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Siberian Mathematical Journal, Vol. 58, No. 3, pp. 467–475, 2017 c 2017 Kondrat ev A.S., Maslova N.V., and Revin D.O. Original Russian Text Copyright 

ON THE PRONORMALITY OF SUBGROUPS OF ODD INDEX IN FINITE SIMPLE SYMPLECTIC GROUPS c A. S. Kondrat ev, N. V. Maslova, and D. O. Revin 

UDC 512.542

Abstract: A subgroup H of a group G is pronormal if the subgroups H and H g are conjugate in H, H g  for every g ∈ G. It was conjectured in [1] that a subgroup of a finite simple group having odd index is always pronormal. Recently the authors [2] verified this conjecture for all finite simple groups other than P SLn (q), P SUn (q), E6 (q), 2 E6 (q), where in all cases q is odd and n is not a power of 2, and P Sp2n (q), where q ≡ ±3 (mod 8). However in [3] the authors proved that when q ≡ ±3 (mod 8) and n ≡ 0 (mod 3), the simple symplectic group P Sp2n (q) has a nonpronormal subgroup of odd index, thereby refuted the conjecture on pronormality of subgroups of odd index in finite simple groups. The natural extension of this conjecture is the problem of classifying finite nonabelian simple groups in which every subgroup of odd index is pronormal. In this paper we continue to study this problem for the simple symplectic groups P Sp2n (q) with q ≡ ±3 (mod 8) (if the last condition is not satisfied, then subgroups of odd index are pronormal). We prove that whenever n is not of the form 2m or 2m (22k + 1), this group has a nonpronormal subgroup of odd index. If n = 2m , then we show that all subgroups of P Sp2n (q) of odd index are pronormal. The question of pronormality of subgroups of odd index in P Sp2n (q) is still open when n = 2m (22k + 1) and q ≡ ±3 (mod 8). DOI: 10.1134/S0037446617030107 Keywords: finite group, simple group, symplectic group, pronormal subgroup, odd index

Introduction By Hall’s definition, a subgroup H of a group G is pronormal if the subgroups H and H g are conjugate in H, H g  for every g ∈ G. Throughout the paper we consider only finite groups and thereby “group” means “finite group.” A subgroup H of a finite group G is a Hall subgroup if its order |H| and its index |G : H| are coprime. In [1], the Hall subgroups of the finite simple groups were proved to be pronormal and further examination of the proof motivated the following: Conjecture 1 [1, Conjecture 1]. All subgroups of odd index are pronormal in finite simple groups. This conjecture was verified by the authors [2, Theorem] for all finite simple groups other than P SLn (q), P SUn (q), E6 (q), 2 E6 (q), where in all cases q is odd and n is not a power of 2, and P Sp2n (q), where q ≡ ±3 (mod 8). But exploiting a criterion for pronormality of supplements to abelian normal subgroups of finite groups which was obtained in [3, Theorem 1], the authors constructed some examples of nonpronormal subgroups of odd index in the groups P Sp6n (q) with q ≡ ±3 (mod 8) (see [3, Theorem 2]). Conjecture 1 was thus refuted. These results lead to the natural extension of Conjecture 1: Problem 1. Classify nonabelian simple groups whose every subgroup of odd index is pronormal. The second author was supported by the President of the Russian Federation (Grant MK–6118.2016.1), the State Maintenance Program for the Leading Universities of the Russian Federation (Agreement 02.A03.21.0006 of 27.08.2013), and the Dynasty Foundation. The third author was supported by the CAS President’s International Fellowship Initiative (Grant 2016VMA078). Ekaterinburg; Novosibirsk; Hefei. Translated from Sibirski˘ı Matematicheski˘ı Zhurnal, Vol. 58, No. 3, pp. 599–610, May–June, 2017; DOI: 10.17377/smzh.2017.58.310. Original article submitted October 17, 2016. 0037-4466/17/5803–0467

467

In this paper we continue to study Problem 1 for symplectic groups. In fact, the result of [3, Theorem 2] enables us to investigate Problem 1 within a much wider family of symplectic groups than that of [3]. We prove the following Theorem 1. Let G = P Sp2n (q), where q ≡ ±3 (mod 8) and n is not of the form 2w or 2w (22k + 1). Then G has a nonpronormal subgroup of odd index. Also, we completely study one of the remaining two cases of symplectic groups that are not covered by Theorem 1; namely, the case of P Sp2n (q) with q ≡ ±3 (mod 8) and n = 2w . Our main result is as follows: Theorem 2. All subgroups of odd index are pronormal in P Sp2n (q), where n = 2w ≥ 2 and q ≡ ±3 (mod 8). Note a consequence of Theorem 2 and the main result of [2]: Corollary. If the dimension or the underlying field order of a simple classical group G is a power of 2, then the subgroups of G of odd index are pronormal. Remark. Problem 1 is still open for P Sp2n (q), where q ≡ ±3 (mod 8) and n = 2m (22k + 1), and also for E6 (q), 2 E6 (q), P SLn (q), and P SUn (q), where in all cases q is odd and n is not a power of 2. Auxiliary Results Our terminology and notation are mostly standard and can be found in [4, 5]. We write H prn G to mean that H is a pronormal subgroup of a group G. For a group G and a subset π of the set of all primes, Oπ (G) and Z(G) denote the π-radical (the largest normal π-subgroup) and the center of G. The set of Sylow p-subgroups of G is denoted by Sylp (G). As usual, π  stands for the set of those primes that do not lie in π. If n is a positive integer, then nπ is the largest divisor of n whose all prime divisors lie in π. Also, for a finite group G, it is common to write O(G) instead of O2 (G). Lemma 1 [1, Lemma 5]. Let G be a finite group and let H ≤ G. Suppose that H contains a Sylow subgroup S of G. Then the following are equivalent: (1) H prn G; (2) H and H g are conjugate in H, H g  for every g ∈ NG (S). To prove that a subgroup H of odd index is pronormal in G, therefore, it suffices to establish that H and H g are conjugate in H, H g  for every nontrivial element g ∈ NG (S) of odd order, where S is a fixed Sylow 2-subgroup of H (and hence of the whole G). Lemma 2. Let p be a prime and let Xp be the class of all finite groups whose Sylow p-subgroups are equal to their normalizers. Then (1) if X  Y , X ∈ Xp and Y /X ∈ Xp , then Y ∈ Xp ; (2) if X ≤ Y ∈ Xp and |Y : X| is coprime to p, then X prn Y . Proof. For (1), suppose that X  Y , Y /X ∈ Xp and X ∈ Xp , and choose S ∈ Sylp (Y ). Then NY (SX) = SX and NX (S ∩ X) = S ∩ X since SX/X ∈ Sylp (Y /X) and S ∩ X ∈ Sylp (X). We argue that NY (S) = S. Indeed, NY (S) ≤ NY (SX) = SX. Hence, NY (S) = NSX (S) = SNX (S),

NX (S) ≤ NX (S ∩ X) = S ∩ X.

Assertion (2) follows from Lemma 1 together with the observation that a subgroup of Y ∈ Xp whose index is coprime to p contains some Sylow p-subgroup S of Y and thus its normalizer in Y .  Lemma 3 [6, Corollary]. Let G ∼ = P Sp2n (q), where q is a power of an odd prime, and let S ∈ Syl2 (G). If q ≡ ±1 (mod 8), then NG (S) = S. If q ≡ ±3 (mod 8), then the factor group NG (S)/S is an elementary abelian 3-group of order 3t , where the number t is defined by the binary expansion n = 2s1 + · · · + 2st , 468

s1 > · · · > st ≥ 0.

Lemma 4 [7, Lemma 1]. Suppose that a Frobenius group F C with kernel F and cyclic complement C = g of order n acts faithfully on a vector space V over a field of nonzero characteristic coprime to |F |. Then the minimal polynomial of g, as a linear transformation of V , is equal to λn − 1. Lemma 5. Let H be a subgroup of G of odd order, where SL2 (q) ≤ G ≤ GL2 (q)

or P SL2 (q) ≤ G ≤ P GL2 (q).

Suppose that (|H|, q) = 1 and g ∈ NG (H) is an element of odd order. Then (1) H is abelian; (2) g ∈ CG (H). Proof. In the case G = P SL2 (q), the assertion follows easily from the list of maximal subgroups of G [8, Theorem II.8.27]. However, we give another proof which covers all cases and relies on representation theory and linear algebra. Let P SL2 (q) ≤ G ≤ P GL2 (q). Since H and H, g are of odd order, these subgroups lie in P SL2 (q). Moreover, these subgroups have isomorphic preimages in SL2 (q) ≤ GL2 (q). Thus it suffices to prove the lemma for G = GL2 (q). We begin with (1). By Maschke’s Theorem [9, Theorem (1.9)], the subgroup H is completely reducible. Hence if H is reducible, then it is conjugate to a subgroup of the group of diagonal matrices, and (1) follows. We may assume therefore that H is irreducible. Furthermore, replacing the field Fq by the splitting field Fqn of H if necessary and exploiting the embedding GL2 (q) → GL2 (q n ), we may assume without loss of generality that the identity embedding of H into G is a faithful absolutely irreducible representation of H. By [9, Theorem (15.13)], the degree of this representation, which is 2, is identical with the degree of some irreducible complex representation. Hence, by [9, Theorem (3.11)] it divides the odd number |H|, which is a contradiction. For (2), denote the characteristic of Fq by p. The element g can be written as g = xy, where x, y ∈ g, x is a p-element, and y is a p -element. Replacing H by the subgroup H, y (whose order divides |H||y| and which is abelian by (1)) and g by x, we may assume that g is a p-element. Furthermore, since a Sylow p-subgroup of G is isomorphic to the additive group of Fq , we conclude that |g| = p. The subgroup H is abelian, so [10, Theorem 4.34] yields H = F × K, where F = [H, g] and K = CH (g). Also we have CF (g) = F ∩ K = 1. Thus, if F is not trivial, then F, g = F g is a Frobenius group with kernel F and cyclic complement g of order p. Applying Lemma 7, we have a contradiction since the minimal polynomial λp − 1 of g must divide its characteristic polynomial det(g − λI) of degree 2 and p = |g| is odd.  Lemma 6 [2, Lemma 3]. Suppose that H is a subgroup and N is a normal subgroup of a group G, and let : G → G/N be the natural epimorphism. In this event (1) If H prn G, then H prn G. (2) If N ≤ H and H prn G, then H prn G. In particular, a subgroup H of odd index is pronormal in G if and only if H/O2 (G) is pronormal in G/O2 (G). Lemma 7 [2, Lemma 5]. Let H and M be subgroups of G, and suppose that H ≤ M . In this event (1) If H prn G, then H prn M. (2) If S ≤ H for some Sylow subgroup S of G, NG (S) ≤ M and H prn M , then H prn G. Lemma 8 [3, Lemma 3]. Let H be a transitive permutation group of degree n and let A be a group. Define the action of H on V = An by putting (x1 , . . . , xn )π = (x1π−1 , . . . , xnπ−1 ) for all (x1 , . . . , xn ) ∈ V and π ∈ H. Then

CV (H) = {(x1 , . . . , xn ) ∈ V | x1 = · · · = xn } ∼ = A. 469

Lemma 9 [3, Theorem 1]. Let H and V be subgroups of a group G such that V is an abelian normal subgroup of G and G = HV . Then the following are equivalent: (1) H prn G; (2) U = NU (H)[H, U ] for every H-invariant subgroup U ≤ V . Lemma 10 [3, Theorem 2]. PSp6n (q), where q ≡ ±3 (mod 8), has a nonpronormal subgroup of odd index. Lemma 11 [3, Lemma 8]. Let H be a permutation group of degree n and let A be a group. Define the action of H on V = An by putting (x1 , . . . , xn )π = (x1π−1 , . . . , xnπ−1 ) for all (x1 , . . . , xn ) ∈ V and π ∈ H. Suppose that H has a transitive subgroup K such that (|A|, |K|) = 1. Then U = CU (H)[H, U ] for every H-invariant subgroup U of V . Lemma 12. Suppose that G satisfies one of the following: (a) G = Sn is the symmetric group of degree n; (b) G is a dihedral group; (c) G = D Sn is a wreath product of a dihedral and a symmetric group. Then (1) each Sylow 2-subgroup of G is equal to its normalizer; (2) each subgroup of odd index is pronormal in G. Proof. It is suffices to prove (1) in cases (a) and (b), since then both (1) in case (c) and (2) will follow by Lemma 2. The case (a) is handled in [11, Lemma 4]. For (b), let G be a dihedral group, S ∈ Syl2 (G) and U = O(G). Then G = SU and hence NG (S) = SNU (S) = SCU (S). We argue that CU (S) = 1 and thus (1) holds. By the definition of a dihedral group, U is cyclic and S contains an involution t such that ut = u−1 for all u ∈ U . Assume that 1 = u ∈ CU (S). Then u is a nontrivial element of odd order and u = u−1 = ut = u, which is a contradiction.  Lemma 13. Let G = A B = V B be the natural permutation wreath product of a cyclic group A of order 3 and the symmetric group B = Sn , with n = 2w . Let V = V1 × · · · × Vn , where Vi ∼ = A, be the base subgroup of G and let H be a subgroup of G of odd index. Then H prn G. Proof. Let S ∈ Syl2 (G) and S ≤ H. By the Sylow Theorem, we may assume that S ≤ B. It follows from Lemma 12(1) that NG (SV ) = SV ≤ HV , and hence NG (S) ≤ NG (SV ) ≤ HV. By Lemma 7, it suffices to show that H prn HV . By Lemma 9, this is equivalent to the assertion that U = NU (H)[H, U ] for every H-invariant subgroup U ≤ V . Let U be an H-invariant subgroup of V . It is clear that NU (H)[H, U ] ⊆ U . Observe that S ≤ B is a transitive permutation group and (|S|, |A|) = 1. By Lemmas 11 and 8, we have U = CU (S)[S, U ] ≤ NU (H)[H, U ]. It follows from Lemma 9 that H prn HV .  Lemma 14. Let G = A B be the natural permutation wreath product of a group A ∼ = A4 and the symmetric group B = Sn , where n = 2w , and let H be a subgroup of G of odd index. Then H prn G. Proof. Observe that G/O2 (G) ∼ = G1 , where G1 is the natural permutation wreath product of a cyclic group of order 3 and the symmetric group Sn . Applying Lemmas 6 and 13 completes the proof.  Lemma 15. Let Q be a subgroup of odd index of a group L = L1 × L2 × · · · × Ln , where Li are finite groups, and let πi be the projection map from L to Li . Suppose that for some i, we have Qπi = Li and Li is a nonabelian simple group. Then Li ≤ Q. Proof. Since Li  L, it follows that Q ∩ Li  Q, and so (Q ∩ Li )πi is a normal subgroup of the group Qπi = Li , which is simple. Thus either (Q ∩ Li )πi = 1 or (Q ∩ Li )πi = Li . In the latter case we have Li ≤ Q. 470

It remains to show that (Q∩Li )πi = 1. Choose S ∈ Syl2 (L) such that S ≤ Q. Then S ∩Li ∈ Syl2 (Li ), and so 1 = S ∩ Li = (S ∩ Li )πi ≤ (Q ∩ Li )πi .  Lemma 16. Let G = A B be the natural permutation wreath product of the group A = A5 and a symmetric group B = Sn , where n = 2w , and let H be a subgroup of odd index in G. Then H prn G. ∼ A. Choose S ∈ Syl (G) Proof. Denote the base subgroup of G by L = L1 ×L2 ×· · ·×Ln , with Li = 2 such that S ≤ H. Observe that G = X/O2 (X), where X is a subgroup of G1 = Sp2n (5) isomorphic to SL2 (5) Sn . By [12, Theorem 9] the index |G1 : X| is odd. By Lemma 3, |NG1 (S1 ) : S1 | = 3, where S1 ∈ Syl2 (G1 ). Hence |NX (S1 ) : S1 | = |G : NG (S)| divides 3. Observe that the element (g, g, . . . , g) ∈ L, where g is an element of A of order 3 normalizing its Sylow 2-subgroup, belongs to NG (S). Thus |NG (S) : S| = 3. Let g be an element of NG (S) of order 3. The above reasoning shows that g ∈ L. We claim that the subgroups H and H g are conjugate in K = H, H g . Let K0 = K ∩ L and let πi be the projection map from L to Li . Since K acts on Li transitively, all subgroups K0πi are pairwise isomorphic. Furthermore, K0πi is a subgroup of odd index in Li , and so only the three possibilities are open: (1) K0πi ∼ = A5 for every i; πi ∼ (2) K0 = A4 for every i; (3) K0πi ∼ = C2 × C2 for every i. In case (1), Lemma 15 implies that L lies K, and so g ∈ K. in n In case (2), K normalizes the subgroup i=1 K0πi and NG (S) ≤ NG

 n i=1

K0πi

 .

 n  K0πi . This is true since Lemma 14 guarantees that By Lemma 7, it suffices to show that H prn NG i=1 n πi ∼ subgroups of odd index are pronormal in NG = A4 Sn . i=1 K0 In case (3), we have K ∩ L = H ∩ L = S ∩ L ∈ Syl2 (L) and L ∩ S  H. Hence H ≤ NG (L ∩ S) and g ∈ NG (L∩S) by the choice of g. Note that subgroups of odd index are pronormal in NG (L∩S)/(L∩S) ≤ B∼ = Sn . Therefore, H and H g are conjugate in H, H g  by Lemma 6. Thus H prn G.  Lemma 17. Let G = A Sn = LH be the natural permutation wreath product of a nonabelian simple group A and a symmetric group H = Sn , where n = 2w , and let L = L1 × · · · × Ln , where Li ∼ = A, be the base subgroup of G. Suppose that K is a subgroup of odd index in G and K0 = K ∩ L. Then for every subgroup M1 of L1 containing the normalizer in L1 of the projection of K0 to L1 , the subgroup K lies in some subgroup of G isomorphic to M1 Sn . Proof. The group H = Sn acts on the set Ω = {1, . . . , n} in a natural way. Denote the stabilizer in H of i ∈ Ω by Xi . Let S be a Sylow 2-subgroup of G contained in K and let T be a regular subgroup of H = Sn . Then |T | = n = 2w and by the Sylow Theorem there is an element g ∈ G such that S g contains T . Hence we may assume that T ≤ K, and then H = T Xi = Xi T for every i. Observe that T acts transitively on {Li | 1 ≤ i ≤ n} by conjugation. Let Ki be the projection of K0 to Li . Then T also acts transitively on {Ki | 1 ≤ i ≤ n} (which yields Ki ∼ = Kj for all i and j), and so  (1) {Ki | 1 ≤ i ≤ n} = K1t | t ∈ T  n   and K0 ≤ ni=1 Ki . Furthermore, we claim that K ≤ NG i=1 Ki .    n  n  n Note that NL i=1 Ki = i=1 NLi (Ki ). We argue that H ≤ NG i=1 Ki . Since Xi centralizes Li , it centralizes Ki as well. Also, T acts regularly on {Ki | 1 ≤ i ≤ n} by (1). For every h ∈ H and every i ∈ {1, . . . , n}, there are x ∈ Xi and t ∈ T such that h = xt. Let Kit = Kj . Then Kih = Kixt = Kit = Kj . 471

 n  Thus we show that H ≤ NG i=1 Ki . Hence     n n n  NG Ki = HNL Ki = H NLi (Ki ) ∼ = NL1 (K1 ) H. i=1

i=1

i=1

Furthermore, if NL1 (K1 ) ≤ M1 ≤ L1 , then K≤H

n  i=1

NLi (Ki ) = H

 t∈T

(NL1 (K1 ))t ≤ H



(M1 )t ∼ = M1 H.



t∈T

Lemma 18. Let G = A B be the natural permutation wreath product of a group A = P SL2 (q) and a symmetric group B = Sn , where q ≡ ±3 (mod 8) and n = 2w , and let H be a subgroup of odd index in G. Then H prn G. Proof. Suppose that the claim is false, and let q be the smallest prime power congruent ±3 modulo 8 such that G has a nonpronormal subgroup H of odd index. Choose a subgroup S ∈ Syl2 (G) so that S ≤ H. Denote the base subgroup of G by L = L1 × L2 × · · · × Ln , where Li ∼ = A. Observe that G = X/O2 (X), where X is a subgroup of G1 = Sp2n (q) isomorphic to SL2 (q) Sn . By [12, Theorem 9] the index |G1 : X| is odd. As in the proof of Lemma 16, we establish that |NG (S) : S| = 3. Let g be an element of NG (S) of order 3. By the above, g ∈ L. Furthermore, we may assume that g ∈ CL (B). We claim that H and H g are conjugate in K = H, H g , and thus H is pronormal in G by Lemma 1. Write H0 = H ∩ L and K0 = K ∩ L. Let πi : L → Li be the projection map. Since H contains a Sylow 2-subgroup of G, it acts transitively on the set {Li | i = 1, . . . , n} by conjugation. Hence the groups Hi = H0πi are pairwise isomorphic, and the same is true for Ki = K0πi . Also note that Hi and Ki are subgroups of odd index in Li because they contain the Sylow 2-subgroup Si = S ∩ Li of Li . The two possibilities are open: Case (i): Ki = Li for every i. Case (ii): Ki < Li for every i. In Case (i), Lemma 15 yields L ≤ K; therefore, g ∈ K and the assertion follows. Consider Case (ii). By Lemma 17, we may assume that K ≤ M1 Sn , where K1 ≤ M1 < L1 and M1 is a maximal subgroup of odd index in L1 . By [12, Theorem 1], one of the following four cases takes place: Case (ii)(1): M1 ∼ = A4 . In this case, H prn M1 Sn by Lemma 14 and NG (S) ≤ M1 Sn , therefore, H prn G by Lemma 7. Case (ii)(2): M1 ∼ = A5 . In this case, H prn M1 Sn by Lemma 16 and NG (S) ≤ M1 Sn . Hence H prn G by Lemma 7. Case (ii)(3): M1 ∼ = P SL2 (q0 ), where q = q0r for some odd prime r. It is easy to see that q0 ≡ ±3 (mod 8), which yields H prn M Sn by the choice of q. Also NG (S) ≤ M Sn , and so H prn G by Lemma 7. Case (ii)(4): M1 is the dihedral group of order q − ε, where ε = ±1 is chosen so that q ≡ ε (mod 4). We argue first that g ∈ NL (H0 ). Observe that in this case M1 and all of its subgroups (in particular, H1 and K1 , and thus all of Hi and Ki ) have normal cyclic 2-complements. Since H0 ≤ H1 , . . . , Hn  ∼ = H1 × · · · × Hn , the subgroup H0 also has the normal 2-complement U = O(H0 ). Define gi = g πi for all i = 1, . . . , n. Also write Ui = O(Hi ) and Vi = O(Ki ) for every i. Then gi Hi = (H g )πi ≤ Ki . Hence   Uigi = O(Hi )gi = O Higi ≤ O(Ki ) = Vi . 472

But Ui is the only subgroup of Vi whose order is equal to |Ui | because Vi is cyclic. Thus Uigi = Ui and by Lemma 5 gi ∈ Li ∼ = P SL2 (q) centralizes Ui . Furthermore, since O(H0 ) = U ≤ U1 , . . . , Un , we have g ∈ g1 , . . . , gn  ≤ CL1 (U1 ), . . . , CLn (Un ) ≤ CL (U1 , . . . , Un ) ≤ CL (U ).

Since U is a 2-complement and S ∩ L is a Sylow 2-subgroup of H0 , we conclude that H0g = (S g ∩ L)U g = (S ∩ L)U = H0 , and g ∈ NL (H0 ), as required. Consider Y = NG (H0 ). By the above, g ∈ Y and it is clear that H ≤ Y , which yields K = H, H g  ≤ H, g ≤ Y . Note that Y is a subgroup of odd index in G and that the projection of Y ∩ L = NL (H0 ) to every factor Li is strictly less than Li (otherwise, it would follow from Lemma 15 that Li ≤ Y , and hence L ≤ Y because Y acts transitively on {Li | i = 1, . . . , n}, and thus H0 would be a nontrivial soluble normal subgroup of L). By Lemma 17, we have Y ≤ M < G, where M ∼ = M1∗ Sn and M1∗ is π ∗ 1 a maximal subgroup of L1 containing NL1 ((Y ∩ L) ). The subgroup M1 , as well as the subgroup M1 above, satisfies the conditions of one of Cases (ii)(1)–(ii)(4). Lemmas 12, 14, and 16 together with the choice of q guarantee that H prn M in all cases, and since g ∈ Y ≤ M , it follows that H and H g are conjugate in H, H g .  Proof of Theorem 1 Let G = Sp2n (q), where q ≡ ±3 (mod 8) and n is not of the form 2w or 2w (22k + 1). Then the 2-adic representing n either has two 1s in positions s1 and s2 of different parity, or three 1s in positions s1 , s2 , and s3 of the same parity. Define m = 2s1 + 2s2 or m = 2s1 + 2s2 + 2s3 , respectively. It is easy to see that m is a multiple of 3. Let V be the n-dimensional vector space over the field Fq with a nondegenerate skew-symmetric bilinear form associated with G. Let H be the stabilizer in G of a nondegenerate m-dimensional subspace of V . Then the index |G : H| is odd by [12, Theorem 1]. Thus if K is a subgroup of odd index in H, then K is a subgroup of odd index in G too. It is easy to see that H = H1 × H2 , where H1 ∼ = Sp2m (q) and H2 ∼ = Sp2(n−m) (q). Since 3 divides m, it follows by Lemma 10 that H1 /O2 (H1 ) has a nonpronormal subgroup K1 /O2 (H1 ) of odd index. By Lemma 6, K1 is not pronormal in H1 , therefore, K1 × H2 is not pronormal in H and hence in G, by Lemma 7. By Lemma 6, (K1 × H2 )/O2 (G) is a nonpronormal subgroup of odd index in the simple group G/O2 (G) = G/Z(G) ∼ = P Sp2n (q).  Proof of Theorem 2 Let G = Spn (q), where n = 2w and q ≡ ±3 (mod 8). Let V be the n-dimensional vector space over the field Fq with nondegenerate skew-symmetric bilinear form associated with G. Suppose that the claim of Theorem 2 is false, and let q be the smallest prime power congruent ±3 modulo 8 such that G has a nonpronormal subgroup H of odd index. Consider the natural homomorphism ¯ : G → G/Z(G). Note that |Z(G)| = 2; therefore, by Lemma 6 and the choice of H, it follows that H is not pronormal in G. Since subgroups of odd index are pronormal in P Sp2 (q) ∼ = P SL2 (q) by a result in [2], we have n ≥ 4. Let S ∈ Syl2 (H) ⊆ Syl2 (G). By Lemma 3, we see that |NG (S) : S| = 3. By Lemma 1, there is an g¯ g¯ element g¯ of NG (S) \ S of order 3 such that H and H are not conjugate in K = H, H . By the choice of g¯, it follows that K is a proper subgroup of G. Hence there is a maximal subgroup M (of odd index) of G such that K ≤ M . By [12, Theorem 1] the following possibilities for M are open: Case (1): M = CG (σ) with σ a field automorphism of odd prime order r of G. In this case M ∼ = P Spn (q0 ), where q = q0r , by [5, Proposition 4.5.4]. Since r is odd, it is clear that q0 ≡ ±3 (mod 8). By the choice of q, it follows that H prn M . Also S ≤ M and |NM (S) : S| = 3 by Lemma 3. Thus g¯ ∈ NG (S) = NM (S) ≤ M . This contradicts the choice of g¯ because H prn M . 473

Case (2): G = P Sp4 (q) and M ∼ = 24 .A5 . By a result of [2], the subgroups of odd index are pronormal in M /O2 (M ). By Lemma 6, this shows that H prn M . Also S ≤ M and |NM (S) : S| = 3 by Lemma 3. Once again we have g¯ ∈ NG¯ (S) = NM (S) ≤ M , which contradicts the choice of g¯ and the fact that H prn M . Case (3): M is the stabilizer in G of the orthogonal decomposition V =

t

Vi

(2)

i=1

of the natural symplectic module V of G, with Vi isometric subspaces of dimension m = 2w1 ≥ 2. Choose M of such type so that m is as small as possible, in other words, so that (2) is unrefinable. By [5, Proposition 4.2.10], we have M ∼ = 2t−1 .(P Spm (q) St ), with n = mt. Furthermore, S ≤ M and M contains an element of order 3 that normalizes S, and so |NM (S) : S| = 3 by Lemma 3. As in the previous cases, g¯ ∈ NG (S) = NM (S) ≤ M . Consider the composition ˜ : M → M /O2 (M ) of the natural homomorphisms M → M and M → ∼ M /O2 (M ). The structure of M shows that M = P Spm (q) St .  corresponding Let L = L1 × L2 × · · · × Lt , where Li ∼ = P Spm (q), be the normal subgroup of M to the base subgroup of P Spm (q) St . We may assume that the nondegenerate subspaces Vi in (2) are associated with the corresponding subgroups Li , which are regarded as projective symplectic groups. ˜ ∈ L. We claim By the choice of g¯, its image g˜ is an element of order 3 of NM (S). It is clear that g g ˜ g ˜ = H, H . This will be a contradiction, since the kernel of ˜ is and H are conjugate in K that H g¯ contained in H and hence H and H will be conjugate in K. ∩ L and let πi be the projection map from L to Li . Note that K acts transitively Write K0 = K πi πi on Li , and so all K0 are pairwise isomorphic. Furthermore, K0 are subgroups of odd index in Li . The two possibilities are open: Case (3)(i): K0πi = Li for every i. Case (3)(ii): K0πi < Li for every i. and g˜ ∈ K. Case (3)(i) is ruled out by Lemma 15 since otherwise L ≤ K Consider Case (3)(ii). ≤Y 2. Also since the subgroup M containing K was chosen so that m was as small as possible, the subgroup R1 of L1 cannot be the stabilizer of an orthogonal direct sum decomposition of the subspace V1 (associated with L1 ) with factors of smaller dimension. Thus by [12, Theorem 1], one of the following cases must arise. Case (3)(ii)(1): R1 = CL1 (σ), where σ is a field automorphism of odd prime order r of L1 regarded as P Spm (q). Then by [5, Proposition 4.5.4], we have R1 ∼ = P Spm (q0 ), with q = q0r . Since r is odd, it is easy to see that q0 ≡ ±3 (mod 8). This shows that Y = X/O2 (X) for a subgroup X of G1 = Spm (q0 ) isomorphic to Spm (q0 ) St . By the choice of q, the subgroups of odd index are pronormal in Spm (q0 ). By [12, Theorem 9] the index |G1 : X| is odd. Lemma 3 yields |NG1 (S1 ) : S1 | = 3, where S1 ∈ Syl2 (G1 ). Thus, the subgroups of odd index are pronormal in X by Lemma 7, and so the subgroups of odd index are pronormal in Y by Lemma 6. Note that the subgroup Y ∼ = R1 St has an element of order 3 that normalizes some of its Sylow 2-subgroups. As above, we conclude that g˜ ∈ NM (S) ≤ Y , and again we have a contradiction ≤K ≤Y. since H Case (3)(ii)(2): m = 4, q is a prime, and R1 ∼ = 24 .A5 . By Lemma 16 the subgroups of odd index are prn Y . Also, S ≤ Y and since Y has an element pronormal in Y /O2 (Y ). By Lemma 6, this shows that H of order 3 that normalizes S, we have g˜ ∈ NM (S) ≤ Y . This is a contradiction.  474

The authors are immensely grateful to A. V. Zavarnitsine for helpful discussions. This work was finished while the second and third authors were visiting the People’s Republic of China, and these authors thank Professors Tasuro Ito and Wenbin Guo for their hospitality. References 1. Vdovin E. P. and Revin D. O., “Pronormality of Hall subgroups in finite simple groups,” Sib. Math. J., vol. 53, no. 3, 419–430 (2012). 2. Kondrat ev A. S., Maslova N. V., and Revin D. O., “On the pronormality of subgroups of odd index in finite simple groups,” Sib. Math. J., vol. 56, no. 6, 1101–1107 (2015). 3. Kondrat ev A. S., Maslova N. V., and Revin D. O., “A pronormality criterion for supplements to abelian normal subgroups,” Proc. Steklov Inst. Math., vol. 296, suppl. 1, S145–S150 (2017). 4. Conway J. H., Curtis R. T., Norton S. P., Parker R. A., and Wilson R. A., Atlas of Finite Groups. Maximal Subgroups and Ordinary Characters for Simple Groups, Clarendon Press, Oxford (1985). 5. Kleidman P. B. and Liebeck M., The Subgroup Structure of the Finite Classical Groups, Cambridge Univ. Press, Cambridge (1990). 6. Kondrat ev A. S., “Normalizers of the Sylow 2-subgroups in finite simple groups,” Math. Notes, vol. 78, no. 3, 338–346 (2005). 7. Mazurov V. D., “On the set of orders of elements of a finite group,” Algebra and Logic, vol. 33, no. 1, 49–55 (1994). 8. Huppert B., Endliche Gruppen. I, Springer-Verlag, Berlin (1967). 9. Isaacs I. M., Character Theory of Finite Groups, Academic Press, New York (1976). 10. Isaacs I. M., Finite Group Theory, Amer. Math. Soc., Providence (2008). 11. Carter R. and Fong P., “The Sylow 2-subgroups of the finite classical groups,” J. Algebra, vol. 1, no. 1, 139–151 (1964). 12. Maslova N. V., “Classification of maximal subgroups of odd index in finite simple classical groups,” Proc. Steklov Inst. Math., vol. 267, suppl. 1, S164–S183 (2009). A. S. Kondrat ev; N. V. Maslova Krasovskii Institute of Mathematics and Mechanics Ural Federal University, Ekaterinburg, Russia E-mail address: [email protected]; [email protected] D. O. Revin Sobolev Institute of Mathematics Novosibirsk State University, Novosibirsk, Russia Department of Mathematics University of Science and Technology of China, Hefei, P. R. China E-mail address: [email protected]

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