Complements of the socle in almost simple groups

Complements of the Socle in Almost Simple Groups.

A. LUCCHINI(*) - F. MENEGAZZO(**) - M. MORIGI(***)

Assume that a finite groupHhas a unique minimal normal subgroup, sayN, and thatNhas a complement inH. We want to bound the number of conjugacy classes of complements ofN inH; in particular we are loo- king for a bound which depends on the order of N. When N4socH is abelian, the conjugacy classes of complements ofNin Hare in bijective correspondence with the elements of the first cohomology group H¹(H/N,N). Using the classification of finite simple groups, Aschbacher and Guralnick [1] proved that NH¹(H/N,N)N E NNN; therefore, when socH4Nis abelian, there are at mostNNNconjugacy classes of comple- ments ofNinH. To study the case whenN4socHis nonabelian we can employ a result proved by Gross and Kovács ([6], Theorem 1): there exists a finite group K containing a (non necessarily unique) minimal normal subgroupS which is simple and nonabelian (indeed Sis isomor- phic to a composition factor ofN) and there is a correspondence between conjugacy classes of complements of N in H and conjugacy classes of complements ofS inK. Using this result it is not difficult to prove that there exists an absolute constantcG4 such that the number of conjuga- (*) Indirizzo dell’A.: Dipartimento di Matematica, Università degli Studi di Brescia, Via Valotti n. 9, 25123 Brescia, Italy. E-mail: lucchiniHing.unibs.it

(**) Indirizzo dell’A.: Dipartimento di Matematica, Pura ed Applicata, Uni- versità di Padova, Via Belzoni n. 7, 35131 Padova, Italy.

E-mail: federicoHmath.unipd.it

(***) Indirizzo dell’A.: Dipartimento di Matematica, Università di Bologna, Piazza di Porta S. Donato n. 5, 40126 Bologna, Italy.

E-mail: mmorigiHdm.unibo.it

Investigation supported by MIUR (project Teoria dei gruppi e applicazioni) and the Universities of Bologna (funds for selected research topics), Brescia and Padova.

cy classes of complements ofNinHis at mostNNN^c(see, for example, [9]

Lemma 2.8). We conjecture that one can takec41 , as occurs when Nis abelian.

In this paper we deal with this conjecture in the case of finite almost simple groups. LetG be a finite simple group. As G`Inn (G), we may identify G with Inn (G). We will prove the following

THEOREM. Let G be a finite non-abelian simple group and assume that HGAut (G) contains G. Then the number of conjugacy classes of complements of G in H is less than NGN.

WhenG4Alt (n) withnc6 orGis a sporadic simple group, it is well known thatNH:GNG2 ; ifHcG, then the complements ofGinHare in bijective correspondence with the involutions ofHwhich are not contai- ned inG; hence the number of complements forGinHis strictly smaller thanNGN. The case G4Alt ( 6 )`PSL ( 2 , 9 ) is dealt with as a group of Lie type.

We may now assume thatGis a finite simple group of Lie type over a fieldK4GF(p^m) of orderp^m, for some primep. We will follow the defi- nitions and notations of the book [4], unless otherwise stated. SoG will be a group of the formG4S_l(q) wherelis the Lie rank ofGandq4p^m, for some prime p.

Also,fdenotes the Frobenius map andGdenotes the group of graph automorphisms of G.

IfG has no complement in Aut (G) there is nothing to prove, so we may assume that there exists CGH such that H4GC and GOC41.

Then we have that C is isomorphic to a subgroup of Out (G), whose structure is well known. In particular,Cis at most 3-generated. Also, if x,y,zare generators ofCandC8is any other complement forGin H, then C8 is generated by three elements of the form xu1,yu2,zu3 sati- sfying the same relations as x,y,z and with u_iG, for i41 , 2 , 3 .

In the whole paper, C will be a fixed complement for G in H.

1. Preliminary results.

We collect in this section some results which will be very useful in the sequel. The first is actually a corollary of Lang’s theorem, in the general form proved by Steinberg.

PROPOSITION 1.1. Let G be an untwisted finite simple group of Lie type over the field K with p^m elements. Let f^raAut (G), with a InnDiag (G)G,and assume thatNf^raN 4m/r.If xInnDiag (G)is such that Nf^raxN 4m/r then f^ra and f^rax are InnDiag (G)-conjugate.

PROOF. Let G4S_l(p^m) and letGbe the connected algebraic group over the algebraic closure K of K such that G is adjoint and G4 4O^p8(C_G(f^m) ) (see [4, Theorem 2.2.6 (e)]). By Lemma 2.5.8. (a) of [4] we have that InnDiag (G)4C_G(f^m).

Let tx be the inner automorphism of G induced by x. There exists aAut (G) such that a is the product of a graph automorphism and an inner automorphism, and a induces a on G. We note that (f^ra)^m/r4 4(f^ratx)^m/r4f^m. Sof^rais a surjective homomorphismcofGwhose set of fixed points in G is finite. By the Lang-Steinberg theorem (see [Theorem 2.1.1] [4]) there exists wG such that x²¹4w²¹w^fra. Let s4^m

r . We have that: f^m4(ctx)^s4c^stxc^s21t_x^cs22Rt_x^ct_x4 4f^mt_x^cs21t^c_x^s22Rt^c_xt_x, so t_x^cs21t_x^cs22Rt^c_xt_x41 . As x4(w²¹)^cw we obtain that (t_w²¹)^cstw41 , so t^f_w^m4tw, that is wInnDiag (G).

It follows that (f^ra)^w4w²¹f^ra w4f^ra(w²¹)^fraw4 4f^ra(w²¹w^fra)²¹4f^rax, as we wanted to prove.

We will also need a lemma proved in [8].

LEMMA 1.2. Let G be a finite simple group of Lie type, and let a Aut (G) then there exists gG such that NaNcNagN.

Our first results are easy consequences of the proposition and lemma above.

PROPOSITION 1.3. Let G be a finite simple group of Lie type, GG GHGAut (G)and assume that a complement C for G in H is cyclic. Then the number of complements for G in H is less than NGN.

PROOF. IfC4aab, then any other complementC8is generated by an element of the form ag, with gG and NagN 4 NaN, and lemma 1.2 applies.

COROLLARY 1.4. Let G be a finite simple group of one of the follo- wing types:³D₄(q),G₂(q),F₄(q),E₈(q),²F₄(q)or²G₂(q)and let GGHG GAut (G). Then the number of complements for G in H is less than NGN.

PROOF. By Theorem 2.5.12 of [4] the groups listed above have cyclic outer automorphism group, so proposition 1.3 applies.

PROPOSITION 1.5. Let G be an untwisted finite simple group of Lie type over the field K. Assume that C4af^ra,bb, with aInnDiag (G), bInnDiag (G)G0Inn (G) and Nf^raN 4 Nf^rN. Then the number of G- conjugacy classes of complements for G in H is less than NGN.

PROOF. IfC8is another complement forGin H, then the first gene- rator ofC8is of the formf^rag, withgGandNf^ragN 4 Nf^raN 4 Nf^rN, so by proposition 1.1 we have at mostd4 NInnDiag (G) :GNchoices for it, up toG-conjugation. Moreover, again by proposition 1.1, we may assume that f^rag4(f^r)^x for some xInnDiag (G). So C8 4af^r, (bv)^x21b^x, for somevG. We now need to count the choices for the second generator, which is of the form (yu)^x, wherey4b^x21andv4u^x21. By lemma 1.2 we have less than NGN choices for u, as NyuN 4 NyN. Moreover, as we are counting G-conjugacy classes of complements, we may count the ele- ments of the formyu up to conjugation by elements of the centralizer of f^rinG. IfG4Sl(q) thenSl(p)GCG(f^r). We have that [yu,Sl(p) ]c1 (see [Lemma 2.5.7] [4]), so thatC_Sl(p)(yu) is a proper subroup ofS_l(p).

As the index of a maximal subgroup ofS_l(p) is at leastd(see Table 5.2 A of [p. 175] [7]) each orbit of the set]yuNuG(under the action ofSl(p) by conjugation has at least d elements. This concludes the proof.

PROPOSITION 1.6. Let G be an untwisted finite simple group of Lie type over the field K. Assume that C4af^ra,bb, with a,b InnDiag (G)G, InnDiag (G)GH andNf^raN 4 Nf^rN.Then the number of G-conjugacy classes of complements for G in H is less than NGN.

PROOF. If C8 is another complement, by proposition 1.1 we may as- sume that the first generator ofC8 is (f^ra)^x, for somexInnDiag (G).

LetC8 4a(f^ra)^x,bub, whereuG. As InnDiag (G)GH4GC8we have thatx4zyfor somezGand someyC8, so thatC8 4a(f^ra)^z, (bu)^y21b isG-conjugate to a complement of the formC9 4af^ra,vb. It follows that the first generator ofC8is uniquely determined, up toG-conjugation. By lemma 1.2 the number of choices for the second generator ofC8are less than NGN, and the conclusion follows.

We recall that ifaH, thena is of one of the following types: inner, inner-diagonal, graph, field or graph-field (see [4], definition 2.5.13).

PROPOSITION 1.7. Let G be a finite simple group of Lie type over the field K.Assume that C4aa,bb,where the type of a is known and b nor- malizesaab.Then the number of conjugacy classes of complements for G in H is bounded by rs,where r is the number of G-conjugacy classes of elements of H of the same type and order as a and s is the order of a ma- ximal subgroup of G.

PROOF. If C8is another complement forG inH, we have that C8 4 4aau,bvb, for someu,vG, whereNauN 4 NaN,NbvN 4 NbNand ifa^b4a^t for some integert, then (au)^bv4(au)^t. There are at most rchoices for au, up to G-conjugacy. Moreover, any two elements bv8 and bv9 such that (au)^bv84(au)^bv94(au)^tsatisfy (bv8)²¹bv9C_G(au), so there are at most NC_G(au)N choices for the second generator, and the conclusion follows.

2. The special linear groups.

LetKbe the finite field with qelements, withq4p^mfor some prime number p. As usual GL(n,q) (resp. SL(n,q)) will denote the general (resp. special) linear group of degreenover the fieldK. In the following we will identify the multiplicative group K³ of K with the subgroup of GL(n,q) consisting of scalar matrices. Then PGL (n,q)4 4GL (n,q) /K³, PSL(n,q)4SL(n,q)K³/K³ and if gGL (n,q) its image in PGL (n,q) will be denoted withg. Also, as usual, det (g) will in- dicate the determinant of a matrix g and diag(a1,R,an) will denote a diagonal matrix, whose entries on the diagonal are those listed between the brackets.

In the whole section, we will considerG4A_n21(q)4PSL (n,q), for nandq fixed. Letfbe the Frobenius automorphism of GL (n,q), given by: (aij)^f4(aijp), for i,j41 ,R,n.

Let t: GL (n,q)KGL (n,q) be the automorphism defined by g^t4 4(g^!)²¹, where g^! denotes the transposed matrix of g.

Bothfandtinduce automorphisms of PGL (n,q), which we will still indicate byfandt.fgenerates the group of field automorphisms,tis a graph automorphism ifnF3 , and it is an inner automorphism if n42 . Also, PGL (n,q) /G is cyclic of order d4(n,q21 ).

We have that C is isomorphic to a subgroup of Out (G)4 4afG,tG,aGb, where aPGL (n,q), (aG)^fG4a^pG, (aG)^tG4a²¹G, [fG,tG]41 and NaGN 4d, NfGN 4m, NtGN 42 .

Case A: C is 3-generated

In this caseChas the groupZ23Z23Z2as an epimorphic image and d is even, so that p is odd and nF4 is even.

We may assume that C4af^rN₁,tM₁,U₁b, where M₁,N₁,U₁ GL (n,q) andrNm. Also we have thatU₁has orderd8, with 2Nd8 Ndand we also have that (f^rN₁)^m/raU₁b.

LEMMA 2.1. In the above setting, we may also assume that [f^rN₁,tM₁]41 and tM₁ has order 2 .

PROOF. AsCis isomorphic to a subgroup of Out (G), it will be isomor- phic to a subgroup T of the group X4aa,b,cNa^d4b²4c^m41 , a^b4 4a²¹,a^c4a^p,b^c4bbwherepis a prime andp^mf1 modd . SinceTis not 2-generated, TO aa,bb and Taab/aab are not cyclic; in particular m is even. Set aa^lb4TO aab. If bT, easy calculations prove that T4 4aa^l,b,c^kbwhere botha^landc^khave even order. Assume thatbTand baT. Note thatC_X(ba)4aa^d/2,ba,ubwhereu4ca²

p21

2 . Similar com- putations prove thatT4aa^l,ba,u^kb, wherel is even, and the orders of a^land ofu^kaabare even. As any subgroup ofXwhich is not 2-generated is aab-conjugate to a subgroup containing either b or ba, the result follows.

OBSERVATION. With the notation of lemma 2.1 we note that it is pos- sible that T does not split over TO aa,bb. Namely, T4aa^l,ba,u^kb is not 2-generated and does not split overTO aa,bbiffpc2 ,l,d,m/kare even, ^p

m21

d is odd, the order of a^l is divisible by 4 , and finally r₂E Emax ( (p^k21 )₂, (p^k11 )₂) where we denote by x₂ the 2-part of the integerx. Also, if Tdoes not split over TO aa,bbwe have that u^m has order 2 .

CASE I: (f^rN₁)^m/r41

We may assume that another complementC8forGinHis generated by f^rN1X, tM,U, with XG, M,UPGL (n,q), satisfying the same relations as f^rN1,tM1,U1. In particular (f^rN1X)^m/r41 , so by pro- position 1.1 there are at most d possibilities for the choice of f^rN₁X, up to conjugation by elements of G. Moreover, again by proposition 1.1, we have that f^rN₁X4(f^r)^S, with SPGL (n,q). Changing no-

tations for the last two generators, we may now assume thatC8 4a(f^r)^S, (tM)^S, (U)^Sb.

We now have to count how many possibilities there are for the other two generators. From the fact that tM has order 2 it follows that M^tM41 , so M^!4aM, with aK and as (M^!)^!4M we have that a²41 , so that M is symmetric or skew-symmetric.

From the fact that [f^r,tM]41 it follows that M^fr4bM, with b K³. This implies thatmijp^r21

4bfor eachi,j41 ,R,nsuch thatmijc c0 . Chooseh,ksuch thatmhkc0 . Thus, for eachi,j41 ,R,nwe have thatm_ijm_hk²¹GF (p^r) , that ism_ij4m_hkm^ij8for somem^ij8GF(p^r). It fol- lows thatM4m_hkM8, withM8GL (n,p^r). ChoosingM8instead ofM as a pre-image of M we may assume that MGL (n,p^r).

As we are counting conjugacy classes of complements, we note that to count the possibilities for the second generator ofC8 we are still free to conjugate it by an elementHofGcentralizingf^r, that isHSL (n,p^r).

Note that in that case we have that (tM)^H4tH^!M H, and by [3] there existsHGL (n,p^r) such thatH^!MH has one of the following forms:

identity, diag (a, 1 ,R, 1 ), where a is a non-square in GF (p^r), or a block-diagonal matrix whose blocks on the diagonal are all equal to

g

h

As we are allowed to conjugate by matrices in PSL (n,q) and not in PGL (n,q), we have at most 3d possibilities for M.

We now count the number of choices for U. We have that (U)^tM4 4(U)²¹, soU^!M4gU, withg²41 , and we have at mostq^{n(n11 ) /2}possibili- ties forUfor each choice ofg. So we have at most 2q^{n(n11 ) /2}/(q21 ) pos- sibilities for U, and thus at most ^6d

3q^{n(n11 ) /2}

(q21 ) E NGN possibilities forC8, as 6qⁿ¹¹E(q³21 )(qⁿ21 ) for nF4 and qF9 .

CASE II: (f^rN1)^m/rc1 In this case ^m

r is even. Actually, if ^m

r is odd, putting x4tM₁, y4f^rN₁, ifm42^ts, with ^m

r Ns, then C4ax,y^2t,y^s,Ub4axy^2t,Ub, as y^say^m/rbaUb. So C is 2-generated, contradicting the assumptions.

Again, we may assume that another complementC8 is generated by f^rN,tM,U, satisfying the same relations asf^rN₁,tM₁,U₁. In particu- lar (tM)²41 . As in Case I, it follows thatMis symmetric or skew-sym- metric, and conjugating by a suitable element of PSL (n,q) we have at

most 3dpossibilities forM. Namely, we may assume thattMis of one of the following types:

i) t^S,

ii) (tA)^S, withA4diag (a, 1 ,R, 1 ), whereais a non-square inK, iii) (tB)^S, where Bis a block-diagonal matrix whose blocks on the diagonal are all equal to

g

1 0

h

Changing notations for the generators, we may assume that C8 4 4a(f^rN)^S, (tM)^S, (U)^Sb, withM]I,A,B(. Also, there is no loss in gene- rality in assuming S41 , as this does not affect calculations.

We now consider the generatorf^rN. Letm4det (N) and (f^rN)^m/r4 4L.

In cases i) and iii) we have that [tM,f^rN]4[tM,N]41 , so that N^tM4N. It follows that (N²¹)^!M4gN, withgK³, and m²Kⁿ(here Kⁿ is the set of elements of K which are n-th powers).

As ^m

r is even and p is odd it follows that 2N^(p

r)^m/r21

p^r21 , so that det (L)4m

(p r)m/r 21

p r21 Kⁿ, which implies that (f^rN)^m/rCOG41 and (f^rN₁)^m/r4(f^rN)^m/r41 , a contradiction.

We now deal with case ii). From [(tA), (f^rN) ]41 it follows that N^tA4A^frN, so N^2!4gA^frNA²¹, with gK³ and m²4a^12prg²ⁿ.

As before, det (L)4m

(p r)m/r 21

p r21 fa^{( 12pr)}

p^m21

2(p^r21 )f21 modulo Kⁿ, so thatL²41 (note that ^q21

d is odd, as it is stated in the observation after lemma 2.1).

We distinguish two subcases:

a) rG^m

4 . We first bound the choices for the generator of the form f^rN. By [p. 52] [5]f^rBandf^rCare conjugate in GL (n,q) if and only if (f^rB)^m/r and (f^rC)^m/r have the same property, so we need to count PGL (n,q)-conjugacy classes of involutions (f^rN)^m/r PGL (n,q)0PSL (n,q). By Table 4.5.1 of [4] there are at mostn/2 choi- ces for (f^rN)^m/r, which means at most ⁿ

2 PGL (n,q)-conjugacy classes of elements of the formf^rN, that is at mostdⁿ

2 choices forf^rN, up to PSL (n,q)-conjugation.

Now once we have chosen an elementtVas a second generator, from the fact that (f^rN)^tV4f^rNit follows that all the other possible choices

for the second generator are of the form tV U, where UC_G(f^rN).

LetKthe algebraic closure ofK. By the Lang-Steinberg theorem [p.

32] [2] we have that f^rN is conjugate to f^r in PGL (n,K), so NCPSL(n,K)(f^rN)N 4 NPGL (n,p^r)N. So we have at most NPGL (n,p^r)N choices for tV.

By our hypothesis, there existsRsuch that (tV)^R21is of the formtA, with A4diag (a, 1 ,R, 1 ), where a is a non-square in K.

We may assume that the third generator is of the form (U)^R. We have thatU^R(tV)4(U)^R(tA)R4(U^tA)^R, and as (U^R)^(tV)4(U²¹)^R, it follows that U^!A4U, that is U^!A4gU, with g]61(.

This means that, fixed g, U is determined by its entries along and above the diagonal, so we have at most 2q

n(n11 )

2 choices forU, and at mo- st ²

q21q

n(n11 )

2 choices for U.

Putting all together, the number of conjugacy classes of complements forGinHis at mostGdn

2 NPGL (n,p^m/4)N ² q21q

n(n11 )

2 E NPSL (n,q)N. (Here we have used that 8Nn, because m is even, so that 8Nq21 and

q21

d is odd).

b) r4^m₂ . We first bound the choices for the generator of the form f^rN.

As (f^m/2N)²4L has order 2 , the canonical form ofL is either a dia- gonal matrix whose entries on the diagonal are in the set]6g(, for some gK³(first type), or it is a block-diagonal matrix, whose blocks on the diagonal are all equal to

g

g

h

[5], by conjugating by a suitable element of GL(n,q) we may assume that N is block-diagonal matrix, whose blocks N_i on the diagonal are of the form

N_i4

. `

` `

` `

´

0 1 0 QQQ 0

. . . QQ Q

1 QQ Q . . .

. . .

QQ Q QQ Q 0

0 QQQ QQQ 0 1

a_{i, 1}

a_{i, 2}

QQQ QQQ ai,m_i

ˆ `

` `

` `

˜

.

So we may assume that alsoLis a block-diagonal matrix, whose blocksL_i on the diagonal have dimension m_i.

We now want to prove that the canonical form of L is diagonal.

Ifm_jF5 for somejit is easy to see thatL_jcannot have order 2 . Also, if the canonical form ofLis of the second type, then 2Nm_jfor eachj. Now assume thatm_j42 for somej. AsL_j²is a scalar matrix,L_jis of the form L_j4

g

y

2x

h

g

b

a

h

g

ab^pm/2

b1a^pm/211

h

42b^pm/211is a square (note that21 is a square) and it follows thatL_jis diagonalizable.

To conclude, assume thatm_j44 for eachj. We have thatL_j is of the form

. `

` `

´

1 1

1 a^pm/2 b^pm/2 c^pm/2 d^pm/2

ˆ

` `

`

˜

. `

´

1 1

1 a b c d

ˆ `

˜

4

.

` `

`

´

1 1

a^pm/2 b^pm/2 c^pm/2 d^pm/2

x x x x

ˆ

` `

`

˜

.

So the first column ofLj2is

. `

´

a^pm/2 b^pm/2 c^pm/2 d^pm/2

ˆ `

˜

, which implies thatb4c4d40 . So

L_j4

. `

´

a

ˆ

` ˜

and L_j²4diag (a^pm/2,a,a^pm/2,a).

AsL²is a scalar matrix it follows thata^pm/24aandais the same for all blocks L_j. We have a4l^{u(pm/211 )}, for some integer number u, and detL4(a²)^n/44l^{u(n/2 )(pm/211 )}, which leads to a contradiction because dNⁿ

2(p^m/211 ).

It follows that L is diagonal.

So we have at most ⁿ

2 choices for Land thus at most ⁿ

2 choices for f^m/2N, up to PGL (n,q)-conjugation. As we are counting PSL (n,q)-con- jugacy classes we have to multiply this number by d.

We may also assume thatL4(L₁,L₂) is a block diagonal matrix with 2 blocks on the diagonal of the formL₁4gI_r1andL₂4 2gI_r2, for someg

inK³, wherer₁1r₂4n. We note thatr₁andr₂are both odd, otherwise det (L)4gⁿ contradicting the fact that LPSL (n,q). Moreover, as 8Nn, we have that r1c ⁿ

2 cr2.

We have thatM,N andU centralizeL, so we may assume that they are all block-diagonal matrices, with M4(M₁,M₂), N4(N₁,N₂) and U4(U₁,U₂). (Note that if L^S4L then L^S4aL for some aK³, but looking at the eigenvalues ofL and keeping in mind that r_icⁿ

2 , it fol- lows that a41 , that is S centralizes L).

By proposition 1.1, we have that f^m/2Ni is conjugate to f in PGL (r_i,q), and so f^m/2N is conjugate to fD in PGL (n,q), with D4 4(I₁,bI₂) for some bK³.

We now work separately on the two blocks, using exactly the same strategy as in case I.

We may assume thatM14jM18, withjK³andM18GL (r1,p^m/2).

MoreoverM1is symmetric (note thatr1is odd). By conjugating with ele- ments of GL (r₁,p^m/2) we find that there are at most 2 choices for M₁8, and at most 2(q21 ) choices up toSL(r₁,p^m/2)-conjugation. So there are at most 2(q21 )²choices forM₁8j. Arguing in the same way forM₂and taking images in PGL (n,q) we obtain that there are at most 4(q21 )³ choices for M.

The number of choices forU_i is now at most q^{ri(ri11 ) /2}(note that the elementgappearing in case I is now forced to be 1 , asr_iis odd). So there are at most q^{r1(r111 ) /2}q^{r2(r211 ) /2}/(q21 ) possibilities for U.

So we have at most ⁿ

2d4(q^{r1(r111 ) /2}q^{r2(r211 ) /2})(q21 )²E NPSL (n,q)N choices for C.

Case B: C is 2-generated

We may assume thatC4af^rN₁,t^ef^sM₁b, whereM₁,N₁GL (n,q) ande]0 , 1(. We may also assume that any other complementC8is ge- nerated by f^rN,t^ef^sM, satisfying the same relations as f^rN1,t^ef^sM1.

CASE I: CGO InnDiag (G)G, (f^rN1)^m/r41 In this case we apply proposition 1.5.

CASE II: CGO InnDiag (G)G, (f^rN₁)^m/r4L₁c1 , nF3

Letu4 NL₁N. We now want to count PSL (n,q)-conjugacy classes of elements of the form f^rN. By [p. 52] [5] f^rAandf^rBare conjugate if and only if (f^rA)^m/r and (f^rB)^m/rare conjugate, so we need to bound the number of PGL (n,q)-conjugacy classes of elements L of order u, and then to multiply this bound byNPGL (n,q) : PSL (n,q)N 4d. AsL^uis a scalar matrix,L is conjugate to a block-diagonal matrix Xwhose blocks X_i have all the same dimension k and are of the form:

Xi4

. `

` `

´

1 1

QQ Q 1

ci

ˆ

` `

` ˜

, (1)

where ci4cei and eid

41 . We may also assume c14c.

Ifk41 then there are at most (q21 )dⁿ²¹choices forXand thus at most dⁿ²¹ choices for L, up to PGL (n,q)-conjugacy.

If kD1 there are at most (q21 )d

n k21

choices for X.

So, summing over all k’s, the choices for L are at most

dⁿ²¹1

!

1EkNd

(q21 )d

n k21

. (2)

Note that dⁿ²¹1

!

1EkNd

(q21 )d

n k21

G(q21 )^d

n21

21 d21 .

We now have that L^tefsM4L^t. Once we have fixed one element M with that property, all the others can be obtained by multiplyingMby an element of the centralizerZ of L^tefsin PSL (n,q), and we may assume without loss of generality that L^tefs has prime order u.

Using theorems 4.8.1, 4.8.2 and 4.8.4 of [4] foruodd and Table 4.5.1 of [4]

foru42 and some easy calculations it is possible to see that an upper bound for the order ofZisNGL (n21 ,q)N. We now have to check that d(q21 )^d

n2121

d21 NGL (n21 ,q)N E NPSL (n,q)N, which is true fornF4 becaused²(q21 )^{2 d}

n2121

d21 E(qⁿ21 )qⁿ²¹. Forn43 we use the more accurate bound (2).

CASE III: CGInnDiag (G)G,nF3.

IfCis cyclic we conclude by proposition 1.3. Otherwise we first choo- se a generator forC8OInnDiag (G), so that the number of possibilities is bounded by (2.6), then we argue as in case II.

CASE IV: n42

IfCis cyclic we conclude by proposition 1.3, otherwise we first choose a generator forC8OInnDiag (G), for which there is at most one possibi- lity, by Table 4.5.1 of [4], and by lemma 1.2 there are less thanNGNchoi- ces for the second generator.

3. The unitary linear groups.

In this section, we will consider the group G4²A_n21(q)4 4PSU (n,q), for n and q fixed.

Let K4GF (q²) be the finite field withq²elements, withq4p^m for some prime numberp. We fix a generatorlof the multiplicative group of the field K³. Then GU (n,q) (resp. SU (n,q)) will denote the general (resp. special) unitary group of degree n, that is GU (n,q)4 ]g GL (n,q²)Ng(g^!)^s41( where s4f^mAut (GL (n,q²) ), and SU (n,q)4 ]gGU (n,q)Ndet (g)41(. All other notations, unless otherwise specified, are as in the previous section.

We may assume thatCis non-cyclic, otherwise we conclude by propo- sition 1.

LetC4af^rN₁,U₁b, withU₁,N₁PGU (n,q). We argue as in case B II of the special linear group.

We have thatUis GL (n,q²)-conjugate to a block-diagonal matrixX whose blocksX_i have all the same dimension kand are of the form (1), where c_i4ce_i, e_i^d41 and we may also assume that c₁4c.

By [10, p. 34] the matrix X as above is conjugate to an element of GU (n,q) if and only if it is similar to the matrix ((X^!)^s)²¹.

So cei4(ce_j)^2q, for some j, which implies that c^q114(e_ieqj)²¹ and c^{(q11 )2}41 . Letc4l^u. We have thatq²21Nu(q11 )², soq21Nu(q11 ).

As (q11,q21)G2, it follows that ^q21

2 Nuand there are at most 2(q11) choices forc. Moreover, again by [10, p. 34] two matrices are conjugate in GU (n,q) if and only if they are conjugate in GL (n,q²), so it

is enough to count the number of choices for the matrix X as above, and then to multiply by d4 NPGU (n,q) : PSU (n,q)N.

As in case B II of the special linear group, the choices for X are at most

dⁿ²¹1

!

1EkNd

2(q11 )d

n k21

. (3)

Note that dⁿ²¹1

!

1EkNd

2(q11 )d

n k21

G2(q11 )^d

n2121 d21 .

To bound the number of choices for the second generator, we look for an upper bound for the order of the centralizerZofUin PGU (n,q). We may assume that U has prime order u.

We first assume that (n,q)]( 3 , 2 ), ( 3 , 5 ), ( 4 , 3 ), ( 8 , 3 )(.

Using theorems 4.8.1, 4.8.2 and 4.8.4 of [4] foruodd and Table 4.5.1 of [4] foru42 and some easy calculations it is possible to see that an upper bound for the order of Z is NGU (n21 ,q)N.

So we have to prove that 2d(q11 )dⁿ²¹21

d21 NGU (n21 ,q)N E ENPSU (n,q)N.

As ^d

d21G2 , this is true because 4(q11 )²dⁿE(qⁿ21 )qⁿ²¹. If (n,q)4( 8 , 3 ) we use the more accurate bound (3) and the fact that NZN G NGU (n21 ,q)N.

We now study the remaining cases.

I: Case (n,q)4( 3 , 2 ),d43 is divided into 2 subcases according asU is diagonalizable or not. For each case, we have to consider the possible canonical forms forUand the order of their centralizers, and the result follows just by counting the possible choices.

II: Case (n,q)4( 3 , 5 ), d43 .

There are at most 15 possibilities for the choice of X and 15Q Q3NGU ( 2 , 5 )N E NPSU ( 3 , 5 )N.

III: Case (n,q)4( 4 , 3 ),d44 is divided into 2 subcases according as NUNis equal to 2 or 4 . For each case, we have to consider the possible ca- nonical forms for U and the order of their centralizers, and the result follows just by counting the possible choices.

4. B_l(q), C_l(q) and E7(q).

LetG]B_l(q),C_l(q),E₇(q)(. We have thatCis isomorphic to a sub- group C of Z₂3Z_m, with Z_m4afGb and Out Diag (G)GZ₂.

Then eitherCis cyclic, and we may apply proposition 1.3, or it is 2-ge- nerated, and it is possible to choose one generator of the formf^rz, with zG and (f^rz)

m

r 41 , so proposition 1.5 applies.

5. D_l(q),lc4.

Case p42

In this case we have thatCis isomorphic to a subgroupCofZ₂3Z_m, with Zm4afGb and Out Diag (G)G4Z2, and we argue as for the case G4Bl(q) or Cl(q).

Case pc2

We have thatC and its image Cin Out (G) are isomorphic to a sub- group of D₈JZ_m, with the following notation: Z_m4afGb and Out Diag (G)GGD8. More precisely, ifl is odd and 4Nq21 or iflis even then Out Diag (G)G4D84awG,tGb, where tis the graph automorphi- sm of order 2 , w4wG has order 4 , w^t4w²¹, [t,f]41 , and w^f4w unless l is odd and 4=p21 , in which case w^f4w²¹.

Iflis odd and 4=q21 then Out Diag (G)G4axG,tGbis elementary abelian,tis the graph automorphism of order 2 , xInnDiag (G). Also f centralizes Out Diag (G).

LetT4COInnDiag (G)G, and letTbe its image in OutG. By propo- sition 1 we may assume thatCis not cyclic, and it is easy to check thatC splits over T.

Let CGO Out Diag (G)G.

I) Assume that it is possible to choose a generator ofCmodulo T of the form f^ra, with aInnDiag (G) and (f^ra)

m r 41 .

IfTis cyclic proposition 1.5 applies, so we may assume that Tis not cyclic.

IfC8is another complement, by proposition 1.1 we may assume that, up to InnDiag (G)-conjugacy, a generator of C8 modulo C8O OInnDiag (G)Gis (f^r)^x, for somexInnDiag (G), and we have at most NInnDiag (G) :GN G4 choices for it, up to G-conjugacy.

Tis generated by two involutionsu^xandv^x, that are of graph type or of inner-diagonal type, depending on which case we are considering. Mo- reover we may assume that u is of the form t^ey, with yG and e ]0 , 1(, and such that [t^ey,f^r]41 , soyD_l(p^r). We note that we may conjugate t^ey by elements of D_l(p^r), which centralize f^r.

From table 4.5.1 of [4] we deduce that both the number ofD_l(p^r)-con- jugacy classes of involutions of graph type and the number of D_l(p^r)- conjugacy classes of involutions of inner-diagonal type are bounded by 2(l13 ). So there are at most 2(l13 ) choices for u. Then we have to count the involutionsv of a fixed type. There are at most 2(l13 ) conju- gacy classes, and each class contains at most NInnDiag (G)G: CInnDiag (G)G(g)N G8NG:C_G(g)Nelements, wheregis any involution in the class considered. We chooseg such that the index of H4CG(g) in G is maximum. So there are at most 2(l13 )NG:HNpossibilities for the choi- ce ofv. So we just have to check that 4Q32(l13 )²NG:HN E NGN, which is true because 128(l13 )²E NHN (the structure of H is also described in table 4.5.1 of [4]).

II) Assume that we are not in the previous case, so thatCdoes not contain OutDiag (G)G; in particularNTN E8 . Letf^rzbe a generator ofC moduloTof order ^m

r , withzInnDiag (G)G0InnDiag (G). We have that m

r is even, otherwise we replacef^rzwith (f^rz)⁴, which is a generator of C modulo T of order ^m

r and of the form f^rx with xG.

IfT4aub has order 2 then we apply proposition 1. By table 4.5.1 of [4] we have at most 2(l13 ) conjugacy classes of involutions of the same type asu; moreover, by Table 5.2 A of [p. 175] [7] the index of a maximal subroup of G is less than 2(l13 ), so in this case the conclusion follows.

IfTis cyclic of order 4 , from the fact that we are not in case I it follo- ws that T4OutDiag (G) and we can conclude by proposition 1.6.

So we may assume that T is elementary abelian of order 4 . Iflis even thenT4OutDiag (G) and as we are not in case I it follows that f^rz does not centralize T, so we conclude by proposition 1.6.

Letlbe odd. Note that we also have that 8Nq21 , becausemis even.

As we are not in case I, one of the following occurs:

– z4t and T4aw²,wtb, or – z4wt and T4aw²,tb.

To deal with these cases we always adopt the same strategy. We first count the number of choices for a generator ofTOInnDiag (G), then we count the number of choices for a generator of T modulo TO OInnDiag (G), and finally we count the number of choices for a generator of C modulo T.

We describe the calculations in detail only for the first case.

LetC8be another complement of ofGinH; then we may assume that it is of the form C8 4af^rtu,w²v,wtxb, with u,v,xG.

By Table 4.5.1 of [4] we have at mostl21 choices forw²v, up to G- conjugacy. Moreover let C*4CInnDiag (G)(w²v) and L*4O^p(C* ). From table 4.5.1 of [4] it follows that

i) either L*4²D_l21(q) and Z4C_C*(L* )4CInnDiag (G)Gk(L* ) has order q11 or

ii) L*4D_i(q)3D_l2i(q) orL*4²D_i(q)32D_l21(q), where 2GiE E ^l

2 and Z4C_C*(L* )4CInnDiag (G)G_k(L* ) has order 2 .

We first deal with case ii). Note that wtxcentralizes w²v, so it nor- malizes L* . Let (y1,y2)Aut (^eDi(q) )3Aut (^eDl2i(q) ) be the image of wtx in Aut (L* ). The number of choices for wtx, up toG-conjugacy, is bounded by NZNr₁r₂, where r₁21 is the number of ^eD_i(q)-conjugacy classes of involutions in InnDiag (^eD_i(q) )G (we have to add one because y₁might be the identity) andr₂21 is the number of^eD_l2i(q)-conjugacy classes of involutions in InnDiag (^eDl2i(q) )G. Again by table 4.5.1 of [4]

we have that r1,r2G6l125 .

Note: Fori42 , 3 it is easy to check thatr₁,r₂G6l125 is still true (see [p. 11] [4] and [p. 43] [7] for the description of D_i in these cases).

So there are at most 2( 6l125 )² choices for wtx.

We now have to choosef^rtu. Note that once we have fixedf^rtuwith the required properties, any other element of the form f^rtu8 is such that (f^rtu)²¹f^rtu8C_G(w²v), so we have at mostNC_G(w²v)N choices for the third generator.

A similar argument applies to case i).

To conclude, we have that the number of complements forGinHis at most (l21 ) 2( 6l125 )²NUN, where U is a maximal subgroup of G, and this number is less thanNGN, as by Table 5.2 A of [p.175] [7] the index of a maximal subgroup ofG is at least ^(q

l21 )(q^l2111 )

q21 and 2(l21 )( 6l1 125 )²E^(q

l21 )(q^l2111 )

q21 (here lF5 and qF9).

Let CGOutDiag (G)G.

Then C is generated by two involutions u and v, that are of graph type or of inner-diagonal type, depending on which case we are conside- ring, and we argue as in Case I above.