so every element w of W1 has a unique decomposition xwyw with xw ∈ D1, yw ∈ M. Since D1 is normal and Abelian, we have CW1 (τw) ≤ CW1 (x¿w) = CW1 (τxw). So |CW;(w)| ≤ 1 |CW1 (τxw) ∩ W | = 2 |CW1 (τxw)| and CW;(w) = CW1 (τxw) ∩ W if and only if w = xw. Since xw ∈ D, τxw contains an odd number of −1; hence τxw 6= ±E, where E is the identity matrix. Assume that τxw has n1 of 1s and n2 of −1s. Then

CW1 (τxw) = D h (Symn1 × Symn2 ) if n1 6= n2 or D h ((Symn1 × Symn2 ) h 2) if n1 = n2 = n, i.e., n is odd. Since n1 and n2 are odd, we have that

1 2n−1 |CW;(w)|2 ≤ |CW (τxw)|2 = 2 · | Sym |2 · | Sym |2 2 1 n1 n2 2n−1 = 2 · | Symn1−1 |2 · | Symn2−1 |2 2n−1 ≤ 2 · | Symn1+n2−2 |2 = |2 o Sym2n−1 |2, if n1 6= n2, and 1 |C (w)| = |C (τx)| = 22n · | Sym | · | Sym | W; 2 2 W1 2 n 2 n 2 2n = 2 · | Symn−1 |2 · | Symn−1 |2 ≤ |2 o Sym2n−1 |2, if n1 = n2. So |N(G, Q)/Q|2 ≤ |N(G, T )/T |2, and we proceed like in previous cases. G = G2(q). If e(2, q) = 1, let T be a Cartan subgroup; if e(2, q) = 2, let T be 2 a torus of order (q + 1) . By [6, Table 7], N(G, T )/T ' W (G2) in both cases. By [6, Propositions 26, 27], it follows that T is unique up to conjugation in G. Now we proceed like in previous cases. G = F4(q). If e(2, q) = 1, let T be a Cartan subgroup; if e(2, q) = 2, let T be 4 a torus of order (q + 1) . By [6, Table 8], N(G, T )/T ' W (F4) in both cases. By 10 D.O.REVIN AND E.P.VDOVIN

[6, Propositions 26, 27], it follows that T is unique up to conjugation in G. Now we proceed like above. G = E6(q). If e(2, q) = 1, let T be a Cartan subgroup; if e(2, q) = 2, let T be a torus of order (q + 1)4(q − 1)2. In both cases, N(G, T )/T contains a Sylow 2-subgroup of W (E6) ([6, Table 9]) and T is unique up to conjugation in G ([6, Propositions 26, 27]). Like above, we obtain the requirement statement. 2 2 G = E6(q ). If e(2, q) = 1 let T be a Cartan subgroup of G, if e(2, q) = 2 6 let T be a torus of order (q + 1) . In the ﬁrst case, N(G, T )/T ' W ' W (F4). In the second case, direct calculations using [10, Proposition 3.3.6] and [46] show that N(G, T )/T ' W (E6). Noting that in view of [14, Table 1, p. 128] T is unique up to conjugation in both cases, we proceed like in previous cases. G = E7(q). Take T to be a Cartan subgroup if e(2, q) = 1 and to be a torus of 7 order (q + 1) if e(2, q) = 2. By [6, Table 10], we have that N(G, T )/T ' W (E7) in both cases. By [6, Propositions 26, 27], T is unique up to conjugation in G. Like in previous cases, we obtain the statement. G = E8(q). If e(2, q) = 1, let T be a Cartan subgroup; if e(2, q) = 2, let T 8 be a maximal torus of order (q + 1) . By [6, Table 11], N(G, T )/T ' W (E8) in both cases and, in view of [6, Propositions 26, 27], it follows that T is unique up to conjugation in G. Like above, we get the statement. 3 3 3 G = D4(q ). If e(2, q) = 1, let T be a Cartan subgroup of order (q −1)(q −1); if e(2, q) = 2, let T be a maximal torus of order (q + 1)(q3 + 1). It follows by [14, Table 7, p. 140] that in the second case, T is contained in a subgroup L = 3 TA1(q ) of G. Since |N(G, L)/L| = 2 and |N(L, T )/T | = 2, Lemma 3.7 implies that |N(G, T )/T | ≥ 4. But |G|2 = 4 · |T |2 in this case, so N(G, T ) contains a Sylow 2-subgroup of G. Noting that, by [14, Table 7, p. 140], T is unique up to conjugation in G, we proceed like in previous cases. (3) Assume by contradiction that G is a group of minimal order with NG(T ) 6= N(G, T ) and S ≤ N(G, T ). Since N(G, T ) ≤ NG(T ), it follows that S ≤ NG(T ). We again may assume G to be universal so that G = G. By the proof of (2), it is clear that T 6≤ Z(G). Clearly, NG(T ) = (NG(T )). In view of rigidity of 0 0 d-groups [26, Corollary to Proposition 16.3], NG(T ) = CG(T ) . By Lemma 3.6, 0 C = CG(T ) is a connected reductive subgroup of maximal rank of G. Since we are assuming NG(T ) 6≤ N(G, T ), it follows that NG(T ) 6≤ NG(T ), so C 6= T . Hence C = T · M, where M is semisimple and nontrivial. Since T 6≤ Z(G) we have that C is a proper subgroup of G. It follows that NG(T ) ≤ N(G, C) and C = TG1∗...∗Gk, where Gi is a subsystem subgroup of G for all i = 1, . . . , k. Since every Gi is subnormal in NG(T ) and since S ≤ NG(T ), it follows that Si = S ∩ Gi is a Sylow 2-subgroup of Gi. In view of (1), we have that Si ≤ N(Gi,Ti) for some maximal torus Ti of Gi. Let Q = Q be a maximal torus of TG1 ∗ ... ∗ Gk, containing T1 ∗ ... ∗ Tk. In view of Lemma 3.7, N (C) = N (C)C, so G NG(Q) (NG(C)) = NN(G;Q)(C) · (G1 ∗ ... ∗ Gk). Clearly, (NG(C)) ≥ NG(T ) ≥ S. Since S ∩ G1 ∗ ... ∗ Gk ≤ N(G1,T1) ∗ ... ∗ N(Gk,Tk), and since we chose Q so that N(G1,T1)∗...∗N(Gk,Tk) ≤ N(G, Q), we have that, up to conjugation in (NG(C)), S ≤ N(G, Q). Now (2) implies that Q and T are conjugate in G, so we may assume that Q = T . Thus we obtain that T ∩ Gi = Ti. Since G is a minimal counter example, NGi (Ti) ≤ N(Gi,Ti). Thus we have NGi (Gi ∩ T ) ≤ N(Gi,Gi ∩ T ) 6= Gi.

But Gi ≤ CG(T ) ≤ NG(T ), so NGi (Gi ∩ T ) = Gi, a contradiction. ¤ HALL SUBGROUPS OF FINITE GROUPS 11

Corollary 3.11. Let G be a universal finite group of Lie type defined over a field of odd characteristic, and G is not a Ree group. Let S be a Sylow 2-subgroup of G and T a maximal torus with S ≤ N(G, T ). Then the following hold. 3 3 k n−k (1) If e(2, q) = 1 and G 6= D4(q ), then |T | = (q − 1) (q + 1) , where k k is the maximal integer with (t − 1) |fG(t). 3 3 k n−k (2) If e(2, q) = 2 and G 6= D4(q ), then |T | = (q + 1) (q − 1) , where k k is the maximal integer with (t + 1) |fG(t). 3 3 3 (3) If G = D4(q ), then |T | = (q − 1)(q − 1) if e(2, q) = 1 and |T | = (q + 1)(q3 + 1) if e(2, q) = 2. Proof. Immediate from the proof of Lemma 3.10. ¤

Lemma 3.12. Let G = G be a universal finite group of Lie type, T a maximal torus such that N(G, T ) contains a Sylow 2-subgroup S of G. Assume that N(G, T ) contains a Sylow r-subgroup R of G for some odd prime r 6= p such that R∩T 6= {1}. Then e(r, q) = e(2, q). In particular , R ∩ T 6≤ Z(G). 3 3 Proof. If G = D4(q ), then the statement follows from the description of 3 3 maximal tori given in [14, Table 7, p. 140]. So we assume that G 6= D4(q ). 2 2 Suppose also that G 6= D2n(q ). In view of Corollary 3.11, r divides either q − 1 or q + 1. In the first case, e(r, q) = 1; in the second case, e(r, q) = 2. Assume by contradiction that e(r, q) 6= e(2, q). Suppose e(r, q) = 1, then e(2, q) = 2. It follows that G is nontwisted, since otherwise we would have |T | = (q +1)n and e(r, q) = e(2, q) = 2. Consider a Cartan subgroup H of G. In view of Corollary 3.11, |T | = (q + 1)k(q − 1)n−k. It is known that |H| = (q − 1)n. From the proof of Lemma 3.10, it follows that k > 0 for all groups. Therefore, |H|r > |T |r. Since N(G, T )/T is isomorphic to a subgroup of a Weyl group W (G) ' N(G, H)/H, we have that |N(G, T )|r = |N(G, T )/T |r ·|T |r < |N(G, H)/H|r · |H|r = |N(G, H)|r, a contradiction to |N(G, T )|r = |G|r. Assume now that e(r, q) = 2, so e(2, q) = 1. Hence G is a twisted group, since otherwise we would have |T | = (q − 1)n and e(r, q) = 1. From the structure of N(G, Q)/Q given in the proof of Lemma 3.10, it follows that for all twisted groups there exists a maximal torus Q such that |Q| = (q + 1)n and N(G, Q)/Q ' W (G). 2 2 (Recall that we are assuming G 6= D2n(q ).) It follows that |G|r ≥ |N(G, Q)|r > |N(G, T )|r, a contradiction. 2 2 Assume now that G = D2n(q ). Assume by contradiction that e(r, q) 6= e(2, q). By Corollary 3.11, |T | = (q − 1)k(q + 1)n−k. If e(r, q) = 2, then e(2, q) = 1 and |T | = (q − 1)n−1(q + 1). Consider Q with |Q| = (q + 1)n−1(q − 1). From the proof of Lemma 3.10, it follows that |N(G, Q)|r > |N(G, T )|r, a contradiction to |N(G, T )|r = |G|r. If e(r, q) = 1, we proceed in the same way. From the structure of T given in Corollary 3.11, we have that R ∩ T is an Abelian r-group of rank at least 2 (if G 6' SL2(q)). Since for all universal groups of Lie type we have O20 (Z(G)) is cyclic, this means that R∩T 6≤ Z(G). If G ' SL2(q), then |Z(G)| ≤ 2; hence in this case, it does not contain elements of odd order. ¤ Lemma 3.13. Let G be a finite group of Lie type over a field of odd characteristic and assume that G is not of Ree type. Let S be a Sylow 2-subgroup of G and T a maximal torus of G with S ≤ N(G, T ). m k Then |NG(S): SCG(S)| = 3 and |NG(S):(NG(S) ∩ N(G, T ))| = 3 for some integers m and k. In particular, if N(G, T ) contains a Sylow 3-subgroup of G, then NG(S) ≤ N(G, T ). 12 D.O.REVIN AND E.P.VDOVIN

Proof. If G is an exceptional group, then, by [32, Theorem 6], NG(S) ≤ N(G, T ) and we have nothing to prove. Moreover, in [32, Theorem 7], it is shown that CG(S) ≤ N(G, T ) for all groups of Lie type; hence it is enough to prove that m |NG(S): SCG(S)| = 3 for all classical groups. Now we use the notations of [1]. Let Ω(G) be the set of fundamental subgroups of G, R(G) be the set of Sylow 2-subgroups of members of Ω(G). It is known that if R ∈ R(G), then R is contained in a unique member K of Ω(G). Denote this member by K(R). If S is a Sylow 2-subgroup of G, let Fun(S) = {K(R): R ∈ R(G) ∩ S}, where R(G) ∩ S is the set of members of R(G) contained in S. Since R(G) is G-invariant, it follows that NG(Fun(S)) ≥ NG(S). Denote NG(Fun(S)) by N, hFun(S)i by ∆, and CG(Fun(S)) by C. Clearly, C E N and C∆EN. Since NG(S) ≤ N, it follows that NN (S) = NG(S), hence it is enough l to prove that |NN (S): SCN (S)| = 3 for some integer l. Now NN (S) maps into NN=C∆(SC∆/C∆). In view of [1, Theorem 2], the structure of N/C∆ is known and a Sylow 2-subgroup of N/C∆ is self normalizing. 0 Since NN (S) is solvable, it contains a Hall 2 -subgroup, say M. In view of the above consideration, we obtain that M ≤ C∆. Now M ∩ C E NN (S), hence M ∩ C ≤ CN (S). Thus it remains to prove that ¯£ ¤ £ ¤¯ ¯ ¯ t (3.1) NC∆=C ((S ∩ C∆)C/C) : (S ∩ C∆)C/C · CC∆=C ((S ∩ C∆)C/C) = 3 . ﬁ But C∆/C ' ∆ is a central product of SL2(q ) ([1, (1.2)]), and (3.1) is evident. The lemma follows. ¤ Lemma 3.14. Let G, M satisfy one of the following conditions:

(a) G ' PSL2(q), (q, 6) = 1, Sym4 ' M < G or Alt4 ' M < G; (b) G ' PSL2(q), (q, 30) = 1, SL2(4) ' M < G; (c) G ' PΩ7(q), (q, 210) = 1, Ω7(2) ' M < G; + + (d) G ' PΩ8 (q), (q, 210) = 1, Ω8 (2) ' M < G. Assume K is chosen so that G < K ≤ Gb. Then K does not contain a subgroup M1 such that M1 ∩ G = M and |M1 : M| = |K : G|. Proof. (a) and (b) follow from [28, Ch. II, § 8], (d) follows from [29, Propo- sition 2.3.8]. As to (c), note that, by [13], the minimal degree of an irreducible nontrivial representation of Ω7(2) is equal to 7 (under assumption of the lemma (|M|, q)= 1) and Out(Ω7(2)) is trivial. Thus NGb(Ω7(2)) = NG(Ω7(2)) = Ω7(2), and the lemma follows. ¤ In later considerations, we shall often meet the situation where a Hall subgroup is contained in the normalizer of some reductive subgroup of maximal rank. Though the structure of such normalizers is known, we need some additional information. Namely, we need to know which automorphisms of simple components of a given reductive subgroup of maximal rank are contained in the whole group. For this purpose, ﬁrst we give below the result by N.A.Vavilov [42, Table 2]. HALL SUBGROUPS OF FINITE GROUPS 13

Table 2 Inner-diagonal automorphisms

Φ ∆ h!(ε)

A‘−1, ` even h!¯2 (ε) B‘ D‘ —

Dk, 3 ≤ k ≤ ` − 1 h!¯1 (ε) C‘ Ck, 1 ≤ k ≤ ` − 1 — 0 00 D‘ A‘−1 and A‘−1, ` even h!¯2 (ε)

Dk, 3 ≤ k ≤ ` − 1 h!¯1 (ε)

E6 A5 h!¯3 (ε)

A7 h!¯4 (ε) 00 E7 A5 h!¯2 (ε)

D6 h!¯6 (ε)

A8 h!¯3 (ε)

E8 D8 h!¯8 (ε) 00 A7 h!¯2 (ε) B4 — F4 D4 —

A3 h!¯2 (ε) G2 A2 —

In the table, Φ is a root system of G, ∆ is a root system of some simple component of a reductive subgroup of maximal rank in G and h!(ε) is a diagonal automorphism of the component that is contained in G. In all groups not mentioned in the table, G contains all inner-diagonal automorphisms. The deﬁnition of h!(ε) is given in [42]. The lemma below is immediate from Table 2.

p0 Lemma 3.15. Let G = O (G) be a ﬁnite group of Lie type over a ﬁeld of odd characteristic p. Assume that R = T (G1 ∗ ... ∗ Gk) is a reductive subgroup of + maximal rank of G. If Gi/Z(Gi) is isomorphic to PSL2(q), or PΩ7(q), or PΩ8 (q), \ then Gi/Z(Gi) < R/CR(Gi) ≤ Gi/Z(Gi), except the following pairs: G = B4(q), + + k = 1, G1/Z(G1) = PΩ8 (q); G = F4(q), k = 1, G1/Z(G1) = PΩ8 (q); and G = B2(q), k = 1, G1/Z(G1) ' PSL2(q). This lemma together with Lemma 2.4 plays an important role in inductive arguments below. But it has three exceptional cases. The next lemma helps us to consider these exceptional cases.

0 Lemma 3.16. Let G = Op (G) be a ﬁnite simple group of Lie type over a ﬁeld of characteristic p. + (i) If G ' B4(q), then there is a subgroup H of G such that H ' 2.Ω8 (2) h 2 ≤ N(G, D4(q)) = CG(t) for some central involution of a Sylow 2- subgroup of G Moreover, H is a Hall {2, 3, 5, 7}-subgroup of G if and 14 5 2 only if |G|{2;3;5;7} = 2 · 3 · 5 · 7. (ii) If G ' F4(q), then for some central involution of a Sylow 2-subgroup of G, we have that CG(t) = N(G, B4(q)) ≥ N(G, D4(q)) does not contain any Hall {2, 3, 5, 7}-subgroup of G. 14 D.O.REVIN AND E.P.VDOVIN

(iii) If G ' B2(q), then N(G, A1(q)) does not contain any Hall subgroup of even order of G.

Proof. (i) By Lemma 3.7, we have that N(G, D4(q)) ' D4(q)hhτi, where τ is + a graph automorphism of order 2 of D4(q) and D4(q) ' 2.PΩ8 (q). By [29, Propo- sition 2.3.8], we see that N(G, D4(q)) contains a subgroup H of the same structure as in the lemma. Noting that, if (q, 210) = 1, then (|G : N(G, D4(q))|, 210) = 1, + we see that H is a Hall subgroup of G if and only if Ω8 (2) is a Hall subgroup of + PΩ8 (q). Thus we are done by Lemma 3.5. (ii) By Lemma 3.7, N(G, B4(q))/B4(q) is isomorphic to a subgroup of NW (W (B4))/W (B4). Now |W : W (B4)| = 3; hence either NW (W (B4)) = W 0 or NW (W (B4)) = W (B4). Using [46], we see that |W : W | = 4; hence W does not contain a normal subgroup of index 3. Therefore, NW (W (B4)) = W (B4) and N(G, B4(q)) = B4(q). But |G : B4(q)|3 ≥ 3, hence B4(q) = N(G, B4(q)) does not contain any Hall {2, 3, 5, 7}-subgroup of G. (iii) By [9, Type C‘], we have 2 ≤ |W : NW (W (A1))|2 ≤ |G : N(G, A1(q))|2, so N(G, A1(q)) does not contain a Sylow 2-subgroup of G. ¤

4. Hall subgroups of alternating groups The following statements are known.

Lemma 4.1. Let G = Symn. (1) G contains a transitive Abelian subgroup of order m if and only if n = m. (2) If E is a transitive Abelian subgroup, then CG(E) = E. (3) If E is a transitive elementary Abelian subgroup of order pﬁ of G, then NG(E)/E is isomorphic to a subgroup of GLﬁ(p). Theorem 4.2. Let π be a set of primes with 3 ∈/ π and 2 ∈ π. A subgroup M of Altn is a Hall π-subgroup of Altn if and only if M is a Sylow 2-subgroup of Altn, i.e., π ∩ π(Altn) = {2}.

Proof. Assume that the theorem is false and A = Altn is a counter example of minimal degree. Assume that M is a Hall π-subgroup of A and M is not a Sylow 2-subgroup. Let S = Symn. Using the description of maximal subgroups in alternating groups of degree less than 9 given in [13], we obtain that n ≥ 9. Note that M is a transitive and primitive subgroup of A. Indeed, suppose that M is not transitive or is not primitive. Then there is a subgroup G of Symn such that M ≤ G and G is either of form Symm × Symk with m + k = n, if M is not transitive, or of form Symm o Symk with mk = n, if M is transitive but is not primitive, and m, k < n in both cases. It follows that M ≤ G ∩ A and every composition factor of G∩A is a 2-group, 3-group, or an alternating group of degree less than n. Minimality of A implies that M is a 2-group, so π ∩ π(A) = {2}. A contradiction. By Jordan’s theorem [38, p. 58], if a primitive permutation group of degree n has minimal degree m > 3 (i.e., the minimum of degrees of its elements), then m2 m m n < log + m(log + 3). 4 2 2 A group M contains neither a transposition nor a cycle of length 3, but contains an element of the form (ij)(kl). Hence the minimal degree of M is 4 and n < 20. HALL SUBGROUPS OF FINITE GROUPS 15

Let P be a minimal normal subgroup of M. Then P is transitive. Assume that P is an Abelian group. In this case, |P | = pfi = n for a prime p. In view of fi Lemma 4.1, n = p and NS(P )/P is isomorphic to a subgroup of GLfi(p). Hence either α = 1 and |M| divides p(p − 1), or α = 2 and p = 2, 3, or α = 3, 4 and p = 2 (recall that n < 20). It is easy to check that in any case M is not a Hall subgroup of Altn. Now assume that P is nonabelian. In this case, P is isomorphic to a group of form T × · · · × T where T is a nonabelian simple group. Since |T | is coprime with 3, T is Suzuki group Sz(22t+1), where t ≥ 1. It follows that Sz(22t+1) has a transitive permutation representation of degree at most n and as a consequence contains a subgroup of index less then 20. But [40, Theorem 9] implies thet the minimal index of subgroups in Sz(22t+1) is 22(2t+1) + 1 ≥ 65. A contradiction. ¤

Theorem 4.3. Let S = Symn, A = Altn. The following statements are equivalent.

(1) S satisfies E…. (2) S satisfies C…. (3) A satisfies E…. (4) A satisfies C…. Proof. (1) ⇒ (2). Follows from the description of Hall π-subgroups in symmetric groups given in [25, 41]. (2) ⇒ (3). Trivial. (4) ⇒ (1). Follows from Lemma 3.1(2). (3) ⇒ (4). Clearly, it may be assumed that 2 ∈ π. In view of Theorem 4.2, we may suppose that 3 ∈ π. Assume by contradiction that the statement is false and A is a minimal counter example. Let M be a Hall π-subgroup of A. Assume that M either is not transitive or is transitive, but is not primitive. Then M is contained in a subgroup K of S such that K is a direct product of nonidentical symmetric groups if M is not transitive, or K is a wreath product of nonidentical symmetric groups if M is transitive but is not primitive. In particular, every nonabelian composition factor of K is an alternating group of degree less then n and K contains a transposition. It is also clear that M is a Hall π-subgroup of K ∩A. It follows that every composition factor of K ∩A satisfies E…. By induction, every composition factor of K ∩ A satisfies C…. Hence, by Lemma 3.1(2), K ∩ A satisfies C… and K satisfies C…. In particular, there is a Hall π-subgroup M0 of K and simultaneously of S and, by Lemma 3.1, M0 ∩ A is a Hall π-subgroup of K ∩A. But K ∩A satisfies C…, so, up to conjugation, M = M0 ∩A. Since S satisfies C…, then all Hall π-subgroups of S are conjugate to M0 under S. Hence all Hall π-subgroups of A are conjugate under S. Since M0 ≤ NS(M) and M0 contains a transposition, we obtain that all Hall π-subgroups are conjugate under A. Therefore, assume that M is transitive and primitive. Since M contains a cycle of order 3, it follows that M coincides with A by Jordan Theorem ([5, Theorem I, p. 207]). Hence A satisfies C…. ¤

Corollary 4.4. Let π be a set of primes. M is a Hall π-subgroup of Altn if and only if M = M0 ∩ Altn for some Hall π-subgroup M0 of Symn.

Proof. If M is a Hall π-subgroup of Altn, then Altn satisﬁes E…. Theorem 4.3 implies that Symn also satisﬁes E…. Let M0 be a Hall π-subgroup of Symn. Then 16 D.O.REVIN AND E.P.VDOVIN

M0 ∩ Altn is a Hall π-subgroup of Altn. Since by Theorem 4.3 Altn satisfies C…, subgroups M and M0 ∩ Altn are conjugate under Altn. Replacing M0 by a suitable conjugate subgroup, we can assume M = M0 ∩ Altn. ¤ 5. Hall π-subgroups of finite groups of Lie type In the sequel, we say that a subgroup A of G acts on a subgroup H of G if A ≤ NG(H) and A acts by conjugation on H. Recall also that a Hall subgroup of G is nontrivial if it is not a Sylow subgroup of G. 2n+1 2 Lemma 5.1. Let q = 3 , G = G2(q), π a set of primes with 3 6∈ π, 2 ∈ π, τ = π \{2}. G contains a nontrivial Hall π-subgroup M if and only if one of the following conditions holds. (1) π∩π(G) ⊆ {2, 7}, M is a normalizer of a Sylow 2-subgroup of G, |M| = 56 and M is a Frobenius group. (2) π ∩ π(G) ⊆ π(q + 1), there exists a torus T of order q + 1 such that M is contained in N(G, T ), and M ∩ T is an Abelian Hall τ-subgroup of G. In both cases, Aut G stabilizes some conjugacy class of Hall π-subgroups. Proof. Since M contains a Sylow 2-subgroup of G, the main result of [33] implies that either M is contained in a subgroup isomorphic to A1(8), or M contains a normal Abelian Hall τ-subgroup. In the first case, M contains a normal Sylow 2-subgroup and do not coincide with this subgroup if and only if π ∩ π(G) = {2, 7}. So we get (1). Clearly, all Hall subgroups with this structure are conjugate under G and Aut G stabilizes this conjugacy class. It is also clear that in both cases a Sylow 2-subgroup of M is elementary Abelian. If M contains a normal Abelian Hall τ-subgroup and M is not contained in A1(8), then M is contained in the normalizer of some τ-subgroup L of G and |NG(L)|2 = |G|2. In view of [33, Theorem 1], L ≤ T , where |T | = q + 1 and M ≤ N(G, T ). ¤

Theorem 5.2. Let G = G be a universal finite group of Lie type over a field 2 GF (q) of odd characteristic p, G 6= G2(q). Let π be a set of primes such that 3, p 6∈ π, 2 ∈ π. Define τ = π \{2}. Then G contains a nontrivial Hall π-subgroup M if and only if one of the following conditions holds:

(1) e(2, q) = 1, π ∩ π(G) ⊆ π(q − 1), |G|… = |N|…, where N is a monomial subgroup of G. In this case, M ≤ N, M contains a normal Abelian Hall τ-subgroup R and R is contained in a Cartan subgroup of G. 3 3 (2) G 6= D4(q ), e(2, q) = 2, π ∩ π(G) ⊆ π(q + 1), |G|… = |N(G, T )|…, where T is a maximal torus of order (q + 1)k(q − 1)m and k is the maximal k integer such that (t + 1) divides fG(t). In this case, M ≤ N(G, T ), M contains a normal Abelian Hall τ-subgroup R and R ≤ T . 3 3 (3) G = D4(q ), e(2, q) = 2, π∩π(G) ⊆ π(q+1), |G|… = |N(G, T )|…, where T is a maximal torus of G of order (q+1)(q3+1). In this case, M ≤ N(G, T ), M contains a normal Abelian Hall τ-subgroup R and R ≤ T .

In particular, in all cases G satisfies C…. Proof. It is easy to see that if one of (1)-(3) holds, then M is a Hall π-subgroup of G. So we need to prove the part “only if”. We prove first that G satisfies C…. Indeed, M is contained in N(G, T ) for some maximal torus T of G. By Lemma 3.10, all such tori are conjugate under HALL SUBGROUPS OF FINITE GROUPS 17

G, so all such N(G, T ) are conjugate under G. Since MT/T is a 2-group, M is unique up to conjugation in N(G, T ). Therefore, every N(G, T ) contains the only conjugacy class of Hall π-subgroups, so G contains the only conjugacy class of Hall π-subgroups. Assume first that G contains a nontrivial Hall π-subgroup M such that there exists a maximal torus T with M ≤ N(G, T ). In particular, |G|… = |N(G, T )|…. Since 2 ∈ π, we have that N(G, T ) contains a Sylow 2-subgroup of G. Hence the structure of N(G, T ) is known and is given in the proof of Lemma 3.10. Now we prove that M, G, and π satisfies to one of (1)-(3) of the theorem. By Lemma 3.1, MT/T is a Hall π-subgroup of N(G, T )/T . If G is not iso- 2 morphic to E6(q), E6(q2), E7(q) or E8(q), then, from the structure of N(G, T )/T given in the proof of Lemma 3.10, it follows that all Abelian composition factors of N(G, T )/T have orders 2 or 3, and all nonabelian composition factors are isomorphic to Altn for some n. Consequently, if N(G, T )/T contains a nontrivial Hall π-subgroup with 3 6∈ π, then Altn also contains a nontrivial Hall π-subgroup with 3 6∈ π. In view of Theorem 4.2, this is impossible. So, every Hall π-subgroup of N(G, T )/T is a Sylow 2-subgroup in this case. It follows that every τ-element of M is contained in T . So M contains a normal Abelian Hall τ-subgroup and this subgroup is contained in T . Moreover, by Lemma 3.12, e(r, q) = e(2, q). If G is 2 isomorphic to E6(q), E6(q2), E7(q) or E8(q), then all Abelian composition factors of N(G, T ) are of order 2 and there is one nonabelian composition factor, isomor- 2 2 2 2 phic to A3(2 ), B3(2) or D4(2), respectively. So, either A3(2 ), B3(2) or D4(2) contains a nontrivial Hall π-subgroup with 2 ∈ π, 3 6∈ π. By [36, Theorem 3.3], this is impossible; hence we again obtain that M contains a normal Abelian Hall τ-subgroup and e(r, q) = e(2, q). Therefore, M, G, and π satisfies one of (1)-(3) of the theorem. Assume that the theorem is false and G is a counter example of minimal order. Let M be a Hall π-subgroup such that M does not satisfy Theorem. As we proved above, M is not contained in N(G, T ) for any maximal torus T of G. By Lemma 3.2, M either normalizes some proper connected σ-stable subgroup R of G, or MZ(G)/Z(G) is almost simple, or MZ(G)/Z(G) is contained in the normalizer of some Jordan subgroup, or G = E8(q) and M satisfies Lemma 3.2(4). In view of Lemmas 3.3, 3.4, and 3.5, M cannot satisfy the last three conditions. Therefore, M normalizes some proper connected σ-stable subgroup R of G. Then R satisfies one of the following conditions.

(a) Ru(R) = {1}, Z(R) 6≤ Z(G). (b) Ru(R) 6= {1}. (c) Ru(R) = {1} and Z(R) ≤ Z(G). Consider these three cases separately. Since G is universal, we may assume G to be simply connected so that G = G. (a) By Lemma 3.9, we may assume that R is a connected reductive σ-stable subgroup of maximal rank. Now R = R = T G1 ∗ ... ∗ Gk, where T is a maximal σ-stable torus of R, groups G1,...,Gk are of Lie type and all of them are subsystem subgroups of G. Since all of Gi are subsystem subgroups of G, it follows that all of them are not of Ree type. Since all Gi are subnormal in R, Lemma 3.1 implies that M ∩ Gi is a Hall π-subgroup of Gi. Since G is a minimal counter example, we obtain that M ∩ Gi is in N(Gi,Ti), where Ti is a maximal torus of Gi and M ∩ R contains a normal Abelian Hall τ-subgroup K. 18 D.O.REVIN AND E.P.VDOVIN

Assume that K 6= {1}. By Lemma 3.12, we have that K 6≤ Z(G). It follows that

K is normal in M and M ≤ (NG(K)). Since K ≤ T1 ∗...∗Tk, we obtain that K is 0 contained in some torus of G. Thus, by Lemma 3.6, C = CG(K) is a reductive (and σ-stable) subgroup of maximal rank of G. Moreover, in view of rigidity of d-groups, we have |NG(K): C| < ∞, hence M ≤ N(G, C). Now C = HX1 ∗ ... ∗ Xm, where H is some maximal σ-stable torus of C and Xi is a subsystem subgroup of G for all i. f τ ∩π(M ∩X1 ∗...∗Xm) 6= ∅, then by induction M ∩X1 ∗...∗Xm contains a normal Abelian Hall τ-subgroup K1 6= {1} and we may substitute K by hK,K1i and consider CG(hK,K1i). So we may assume that π(M ∩ X1 ∗ ... ∗ Xm) = {2}. Let S be a Sylow 2-subgroup of M. Precisely the same arguments as in the proof of [19, Theorem 4.10.2] show that S ≤ N(N(G, C),Q) for some maximal torus Q of C. Since K ≤ Z(C), we have that K is contained in all maximal tori of C. In particular, K ≤ Q; therefore, M ≤ N(N(G, C),Q) ≤ N(G, Q). As we proved above, in this case M satisfies to one of (1)–(3) of the theorem. Assume now that K = {1}, i.e., M ∩ R is a Sylow 2-subgroup of R. By Lemma 3.7(3), we have that M ≤ N(G, R) ≤ N(G, T )G1 ∗ ... ∗ Gk. Since M is a Hall π-subgroup of G and 2 ∈ π we have that N(G, R) contains a Sylow 2-subgroup of G. Thus we may choose T so that N(G, T ) contains a Sylow 2- subgroup of G. Therefore, MT/T ≤ N(G, T )/T is a Hall π-subgroup of N(G, T )/T . As we noted above, N(G, T )/T does not contain nontrivial Hall π-subgroups, so π ∩ π(N(G, T )) = {2}. Now M ≤ N(G, T )R and M ∩ R is a 2-group, hence M is a 2-group and π ∩ π(G) = {2}. By Lemma 3.10, we obtain that M satisfies one of (1)–(3) of the theorem. (b) In view of Lemma 3.8, we may suppose that R is a parabolic subgroup of G. So M is contained in R = R, a parabolic subgroup of G. Since p 6∈ π, we have that M is contained in a Levi factor L of R. Now O2(Z(L)) 6≤ Z(G), so Z(M) 6≤ Z(G) and we are done by case (a). (c) Consider R = T G1∗...∗Gk, where T is some maximal σ-stable torus of R, G1,...,Gk are finite groups of Lie type. Assume that one of Gi is not isomorphic to 2 2n+1 G2(3 ). We may suppose that it is G1. Consider the orbit of G1 under NG(R). Renumbering factors, if necessary, we may assume that G1 ∗ G2 ∗ ... ∗ Gm is the p0 subgroup generated by the orbit of G1 under NG(R). Since G1∗...∗Gk = O (R), it follows that G1 ∗ G2 ∗ ... ∗ Gm is normal in NG(R). Clearly, M ∩ G1 ∗ ... ∗ Gm is a Hall π-subgroup of G1 ∗ ... ∗ Gm. Hence, by induction, either M ∩ G1 ∗ ... ∗ Gm contains a normal Abelian Hall τ-subgroup K 6= {1} or M ∩ G1 ∗ ... ∗ Gm is a 2-group. Therefore, M contains an Abelian normal subgroup A 6≤ Z(G) such that 0 A ≤ S for some torus S of G1 ∗...∗Gm (hence of G), so M normalizes CG(A) = C. In both cases, we proceed as in case (a). 2 2ni+1 Now assume that all Gi are isomorphic to G2(3 ). Since all Gi are normal in G1 ∗ ... ∗ Gk and M normalizes G1 ∗ ... ∗ Gk, Lemma 3.1 implies that M ∩ Gi is a Hall π-subgroup of Gi for all i. Assume that Gi ∩ M contains a nontrivial normal Abelian Hall τ-subgroup, i.e., satisfies to Lemma 5.1 (2) for some i. We may suppose that i = 1. Consider the subgroup G1 ∗ ... ∗ Gm generated by the orbit of G1 under M. Again M ∩ G1 ∗ ... ∗ Gm contains a normal Abelian Hall 0 τ-subgroup K 6= {1}. So M normalizes CG(K) and we are done by case (a). Therefore, we may assume that M ∩ Gi contains a normal elementary Abelian 2-subgroup for all i. Denote this subgroup by Ji. Then J = h J1,...,Jm i is a normal elementary Abelian 2-subgroup of M. So M is contained in NG(J). Since HALL SUBGROUPS OF FINITE GROUPS 19 the rank of Z(G) is at most 2 and |Ji| = 8 for all i, we have that J 6≤ Z(G). 0 Therefore, NG(J) is trivial, since otherwise we would obtain the first case. By Lemma 3.2, it follows that NG=Z(G)(JZ(G)/Z(G)) is contained in the normalizer of a Jordan subgroup of G/Z(G). Therefore, MZ(G)/Z(G) is contained in the normalizer of a Jordan subgroup of G. A contradiction to Lemma 3.3. ¤ Corollary 5.3. Let G be a finite group of Lie type over a field of characteristic p and let K be a subsystem subgroup of G. Assume that G satisfies E… with 2 ∈ π, 3, p 6∈ π. Then K satisfies E…. Proof. Immediate from Lemma 5.1 and Theorem 5.2. ¤ Corollary 5.4. Let π be a set of primes such that 3 6∈ π. Assume that G is a finite group of Lie type and G is not a Ree group. Then G is an E…-group if and only if G is a C…-group.

Proof. Assume that G is an E…-group and M is a nontrivial Hall π-subgroup of G. If p ∈ π, we are done by [23, Theorem 3.1]. If 2 6∈ π, the corollary follows from [22, Theorem A]. So, we may assume that 2 ∈ π and p 6∈ π. In this case, our statement follows from Theorem 5.2. ¤ This corollary gives an affirmative answer to Conjecture 1.2. Combining Lemma 5.1 and Theorem 5.2, we obtain an affirmative answer to Conjecture 1.4. If G is a known finite simple group and G satisfies E…, with 3 6∈ π, then Theorems 4.2, 5.2, Lemma 5.1, and Corollary 5.4 imply that Aut G satisfies E…. Thus [21, Corollary 3.6] implies (modulo CFSG) that a finite group satisfies E… if and only if all its composition factors satisfy E…. Now we prove a general theorem about the structure of Hall π-subgroups in finite groups of Lie type with 2, 3 ∈ π, p 6∈ π.

0 Theorem 5.5. Let G = Op (G) be a ﬁnite simple group of Lie type with a base ﬁeld GF (q) of characteristic p, π a set of primes such that 2, 3 ∈ π and p 6∈ π. Then G contains a Hall π-subgroup M if and only if G, π, and M satisfy one of the following conditions.

(a) π ∩ π(G) ⊆ π(q − 1) ∪ π(W (G)), |N|… = |G|…, and M is a Hall π-subgroup of N, where N is a monomial subgroup of G. (b) π ∩ π(G) ⊆ π(q + 1) ∪ π(N(G, T )/T ), |N(G, T )|… = |G|…, and M is a Hall π-subgroup of N(G, T ). Here T is a maximal torus either of order 1/δ · (q − 1)n−k(q + 1)k, where k is the maximal integer such that (t + 1)k 3 3 3 3 3 divides fG(t) if G 6' D4(q ), or of order (q + 1)(q + 1) if G ' D4(q ). (c) G = PSL2(q), π∩π(G) = {2, 3}, |G|… = 24 (resp., |G|… = 12), M ' Sym4 (resp., M ' Alt4). (d) G = PSL2(q), π ∩ π(G) = {2, 3, 5}, |G|… = 60, M ' SL2(4). (e) G = PΩ7(q), π ∩ π(G) = {2, 3, 5, 7}, M ' Ω7(2), |G|… = |M|. + + (f) G = PΩ8 (q), π ∩ π(G) = {2, 3, 5, 7}, M ' Ω8 (2), |G|… = |M|. + (g) G = PΩ9(q), π ∩ π(G) = {2, 3, 5, 7}, M ' 2.Ω8 (2) h 2, |G|… = |M|. Proof. If π, G, and M satisfy the conditions (a)–(g), then M is a Hall subgroup of G. Indeed, note that the ﬁrst two cases give us most of Hall subgroups and they appear from the normalizer of maximal tori. Case (c) appears from Jor- dan subgroups (Lemma 3.3), cases (d)–(f) appear from almost simple subgroups 20 D.O.REVIN AND E.P.VDOVIN

(Lemma 3.5), and case (g) appears from (Lemma 3.16(i)). Thus we need to prove the “only if” part. Assume that G is a counter example of minimal order to the theorem. This means that M < G is a Hall π-subgroup of G such that π, G, and M do not satisfy one of (a)–(g). Note ﬁrst that in this case there does not exist a maximal torus T of G with M ≤ N(G, T ). Indeed, if such a torus T exists, then N(G, T ) contains a Sylow 2-subgroup of G, as 2 ∈ π. In view of Lemma 3.10, T is unique up to conjugation in G. So in this case π, G, and M satisfy either (a) or (b). Since G is simple, it follows that G can be chosen of adjoint type. By Lemma 3.2, M normalizes some proper connected closed σ-stable subgroup of G, or M is contained in the normalizer of some Jordan subgroup of G, or M is almost simple and does not normalize any proper connected σ-stable subgroup, or M and G satisfy (4) of Lemma 3.2. In view of Lemmas 3.3, 3.4, and 3.5, M can not satisfy the last three conditions. So, M normalizes some proper connected closed σ-stable subgroup R of G. There can be three cases:

(i) Ru(R) = {1} and either Z(R) 6= {1} or R is of maximal rank. (ii) Ru(R) 6= {1}. (iii) Z(R) = {1}, Ru(R) = {1}, and R is not of maximal rank. Consider these three cases separately. (i) In view of Lemma 3.9, we may suppose that R is of maximal rank in both cases. So R is a subsystem subgroup of G and R = G ∩ R = T (G1 ∗ ... ∗ Gk), where all Gi are subsystem subgroups of G and T is a maximal torus of G. Since all Gi are subnormal in N(G, R), Lemma 3.1(4) implies that Mi = M ∩ Gi is a Hall π-subgroup of Gi. By induction, π, Gi, and Mi satisfy one of the conditions (a)–(g) of the theorem. Assume, for some i that π, Gi and Mi satisfy one of the conditions (c)–(g). Note that if π, Gi, and Mi satisfies (g), then Mi has a central involution s and, fi fi fi hence, we may substitute Gi ' B4(q )(α ≥ 1) by CGi (s) = N(B4(q ),D4(q )). So we may assume that π, Gi, and Mi satisfy one of the conditions (c)–(f). In view \ of Lemma 3.15, it follows that Gi/Z(Gi) < R/CR(Gi) ≤ Gi/Z(Gi), except three cases mentioned in the lemma. By Lemma 3.1, it follows that R/CR(Gi) contains a Hall π-subgroup M˜ i such that |M˜ i : MiZ(Gi)/Z(Gi)| = |R/CR(Gi): Gi/Z(Gi)| and M˜i ∩Gi/Z(Gi) = MiZ(Gi)/Z(Gi). But in view of Lemma 3.14 it is impossible. The three exceptional cases can be excluded using Lemma 3.16. So we may assume that π, Gi, and Mi satisfy either (a) or (b). This means that, for all i, Mi ≤ N(Gi,Ti) for some maximal torus Ti of Gi. Now we want to show that M ≤ N(G, T ) for some maximal torus T of G. Let T be a maximal torus of R with T1 ∗ ... ∗ Tk ≤ T (here all Ti are chosen in order that the condition Mi ≤ N(Gi,Ti) holds for all i). In view of Lemma 3.7, N(G, R) = N(N(G, R),T ) · (G1 ∗ ... ∗ Gk). So every element x ∈ N(G, R) has a decomposition x = yz, where y ∈ N(N(G, R),T ), z ∈ G1 ∗ ... ∗ Gk. Since G1 ∗ ... ∗ Gk E N(G, R), it follows that M ≤ NN(G;R)(M1 ∗ ... ∗ Mk). Assume that Si is a Sylow 2-subgroup of Mi, hence of Gi. In view of Frattini’s Lemma, NN(G;R)(M1 ∗ ... ∗ Mk) ≤ M1 ∗ ... ∗ Mk · NN(G;R)(S1 ∗ ... ∗ Sk). Since, under our assumption, M1 ∗ ... ∗ Mk ≤ N(N(G, R),T ), it is enough to show that NN(G;R)(S1 ∗...∗Sk) ≤ N(N(G, R),T ). To this end, let x ∈ NN(G;R)(S1 ∗...∗Sk). As we noted above, x has a decomposition x = yz with y ∈ N(N(G, R),T ), z ∈ G1∗...∗Gk. Since S1∗...∗Sk is a Sylow 2-subgroup of N(G1,T1)∗...∗N(Gk,Tk) HALL SUBGROUPS OF FINITE GROUPS 21

y and N(G1,T1) ∗ ... ∗ N(Gk,Tk) E N(N(G, R),T ) we obtain that (S1 ∗ ... ∗ Sk) z is a Sylow 2-subgroup of N(G1,T1) ∗ ... ∗ N(Gk,Tk). Hence (S1 ∗ ... ∗ Sk) is a Sylow subgroup of N(G1,T1) ∗ ... ∗ N(Gk,Tk). Thus there is an element z g−1 g ∈ N(G1,T1) ∗ ... ∗ N(Gk,Tk) with (S1 ∗ ... ∗ Sk) = (S1 ∗ ... ∗ Sk) . Therefore, zg ∈ NG1∗:::∗Gk (S1 ∗ ... ∗ Sk). Since N(Gi,Ti) contains a Sylow 3-subgroup of Gi for all i, Lemma 3.13 implies NG1∗:::∗Gk (S1 ∗ ... ∗ Sk) ≤ N(G1,T1) ∗ ... ∗ N(Gk,Tk). Hence z ∈ N(G1,T1) ∗ ... ∗ N(Gk,Tk) ≤ N(N(R,T ),T ). Thus x ∈ N(N(G, R),T ) and M ≤ NN(G;R)(M1 ∗ ... ∗ Mk) ≤ N(N(G, R),T ) ≤ N(G, T ). A contradiction to the assumption M 6≤ N(G, T ). (ii) Lemma 3.8 implies that we may assume R to be parabolic. So R = R ∩ G is a parabolic subgroup of G and R = L i Op(R), where L is a Levi factor of R. Consider a homomorphism ϕ : R → R/Op(R) ' L. We have that Z(L) is nontrivial and, since p is odd, its order is even. Hence L ' L’ has a central involution. Since M ’ ≤ L’ and M ’ contains a Sylow 2-subgroup of L’, we obtain that M ’ contains a nontrivial central involution as well. Now M ’ ' M, thus there exists a nontrivial 0 central involution, say s, of M. Therefore, M ≤ CG(s) ≤ NG(CG(s) ). Since 0 CG(s) is a proper connected σ-stable subgroup of maximal rank of G we are done by the first case. (iii) Clearly, M is contained in some maximal subgroup of odd index of G. The list of such subgroups is given in [34, Main Theorem]. Here we shall not give a complete statement of the theorem and assume that the reader has it. Since p 6∈ π it follows that M is contained in one of subgroups from the list (III)(A). Denote this subgroup by K. c Assume first that K = NG(L(q0)), where q = q0 and c is an odd prime. Then \ p0 c K ≤ L(q0). Indeed, L(q0) = O (G¿ ) for some automorphism τ with τ = σ (see [18, (7-2)]). Since G does not contain graph, field, and graph-field automorphisms p0 p0 of O (G¿ ) (in view of [39, Theorem 30] every automorphism of O (G¿ ) can be extended to an automorphism of G), it follows that NG(L(q0)) ≤ NG(L(q0)) ≤ \ L(q0). Using induction we obtain that M satisfies to one of the conditions (a)–(g) of the theorem. Suppose now that G is a classical group. Clearly, if K satisfies (III)(A)(ii)(1)– (III)(A)(ii)(3), then K normalizes some connected group of maximal rank, hence, M also does. Therefore, M is contained either in some parabolic subgroup or in the normalizer of some reductive subgroup of maximal rank. In both cases, we are done by (i) or by (ii). In cases (III)(A)(ii)(4)–(III)(A)(ii)(7), we have that K is a simple group of Lie type over a field of even characteristic; or K is isomorphic to one of the groups: 1=2 M10, dihedral, Alt4, Sym4, Alt5, PGL2(q ). If K = Alt5, then either M = Alt4 or M = Alt5 and M again is not in contrast with the theorem. So, suppose that 1=2 G = PSL2(q), K = PGL2(q ), and M is a Hall π-subgroup of G and K. Moreover, 1 1 M ∩ PSL2(q 2 ) is a Hall π-subgroup of PSL2(q 2 ). By induction, M is contained in N(K,TK ). So M contains an Abelian subgroup of index 2. Hence M is contained in the normalizer of this subgroup and this normalizer is contained in N(G, T ) for some maximal torus T of G. But we already noted that M is not contained in N(G, T ), a contradiction. If G is an exceptional group, then [34, Table 1] implies that either K normalizes some proper connected subgroup of maximal rank of K or K is the normalizer of a 22 D.O.REVIN AND E.P.VDOVIN

3 3 Jordan subgroup, or G = D4(q ) and K = G2(q), or G = G2(q) and K = G2(2). In the ﬁrst case, we are done by (i) or (ii); in the second case, we apply Lemma 3.3. In the third case, we use induction; the fourth case is considered in the proof of Lemma 3.5. Anyway, G, M, and π are not counterexamples to the theorem. ¤

6. Structure of Hall subgroups in finite groups of Lie type The main technical tool of the section is Theorem 5.5. Using this theorem, we find the structure of Hall π-subgroups such that 2, 3 ∈ π, p 6∈ π in all finite simple groups of Lie type. The information about the structure of N(G, T ) (with |N(G, T )|2 = |G|2) obtained in the proof of Lemma 3.10 is also useful. Following [36], we call a nonabelian simple group S bad for π if S satisfies the following conditions. (1) S satisfies E…. (2) every π-subgroup is contained in some Hall π-subgroup. (3) any two Hall π-subgroup are conjugate under Aut(S); (4) the number of conjugacy classes of Hall π-subgroups is not a π-number. If a nonabelian finite simple group is not bad, then it is called good. Note that if S satisfies C…, then S is good.

6.1. Linear and unitary groups.

p0 Lemma 6.1. Let G = O (G) be a finite simple group of Lie type over a field GF (q) of characteristic p and G = PGLn(F), where F is the algebraic closure of GF (q). Assume that π is a set of primes such that 2, 3 ∈ π and p 6∈ π. Then G satisfies E… if and only if π and G satisfy one of the following conditions.

(i) G = PSL2(q), |G|… = 24 (resp., |G|… = 12). In this case, a Hall π- subgroup M is isomorphic to Sym4 (resp., Alt4). (ii) G = PSL2(q), |G|… = 60. In this case, a Hall π-subgroup M is isomorphic to SL2(4). 2 (iii) G ' PSLn(q) (resp., G = PSUn(q )), e(2, q) = 1 (resp., e(2, q) = 2), π ∩ π(G) ⊆ π(q − 1) ∪ π(n!) (resp., π ∩ π(G) ⊆ π(q + 1) ∪ π(n!)), Symn satisfies E…, and, if a prime r belongs to π(n!) ∩ π \ π(q − 1) (resp., π(n!) ∩ π \ π(q + 1)), then (n!)r = | Symn |r = |G|r. In this case, a Hall π- subgroup M is contained in N(G, T ), where T is a maximal torus of order 1 n−1 1 n−1 (n;q−1) (q − 1) (resp., (n;q+1) (q + 1) ). Moreover, G satisfies C…. 2 (iv) G ' PSLn(q) (resp., G = PSUn(q )), e(2, q) = 2 (resp., e(2, q) = 1), π ∩ π(G) ⊆£ π¤(q + 1) (resp., π ∩ π(G) ⊆ π(q − 1)), Symk satisfies E…, n where k = 2 . In this case, a Hall π-subgroup M is contained in N(G, T ), 1 n−k−1 k where T is a maximal torus of order (n;q−1) (q − 1) (q + 1) (resp., 1 k n−k−1 (n;q+1) (q − 1) (q + 1) ). Further G satisfies C….

In all cases, G does not satisfy D… and G is good. Proof. In view of Theorem 5.5, we may assume that M is contained in N(G, T ) for some maximal torus T of G. By Lemma 3.10, this torus is unique up to conjugation in G and the structure of N(G, T )/T is known. 2 If we have that either G = PSLn(q) and e(2, q) = 1, or G = PSUn(q ) and 1 n−1 e(2, q) = 2, then N(G, T )/T ' Symn; and |T | is known to equal (n;q−1) (q − 1) HALL SUBGROUPS OF FINITE GROUPS 23 or 1 (q + 1)n−1, respectively. If either G = PSL (q) and e(2, q) = 2, or (n;q+1) n £ ¤ 2 n G = PSUn(q ) and e(2, q) = 1, then N(G, T )/T ' 2 o Symk with k = 2 . Further 1 n−k−1 k 1 k n−k−1 |T | is equal to (n;q−1) (q−1) (q+1) or (n;q+1) (q−1) (q+1) , respectively. Lemma 3.1(1) implies that N(G, T )/T satisfies E…; therefore, we obtain statements (i)–(iv) of the lemma. Note that in cases (iii) and (iv) T is unique up to conjugation in G (see Lemma 3.10). The only nonabelian composition factor of N(G, T ) is isomorphic to Altm for some m; hence, by Theorem 4.2, this composition factor satisfies C…. Therefore, N(G, T ) satisfies C… and in cases (iii), (iv), G satisfies C…. Thus it remain to prove that G does not satisfy D…, and G is good. Consider all cases separately. In cases (i) and (ii), M does not contain an element of order 6, but G does. So G does not satisfy D…, and G is good in these two cases. 2 (iii) Note first that, by Lemma 3.1, PSLn(q) (resp., PSUn(q )) satisfies D… if 2 and only if SLn(q) (resp., SUn(q )) satisfies D…. To simplify our consideration in 2 this case, we assume G to be equal to SLn(q) or SUn(q ), respectively. In order " + to unify notations, denote G by SLn(q), where ε ∈ {−, +} and SLn(q) = SLn (q), 2 − SUn(q ) = SLn (q). We have that M is contained in N(G, T ), where T is a maximal n−1 torus of order (q−(ε1)) and N(G, T )/T ' Symn. Further M/(M ∩T ) ' MT/T is a Hall π-subgroup of Symn. Assume first that MT/T 6= Symn. From [25, Theorem A4] and [41], it follows that either n is a prime and MT/T = Symn−1, or n ∈ {7, 8} and π ∩ π(Symn) = {2, 3}. In the first case, a π-element x1 of order 2(n− 2) does not belong to MT/T , and in the second case a π-element x2 of order 6 does not belong to MT/T . Let x be a preimage of xi in N(G, T ) under the natural homomorphism ϕ : N(G, T ) → N(G, T )/T . Evidently, we may choose x to be a π-element. Denote MT = M ∩ T . It is clear that hx, MT i is a π-group and it is not isomorphic to any subgroup of M. m1 mk Assume now that MT/T = Symn. Let n = 2 + ... + 2 be the 2-adic " decomposition of n and m1 > m2 > . . . > mk. Consider L = SL2m1 (q) × ... × " SL2mk (q) ≤ G. Here L is the group of block-diagonal matrices diag(A1,...,Ak), " where Ai ∈ SL2mi (q). Clearly, we may assume that T normalizes L. So TL is " a reductive subgroup of maximal rank and T ∩ SL2mi (q) is a maximal torus of " " SL2mi (q) for all i. In view of [31, Tables 3.5.A and 3.5.B], we have that SL2m (q) contains a symplectic-type 2-subgroup S = 4 ∗ 21+2m; moreover, S is absolutely irreducible ([31, § 4.6]) and NG(S) = NG(S) = S. Sp2m(2). Let S1,...,Sk be " " + such subgroups of SL m (q),..., SL m (q), respectively. Let P ' Ω (2) be a 2 1 2 k i 2mi subgroup of Sp (2) and let Ri be the complete preimage of Pi in NSLε (q)(Si) 2mi 2mi under the natural homomorphism NSLε (q)(Si) → NSLε (q)(Si)/Si ' Sp (2). 2mi 2mi 2mi Denote by R the subgroup R1 × ... × Rk. Clearly, R is a π-subgroup. We want to show that R is not conjugate to any subgroup of N(G, T ), hence of M. Since Si is absolutely irreducible for all i, it is enough to show that Ri does not belong " " to N(SL2mi (q), SL2mi (q) ∩ T ) for all i. Thus we may assume that k = 1, i.e., n = 2m. Assume that R is conjugate to a subgroup of N(G, T ). Clearly, we may suppose that R ≤ N(G, T ). So RT/T ≤ N(G, T )/T = Symn and RT/T is clearly nontrivial. But in view of [31, Lemma 7.6.1(i)], the degree of the minimal nontrivial permutation representation of R is greater than n, a contradiction. 1 n−1 (iv) Consider N(G, S), where S is a maximal torus of order (n;q−1) (q − 1) if 1 n−1 2 G = PSLn(q) and of order (n;q+1) (q + 1) if G = PSUn(q ). Then N(G, S)/S ' 24 D.O.REVIN AND E.P.VDOVIN

Symn and |S|3 = 1. Let P1 be a Sylow 2-subgroup of S and P2 be a Sylow 3-subgroup of N(G, S). Consider L = hP1,P2i = P1 h P2. Assume that L ≤

N(G, T ). Easy calculations shows that CP2 (P1) is trivial. Hence L ∩ T ≤ P1. ’ Thus, for a natural homomorphism ϕ : N(G, T ) → N(G, T )/T , we have P2 ' P2. Therefore, P2 is isomorphic to a subgroup of N(G, T )/T . But |P2| = | Symn |3 > |N(G, T )/T |3 = | Sym n |3, a contradiction. [ 2 ] Since in cases (iii) and (iv) G satisﬁes C…, we have that G is good. ¤

6.2. Groups of type Bn. p0 Lemma 6.2. Let G = O (G) be a finite simple group of Lie type over a field GF (q) of characteristic p and let G = Bn(F) be of adjoint type, where F is the algebraic closure of GF (q). Assume that π is a set of primes with 2, 3 ∈ π, p 6∈ π. Then G satisfies E… if and only if π and G satisfy one of the following conditions. 9 4 (i) G = PΩ7(q), π ∩ π(G) = {2, 3, 5, 7}, |G|… = 2 · 3 · 5 · 7. In this case, a Hall π-subgroup M is isomorphic to Ω7(2). 14 5 2 (ii) G = PΩ9(q), π ∩ π(G) = {2, 3, 5, 7}, |G|… = 2 · 3 · 5 · 7. In this case, a + Hall π-subgroup M is isomorphic to 2.Ω8 (2) h 2. (iii) e(2, q) = 1 (resp., e(2, q) = 2), π ∩ π(G) ⊆ π(q − 1) ∪ π(n!) (resp., π ∩ π(G) ⊆ π(q +1)∪π(n!)), Symn satisfies E…, and, if r ∈ π(n!)∩π \π(q −1) (resp., if r ∈ π(n!) ∩ π \ π(q + 1)), then (n!)r = | Symn |r = |G|r. In this case, a Hall π-subgroup M is contained in N(G, T ), where T is a maximal 1 n 1 n torus of order 2 (q − 1) (resp., 2 (q + 1) ). Further G satisfies C…. In all cases, G does not satisfy D… and G is good. Proof. In view of Theorem 5.5, we may assume that M is contained in N(G, T ) for some maximal torus T of G. In particular, N(G, T ) contains a Sy- 1 n low 2-subgroup of G. By Lemmas 3.10 and 3.11, we have that |T | = 2 (q ± 1) (the sign depends on e(2, q)) and N(G, T )/T ' 2 o Symn. Thus we obtain statements (i)–(iii) of the lemma. Now we prove that G does not satisfy D…. We proceed case by case. 1 3 (i) If e(3, q) = 1, consider a maximal torus T of order 2 (q−1) ; and if e(3, q) = 2, 1 3 consider a maximal torus T of order 2 (q+1) . In both cases, we have that T contains an Abelian (since T is Abelian) subgroup S = 22 × 33. In view of [13], we have that M does not contain any Abelian subgroup of such structure, hence S 6≤ M. Therefore, G does not satisfy D… and G is good. Using analogous arguments, we see that G does not satisfy D… and G is good in case (ii). (iii) In view of [31, Table 3.5.D], we have that G contains a subgroup R = 2n 2n 2 . Sym2n+1. Let S1 be a Sylow 3-subgroup of R. Consider R1 = 2 h S1 ≤ R. We want to show that R1 is not conjugate to any subgroup of N(G, T ). Indeed, 2n assume that R1 ≤ N(G, T ) (up to conjugation). Clearly, CS1 (2 ) = 1, so R1 ∩ T is a 2-group. Thus |R1T/T |3 = | Sym2n+1 |3. But |N(G, T )/T |3 = |2 o Symn |3 = | Symn |3 < | Sym2n+1 |3, a contradiction. ¤ 6.3. Symplectic groups.

p0 Lemma 6.3. Let G = O (G) be a ﬁnite simple group of Lie type over a ﬁeld GF (q) of characteristic p and G = PSp2n(F), where F is the algebraic closure of GF (q) and n ≥ 2. Assume that π is a set of primes with 2, 3 ∈ π, p 6∈ π. HALL SUBGROUPS OF FINITE GROUPS 25

Then G satisfies E… if and only if either e(2, q) = 1 or e(2, q) = 2, π ∩ π(G) ⊆ π(q − 1) ∪ π(n!) (resp., π ∩ π(G) ⊆ π(q + 1) ∪ π(n!)); Symn satisfies E…; and, if r ∈ π(n!) ∩ π \ π(q −1) (resp., r ∈ π(n!) ∩ π \ (q + 1)), then (n!)r = | Symn |r = |G|r. 1 n Moreover, M is contained in N(G, T ), where T is a maximal torus of order 2 (q−1) 1 n (resp., 2 (q + 1) ), G satisfies C…, does not satisfy D… and G is good. Proof. By Theorem 5.5, we obtain that a Hall subgroup of G is contained in N(G, T ) for some maximal torus T and N(G, T ) contains a Sylow 2-subgroup of G. 1 n In view of Lemma 3.10, we have that |T | = 2 (q ± 1) (the sign depend on e(2, q)) and N(G, T )/T ' 2 o Symn. Thus we only need to prove that G satisfies C… and does not satisfy D…. The first part is clear, since T is unique up to conjugation and Symn satisfies C…. Since G satisfies C…, G is good. Now we prove that G does not satisfy D…. If a Hall π-subgroup of Symn does not coincide with Symn, we easily obtain that G does not satisfy D…. So assume that Symn is a π-group. In view of Lemma 3.1(3), PSp2n(q) satisfies D… if and only m if Sp2n(q) satisfies D…. So we may assume that G = Sp2n(q). If 2n = 2 , then G contains an absolutely irreducible symplectic-type 2-subgroup S = 21+2m such − − that NG(S) = NG(S) = S.Ω2m(2) ([31, Table 3.5.C]). Clearly, Ω2m(2) contains a subgroup isomorphic to Ω2m−1(2). Let R = S.Ω2m−1(2) be a subgroup of NG(S). Evidently, π(R) ⊆ π. Now consider the 2-adic decomposition 2n = 2m1 + ... + 2mk .

Thus there is a subgroup L = Sp2m1 (q) × ... × Sp2mk (q) (the subgroup of block- diagonal matrices) of G, and we may assume that T normalizes L. Let R1 ×...×Rk be a subgroup of L, where Ri ≤ Sp2mi (q) is deﬁned above. Using [31, Table 5.2.A], it is easy to see that the degree of any nontrivial permutation representation of 2mi Ri is not less than 2 − 1. As in the proof of Lemma 6.1(iii), we obtain a contradiction. ¤ 6.4. Orthogonal groups of even dimension.

p0 Lemma 6.4. Let G = O (G) be a finite simple group of Lie type over a field GF (q) of characteristic p, n ≥ 4, and let G = Dn(F) be of adjoint type, where F is the algebraic closure of GF (q). Assume that π is a set of primes with 2, 3 ∈ π, p 6∈ π. Then G satisfies E… if and only if π and G satisfy one of the following conditions. + 12 5 2 (i) G = PΩ8 (q), π ∩ π(G) = {2, 3, 5, 7}, |G|… = 2 · 3 · 5 · 7. In this case, + a Hall π-subgroup M is isomorphic to Ω8 (2). (ii) e(2, q) = 1, π ∩ π(G) ⊆ π(q − 1) ∪ π(n!) (π ∩ π(G) ⊆ π(q − 1) ∪ π((n − 1)!) if G is twisted), Symn (Symn−1, if G is twisted) satisfies E…, and if r ∈ π(n!) ∩ π \ π(q − 1) (r ∈ π((n − 1)!) ∩ π \ π(q − 1) if G is twisted), then (n!)r = |G|r (((n − 1)!)r = |G|r if G is twisted). In this case, a Hall π-subgroup M is contained in a monomial subgroup N(G, T ), where T is a Cartan subgroup of G. + − (iii) Either G = PΩn (q) and n even, or G = PΩn (q) and n odd; e(2, q) = 2, π∩π(G) ⊆ π(q+1)∪π(n!), Symn satisfies E…, and if r ∈ π(n!)∩π\π(q+1), then (n!)r = |G|r. In this case, a Hall π-subgroup M is contained in 1 n N(G, T ), where T is a maximal torus of order – (q+1) and N(G, T )/T ' W (Dn). + − (iv) Either G = PΩn (q) and n odd, or G = PΩn (q) and n even; e(2, q) = 2, π ∩ π(G) ⊆ π(q + 1) ∪ π((n − 1)!), Symn−1 satisfies E…, and if r ∈ 26 D.O.REVIN AND E.P.VDOVIN

π((n − 1)!) ∩ π \ π(q + 1), then ((n − 1)!)r = |G|r. In this case, a Hall π-subgroup M is contained in N(G, T ), where T is a maximal torus of 1 n−1 order – (q − 1)(q + 1) and N(G, T )/T ' 2 o Symn−1. In the last three cases, G satisfies C…; in all the cases, G does not satisfy D… and G is good. Proof. As above, (i)–(iv) follow from Theorem 5.5 and Lemma 3.10. Since N(G, T )/T satisfies C… in cases (ii)–(iv) (see Theorem 4.2) and since T is unique up to conjugation in G (see Lemma 3.10), we easily obtain that G satisfies C… in cases (ii)–(iv). Thus we only need to prove that G does not satisfy D… and G is good. Since in cases (ii)–(iv) G satisfies C…, G is good in these cases. In view of [31, 2n−2 Tables 3.5.E and 3.5.F], G contains a subgroup 2 h Sym2n. As in the proof of Lemma 6.2, we obtain that G does not satisfy D… in these cases. Now in case (i) 3 4 G contains a subgroup 2 × 3 , but M does not. Hence G does not satisfy D…, and G is good in this case. ¤

2 6.5. Groups of type E6 and E6.

p0 Lemma 6.5. Let G = O (G) be a finite simple group of Lie type over a field GF (q) of characteristic p and let G = E6(F) be of adjoint type, where F is the algebraic closure of GF (q). Assume that π is a set of primes such that 2, 3 ∈ π and p 6∈ π. Then G satisfies E… if and only if π and G satisfy one of the following conditions. 2 2 (i) G = E6(q), e(2, q) = 1 (resp., G = E6(q ), e(2, q) = 2); π ∩ π(G) ⊆ π(q − 1) (resp., π ∩ π(G) ⊆ π(q + 1)), 5 ∈ π. In this case, a Hall π- subgroup M is contained in N(G, T ), where T is a maximal torus of order 1 6 1 6 3 (q − 1) (resp., 3 (q + 1) ) and N(G, T )/T ' MT/T ' W (E6). 2 2 (ii) G = E6(q), e(2, q) = 2 (resp., G = E6(q ), e(2, q) = 1); π ∩ π(G) ⊆ π(q +1) (resp., π ∩π(G) ⊆ π(q −1)). In this case, a Hall π-subgroup M is contained in N(G, T ), where T is a maximal torus of order (q−1)2(q+1)4 4 2 (resp., (q − 1) (q + 1) ) and N(G, T )/T ' MT/T ' W (F4).

In all the cases, G satisfies C…, does not satisfy D…, and G is good. Proof. By Theorem 5.5, we have that M is contained in N(G, T ) for some maximal torus T and, moreover, N(G, T ) contains a Sylow 2-subgroup of G. Ap- plying Lemma 3.10 and noting that, in view of [13], W (E6) ' Bb2(3) does not contain a Hall {2, 3}-subgroup, we obtain (i) and (ii). By Lemma 3.10, it follows that T is unique up to conjugation in G. Since M maps onto N(G, T )/T under the natural homomorphism we obtain that N(G, T ) satisfies C…, hence G satisfies C…. It also follows that G is good. Thus we need to prove that G does not satisfy D…. Assume first that (i) holds. Then in view of [12, Theorem 1(II) and Table 1] 3 3 3 there is a subgroup J = 3 of G such that NG(J) ' 3 .3 . SL3(3). This normalizer 3 3 contains a subgroup L = 3 .3 .P , where P is a Sylow 2-subgroup of SL3(3). Thus the maximal normal Abelian subgroup of L has order 33. If L would be contained in N(G, T ), it would imply that L maps onto 33.P under the natural homomorphism N(G, T ) → N(G, T )/T = W (E6). But W (E6) ' PSp4(3).2 clearly does not contain a subgroup of such structure, a contradiction. HALL SUBGROUPS OF FINITE GROUPS 27

6 Assume now that (ii) holds. Let S be a torus of order (q − 1) if G ' E6(q) 6 2 2 and of order (q + 1) if G ' E6(q ). Then N(G, S)/S ' W (E6); hence N(G, S) contains a subgroup K = 26 h Q, where Q is a Sylow 3-subgroup of N(G, S) and 6 6 2 is a Sylow 2-subgroup of S. Note that CQ(2 ) = 1. Indeed, consider the natural ’ homomorphism ϕ : N(G, S) → N(G, S)/S ' W (E6). Then Q ' Q. Moreover, this automorphism shows how Q acts on S: if α ∈ Φ(E6), x ∈ Q, hfi(z) ∈ S, then x (hfi(z)) = hfixϕ (z). Since δ = 1 in our case, we may assume that G is universal.

Hence S = {hp1 (ε1) · hp2 (ε2) · hp3 (ε3) · hp4 (ε4) · hp5 (ε5) · hp6 (ε6)}, where εi = ±1 for all i and p1, p2, p3, p4, p5, p6 form a fundamental system of Φ(E6). Now, since ’ xϕ Q is a subgroup of W (E6), for any x ∈ Q there exists pi with pi 6= pi. Hence x 6 hpi (−1) 6= hpi (−1); therefore, CQ(2 ) = 1. If L is contained in N(G, T ), then 6 we obtain that L ∩ T is a normal Abelian subgroup of L. Since CQ(2 ) is trivial, L ∩ T is a 2-group. For the natural homomorphism ψ : N(G, T ) → N(G, T )/T , we obtain that Qˆ ' Q. Hence Qˆ is contained in a Sylow 3-subgroup of N(G, T )/T . 4 2 But |Q|3 = 3 > |N(G, T )/T |3 = 3 , a contradiction. Therefore, L is not contained in M. ¤

6.6. Groups of type E7.

p0 Lemma 6.6. Let G = O (G) be a finite simple group of Lie type over a field GF (q) of characteristic p and let G = E7(F) be of adjoint type, where F is the algebraic closure of GF (q). Assume that π is a set of primes such that 2, 3 ∈ π and p 6∈ π. Then G satisfies E… if and only if 5, 7 ∈ π and, either e(2, q) = 1 and π∩π(G) ⊆ π(q − 1), or e(2, q) = 2 and π ∩ π(G) ⊆ π(q + 1). A Hall π-subgroup M is contained 1 7 1 7 in N(G, T ), where T is a maximal torus of order 2 (q − 1) (resp., 2 (q + 1) ) and N(G, T )/T ' MT/T ' W (E7) ' 2.Ω7(2). Moreover, G satisfies C…, does not satisfy D…, and G is good.

Proof. In view of [13], W (E7) does not contain a nontrivial Hall subgroup of even order. Now, to obtain necessary and suﬃcient conditions and the C…-property, we proceed like in previous lemmas. Since G satisﬁes C…, G is good. In view of [12, Theorem 1 and Table 1], we have that there exists a subgroup J = 22 such that 2 + 2 NG(J) = 2 × (PΩ8 (q).2 ). Sym3. It follows that NG(J) contains a π-subgroup 2 + L = 2 × (Ω8 (2).2 × 2). But N(G, T ) clearly does not contain such a subgroup. ¤

6.7. Groups of type E8.

Lemma 6.7. Let G = G be a finite simple group of Lie type over a field GF (q) of characteristic p and G = E8(F), where F is the algebraic closure of GF (q). Assume that π is a set of primes such that 2, 3 ∈ π and p 6∈ π. Then G satisfies E… if and only if 5, 7 ∈ π and, either e(2, q) = 1 and π∩π(G) ⊆ π(q − 1), or e(2, q) = 2 and π ∩ π(G) ⊆ π(q + 1). A Hall π-subgroup M is contained in N(G, T ), where T is a maximal torus of order (q − 1)8 (resp., (q + 1)8) and + N(G, T )/T ' MT/T ' W (E8) ' 2.Ω8 (2).2. Moreover, G satisfies C…, does not satisfy D…, and G is good.

Proof. By [13], we obtain that W (E8) does not contain nontrivial Hall subgroups of even order. As in previous lemmas, we obtain necessary and suﬃcient conditions, the C…-property, and the fact that G is good. In view of [12, The- orem 1 and Table 1], we obtain that G contains a subgroup J = 53 such that 28 D.O.REVIN AND E.P.VDOVIN

3 NG(J) = 5 . SL3(5). Let P be a parabolic subgroup of SL3(5) with Levi factor 3 GL2(5). Then we obtain that NG(J) contains a π-subgroup L = 5 .P . It is easy to see that the order of a maximal normal Abelian 5-subgroup of L is at most 53. 3 Assume that L ≤ N(G, T ). Hence |L ∩ T |5 ≤ 5 . Now L/(L ∩ T ) is isomorphic to 3 2 a subgroup of N(G, T )/T ' W (E8). But |L/(L ∩ T )|5 ≥ 5 > |W (E8)|5 = 5 , a contradiction. ¤

6.8. Groups of type F4.

Lemma 6.8. Let G = G be a finite simple group of Lie type over a field GF (q) of characteristic p and G = F4(F), where F is the algebraic closure of GF (q). Assume that π is a set of primes such that 2, 3 ∈ π and p 6∈ π. Then G satisfies E… if and only if either e(2, q) = 1 and π ∩ π(G) ⊆ π(q − 1) or e(2, q) = 2 and π ∩π(G) ⊆ π(q +1). A Hall π-subgroup M is contained in N(G, T ), where T is a maximal torus of order (q − 1)4 (resp., (q + 1)4) and N(G, T )/T ' MT/T ' W (F4). Moreover, G satisfies C…, does not satisfy D…, and G is good.

Proof. Necessary and suﬃcient conditions, the C…-property, and the fact that G is good can be derived as above. In view of [12, Theorem 1 and Table 1], there 3 3 exists a subgroup J = 3 of G such that NG(J) = 3 . SL3(3). Let P be a parabolic 3 subgroup of SL3(3) with Levi factor GL2(3). It is easy to see that L = 3 .P is not contained in N(G, T ). ¤

6.9. Groups of type G2.

Lemma 6.9. Let G = G be a finite simple group of Lie type over a field GF (q) of characteristic p and G = G2(F), where F is the algebraic closure of GF (q). Assume that π is a set of primes such that 2, 3 ∈ π and p 6∈ π. Then G satisfies E… if and only if either e(2, q) = 1 and π ∩ π(G) ⊆ π(q − 1) or e(2, q) = 2 and π ∩π(G) ⊆ π(q +1). A Hall π-subgroup M is contained in N(G, T ), where T is a maximal torus of order (q − 1)2 (resp., (q + 1)2) and N(G, T )/T ' MT/T ' W (G2). Moreover, G satisfies C…, does not satisfy D…, and G is good.

Proof. Necessary and sufficient conditions, the C…-property, and the fact that G is good can be obtained as above. By [12, Theorem 1 and Table 1], there exists 3 3 a subgroup J = 2 of G such that NG(J) = 2 . SL3(2). Let P be a parabolic 3 subgroup of SL3(2) with Levi factor isomorphic to SL2(2). It is clear that 2 .P is not contained in N(G, T ). ¤ 3 6.10. Groups of type D4. 3 3 Lemma 6.10. Let G = D4(q ) = G be a finite simple group of Lie type over a field GF (q) of characteristic p and G = D4(F), where F is the algebraic closure of GF (q). Assume that π is a set of primes such that 2, 3 ∈ π and p 6∈ π. Then G satisfies E… if and only if either e(2, q) = 1 and π ∩π(G) ⊆ π(q −1), or e(2, q) = 2 and π ∩π(G) ⊆ π(q +1). A Hall π-subgroup M is contained in N(G, T ), where T is a maximal torus of order (q − 1)(q3 − 1) (resp., (q + 1)(q3 + 1)) and N(G, T )/T ' MT/T ' W (G2). Moreover, G satisfies C…, does not satisfy D…, and G is good.

Proof. Necessary and suﬃcient conditions, the C…-property, and the fact that G is good can be obtained as above. In view of [30], G contains a subgroup 3 PSL2(q ) × PSL2(q), hence, G contains a π-subgroup K = Alt4 × Alt4. Clearly, M does not contain a subgroup of such structure. ¤ HALL SUBGROUPS OF FINITE GROUPS 29

7. D…-theorems Theorem 7.1. Let G be a finite group of Lie type defined over a field of characteristic p. Let π be a set of primes such that 2 ∈ π and 3, p 6∈ π, τ = π \{2}. If G is an E…-group, then the following statements are equivalent. (1) Every π-subgroup of G contains a normal Abelian Hall τ-subgroup. (2) G is a D…-group. 2 Proof. (2) ⇒ (1). Indeed, if G 6= G2(q) or π∩π(G) 6= {2, 7}, then Lemma 5.1 and Theorem 5.2 imply that every Hall π-subgroup of G contains a normal Abelian Hall τ-subgroup. So every π-subgroup contains a normal Abelian Hall τ-subgroup. 2 Assume that G = G2(q) and π ∩ π(G) = {2, 7}. Then some Hall π-subgroup M of G is isomorphic to a Borel subgroup of A1(8). So, if s is an element of order 7 of M, then NM (h s i) = h s i. But NG(h s i) contains an element of order 2. So G contains a subgroup h t i i h s i, where t is an involution. Clearly, this subgroup is not isomorphic to any subgroup of M. So G does not satisfy D…. (1) ⇒ (2). Lemma 5.1, Theorem 5.2, and (1) imply that G contains an Abelian Hall τ-subgroup K 6= {1} such that K is contained in some maximal torus T of G. Since a Hall τ-subgroup of G is Abelian, [25, Theorem D5] implies that G is a D¿ -group. In particular, every τ-subgroup of G is Abelian and is contained in some Hall τ-subgroup, hence, up to conjugation, we may assume that every τ-subgroup of G is contained in T . Moreover, since G satisfies E…, Lemma 5.1 and Theorem 5.2 imply that T can be chosen so that N(G, T ) contains a Sylow 2-subgroup of G. Lemma 3.10 implies that such T is unique within to conjugation in G. By Lemma 3.1, we may assume G to be simply connected so that G = G. Assume that M is a maximal π-subgroup of G. Since G is a E…-group, then M cannot be a 2-group. In view of (1), M contains a normal Abelian Hall τ-subgroup, say L, and, further, L 6≤ Z(G). Therefore, M is contained in NG(L). In view 0 0 of Lemma 3.6 and rigidity of d-groups, R = NG(L) = CG(L) is a proper σ- stable connected reductive subgroup of maximal rank of G. Consequently, R = T G1 ∗ ... ∗ Gk, where T = T is any maximal torus containing L and G1,...,Gk are subsystem subgroups of G. As we noted above, T can be chosen so that N(G, T ) contains a Sylow 2-subgroup of G. Since L E NG(L) and in view of maximality of M, we have that M contains a Sylow 2-subgroup of (NG(L)) = N(G, L). By Lemma 3.7, N(G, L) = N((NG(L)),T ) · G1 ∗ ... ∗ Gk. Since we chose T so that a Sylow 2-subgroup of G is contained in N(G, T ), arguing as in the proof of Lemma 3.10(3), we may show that a Sylow 2-subgroup of Gi is contained in N(Gi,T ∩Gi) ≤ N((N(G, L),T ) for all i. Therefore, up to conjugation in N(G, L), we may assume that M ≤ N((NG(L)),T ) ≤ N(G, T ). Since N(G, T ) satisfies D… and contains a Hall π-subgroup of G, the maximality of M implies that M is a Hall π-subgroup of G. ¤

Corollary 7.2. Let π be a set of primes with 3 6∈ π. Then every normal subgroup of a D…-group is a D…-group. Proof. It is suﬃcient to prove that every nonabelian simple group S is good if 3 6∈ π ([37, Theorem 1.3]). If 2 6∈ π, we are done by [22, Theorem A]. If S is an alternating or a sporadic group or a simple group of Lie type with characteristic belonging to π, then S is good ([37]). So we can assume that S is a group of Lie type of characteristic p with p 6∈ π and 2 ∈ π. Now S is good by Theorem 7.1. ¤ 30 D.O.REVIN AND E.P.VDOVIN

Lemma 7.3. Let G be a connected simple algebraic group of adjoint type and p0 let O (G) be a finite simple group of Lie type defined over a field of characteristic p. Let π be a set of primes with 3, p 6∈ π, 2 ∈ π, define τ = π \{2}. Assume also that G satisfies E…. Then every π-subgroup of G has a normal Abelian Hall τ-subgroup if and only if every π-subgroup of Gb = G has a normal Abelian Hall τ-subgroup. Proof. Clearly, if every π-subgroup of Gb contains a normal Abelian Hall τ- subgroup, then every π-subgroup of G also contains a normal Abelian Hall τ- subgroup. So assume that every π-subgroup of G contains a normal Abelian Hall π-subgroup. If G is not of type An, then the statement is evident, since π(Gb : G) ∩ τ = ∅. So assume that G is of type An. Then G is isomorphic to either An(q) or 2 2 An(q ). So, |Gb : G| divides (n + 1, q − 1) in the first case and (n + 1, q + 1) in the second case. Assume that π(Gb : G) ∩ τ 6= ∅. Since G satisfies E…, Theorem 5.2 implies that a Hall π-subgroup of G is contained in N(G, T ), where T is a maximal torus of order (q − 1)n in the first case and of order (q + 1)n in the second case. Moreover, by the same theorem, it follows that τ ∩ π(N(G, T )/T ) = ∅. In view of the structure of N(G, T )/T given in the proof of Lemma 3.10, we have that N(G, T )/T ' Symn+1 in both cases. In particular, π(n + 1) ⊆ π(N(G, T )/T ). Therefore, ∅ = τ ∩ π(N(G, T )/T ) ⊇ τ ∩ π(n + 1) ⊇ τ ∩ π(Gb : G). ¤

p0 Theorem 7.4. Let G = O (G) be a finite group of Lie type, G not a Ree group. Let π be a set of primes such that 2 ∈ π, 3, p 6∈ π, define τ = π \{2}. Assume that G is an E…-group. Then the following statements are equivalent: (1) every π-subgroup of G contains a normal Abelian Hall τ-subgroup; (2) Aut G is a D…-group; (3) if L is a group such that G E L and L/G is a D…-group, then L is a D…-group; (4) G is a D…-group. Proof. (2) ⇒ (3). Follows from [35, Corollary to Theorem 1], Theorem 7.1 and Corollary 7.2. (3) ⇒ (4). Evident. (4) ⇒(1). Immediate from Theorem 7.1. (1) ⇒ (2). Assume by contradiction that G0 is a minimal counter example to p0 the statement. In particular, Z(G0) is trivial and G0 = O (G), where G is of adjoint type. Denote G1 = Aut G0. Let M1 be a maximal π-subgroup of G1, which is not a Hall π-subgroup of G1. Let G = Gb0 = G. By Lemma 7.3, G0 satisfies (1) if and only if G satisfies (1). Consider M = M1 ∩ G. There can be three cases. Case (a). M contains a nonidentical τ-subgroup. Case (b). M is a nontrivial 2-group. Case (c). M is trivial. Consider these cases separately. (a) In view of (1), M contains a normal Abelian Hall τ-subgroup T 6= {1}. Con- sider CG(T ). Since G contains an Abelian Hall τ-subgroup, by [25, Theorem D6] G satisfies D¿ . Moreover, by Theorem 5.2, there exists a maximal torus S of G containing a Hall τ-subgroup of G. Thus we may assume T ≤ S. By Lemma 3.6, HALL SUBGROUPS OF FINITE GROUPS 31

CG(T ) is a reductive (but not necessarily connected) σ-stable subgroup of maximal 0 rank of G. Moreover, CG(T ) = C is a characteristic connected σ-stable subgroup p0 of CG(T ). So O (C) = G1 ∗ ... ∗ Gk, where all Gi are subsystem subgroups of G. By Corollary 5.3, Gi satisﬁes to the conditions of the theorem for i = 1, . . . , k. Since G is a minimal counter example, Aut G1/Z(G1),..., Aut Gk/Z(Gk) are D…- groups. Then, by [35, Corollary to Theorem 1], C is a D…-group and every extension of C by a D…-group is a D…-group. Since M1 normalizes C and M1 is a D…-group, M1C is a D…-group. Since M1 is maximal, M1 is a Hall π-subgroup of M1C. Noting that C contains a Hall τ-subgroup of G we have that T is a Hall τ-subgroup of G. Since G is an E…-group, Lemma 5.1 and Theorem 5.2 imply that NG(T ) contains a Hall π-subgroup of G. Clearly, a Hall π-subgroup of every nonabelian composition factor of NG(T ) coincides with its Sylow 2-subgroup and hence is nilpotent. By [25, Theorem D5], [35, Corollary to Theorem 1] and Lemma 3.1, every extension of NG(T ) by a D…-group is again a D…-group. In particular, M1NG(T ) satisﬁes D…. Since M1 is maximal, M1 is a Hall π-subgroup of M1NG(T ). Therefore, M = M1 ∩ G = M1 ∩ NG(T ) is a Hall π-subgroup of G.

Now consider NG1 (M). Since M contains a Sylow 2-subgroup of G and M is soluble, and since G1/G is soluble, we have that NG1 (M) is soluble. Hence NG1 (M) is a D…-group, in particular M1 is a Hall π-subgroup of NG1 (M). By Lemma 3.1,

G1 is a C…-group, thus NG1 (M) contains a Hall π-subgroup of G1. It follows that M1 is a Hall π-subgroup of G1. (b) M is a nontrivial 2-group. Prove ﬁrst the following technical proposition.

Proposition 7.5. Assume that M1, G1, M, and G are chosen as above and M is a Sylow 2-subgroup of MCG(M). Then M is a Sylow 2-subgroup of G.

Proof. Consider NG(M). Clearly, all τ-elements of NG(M) are in C = CG(M). Indeed, assume that x is a τ-element and x ∈ NG(M). Consider h x, M i. This is a π-subgroup of G, so, by (1), it must contain a normal Abelian Hall τ-subgroup. Since h x i is a Hall τ-subgroup of h x, M i, we have that h x i is a normal subgroup of h x, M i. But M is also normal, so h x, M i = h x i ×M. Thus x ∈ CG(M). Now we note that C/(C ∩ M) ' CM/M is 20-group. Hence C is solvable by [15] and every extension of C by a D…-group satisfy D…. In particular, M1C is a D…-group with a Hall π-subgroup M1, and C ∩ M is a Hall π-subgroup of C. So, 0 C and NG(M) are τ -groups. It follows that every nonabelian composition factor of NG(M) contains a nilpotent Hall π-subgroup, which is its Sylow 2-subgroup. By [25, Theorem D5] and [35, Corollary to Theorem 1], M1NG(M) is a D…-group. Since M1 is maximal, M1 is a Hall π-subgroup of M1NG(M). In particular, M is a Sylow 2-subgroup of NG(M). It follows that M is a Sylow 2-subgroup of G. ¤ Now continue the proof of Theorem 7.4. Since M is nontrivial, Z(M) is nontrivial as well. So Z = Ω1(Z(M)) is nontrivial. Consider C = CG(Z). Since Z is a closed σ-stable and M1-invariant subgroup, C is also a closed σ-stable and 0 M1-invariant subgroup. So C is a connected, σ-stable and M1-invariant subgroup 0 0 of G. Since Z(G) is trivial, C is a proper subgroup of G. Let U = Ru(C ) be a unipotent radical of C0. Since U is a characteristic subgroup of C0, U is σ-stable and M1-invariant. Suppose that U is nontrivial. Then, by Lemma 3.8, we may take a proper σ- stable invariant under M1 parabolic subgroup P of G with M ≤ NG(P ). Moreover, 32 D.O.REVIN AND E.P.VDOVIN since NG(P ) = P , we have that M ≤ P . Thus, P = P is a proper parabolic subgroup of G. It follows that P has a Levi decomposition LV , where V = Op(P ), L = S(G1 ∗...∗Gk) is a Levi factor of P . Further all Gi are subsystem subgroups of G, and S is a maximal torus of G. So all composition factors of P are either Abelian or satisfy the conditions of the theorem. Hence, by [35, Corollary to Theorem 1], P and M1P are D…-groups. Since M1 is maximal, it is a Hall π-subgroup of M1P . In particular, it contains a Hall π-subgroup of P . By Lemma 5.1 and Theorem 5.2, we have that π∩π(G) ⊆ π(q−1) or π∩π(G) ⊆ π(q + 1). So, if k ≥ 1, then ∅ 6= τ ∩ π(G1 ∗ ... ∗ Gk) ⊆ τ ∩ π(P ). Therefore, M contains a τ-subgroup T 6= {1}, a contradiction. If G1 ∗ ... ∗ Gk is trivial, we have that P is a Borel subgroup of G; hence S is a Cartan subgroup of G and M is a Sylow 2-subgroup of S. Since CG(M) ≤ CG(Z), we obtain that M is a Sylow 2-subgroup of MCG(M) in this case. Hence, by Proposition 7.5, M is a 2 2n+1 Sylow 2-subgroup of G. Using Lemma 3.10 if G 6= G2(3 ), and using [33] if 2 2n+1 G = G2(3 ), we obtain that a Sylow 2-subgroup of G is not contained in any maximal torus. A contradiction to the fact that M is a Sylow 2-subgroup of G. Assume that U is trivial. Then C is a reductive (but not necessarily connected) subgroup of G. There can be three cases. First case: dim(C) = 0. Since Z is finite, it is a d-group. In view of rigidity of d- groups [26, Corollary to Proposition 16.3], NG(Z) is a finite group. By Lemma 3.2, NG(Z) is contained in the normalizer of a Jordan subgroup J of G. Since J is a minimal normal subgroup of NG(J)[3, Condition b], then NG(Z) ≤ NG(J) implies Z ≥ J. In view of Table 2, J is a maximal elementary Abelian 2-group of NG(J), hence Z = J. If |NG(Z)|¿ > 1, then there exists a τ-element x in NG(Z). Since CG(Z) = Z, we have x 6∈ CG(Z). It follows that h x, Z i is a π-subgroup and h x, Z i does not contain a normal Hall τ-subgroup. A contradiction to (1). Therefore, a Hall π-subgroup of NG(Z) coincides with its Sylow 2-subgroup; hence a Hall π- subgroup of NG(Z) is nilpotent. It follows that M1NG(Z) is a D…-group. In view of maximality of M1, M1 is a Hall π-subgroup of M1NG(Z), hence M is a Sylow 2-subgroup of NG(Z). But a Sylow 2-subgroup of NG(Z) does not centralize Z, a contradiction to the assumption Z ≤ Z(M). Second case. Z(C0) = 1, dim(C) > 0, and rank(C) < rank(G). Then C0 is a σ-stable and M1-invariant connected semisimple subgroup of G, so all composition 0 factors of (C ) are either simple groups of Lie type over a field of characteristic p, or elementary Abelian. By induction and [35, Corollary to Theorem 2], we obtain 0 0 that M1(C ) is a D…-group. If τ ∩ π(C ) 6= ∅, we obtain that M contains a nonidentical τ-subgroup, so we are done by (a). Otherwise a Hall π-subgroup of 0 M(C ) coincides with its Sylow 2-subgroup. Maximality of M1 implies that M is 0 0 0 0 a Sylow 2-subgroup of M(C ). Since (C ) is normal in M(C ) and M ∩ (C ) 0 is nontrivial, then M ∩ (C ) is a nontrivial normal subgroup of M. Hence Z(M) ∩ 0 0 0 0 (C ) is nontrivial, so Z ∩ (C ) is nontrivial. But Z ∩ (C ) ≤ Z(C ) = 1, a contradiction. Third case. Either Z(C0) 6= 1 or C is of maximal rank. In view of Lemma 3.9, 0 we may assume that C is of maximal rank in both cases. Then (C ) = T G1 ∗ 0 ... ∗ Gk, where T is some σ-stable maximal torus of C and all Gi are subsystem subgroups of G. By induction and [35, Corollary to Theorem 2], we have that 0 0 0 M1(C ) is a D…-group. So M ∩(C ) is a Hall π-subgroup of (C ). If G1 ∗...∗Gk 0 is nontrivial, then τ ∩ π((C )) is nontrivial; so M contains a τ-subgroup S 6= {1}, HALL SUBGROUPS OF FINITE GROUPS 33 and we are done by (a). Hence we may assume that C0 = T is a maximal torus and |T |¿ = 1. Then M ≤ N(G, T ) and M1 normalizes N(G, T ). Moreover, by Theorem 5.2, it follows that |N(G, T )|¿ = 1. Therefore, a Hall π-subgroup of N(G, T ) coincides with its Sylow 2-subgroup, hence is nilpotent. By [25, Theorem D5], M1N(G, T ) satisfies D…. Therefore, in view of the maximality of M1, M1 is a Hall π-subgroup of M1N(G, T ). In particular, M is a Sylow 2-subgroup of N(G, T ); hence M is a Sylow 2-subgroup of N(G, T ) ≥ MNG(Z) ≥ MCG(Z) ≥ MCG(M). By Proposition 7.5, N(G, T ) contains a Sylow 2-subgroup of G. Since G is an E…-group, Lemmas 3.10, 3.12 and 5.1, and Theorem 5.2 imply that T contains a Hall τ-subgroup L 6= {1}.A contradiction to the fact that |T |¿ = 1. (c). M1 ∩ G is trivial. Every element of G1 has a unique decomposition into the product of inner-diagonal, field, and graph automorphisms. Assume first that M1 is contained in the subgroup of G1 generated by inner-diagonal and field automorphisms. Denote this subgroup by G2. It is known that G2/G is cyclic. Since M1 ∩ G = 1, M1 is also cyclic. Thus M1 = h h i. By [18, (7-2)], we may assume p0 that h is a field automorphism of G. Then K = O (Gh) is a finite group of Lie type. Clearly, K ≤ CG(M1) and a Sylow 2-subgroup of K is nontrivial. Since M1 is maximal, M1 contains a Sylow 2-subgroup of K, hence a nontrivial 2-subgroup of G, a contradiction to the assumption that M1 ∩ G is trivial. Assume now that M1 6≤ G2. Then M1 ∩ G2 is a normal cyclic subgroup of M1. If it is nontrivial, as above we obtain that M1 ∩ G is nontrivial, a contradiction to our assumption. If it is trivial, then M1 is a 2-group, hence it is contained in a Sylow 2-subgroup of G1. Since M1 is maximal, it is a Sylow 2-subgroup of G1 and so M1 ∩ G is nontrivial. This final contradiction completes the proof. ¤ Corollary 7.6. Let G be a finite group, H E G, and π a set of primes with 3 6∈ π. If H and G/H satisfy D…, then G also does. Proof. Let 3 6∈ π. In view of [35, Corollary to Theorem 1], it is enough to show that if a finite nonabelian simple group S satisfies D…, then Aut S does. If 2 6∈ π, the result follows from [43]. So assume that 2 ∈ π. If S is alternating, sporadic or of Lie type and p ∈ π, the result follows from [37]. So, by the classification of finite simple groups, G is of Lie type with p 6∈ π. In this case, the corollary follows from Theorem 7.4. ¤ Now we give affirmative answers to Problems 3.62 and 13.33 posed by L.A. Shemetkov and V.D. Mazurov in “The Kourovka Notebook” [45].

Theorem 7.7. Let G be a finite group, AEG. Then G satisfies D… if and only if G/A and A satisfy D…. Proof. If 2 6∈ π or 3 6∈ π, then the result follows from [35], [43, Theorem 2], and Corollaries 7.2 and 7.6. Thus we may assume that 2, 3 ∈ π. In view of [37, Theorem 1.3], for the part “if,” it is sufficient to prove that every nonabelian simple group S is good. If S is either an alternating group, a sporadic group, or a simple group of Lie type with characteristic in π, then S is good ([37]). So we may assume that S is a group of Lie type over a field of characteristic p with p 6∈ π and 2, 3 ∈ π. By Lemmas 6.1–6.9, S is good. In view of [35, Theorem 1], for the part “only if,” it is sufficient to prove that if a finite simple group S satisfies D…, then any group G with S ≤ G ≤ Aut(S) 34 D.O.REVIN AND E.P.VDOVIN satisfies D…. If G is alternating, the result follows from [25, Theorem A4], [41], and Corollary 4.4. If G is sporadic or of Lie type and p ∈ π, the result follows from [37]. If G is of Lie type, 2, 3 ∈ π, and p 6∈ π, then, by Lemmas 6.1–6.9, we have that S does not satisfy D… and the theorem follows. ¤ Note that we can state this theorem without using the classification of finite simple groups. A finite group is called a K-group if all its composition factors are known simple groups.

Theorem 7.8. Let G be a K-group and assume that G satisfies D…. Then every normal subgroup of G and every extension of G by a D…-group satisfy D…. References [1] M. Aschbacher, On finite groups of Lie type and odd characteristic, J. Algebra 66 (1980), no. 2, 400-424. [2] A. Borel and J. de Siebental, Les-sous-groupes fermésde rang maximum des groupes de Lie clos, Comment.Math.Helv. 23 (1949), 200-221. [3] A.V. Borovick, The structure of Jordan subgroups of simple algebraic groups, Algebra and Logic 28 (1989), no. 2, 97-108. [4] A.V. Borovick, The structure of finite subgroups of simple algebraic groups, Algebra and Logic 28 (1989), no. 3, 163-183. [5] W. Burnside, Theory of Groups of Finite Order, 2nd ed., Dover Publications, NY, 1955. [6] R.W. Carter, Conjugacy classes in the Weyl group, Compositio Mathematica 25 (1972), no. 1, 1-59. [7] R.W. Carter, Simple Groups of Lie Type, John Wiley and Sons, London, 1972. [8] R.W. Carter, Centralizers of semisimple elements in finite groups of Lie type, Proc. London Math. Soc. (3) 37 (1978), no. 3, 491-507. [9] R.W. Carter, Centralizers of semisimple elements in the finite classical groups, Proc. London Math. Soc. (3) 42 (1981), no. 1, 1-41. [10] R.W. Carter, Finite groups of Lie type, conjugacy classes and complex characters, John Wiley and Sons, 1985. [11] R.W. Carter and P. Fong, The Sylow 2−subgroups of the finite classical groups, J. Algebra 1 (1964), 139-151. [12] A.M. Cohen, M.W. Liebeck, J. Saxl, and G.M. Seitz, The local maximal subgroups of exceptional groups of Lie type, finite and algebraic, Proc. London Math. Soc. (3) 64 (1992), no. 1, 21-48. [13] J.H. Conway, R.T. Curtis, S.P. Norton, R.A. Parker, and R.A. Wilson, Atlas of Finite Groups, Clarendon Press, Oxford, 1985. [14] D. Deriziotis, Conjugacy classes and centralizers of semisimple elements in finite groups of Lie type, Vorlesungen aus dem Fachbereich Mathematik der UniversitätEssen, 1984, Heft 11. [15] W. Feit and J. Thompson, Solvability of groups of odd order, Pacif. J. Math. 13 (1963) no. 3, 775-1029. [16] P.A. Ferguson, P. Kelley, and P. Hall, Hall π-subgroups which are direct product of nonabelian simple groups, J. Algebra 120 (1989), no. 1, 40-46. [17] D. Gorenstein, Finite Groups, Harper and Row, New York, 1968. [18] D. Gorenstein and R. Lyons, The Local Structure of Finite Groups of Characteristic 2 Type, Mem. AMS, 42 (1983), no. 276. [19] D. Gorenstein R. Lyons, and R. Solomon, The Classification of the Finite Simple Groups, Number 6. Part IV, Mathematical Surveys and Monographs, American Mathematical Society, Providence, RI, 2005. [20] F. Gross, On a conjecture of Philip Hall, Proc. London Math. Soc. (3) 52 (1986), no. 3, 464-494. [21] F. Gross, On the existence of Hall subgroups, J. Algebra 98 (1986), no. 1, 1-13. [22] F. Gross, Conjugacy of odd order Hall subgroups, Bull. London Math. Soc. 19 (1987), no. 4, 311-319. [23] F. Gross, Hall subgroups of order not divisible by 3, Rocky Mountain J. Math. 23 (1993), no. 2, 569-591. HALL SUBGROUPS OF FINITE GROUPS 35

[24] F. Gross, Odd order Hall subgroups of the classical linear groups, Math. Z. 220 (1995), no. 3, 317-336. [25] P. Hall, Theorems like Sylow’s, Proc. London Math. Soc. 6 (1956), 286-304. [26] J.E. Humphreys, Linear Algebraic Groups, Springer-Verlag, New York, 1972. [27] J.E. Humphreys, Conjugacy Classes in Semisimple Algebraic Groups, American Mathemat- ical Society, Providence, Rhode Island, Mathematical Survey and Monographs 43, 1995. [28] B. Huppert, Endliche Gruppen I, Springer, Berlin, 1967. + [29] P.B. Kleidman, The maximal subgroups of the finite 8-dimensional orthogonal groups P Ω8 (q) and of their automorphism groups, J. Algebra 110 (1987), no. 1, 173-242. 3 3 [30] P.B. Kleidman, The maximal subgroups of the Steinberg triality groups D4(q ) and of their automorphism groups, J. Algebra 115 (1988), no. 1, 182-199. [31] P.B. Kleidman and M. Liebeck, The Subgroups Structure of Finite Classical Groups, Cam- bridge Univ. Press, 1990. [32] A.S. Kondratiev and V.D. Mazurov, On 2-indicators of finite simple groups, Algebra and Logic 42 (2003), no. 6, 333-348. [33] V.M. Levchuk and Ja.N. Nuzhin, Structure of Ree groups, Algebra and Logic 24 (1985), no. 1, 16-26. [34] M.W. Liebeck and J. Saxl, The primitive permutation groups of odd degree, J. London Math. Soc. (2) 31 (1985), no. 2, 250-264. [35] V.D. Mazurov and D.O. Revin, On Hall Dπ-property for finite groups, Sib. Math. J. 38 (1997), no. 1, 106-113. [36] D.O. Revin, Hall π-subgroups of finite Chevalley groups with characteristic in π, Siberian Adv. Math. 9 (1999), no. 2, 25-71. [37] D.O. Revin, A Dπ-property in a class of finite groups, Algebra and Logic 41 (2002), no. 3, 187-206. [38] J.-A. de Seguir, Groupes de Substitutions, Paris, Gautier-Villars, 1912. [39] R. Steinberg, Lectures on Chevalley Groups, Yale University, 1967. [40] M. Suzuki, On a class of doubly transitive groups, Ann. Math. 75 (1962), no. 1 105-145. [41] J. Thompson, Hall subgroups of symmetric groups, J. Combin. Theory, Ser. A 1 (1966), 271-279. [42] N.A. Vavilov, Weight elements of Chevalley groups, preprint. [43] E.P. Vdovin and D.O. Revin, Hall subgroups of odd order of finite groups, Algebra and Logic 41 (2002), no. 1, 8-29. [44] V.A. Vedernikov, Subdirect products of finite groups with Hall π-subgroups, Math. Notes 59 (1996), no. 2, 311-313. [45] The Kourovka Notebook. Unsolved Problems in Group Theory, Novosibirsk, 1999. [46] The GAP Group, GAP — Groups, Algorithms, and Programming, Version 4.2, 2000. http://www.gap-system.org

Institute of Mathematics, pr-t Acad. Koptiug, 4, Novosibirsk, 630090, Russia E-mail address: [email protected]