ON THE PRONORMALITY OF SUBGROUPS OF ODD INDICES IN FINITE SIMPLE - - PowerPoint PPT Presentation

ON THE PRONORMALITY OF SUBGROUPS OF ODD INDICES IN FINITE SIMPLE GROUPS

Natalia V. Maslova

IMM UB RAS, UrFU Joint work with Anatoly Kondrat’ev and Danila Revin

Yekaterinburg, August 11, 2015

Natalia V. Maslova On the finite groups

DEFINITIONS

• AGREEMENT. If another isn’t established, we consider

finite groups only. Let G be a group.

• DEFINITION. A subgroup H of a group G is pronormal in

G if H and Hg are conjugate in H, Hg for every g ∈ G.

• REMARK. H is a normal subgroup of G ⇒ H is pronormal

in G.

• QUESTION. What are pronormal subgroups of G?
• EXAMPLES. The following subgroups are pronormal in

finite groups:

• Normal subgroups;
• Maximal subgroups;
• Sylow subgroups.

Natalia V. Maslova On the finite groups

PRONORMAL SUBGROUPS

PROPOSITION 1 (Evgeny Vdovin and Danila Revin, 2012). Let G be a group, H ≤ G and S ≤ H for some pronormal (possibly, Sylow) subgroup S of G. Then the following conditions are equivalent:

(1) H is pronormal in G; (2) subgroups H and Hg are conjugate in H, Hg for every g ∈ NG(S).

• PROOF. Prove (2) ⇒ (1). Let g ∈ G and note

S, Sg ∈ H, Hg. S in pronormal, so, there exists y ∈ S, Sg such that S = Sgy. In particular, gy ∈ NG(S). In view of (2) the subgroups H and Hgy are conjugate in H, Hgy. Let t ∈ H, Hgy and Ht = Hgy. Then t−1Ht = y−1g−1Hgy and (yt−1y−1)yHy−1(yty−1) = g−1Hg. Moreover, it’s easy to see, yty−1 ∈ Hy−1, Hg. Thus, the subgroups Hy−1 and Hg are conjugate in Hy−1, Hg.

Natalia V. Maslova On the finite groups

PRONORMAL SUBGROUPS

PROPOSITION 1 (Evgeny Vdovin and Danila Revin, 2012). Let G be a group, H ≤ G and S ≤ H for some pronormal (possibly, Sylow) subgroup S of G. Then the following conditions are equivalent:

(1) H is pronormal in G; (2) subgroups H and Hg are conjugate in H, Hg for every g ∈ NG(S).

• PROOF. Recall, y ∈ S, Sg such that S = Sgy. We have

proved, the subgroups Hy−1 and Hg are conjugate in Hy−1, Hg. The following Lemma completes the proof. LEMMA (EXERCISE). Let H be a subgroup of G, g ∈ G and y ∈ H, Hg. If the subgroups Hy and Hg are conjugate in Hy, Hg then the subgroups H and Hg are conjugate in H, Hg.

• REMARK. Let G be a group, H ≤ G and S be a pronormal

subgroup of G. If NG(S) ≤ H then H is pronormal in G.

Natalia V. Maslova On the finite groups

EXAMPLES

• DEFINITION. H is a Hall subgroup of G if

(|H|, |G : H|) = 1. Let G be a group and H ≤ G.

• EXAMPLES. H is a pronormal subgroup of G in the

following cases:

• H is a normal subgroup of G;
• H is a maximal subgroup of G;
• H is a Sylow subgroup of G;
• G is solvable and H is a Hall subgroup of G;
• S is a pronormal subgroup of G and NG(S) ≤ H.

Natalia V. Maslova On the finite groups

PROBLEMS

• QUESTION. What are pronormal subgroups of finite

simple groups?

Natalia V. Maslova On the finite groups

CONJECTURE

• DEFINITION. H is a Hall subgroup of G if

(|H|, |G : H|) = 1. THEOREM (Evgeny Vdovin and Danila Revin, 2012). All Hall subgroups are pronormal in finite simple groups. The following conjecture was formulated by E. P. Vdovin and D. O. Revin.

• CONJECTURE. Subgroups of odd indices are pronormal in

finite simple groups.

Natalia V. Maslova On the finite groups

SYLOW p-SUBGROUPS, WHERE p > 2

EXAMPLE 1. Let p doesn’t divide |G|. In view of Glauberman’s Z∗-Theorem in every finite simple group G there exist two conjugate involutions u and v such that uv = vu. The subgroups u and v are conjugate in G and every of them contain a (trivial) Sylow p-subgroup of G. But u and v are not pronormal since the subgroup u, v is abelian.

Natalia V. Maslova On the finite groups

SYLOW p-SUBGROUPS, WHERE p > 2

EXAMPLE 2. Let p > 3 and G = Ap+4. Consider conjugate subgroups H1 = (1, ..., p)(p + 1, p + 2)(p + 3, p + 4) and H2 = (1, ..., p)(p + 1, p + 3)(p + 2, p + 4) containing a Sulow p-subgroup of G. The subgroup H1, H2 ∼ = Cp × V4 is abelian, so H1 and H2 are not pronormal in G.

Natalia V. Maslova On the finite groups

SYLOW p-SUBGROUPS, WHERE p > 2

EXAMPLE 3. Let p = 3, G = M23, H ∼ = PSL3(4)τ is a subgroup of index 253 of G, H1 ∼ = A6 ≤ PSL3(4), H2 = Hτ

1

and H1 and H2 are not conjugate in PSL3(4). H1 and H2 contain Sylow 3-subgroups of G and H1, H2 = PSL3(4). So, H1 and H2 are not pronormal in G.

Natalia V. Maslova On the finite groups

PRONORMAL SUBGROUPS

PROPOSITION 1. Let G be a group, H ≤ G and S ≤ H for some pronormal (possibly, Sylow) subgroup S of G. Then the following conditions are equivalent:

(1) H is pronormal in G; (2) subgroups H and Hg are conjugate in H, Hg for every g ∈ NG(S).

• REMARK. If S is a Sylow subgroup of G and NG(S) = S

then H is pronormal in G for any subgroup H ≥ S.

Natalia V. Maslova On the finite groups

ON THE FINITE SIMPLE GROUPS

Recall, a non-trivial group G is simple if it doesn’t contain nontrivial proper normal subgroups. Finite simple groups were classified. With respect to this classification, finite simple groups are:

• Alternating groups An for n ≥ 5;
• Classical groups PSLn(q) = Ln(q),

PSUn(q) = Un(q) = PSL−

n (q) = L− n (q), PSp2n(q) = S2n(q),

PΩn(q) = On(q) (n is odd), PΩ+

n (q) = O+ n (q) (n is even),

PΩ−

n (q) = O− n (q) (n is even);

• Exceptional groups of Lie type: E8(q), E7(q), E6(q),

2E6(q) = E− 6 (q), 3D2n(q), F4(q), 2F4(q), G2(q), 2G2(q) (q is

a power of 3), 2B2(q) (q is a power of 2);

• 26 sporadic groups.

Natalia V. Maslova On the finite groups

PRONORMAL SUBGROUPS

LEMMA (A. S. Kondrat’ev, 2005). Let G be a finite nonabelian simple group and S ∈ Syl2(G). Then NG(S) = S excluding the following cases:

(1) G ∼ = J2, J3, Suz or HN and |NG(S) : S| = 3; (2) G ∼ = 2G2(32n+1) or J1 and NG(S) ∼ = 23.7.3 < Hol(23); (3) G is a group of Lie type over field of characteristic 2 and NG(S) is a Borel subgroup of G; (4) G ∼ = L2(q) where 3 < q ≡ ±3 (mod 8) and NG(S) ∼ = A4; (5) G ∼ = Eη

6(q) where η = ± and q is odd and

|NG(S) : S| = (q − η1)2′/(q − η1, 3)2′ = 1; (6) G ∼ = S2n(q), where n ≥ 2, q ≡ ±3 (mod 8), n = 2s1 + · · · + 2st for s1 > · · · > st ≥ 0 and NG(S)/S is the elementary abelian group of order 3t; (7) G ∼ = Lη

n(q), where n ≥ 3, η = ±, q is odd, n = 2s1 + · · · + 2st

for s1 > · · · > st > 0 and NG(S) ∼ = S × C1 × · · · × Ct−1, where C1, . . . Ct−2, Ct−1 are cyclic subgroup of orders (q − η1)2′, . . . , (q − η1)2′, (q − η1)2′/(q − η1, n)2′ respectively.

Natalia V. Maslova On the finite groups

PRONORMAL SUBGROUPS

THEOREM (A. Kondrat’ev, N.M., D. Revin, 2015). All subgroups of odd indices are pronormal in the following finite simple groups:

(1) An, where n ≥ 5; (2) sporadic groups; (3) groups of Lie type over fields of characteristic 2; (4) L2n(q); (5) U2n(q); (6) S2n(q), where q ≡ ±3 (mod 8); (7) Oε

n(q), where ε ∈ {+, −, epmty symbol};

(8) exeptional groups of Lie type not isomorphic to E6(q) or

2E6(q).

Natalia V. Maslova On the finite groups

PROBLEM

• PROBLEM. Are all subgroups of odd indices pronormal in

the following finite simple groups:

(1) Ln(q), where n = 2w and q is odd; (2) Un(q), where n = 2w and q is odd; (3) S2n(q), where q ≡ ±3 (mod 8); (4) exeptional groups of Lie type E6(q) and 2E6(q), where q is

• dd?

Natalia V. Maslova On the finite groups

COUNTEREXAMPLE TO CONJECTURE

LEMMA (A. Kondrat’ev, N.M., D. Revin, 2015). 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 statements are equivalent:

(1) H is pronormal in G; (2) U = NU(H)[H, U] for any H-invariant subgroup U ≤ V .

• COROLLARY. Let G = A ≀ Sn = HV , where A is abelian,

H = Sn and V = An. Then the following statements are equivalent:

(1) H is pronormal in G; (2) (|A|, n) = 1.

Natalia V. Maslova On the finite groups

COUNTEREXAMPLE TO CONJECTURE

Let q ≡ ±3 (mod 8) be a prime power and n be a positive

• integer. It’s well known, Sylow 2-subgroup S of a group

T = Sp2(q) = SL2(q) is isomorphic to Q8 and NT (S) ∼ = SL2(3) = Q8 : 3. We have H = Q8 ≀ S3n ≤ X = SL2(3) ≀ S3n ≤ Y = Sp2(q) ≀ S3n ≤ G = Sp6n(q). PROPOSITION (N.M., 2008). If L = Spn(q), where n ≥ 4 and P ∼ = Spm(q) ≀ St ≤ L then |L : P| is odd if and only if m = 2w ≥ 2. THEOREM (A. Kondrat’ev, N.M., D. Revin, 2015). The index |G : H| is odd and H isn’t pronormal in G, so H/Z(G) is a subgroup of odd index in G/Z(G) ∼ = PSp6n(q) which isn’t pronormal.

Natalia V. Maslova On the finite groups

PRONORMAL SUBGROUPS

THEOREM (A. Kondrat’ev, N.M., D. Revin, 2015). All subgroups of odd indices are pronormal in finite simple groups PSp2n(q).

Natalia V. Maslova On the finite groups

Thank you for your attention!

Natalia V. Maslova On the finite groups