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BY 4.0 license Open Access Published by De Gruyter December 15, 2020

TI subgroups and depth 3 subgroups in simple Suzuki groups

  • Hayder Abbas Janabi , László Héthelyi and Erzsébet Horváth EMAIL logo
From the journal Journal of Group Theory

Abstract

In this paper, we determine the TI subgroups of the simple Suzuki groups Sz(q). More generally, we determine those nontrivial subgroups that are disjoint from some of their conjugates. It turns out that the latter are exactly those subgroups that have ordinary depth 3. The Sylow 2-subgroups of simple Suzuki groups belong to the class of so-called Suzuki 2-groups, which have been studied extensively by Higman. These results were extended later by Goldschmidt, Shaw, Shult, Gross, Wilkens and Bryukhanova. As a corollary of our investigations, we get some interesting results for the Sylow 2-subgroups of Suzuki groups, as well. We relate this to an open problem on Suzuki 2-groups, and we ask a question concerning that. We also give some characterization of Suzuki groups.

1 Motivation and results

In an earlier paper [14], we determined the ordinary and combinatorial depth of several subgroups of the simple Suzuki groups Sz(q). For the definitions of these notions, see Definition 2.5 and Definition 2.8 in Section 2. There are several ways to define ordinary depth; a good summary of these can be found in [18]; see also [6, Definition 3.5] and the introduction of [8]. The notion of combinatorial depth was introduced in [2, Definition 3.2]. We use [2, Theorem 3.9] as an equivalent definition. A subgroup LG is called TI if, for every xG, from LxL{1}, it follows that xNG(L). In an arbitrary group, non-normal TI subgroups are always of combinatorial depth three, hence also of ordinary depth three. Moreover, nontrivial subgroups having a disjoint conjugate are always of ordinary depth three in every group. However, in general, the converse is not true, e.g. L=A5 has in G=A7 ordinary depth 3, but there is no element xG such that LLx={1}. (A similar example is A6 in A10.)

It is an open problem how to characterize subgroups of ordinary depth 3 in a group-theoretical way in an arbitrary group.

We will show in Theorem 1.3 that, in the case of simple Suzuki groups, subgroups of ordinary depth 3 are exactly those nontrivial subgroups that have a disjoint conjugate.

In Theorem 1.1, we also characterize those nontrivial TI subgroups of Sz(q) which are non-cyclic elementary abelian 2-subgroups. These are exactly those subgroups that are conjugate to the center of a Sylow 2-subgroup of a smaller Suzuki subgroup Sz(s)Sz(q). This property can help the recognition of those subgroups of Sz(q) which are isomorphic to a Suzuki subgroup Sz(s)Sz(q). Recognition of Suzuki groups in GL(4,q) is considered in some recent papers; see [1, 4]. In another paper [23], Suzuki groups were used to construct some block designs. So results about intersections in Suzuki groups might also be helpful in combinatorial investigations.

In this paper, we will use the following notation. Let 𝐺 be the simple Suzuki group Sz(q), where q=22m+1. Let 𝐹 be a fixed Sylow 2-subgroup of 𝐺. Then NG(F)=FH, where |H|=q-1. Let A1 and A2 be Hall subgroups of 𝐺 of orders q+2r+1 and q-2r+1, respectively, where r=2m; see Theorem 2.1 and also [17, Theorem 3.10, Chapter XI] and [22, Theorem 4.12].

We will denote by K2n an elementary abelian subgroup of order 2n. The center Z(F) of the Sylow 2-subgroup 𝐹 of the Suzuki group Sz(q) will be of order q=22m+1, and we will denote 2m+1 by 𝑓. We also suppose that m>0. The main results of this paper are the following theorems.

Theorem 1.1

If G=Sz(q) is a simple Suzuki group, then 𝐺 has the following TI subgroups.

  1. Cyclic subgroups of prime order and the trivial subgroup.

  2. Subgroups F,H,A1,A2, their characteristic subgroups and the conjugates of these.

  3. An elementary abelian subgroup K2n of order 2n>2 is a TI-subgroup if and only if it is the center of a Sylow 2-subgroup of a simple Suzuki subgroup G1G, or a conjugate to it. This holds if and only if n>1 and nf. (Remember |Z(F)|=2f.) These subgroups are exactly those non-cyclic elementary abelian 2-subgroups of 𝐺 that have combinatorial depth 3.

All other nontrivial subgroups are not TI.

Remark 1.2

This classification of TI subgroups shows that [1, Theorem 2.1 (7)] is wrong; the centralizer of an element of order 4 cannot be TI.

Theorem 1.3

Subgroups of ordinary depth 3 of a simple Suzuki group G=Sz(q) are the following.

  1. Every nontrivial subgroup contained in a maximal subgroup different from a conjugate of NG(F).

  2. 𝐹 and all its nontrivial subgroups, and the conjugates of these.

  3. All nontrivial subgroups U=F1K of NG(F), where F1<F and KH, and the conjugates of these subgroups.

Moreover, a nontrivial subgroup LG=Sz(q) is of ordinary depth 3 if and only if there exists an element xG with LLx={1}.

As a corollary, we get a characterization of Suzuki groups.

Corollary 1.4

Let 𝐺 be a simple Zassenhaus group acting on q+1 points, where 𝑞 is a 2-power. Then the following are equivalent.

  1. GSz(q).

  2. If FSyl2(G), then every subgroup not containing a conjugate of the subgroup 𝐹 has a disjoint conjugate.

Proof

By Theorem 1.3 and by the description of subgroups of Sz(q) in Theorem 2.1, we have that (i) implies (ii). For the other direction, observe that the only simple Zassenhaus groups with 𝑞 even are Sz(q) and PSL(2,q). In PSL(2,q), there are dihedral subgroups of order 2(q+1), that are not disjoint from their conjugates; see [7]. ∎

We also have a wider class of groups where ordinary depth 3 is equivalent to having a disjoint conjugate.

Corollary 1.5

Let 𝐺 be a simple Zassenhaus group acting on q+1 points, where 𝑞 is an odd power of 2. Then the following are equivalent for a nontrivial subgroup LG.

  1. The subgroup LG has ordinary depth 3.

  2. The subgroup LG has a disjoint conjugate.

Proof

Since these Zassenhaus groups are Sz(q) and PSL(2,q), we have to prove the equivalence for these groups. In Theorem 1.3, we proved the equivalence for Sz(q). In [7], this is proved for PSL(2,22n+1). ∎

We also get some information about Sylow 2-subgroups of Suzuki groups.

Corollary 1.6

Every Sylow 2-subgroup of G=Sz(q) is the union of conjugates of the Sylow 2-subgroups of a smallest simple Suzuki subgroup G1 contained in 𝐺. If FSyl2(G) and F1Syl2(G1) are contained in 𝐹, then F=xNG(F)F1x. Every element of order 4 of 𝐹 is in exactly one conjugate of F1, and any two of the conjugates of F1 in this union either have trivial intersection or their intersection is their center, which is a conjugate of Ω1(F1).

We give a proof of this in Section 3.

Remark 1.7

It is easy to see that F=xNG(F)F2x, where F2C4, and in this union, any two conjugates of F2 either have trivial intersection, or the intersection is a conjugate of Ω1(F2). This is because there are two conjugacy classes of order 4 elements in 𝐺. Let they be KG(a) and KG(a-1), where aF. Consider xNG(F)ax. This will contain all elements of order 4 in 𝐹, i.e. all elements of FZ(F). However, each involution in 𝐹 is a square of an element of order 4, since this is true for at least 2 involutions and the other involutions are conjugate by elements of HNG(F). Hence we are done.

About the structure of the paper: In Section 2, we give some properties of the Suzuki groups that are needed in the proofs; we also give the definitions of combinatorial and ordinary depth of a subgroup. In Section 3, we give the proof of Theorem 1.1 and of Corollary 1.6. In Section 4, we prove Theorem 1.3. At the end of the paper, we formulate two questions.

2 Preliminaries

Suzuki groups Sz(q) are twisted groups of Lie type B22(q), where q:=22m+1. If m>0, then they are simple. Suzuki groups are also doubly transitive permutation groups on q2+1 points; they belong to the class of Zassenhaus groups. Suzuki groups also can be defined as subgroups of GL(4,q). The order of Sz(q) is (q2+1)(q-1)q2. The order of Sz(q) is not divisible by 3; however, it is always divisible by 5. For further information, see [14].

We will use the following results on the Suzuki groups; see [17, Theorem 3.10, Chapter XI] and [22, Theorem 4.12].

Theorem 2.1

Theorem 2.1 (Suzuki)

Let G=Sz(q), where q=22m+1 for some positive integer 𝑚. Then 𝐺 has the following subgroups.

  1. The Hall subgroup NG(F)=FH, a Frobenius group of order q2(q-1), where FSyl2(G) and 𝐻 is cyclic of order q-1.

  2. The dihedral group B0=NG(H) of order 2(q-1).

  3. The cyclic Hall subgroups A1,A2 of orders q+2r+1, q-2r+1, respectively, where r=2m and |A1||A2|=q2+1.

  4. The Frobenius subgroups B1=NG(A1) and B2=NG(A2) of orders 4|A1| and 4|A2|, respectively.

  5. Subgroups of the form Sz(s), where 𝑠 is an odd power of 2, s8, and q=sn for some positive integer 𝑛. Moreover, for every odd 2-power 𝑠, where sn=q for some positive integer 𝑛, there exists a subgroup isomorphic to Sz(s).

  6. Subgroups (and the conjugates of the subgroups) of the above groups.

Remark 2.2

The group B22(2):=Sz(2) is not simple; it is a Frobenius group of order 20. It is a subgroup of each simple Sz(q). Since 5 divides either |A1| or |A2|, the group Sz(2) is a subgroup of B1 or B2 up to isomorphism.

Remark 2.3

By [25, Theorem 10], if Sz(s)Sz(q), then every subgroup of Sz(q) isomorphic to Sz(s) is also conjugate to it in Sz(q). Later in the paper, we will call the subgroups Sz(s)Sz(q)Suzuki subgroups of Sz(q). In the case of matrix generators of Sz(q) in GL(4,q), it is the subgroup generated by the same matrices when the matrix entries are restricted to GF(s).

Theorem 2.4

Let q=22m+1, m>0, r=2m and G=Sz(q).

  1. Let i{1,2}, and let uiAi, ui1. Then CG(ui)=Ai. If Bi=NG(Ai), then Bi=Ai,ti, where ti is an element of order 4, and uti=uq for all uAi. Moreover, NG(Ai) is a Frobenius group with kernel Ai.

  2. Let F,H,A1,A2 be as in Theorem 2.1. Then the conjugates of F,H,A1,A2 form a partition of 𝐺. In particular, F,H,A1,A2, their conjugates and the conjugates of their characteristic subgroups are TI sets in 𝐺.

The Sylow 2-subgroup 𝐹 of 𝐺 is a Suzuki 2-group. This means that it is a non-abelian 2-group, having more than one involution, and having a solvable group of automorphisms which permutes the set of involutions of 𝐹 transitively. See [16, p. 299] for details.

The group 𝐹 is a class 2 group of order q2 and exponent 4. Moreover, its center Z(F)=F=Φ(F) is of order 𝑞. The involutions in 𝐹 together with the identity element constitute Z(F), and 𝐹 does not contain any quaternion subgroups. A nontrivial element of 𝐹 is real in 𝐺 if and only if it is an involution. (An element of a group 𝐺 is called a real element in 𝐺 if it is conjugate in 𝐺 to its inverse; see [10, p. 303].)

The subgroup 𝐻 acts sharply 1-transitively on the involutions of 𝐹, and on the nontrivial elements of F/Z(F). The centralizer in 𝐺 of every nontrivial element of 𝐹 is a subgroup of 𝐹. For m>0, Suzuki groups are simple and their orders are congruent to 2mod3.

Zassenhaus groups are doubly transitive permutation groups without any regular normal subgroup, where any non-identity element has at most two fixed points. Zassenhaus groups are always of degree q+1, where 𝑞 is a prime power. If 𝑞 is even, then there are two series of Zassenhaus groups PSL(2,2n) for n>1 and Sz(22m+1) for m>0. If q=pf is odd and 𝑓 odd, then Zassenhaus groups are PGL(2,pf) and PSL(2,pf) if pf3. If f=2m, then beside PGL(2,pf) and PSL(2,pf), there exists a third type of Zassenhaus group M(pf). It can be constructed in the following way: let H=PGL(2,pf)αPΓL(2,pf), where αAut(GF(pf)) of order 2, raising elements of GF(pf) to the 𝑚-th power. There are 3 subgroups of 𝐻 of index 2 containing PSL(2,pf): PGL(2,pf), PSL(2,pf)α and M(pf). See [17, Chapter XI].

The notion of ordinary depth was originally defined for von-Neumann algebras; see [9]. Later, it was also defined for Hopf algebras; see [21]. For some recent results in this direction, see [13, 19, 12]. In [20] and later in [6], the depth of semisimple algebra inclusions was studied. The ordinary depth of a group inclusion LG (denoted by d(L,G)) is defined as the minimal depth of the group algebra inclusion CLCG, studied in [2]. We will use an equivalent definition which is established in [6]. For the case of depth one and two, see [3, 20].

Let 𝐺 be a finite group, LG a subgroup. We introduce the ordinary depth as follows.

Two irreducible characters α,βIrr(L) are related (denoted by αGβ) if they are constituents of the restriction χ|L for some χIrr(G). The distancedG(α,β)=m is the smallest integer 𝑚 such that there is a chain of irreducible characters of 𝐿 such that α=ψ0Gψ1GGψm=β. If there is no such chain, then dG(α,β)=-, and if α=β, then the distance is zero. If 𝑋 is the set of irreducible constituents of χ|L, then m(χ):=maxαIrr(L)minψXdG(α,ψ).

Definition 2.5

Definition 2.5 ([6, Theorem 3.6, Theorem 3.10], [20, Corollary 3.2], [3, Theorem 1.7])

Let 𝐿 be a subgroup of a finite group 𝐺. The ordinary depthd(L,G) is the minimal possible positive integer which can be determined from the following upper bounds.

  1. For m1, d(L,G)2m+1 if and only if the distance between two irreducible characters of 𝐿 is at most 𝑚.

  2. For m2, d(L,G)2m if and only if m(χ)m-1 for all χIrr(G).

  3. d(L,G)2 if and only if 𝐿 is normal in 𝐺, d(L,G)=1 if and only if G=LCG(x) for all xL.

Theorem 2.6

Theorem 2.6 ([6, Theorem 6.9])

Suppose that 𝐿 is a subgroup of a finite group 𝐺 and N=CoreG(L) is the intersection of 𝑚 conjugates of 𝐿. Then d(L,G)2m. If NZ(G) holds, then d(L,G)2m-1.

Remark 2.7

It follows from the first part of Theorem 2.6 that if a non-normal subgroup LG has a disjoint conjugate, i.e. LLx={1} for some xG, then d(L,G)4. Since in this case CoreG(L)={1}, it follows from the second part of Theorem 2.6 that d(L,G)3. But since 𝐿 is not normal in 𝐺, d(L,G)=3. However, the converse is not true, e.g. d(A5,A7)=3, but A5 does not have a disjoint conjugate in A7.

The combinatorial depth can be defined as follows; see [2].

Definition 2.8

Let 𝐿 be a subgroup of the finite group 𝐺, and let i1. Then the combinatorial depthdc(L,G) of the subgroup 𝐻 in 𝐺 is defined in the following way.

  1. dc(L,G)2i if and only if, for every x1,,xiG, there exist some y1,,yi-1G with LLx1Lxi=LLy1Lyi-1.

  2. Let i>1. Then dc(L,G)2i-1 if and only if, for every x1,,xiG, there exist some y1,,yi-1G with

    LLx1Lxi=LLy1Lyi-1

    and x1hx1-1=y1hy1-1 for all hLLx1Lxi.

  3. dc(L,G)=1 if and only if, for every xG, there exists some yL with xhx-1=yhy-1 for all hL. This holds if and only if G=LCG(L).

Remark 2.9

It is easy to see from Definition 2.8 that if a non-normal subgroup LG is TI, then dc(L,G)=3. The converse is not true; we will see later that C4 is not TI in Sz(q); however, dc(C4,Sz(q))=3 by [14, Theorem 4.1]. By Theorem [2, Theorem 4.1], we also know that d(L,G)dc(L,G) for every subgroup LG. For a non-normal subgroup LG to have a disjoint conjugate is a weaker property than being TI, e.g. in the case of simple Suzuki groups, every nontrivial 2-subgroup has a disjoint conjugate; however, by Theorem 1.1, only some of them are TI.

3 Proofs of Theorem 1.1 and of Corollary 1.6

Proof of Theorem 1.1

We prove Theorem 1.1 in eight steps. It is obvious that every subgroup of prime order is TI. Now we examine the subgroups of maximal subgroups of 𝐺 to establish which are TI and which are not TI.

(1) The subgroups 𝐹, 𝐻, A1, A2, their characteristic subgroups and their conjugates are TI. In particular, all subgroups of H,A1,A2 are TI; however, not every subgroup of 𝐹 is TI.

The first part follows from Theorem 2.4. Since H,A1,A2 are cyclic and TI, each subgroup of them is characteristic, hence TI. We will see below that not every subgroup of 𝐹 is TI.

(2) If UNG(F) and U=F1H1, where F1F, H1H are nontrivial subgroups, then 𝑈 is not TI.

Let xNG(H)FH. Then H1x=H1; hence UxHH1{1}. Since xNG(F), thus F1xF since 𝐹 is TI. Hence UxU is a proper subgroup of 𝑈; thus 𝑈 is not TI.

(3) For every nontrivial subgroup H1H and uNG(H)H, the subgroup u,H1 is not TI.

Let uNG(H)H; then this is an element of order 2 by Theorem 2.1 (2). So we may suppose that uF1 for some subgroup F1Syl2(G). Since we know, see e.g. [22, Theorem 4.1 (b)], that all involutions in a Sylow 2-subgroup of Sz(q) are in the center of the Sylow 2-subgroup, thus we have that uZ(F1). Let xF1NG(H). Then (H1u)xH1u contains 𝑢. However, 𝑥 does not normalize H1. Otherwise, it would also normalize 𝐻 since 𝐻 is TI. Hence u,H1 is not TI.

(4) If we take a nontrivial subgroup A~ of Ai, and a nontrivial subgroup C1C of a cyclic complement of order 4 in the Frobenius group NG(Ai), then C1A~ is not TI; in particular, Sz(2) is not TI.

Let A~Ai, CF1, where F1Syl2(G). Let

nNF1(C)C=NF1(C)NG(Ai)

be an involution. Then (C1A~)nC1A~{1} since C1 is characteristic in 𝐶. However, it does not contain A~ since then nNG(A~)NG(Ai), which is not the case. Hence C1A~ is not TI.

(5) Let G1:=Sz(s) be a simple smaller Suzuki subgroup in 𝐺. The subgroups 𝑈 of G1, whose order is divisible both by 2 and by some odd integer greater than 1, are not TI.

Let U2Syl2(U), and let U2P2Syl2(G). Then Z(P2)NG(U2); however, since Z(P2)G1, we have that Z(P2)U. If NP2(U2)NG(U), then UZ(P2) is a subgroup of 𝐺. It cannot be a subgroup of any maximal subgroup of type NG(Ai), NG(H) and Sz(s1) since the orders of these subgroups are not divisible by 𝑞. If UNG(S) for some SSyl2(G), then 𝑈 is conjugate to a subgroup discussed in point (2); hence it is not TI. The subgroup UZ(P2) cannot be 𝐺 since 𝑈 is normal in it. So we can suppose that NP2(U2)NG(U). Let xNP2(U2)NG(U). Then U2UxU<U; hence 𝑈 is not TI.

(6) Let C:=c be a cyclic subgroup of 𝐹 of order 4. This is not TI.

By the proof of [17, Lemma 5.9, Chapter XI], we have that CF(C)=Z(F)C is of order 2q. Since 𝑐 is not real in 𝐹, we have that NF(C)=CF(C). Thus there is an element xFNF(C). Let u:=c2. Then CG(u)=F, and hence CxC contains 𝑢. Hence 𝐶 is not TI.

(7) Let us denote an elementary abelian subgroup of order 2n in 𝐹 by K2n. The main results in this step are (e) and (f). This is exactly the content of (iii) in Theorem 1.1, which we will prove with the help of statements (a)–(d).

(a) Let K2n=Z(S1), where S1Syl2(G1) and G1 is a simple Suzuki subgroup of 𝐺. Then K2n is TI in 𝐺.

We may assume that S1F. Suppose that Z(S1)xZ(S1){1} for some xG. Then there exist involutions a,bZ(S1) with ax=b. Since 𝐹 is TI, thus xNG(F)=FH. Since 𝐹 acts trivially on Z(S1), we may suppose that xH. Moreover, 𝐻 acts sharply 1-transitively on the involutions of 𝐹. We also have that NG1(S1)NG(F)=FH since 𝐹 is TI. For some complement H1 in NG1(S1), H1H. However, H1 also acts sharply 1-transitively on the involutions of Z(S1). Thus xH1NG1(S1)NG1(Z(S1)). Hence Z(S1) is TI in 𝐺.

(b) If a non-cyclic elementary abelian subgroup of 𝐺 is TI, and its order is equal to |Z(S1)| for a Sylow subgroup S1Syl2(G1) for some simple Suzuki subgroup G1 of 𝐺, then this elementary abelian subgroup is the center of a Sylow 2-subgroup of a subgroup conjugate to a simple Suzuki subgroup of 𝐺.

Let us suppose that a non-cyclic elementary abelian subgroup K2nF of order 2n in 𝐺 is a TI set. We suppose that 2n=|Z(S1)| for S1Syl2(G1), where G1 is a simple Suzuki subgroup of 𝐺. The involutions of K2n and their 𝐻-conjugates form blocks in Z(F) since K2n is TI in 𝐺. Since 𝐻 acts sharply 1-transitively on the involutions of Z(F), hence Z(F) is the disjoint union of different 𝐻-conjugates of K2n. We claim that the normalizers in 𝐻 of the elementary abelian TI subgroups K~2n of order 2n of 𝐺 contained in Z(F) are the same. To see this, note that no element of 𝐻 fixes any element in 𝐹; hence the elements of the cyclic subgroup NG(K~2n)H move the nonunit elements of K~2n sharply 1-transitively; hence NG(K~2n)H has order 2n-1. Since 𝐻 is cyclic, it has only 1 subgroup of this order; thus each elementary abelian TI-subgroup of order 2n in Z(F) has the same normalizer in 𝐻. We know that K~2nhHK2nh. We claim that K~2n is one of the conjugates of K2n. Suppose that aK~2nK2nh1 and bK~2nK2nh2 are two different involutions and h1,h2H. Then there exists a unique hHNG(K~2n)=HNG(K2nh1) with ah=bK2nh1hK2nh2. However, then K2nh1=(K2nh1)h=K2nh2. Hence K~2n=K2nh1. Thus all elementary abelian TI subgroups of order 2n in 𝐹 are conjugate to Z(S1) for a Sylow subgroup S1Syl2(G1) for some Suzuki subgroup G1G.

(c) Let K2rF be an elementary abelian subgroup of order 2r in 𝐺. Then it is TI if and only if NH(K2r)=H1 is of order 2r-1.

Let K2rF be an elementary abelian subgroup of order 2r with the property that NH(K2r)=H1 and |H1|=2r-1. Then H1 permutes the elements of K2r{1} sharply 1-transitively, and every hHH1 transports each involution of K2r outside this group. Thus K2r is TI.

Conversely, let K2rF be TI. Since by [22, Theorem 4.1 (e), (f)] 𝐻 acts sharply 1-transitively on the involutions of 𝐹, if aK2r{1}, then for every bK2r{1}, there exists a unique hH with ah=b. Then bK2rhK2r. Since K2r is TI, then hNH(K2r). Thus NH(K2r) is regular on K2r{1}. Hence it follows that |NH(K2r)|=2r-1.

(d) Let r>1, and let NH(K2r)=H1 be of order 2r-1. Then 2r-1 divides q-1=2f-1, which happens if and only if |K2r|=|Z(S1)| for the Sylow subgroup S1Syl2(G1) for a simple Suzuki subgroup G1G. This happens if and only if rf and r>1.

Suppose that r>1 and |H1|=2r-1. This divides |H|=q-1=22m+1-1. Then r2m+1, and hence (2r)k=22m+1 for some positive integer 𝑘. Thus if G1=Sz(2r), then S1Syl2(G1) has center Z(S1) of order 2r. Conversely, if 2r is the size of Z(S1) for some S1Syl2(G1) for a simple Suzuki subgroup G1G, then r>1 and (2r)k=22m+1 for some positive integer 𝑘; hence 2r-122m+1-1.

(e) A non-cyclic elementary abelian 2-subgroup of 𝐺 is TI if and only if it is the center of a Sylow subgroup S1Syl2(G1) for some simple Suzuki subgroup G1G or conjugate to it; in particular, Klein four subgroups of 𝐺 are not TI.

One direction follows from (a); the other direction follows from (b) using (c) and (d).

(f) A non-cyclic elementary abelian 2-subgroup of 𝐺 is TI if and only if it is of combinatorial depth 3.

If K2n is TI, then it is of combinatorial depth 3 by Remark 2.9. Let n>1. If K2nF is not TI, then we have that there exists an element x2NG(K2n) such that K2nx2K2n{1}. Let x1F. Then x1 centralizes K2n and we cannot find an element 𝑦 with K2nK2nx1K2nx2=K2nK2ny such that 𝑦 also centralizes the intersection since then yF and it also centralizes K2n. Hence the combinatorial depth of K2n in 𝐺 is bigger than 3.

(8) If L<F is not elementary abelian, then 𝐿 is not TI.

Suppose by contradiction that 𝐿 is TI and not elementary abelian. Let 𝐼 be the subgroup generated by the involutions of 𝐿. If |I|=2, then since 𝐹 does not contain quaternion subgroups, LC4; hence it is not TI by step (6). Let |I|>2.

Suppose that IIx{1}; then LLx{1} and xNG(L)NG(I). Thus 𝐼 is also TI; hence it is the center of a Sylow 2-subgroup of a simple Suzuki subgroup of 𝐺 or conjugate to it. If xNG(I)NG(L), then

Ix=ILxL{1},

and hence 𝐿 is not TI. So we may assume that NG(I)=NG(L). Since IZ(F), FNG(I)=NG(L). Thus 𝐿 is normal in 𝐹 and Z(F)LF. If xL with o(x)=4, then by the proof of [17, Lemma 5.9, Chapter XI, p. 216], the number of conjugates of 𝑥 in 𝐹 is |F:CF(x)|=|F|/(2|Z(F)|)=q/2. Since F/Z(F) is abelian, all 𝐹-conjugates of 𝑥 are in the coset Z(F)x; moreover, using that LF, we have that they are also in (Z(F)L)x. Thus |Z(F)L|q/2. Z(F)L cannot be the center of a Sylow 2-subgroup of a smaller Suzuki subgroup since the order of that subgroup was smaller than q/2. Thus Z(F)L. Since Z(F) is normalized by 𝐻, we have that LLh{1} holds for every hH. Hence 𝐻 must normalize 𝐿. However, 𝐻 acts fixed point freely on L/Z(F). Hence |L/Z(F)|=q, and L=F. We are done.

Thus we have considered all the possible subgroups up to conjugacy; see Theorem 2.1

Remark 3.1

Let |Z(F)|=2f. Not every elementary abelian subgroup of order 2n, where 1<nf, is conjugate to the center of a Sylow 2-subgroup of a simple Suzuki subgroup of 𝐺. The number of subgroups of order 2n in Z(F) is

(2f-1)(2f-2)(2f-2n-1)(2n-1)(2n-2)(2n-2n-1).

If S1Syl2(G1) for a Suzuki subgroup G1G and |Z(S1)|=2n, then the number of 𝐻-conjugates of Z(S1) is |H:NH(Z(S1))|=(2f-1)(2n-1), which is strictly smaller than the above number if n<f. Thus there are some subgroups, Z(S1) and its 𝐻-conjugates, which are TI, and the remaining subgroups of Z(F) of order 2n are not TI.

Proof of Corollary 1.6

By part (7) (b) in the proof of Theorem 1.1, we have that hHZ(F1)h=Z(F), and the subgroups Z(F1)h for hH intersect trivially. Let aFZ(F) such that aF1. Since 𝐻 acts sharply 1-transitively on the nontrivial elements of F/Z(F), each coset of Z(F) contains an element of a conjugate of F1. By the proof of [17, Lemma 5.9, Chapter XI], the element 𝑎 has q/2=|F:CG(a)|=|F:Z(F)a| conjugates and a-1 also. The 𝐹-conjugates of these elements all lie in Z(F)a. Thus these give all the elements of Z(F)a since this coset has 𝑞 elements. Hence each element of 𝐹 is in a conjugate of F1. Since 𝐹 is a TI set, the conjugating elements are in NG(F).

For the second part, see [14, Proposition 4.13] or the following shorter argument. Suppose that xGNG(F1) and aF1F1x is an element of order 4. Then there exists an element bF1 such that bx=a. By [17, Lemma 11.7, Chapter XI], we have that, in 𝐺, there are two conjugacy classes of elements of order 4. We know that KG(a)KG(a-1); otherwise, an element of 𝐻 would centralize a2; similarly, KG(b)KG(b-1). They are also in different conjugacy classes in G1. Thus 𝑎 and 𝑏 are conjugate also in G1. Hence they are conjugate in NG1(F1). Thus there exists an element yNG(F1) such that by=a. Hence we have xy-1CG(b). Since xNG(F1) and yNG(F1), then xy-1NG(F1). However, xy-1CG(b)=Z(F)bNG(F1). This contradiction shows that there cannot be an order 4 element in two conjugates of F1. If F1F1x{1}, then they have common elements of order 2; hence Z(F1)Z(F1)x{1}. We have seen that Z(F1) is a TI subgroup of 𝐺; hence Z(F1)=Z(F1)x. Since F1 and F1x do not have common elements of order 4 and all their elements of order 2 are central, hence F1F1x=Z(F1). ∎

4 Proof of Theorem 1.3

Proof of Theorem 1.3

(i) As by [14], except for NG(F), all maximal subgroups have a disjoint conjugate, this holds also for their subgroups; thus any such nontrivial subgroup has ordinary depth 3.

(ii) This holds since 𝐹 is TI.

(iii) Now we have to consider the subgroups of NG(F)=FH. We want to prove that if the subgroup does not contain 𝐹, then it has a disjoint conjugate.

Let 𝑈 be such a subgroup. Let us suppose now that U=F1K, where F1<F and KH. Then by [24, Theorem 17.3], we have that 𝑈 is a Frobenius group with kernel F1=FU and complement 𝐾. Moreover, since 𝐾 is a characteristic subgroup in the cyclic group 𝐻, and 𝐻 is a TI set in 𝐺, so 𝐾 is also a TI set in 𝐺. Let xFH=NG(F). If UxU={1}, then we are done.

Otherwise, UxU{1}. The subgroup UxU cannot contain a nontrivial element f1F1 since then f1F1F1xFFx. Since 𝐹 is TI, we have that xNG(F), which is not the case. Then |UxU| is a divisor of |K|. Since 𝑈 is solvable, by Hall’s theorem, UxU is contained in a complement of F1, and it can be conjugated in 𝑈, moreover, in F1, to a subgroup of any other complement of F1 in 𝑈. We may suppose that UxUKH. Otherwise, let sF1 be an element such that (UxU)sK. Then K(UxU)s=UxsU and xsNG(F). Thus we can, if necessary, exchange 𝑥 to xs to get UxUK. Let K1:=UxU. So we may suppose that K1K.

The Frobenius complements of Ux are of the form Kf1x for some f1F1. Since Ux=F1x(f1F1Kf1x), for some element f1F1, we have that

{1}Kf1xKHf1xH.

Since 𝐻 is TI, we have that f1xNG(H)NG(K), as 𝐾 is characteristic in the cyclic group 𝐻. Thus Kf1x=KUxU=K1. Hence UxU=K. Let lNFx(F1x)F1x. Then (F1xK)l=F1xKlFf1xK=(FK)f1x since lFx and f1xNG(K).

Now, since lNFx(F1x) and f1xNG(K), moreover (F1xK)l(FK)f1x, we have that F1K(F1K)f1xl=F1KF1xKlFK(FK)f1x. The subgroup FK(FK)f1x contains 𝐾; however, it cannot contain any elements of 𝐹; otherwise, as 𝐹 is TI, this would imply that f1xNG(F), which is not the case. So FK(FK)f1x=K. If K(F1xKl){1}, then since F1xKl=(F1xK)l, for some element f2xF1x, we have that KKf2xl{1}. Hence f2xlNG(K) since 𝐾 is a TI set in 𝐺. However, lFxF1x so f2xlFxF1x. Thus a nontrivial element of Fx normalizes a Frobenius complement of FxK, which cannot happen. Thus K(F1xKl)={1}, and so F1K(F1K)f1xl={1}. Hence d(F1K,G)=3. (If F1=F, this proof does not work since then such 𝑙 does not exist.)

Let us suppose now that U=FK, where K<H. We want to prove that d(U,G)=5. We know, see the proof of [14, Proposition 4.3], that there exist elements x1,x2G with FH(FH)x1(FH)x2={1}. Hence, by the second part of Theorem 2.6, we have that d(U,G)5. Suppose by contradiction that d(U,G)4. Then, by Definition 2.5, we have that m(χ)1 for each irreducible character χIrr(G). We will prove that, for χ=1G, this is not true. For let us take a nontrivial irreducible character ψIrr(FK/F). We will prove that d(ψ,1FK)=2.

Since 𝜓 can be extended to a nontrivial irreducible character 𝜃 of NG(F) containing 𝐹 in its kernel, by the proof of [14, Corollary 5.1], d(θ,1NG(F))=2. Hence d(ψ,1FK)2.

Suppose by contradiction that this distance was 1. Then (ψG,1FKG)0. However, 𝜓 can be extended in |H:K| ways to FH=NG(F). Let us suppose that these extensions are ψ1,,ψ|H:K|. These are all the constituents of ψFH. Then

(ψG,1FKG)=((ψFH)G,1FKG)=(ψiG,1FKG).

However, 1FKG=(1FKFH)G, and the irreducible constituents of 1FKFH are exactly those characters ϕIrr(FH) whose kernel contains FK. Thus if (ψG,1FKG)0, then for some ϕIrr(FH/FK), (ψiG,ϕG)0. However, by [17, Chapter XI, Lemma 5.3], (ϕG,ψiG)0 if and only if ϕ=ψi or ϕ=ψ¯i. We know that ψi is an extension of a nontrivial irreducible character of FK/F; hence neither ψi nor ψi¯ can contain 𝐾 in its kernel. Thus (ψiG,1FKG)=0 for all 𝑖 in the above sum; hence (ψG,1FKG)=0, and d(U,G)4. Thus it is 5, and we are done.

In [14], we proved that subgroups of type (i) and (ii) have disjoint conjugates. In (iii), we proved above this property for subgroups of type (iii). Conversely, if a non-normal subgroup has a disjoint conjugate, then by the second part of Theorem 2.6, it is of ordinary depth 3. ∎

5 Final remarks

Since for subgroups 𝐿 of ordinary depth 3 in an arbitrary group 𝐺, it is not true in general that 𝐿 has a disjoint conjugate, it is natural to ask the following.

Question 5.1

Let LG be a subgroup of a finite group 𝐺 of ordinary depth 3. Is it true that CoreG(L)=LLx1Lx2 for suitable elements x1 and x2?

We are also interested in a possible converse of Corollary 1.6.

The following group is defined in [16, Example 6.7, Chapter VIII].

Definition 5.2

Let F=GF(2n), and let 𝜃 be an automorphism of 𝐹. Let A(n,θ) be the set of matrices of the form

u(a,b)=(100a10baθ1)

with a,bF. Then

u(a,b)u(a,b)=u(a+a,b+b+a(aθ)),
u(a,b)-1=u(a,b+a(aθ)).
Then A(n,θ) is a group of order 22n with unit element u(0,0).

The following theorem characterizes Suzuki 2-groups; see [16, Theorem 7.9, Chapter VIII].

Theorem 5.3

Let 𝑆 be a Suzuki 2-group; then

  1. S=Φ(S)=Z(S)={xSx2=1};

    1. either 𝑆 is isomorphic to a group A(n,θ) for some non-identity automorphism 𝜃 of GF(2n) of odd order and |S|=|Z(S)|2=22n,

    2. or |S|=|Z(S)|3.

Remark 5.4

Sylow 2-subgroups of simple Suzuki groups Sz(q) are isomorphic to certain A(2m+1,θ), where θAut(GF(22m+1)) (acting as xθ=x2m+1), which is of odd order, thus falls into the category (i) in the previous theorem. However, not every A(n,θ) is isomorphic to a Sylow 2-subgroup of a simple Suzuki group.

Question 5.5

Let 𝑆 be a 2-group, and let HAut(S). Suppose that S1S is a Suzuki 2-group, which is isomorphic to a Sylow 2-subgroup of a simple Suzuki group Sz(q). Let H1Aut(S1) be a solvable group of automorphisms, acting on the involutions of S1 transitively, and let H1H. Let us consider the semidirect product SH. Suppose further that S=xSHS1x and S1S1x is either trivial or the common center of them. Is it true that 𝑆 is a Suzuki 2-group?

Remark 5.6

From Theorem 5.3, it follows that, in Question 5.5, Aut(S) acts transitively on the involutions of 𝑆, so this conjecture is a weakened form of a conjecture of Gross on finite 2-groups 𝑆 with more than 1 involution admitting a group of automorphisms that transitively permutes the involutions of 𝑆, called 2-automorphic 2-groups; see [11]. According to the results of Gross, these groups fall into three classes: (a) homocyclic, (b) of exponent 4 and class 2 of order |Z(S)|2 or |Z(S)|3 and S=1(S)=Z(S) has exponent 2, or (c) of exponent 8 and class 3, S is homocyclic of order |Z(S)|2 and exponent 4, and S=1(S)=Z(S)=[S,S] has exponent 2. The conjecture was that the groups of type (b) are Suzuki 2-groups and those of type (c) do not occur. In the paper [26], it was proved that 2-automorphic 2-groups are of class at most 2, so case (c) does not occur. In the paper [5], it was proved that 2-automorphic 2-groups of type (b) are Suzuki 2-groups if |Z(S)|3=|S|. The other case in type (b) is still open. Thus, in Question 5.5, the group 𝑆 must be of type (b), and in the case |Z(S)|3=|S|, the question has a positive answer. In [5], it was also proved that the automorphism group of a Suzuki 2-group is always solvable. In [27], it was proved that, in Suzuki 2-groups 𝑆 of order |Z(S)|2, any two elements of the same order are conjugate in Aut(S).

Now we summarize the conjectures. Taking into consideration the results of Higman, Shult, Gross, Bryukhanova and Wilkens, the reformulated conjecture is the following: if a finite group 𝐺 acts on a non-abelian finite 2-group 𝑆 with more than 1 involution and 𝐺 is transitive on the involutions of 𝑆, then 𝐺 is solvable.

So far, it has been proved that if a finite group 𝐺 acts on a non-abelian finite 2-group 𝑆 with more than 1 involution, then if there is a solvable subgroup G1G that acts on the involutions of 𝑆 transitively, then 𝐺 is solvable. We ask that if 𝑆 has some more properties, namely we suppose that 𝑆 has a special kind of covering of Question 5.5, then 𝐺 is solvable.

Award Identifier / Grant number: 115288

Award Identifier / Grant number: 115799

Funding statement: The first author was supported by the Stipendium Hungaricum PhD fellowship at the Budapest University of Technology and Economics. The second and the third author were supported by the NKFI Grants No. 115288 and 115799.

Acknowledgements

We are grateful for the referee for the helpful comments and suggestions.

  1. Communicated by: Michael Giudici

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Received: 2020-03-23
Revised: 2020-11-24
Published Online: 2020-12-15
Published in Print: 2021-05-01

© 2020 Janabi, Héthelyi and Horváth, published by De Gruyter

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