Algebra and Logic, Vol. 41, No. 2, 2002

RECOGNITION OF FINITE SIMPLE GROUPS S4 (q) BY THEIR ELEMENT ORDERS V. D. Mazurov∗

UDC 512.542

Key words: finite simple groups, recognizability of groups by their element orders. It is proved that among simple groups S4 (q) in the class of finite groups, only the groups S4 (3n ), where n is an odd number greater than unity, are recognizable by a set of their element orders. It is also shown that simple groups U3 (9), 3 D4 (2), G2 (4), S6 (3), F4 (2), and 2 E 6 (2) are recognizable, but L3 (3) is not. For a finite group G, we denote by ω(G) a set of orders of elements of G. This set is closed under divisibility and hence is uniquely determined by a set µ(G) of elements in ω(G) which are maximal under the divisibility relation. A group G is said to be recognizable by ω(G) (shortly, recognizable) if every finite group H with ω(H) = ω(G) is isomorphic to G. In other words, G is recognizable if h(G) = 1 where h(G) is the number of pairwise non-isomorphic groups H with ω(H) = ω(G). A list of finite simple groups which are presently known to be or not to be recognizable is given under Sec. 7. In [1], in particular, it was proved that, for every simple symplectic group S = S4 (2m ) of dimension 4 over a field of order 2n , h(S) is infinite. The goal of this article is to prove that among S4 (q) with q odd, only the groups S4 (3n ), where n > 1 is odd, are recognizable. We also show that simple groups U3 (9), 3 D4 (2), G2 (4), S6 (3), F4 (2), and 2 E 6 (2) are recognizable but L3 (3) is non-recognizable. MAIN THEOREM. Let S = S4 (q) be a finite simple symplectic group of dimension 4 over a field of order q. If q = 32n+1 > 3 then S is recognizable. In all other cases, h(S) = ∞. The set ω(H) of a finite group H defines the Gruenberg–Kegel graph GK(H) whose vertices are prime divisors of the order of H, and two primes p and q are adjacent if H contains an element of order pq. Denote by s = s(H) the number of connected components in GK(H), and by πi = πi (H) an ith connected component, i = 1, . . . , s. For a group H of even order, let 2 ∈ π1 . Denote by µi = µi (H) (resp., by ωi = ωi (H)) a set consisting of n ∈ µ(H) (resp., of n ∈ ω(H)) such that every prime divisor of n lies in πi . 1. PRELIMINARY RESULTS LEMMA 1 [2]. If G is a finite group with disconnected graph GK(G) then one of the following holds: (1) s(G) = 2, G is a Frobenius group. (2) s(G) = 2, G = ABC, where A and AB are normal subgroups of G, B is a normal subgroup of BC, and AB and BC are Frobenius groups. ∗ Supported by RFFR grant No. 99-01-0550, by RF Ministry of Education grant No. E00-1.0-77, and by FP “Universities of Russia” grant No. UR.04.01.031.

Translated from Algebra i Logika, Vol. 41, No. 2, pp. 166-198, March-April, 2002. Original article submitted November 29, 2000. c 2002 Plenum Publishing Corporation 0002-5232/02/4102-0093 $27.00

93

(3) There exists a non-Abelian simple group P such that P ≤ G = G/K ≤ Aut(P ) for some nilpotent normal π1 (G)-subgroup K of G, and G/P is a π1 (G)-group. Moreover, GK(P ) is disconnected, s(P ) > s(G), and for every i, 2 6 i 6 s(G), there exists j, 2 6 j 6 s(P ), such that ωi (G) = ωj (P ). LEMMA 2 [3]. Let P be a finite simple group with disconnected graph GK(P ). Then |µi (P )| = 1 for 2 6 i 6 s(P ). Denote by ni a unique element in µi (P ), i > 2. Then P , π1 (P ), and ni , 2 6 i 6 s(P ), are as in Tables 1a-1c. In Tables 1a-c, p is an odd prime. Notice that there are three misprints in [3]: Table 1b, rows 2 G2 (q) and F4 (q); Table 1c, row F i024 . LEMMA 3. Let p be a prime and s a natural number, s > 2. Then one of the following holds: (a) there exists a prime q such that q divides ps − 1 and q does not divide pt − 1 for all natural t < s; (b) s = 6 and p = 2; (c) s = 2 and p = 2t − 1 for some natural t. Proof. See [8]. A prime q satisfying (a) is called a primitive prime divisor of ps − 1. LEMMA 4. (a) Let x = 3a < 3b = y < 32s , where a, b, and s are non-negative integers. If one of the numbers x ± y is divisible by 3s + 1 then b = a + s. (b) If s ∈ N and a, b, c < 2s are pairwise distinct non-negative integers then either some number of the form 2 · 3a ± 3b ± 3c is divisible by 5, or each of these numbers is not divisible by 2(1 + 3s ). (c) If a1 , a2 , . . . , at are integers not divisible by 5 and t > 4 then there exist numbers εi = ±1, i = t P 1, 2, . . . , t, such that εi ai is divisible by 5. i=1

Proof. Part (a) is obvious. To prove (b), we can assume that a = 0 or c = 0. In this case only numbers 2 + 3s+1 − 3s and 2 · 3s−1 − 32s−1 + 1 in the desired form are divisible by 2(1 + 3s ). Since 32s−1 ≡ ±3 (mod 5) and numbers of the form ±3 ± 1 (resp., of the form ±3s+1 ± 3s ) compose a full system of non-zero residues modulo 5, one of the numbers 2 · 3s−1 ± 32s−1 ± 1 (resp., one of 2 ± 3s+1 ± 3s ) is divisible by 5.  tTo prove (c), notice that a  more general assertion is true, namely, that the set P εi ai + 5Z εi = ±1, i = 1, 2, . . . , t coincides with Z/5Z. This is easy to prove for t = 4 (we i=1 need only examine tetrads (a1 , a2 , a3 , a4 ) in the set {(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 2, 2)}). For t > 4, it suffices 4 P to consider cosets of the form εi ai + a5 + . . . + at + 5Z. i=1

2. ELEMENT ORDERS IN COVERINGS OF FROBENIUS GROUPS LEMMA 5. Let G be a Frobenius group with core H and complement C. Then GK(H) and GK(C) are connected components of GK(G). Moreover, the following hold: (a) H is nilpotent. In particular, |µ(H)| = 1 and GK(H) is complete, that is, every two vertices of GK(H) are adjacent. (b) Every subgroup of C of order pq, where p and q are (not necessarily distinct) primes, is cyclic. In particular, every Sylow subgroup of C of odd order is cyclic, and a Sylow 2-subgroup of C either is cyclic or a (generalized) quaternion group. If |C| is even then the center of C is of even order. If C contains an element of order 16 then C has a normal 2-complement. The subgroup C0 of C generated by all elements of prime orders is isomorphic to Z × C1 , where Z is a cyclic group and C1 either is trivial or isomorphic to one of the groups SL2 (3) or SL2 (5). Moreover, either C is soluble and GK(C) is complete, or C contains a normal subgroup L ' SL2 (5) such that (|L|, |C : L|) 6 2 and GK(C) can be obtained from the complete graph on π(C) by deleting the edge (3, 5). 94

TABLE 1a. Finite Simple Groups P with s(P ) = 2 P

Restrictions on P

π1 (P )

n2

An

6 < n = p, p + 1, p + 2; one of n, n − 2 is not a prime

π((n − 3)!)

p

Ap−1 (q)

(p, q) 6= (3, 2), (3, 4)

π(q

p−1 Q

(q i − 1))

(q p − 1)/(q − 1)(p, q − 1)

i=1

Ap (q)

(q − 1)|(p + 1)

π(q(q p+1 − 1)

p−1 Q

(q i − 1))

(q p − 1)/(q − 1)

i=1 2

Ap−1 (q)

π(q

p−1 Q

(q i − (−1)i ))

(q p + 1)/(q + 1)(p, q + 1)

i=1 2

Ap (q)

(q + 1)|(p + 1),

π(q(q p+1 − 1)

p−1 Q

(q i − (−1)i ))

(q p + 1)/(q + 1)

i=1

(p, q) 6= (3, 3), (5, 2) 2

A3 (2)

Bn (q)

m

n=2

> 4, q odd

{2, 3} n−1 Q 2i π(q (q − 1))

5 (q n + 1)/2

i=1

π(3(3p + 1)

Bp (3)

p−1 Q

(32i − 1))

(3p − 1)/2

i=1

Cn (q)

n = 2m > 2

π(q

n−1 Q

(q 2i − 1))

(q n + 1)/(2, q − 1)

i=1

Cp (q)

q = 2, 3

π(q(q p + 1)

p−1 Q

(q 2i − 1))

(q p − 1)/(2, q − 1)

i=1

Dp (q)

p > 5, q = 2, 3, 5

π(q

p−1 Q

(q 2i − 1))

(q p − 1)/(q − 1)

i=1

Dp+1 (q)

q = 2, 3

π(q(q p + 1)

p−1 Q

(q 2i − 1))

(q p − 1)/(2, q − 1)

i=1 2

Dn (q)

n = 2m > 4

π(q

n−1 Q

(q 2i − 1))

(q n + 1)/(2, q + 1)

i=1 2

Dn (2)

n = 2m + 1 > 5

π(2(2n + 1)

n−2 Q

(22i − 1))

2n−1 + 1

i=1 2

Dp (3)

5 6 p 6= 2m + 1

π(3

p−1 Q

(32i − 1))

(3p + 1)/4

i=1 2

Dn (3)

G2 (q) 3 D4 (q) F4 (q) 2 F4 (2)0 E6 (q) 2 E6 (q) M12 J2 Ru He M cL Co1 Co3 F i22 HN

9 6 n = 2m + 1 6= p 2 < q ≡ ε (mod 3), ε = ±1 q odd

q>2

π(3(3n + 1)

n−2 Q i=1 3

(32i − 1))

π(q(q 2 − 1)(q − ε)) π(q(q 6 − 1)) π(q(q 6 − 1)(q 8 − 1)) {2, 3, 5} π(q(q 5 − 1)(q 8 − 1)(q 12 − 1)) π(q(q 5 + 1)(q 8 − 1)(q 12 − 1)) {2, 3, 5} {2, 3, 5} {2, 3, 5, 7, 13} {2, 3, 5, 7} {2, 3, 5, 7} {2, 3, 5, 7, 11, 13} {2, 3, 5, 7, 11} {2, 3, 5, 7, 11} {2, 3, 5, 7, 11}

(3n−1 + 1)/2 q 2 − εq + 1 q4 − q2 + 1 q4 − q2 + 1 13 (q 6 + q 3 + 1)/(3, q − 1) (q 6 − q 3 + 1)/(3, q + 1) 11 7 29 17 11 23 23 13 19

95

TABLE 1b. Finite Simple Groups P with s(P ) = 3 P

Rectrictions on P

π1 (P )

n2

n3

An

n > 6, n = p, p − 2 are primes 3 < q ≡ ε (mod 4), ε = ±1 q > 2, q even

π((n − 3)!)

p

p−2

π(q − ε)

π(q)

(q + )/2

{2} {2, 3, 5}

q−1 7

q+1 11

(3p−1 + 1)/2

(3p + 1)/4

q2 − q + 1 √ q − 3q + 1 q4 + 1 p q 2 − 2q 3 + q− √ 2q + 1 73 757 5 11 11 17 7 11 11 17 19 31

q2 + q + 1 √ q + 3q + 1 q4 − q2 + 1 p q 2 + 2q 3 + q+ √ 2q + 1 127 1093 11 23 23 19 11 13 23 23 31 47

A1 (q) A1 (q) A5 (2)

2 2

Dp (3)

p = 2m + 1

π(3(3p−1 − 1)

p−2 Q

(32i − 1))

i=1

G2 (q) G2 (q) F4 (q) 2 F4 (q) 2

E7 (2) E7 (3) M11 M23 M24 J3 HiS Suz Co2 F i23 F3 F2

q q q q

≡ 0 (mod 3) = 32m+1 > 3 even = 22m+1 > 2

π(q(q 2 − 1)) π(q(q 2 − 1)) π(q(q 4 − 1)(q 6 − 1)) π(q(q 3 + 1)(q 4 − 1)) {2, 3, 5, 7, 11, 13, 17, 19, 31, 43} {2, 3, 5, 7, 11, 13, 19, 37, 41, 61, 73, 547} {2, 3} {2, 3, 5, 7} {2, 3, 5, 7} {2, 3, 5} {2, 3, 5} {2, 3, 5, 7} {2, 3, 5, 7} {2, 3, 5, 7, 11, 13} {2, 3, 5, 7, 13} {2, 3, 5, 7, 11, 13, 17, 19, 23}

Proof. Part (a) is the Thompson theorem in [4]. Part (b) is an easy consequence of the classification of Frobenius complements in [5, 6]. LEMMA 6. Let G be a finite group, N / G, and G/N be a Frobenius group with core F and cyclic complement C. If (|F |, |N |) = 1 and F is not contained in N CG (N )/N , then p|C| ∈ ω(G) for some prime divisor p of |N |. Proof. This is part of Lemma 1 in [7]. LEMMA 7. Suppose that the conditions of case (2) of Lemma 1 are met. For t = |C|, we then have the following: Q (1) A is nilpotent, B is cyclic of odd order, C is cyclic, and G contains an element of order t p. p∈π(A)

In particular, GK(B) and GK(AC) are connected components of GK(G) and are both complete graphs. (2) If A is Abelian and m is the least common multiple of element orders in A then, for any generator c of C, there exists an element a ∈ A such that (ac−1 )t is an element of order m in CA (C). In particular, ac−1 is an element of order mt. (3) If t = 3, |A| is not divisible by 3, and CA (C) is elementary Abelian, then factors of the lower central series of A are elementary Abelian and the nilpotency class of A is at most 2. Proof. For part (1), see [1]. To prove (2), it suffices to consider the case m = pn , where p is a prime. n−1 | a ∈ A}. It is known that A1 , if treated as a C-module over a field Fp of order p, contains Let A1 = {ap 96

TABLE 1c. Finite Simple Groups P with s(P ) > 3 s(P )

P

4

A2 (4) 2 B2 (q) 2

Rectrictions on P

π1 (P )

n2

n3

n4

q = 22m+1 > 2

{2} {2}

3 q−1 13

5 √ q − 2q+ 1 17

7 √ q + 2q+ 1 19

q 10 −q 5 +1 q 2 −q+1

q 10 +q 5 +1 q 2 +q+1

q8 − q4 +

{2, 3, 5, 7, 11}

E6 (2)

E8 (q)

q ≡ 2, 3 (mod 5)

8

π(q(q − 1)(q 12

E8 (q)

6

− 1)·

18

(q − 1)(q − 1)· (q 20 − 1)) {2, 3} {2, 3, 5} {2, 3, 5, 7} {2, 3, 5, 7, 11} {2, 3, 5, 7, 11, 13} {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 47}

M22 J1 O0 N LyS F i024 F1 5

14

q ≡ 0, 1, 4 (mod 5)

π(q(q 8 − 1)(q 10 − 1)· (q 12 − 1)(q 18 − 1)· (q 18 − 1)) {2, 3, 5, 7, 11}

J4

n5

n6

1 5 7 11 31 17 41

7 11 19 37 23 59

11 19 31 67 29 71

q 10 −q 5 +1 q 2 −q+1

q 10 +q 5 +1 q 2 +q+1

q8 − q4 +

q 10 +1 q 2 +1

1 23

29

31

37

43

a regular C-submodule (see, e.g., [7, proof of Lemma 1]); in other words, there exists x ∈ A1 such that i n−1 elements xc , i = 0, 1, . . . , t are linearly independent. Let x = ap , a ∈ A. We argue that i

t−1

hac | i = 0, 1, . . . , t − 1i = hai × hac i × . . . × hac t−1

Suppose that am0 acm1 · · · ac

mt−1

i.

(1)

= 1, where mi are integers, i = 0, 1, . . . , t − 1. Then t−1

xm0 xcm1 · · · xc

mt−1

t−1

= (am0 acm1 · · · ac

mt−1 pn−1

)

= 1, t−1

and hence mi = pli , where li are integers, i = 0, 1, . . . , t − 1. By the same argument, xl0 xcl1 · · · xc lt−1 = 1, and hence li = pki , where ki are integers, i = 0, 1, . . . , t − 1. After n − 1 such steps we see that mi , t−1 i = 0, 1, . . . , t − 1, are divisible by pn , that is, am0 = acm1 = . . . = ac mt−1 = 1, proving (1). It then t−1 follows from (1) that y = aac · · · ac = (ac−1 )t is an element of order pn , and y ∈ CA (C). We embark on (3). We may suppose that q = |B| is a prime. Let B = hbi, C = hci, and bc = br for an integer r. Notice that r2 + r + 1 is divisible by q. By (2), A is a p-group for some prime p and all factors of the lower central series of A are elementary Abelian. Thus an associated Lie ring L of A (for definition and relevant facts, see [13]) is an algebra over a field Fp of order p, BC acts on L, and CL (C) is an Abelian subalgebra. Moreover, B acts regularly on L. b = L N F , where F is an extension of Fp containing a non-trivial The same holds true if we replace L by L Fp qth root of identity. To prove this, we need the following: b be such that xb = αx and yb = βy, where α, β ∈ F and [x, y] 6= 0. Then β = αr , or (7.1) Let x, y ∈ L 2

2

β = αr and [x, y]b = α−r [x, y], or [x, y]b = α−r [x, y], respectively. 2 2 To prove (7.1), notice that α 6= 1 6= β, αq = β q = 1, and hence αr +r+1 = β r +r+1 = 1. Moreover, 2

2

(xc)b = xbr c = αr (xc), (xc2 )b = αr (xc2 ), 2 (yc)b = β r (xc), (yc2 )b = β r (xc2 ).

(2)

97

Thus the vectors x, xc, and xc2 are linearly independent, and so are y, yc, and yc2 . Vectors x + xc + xc2 and y + yc + yc2 are in CLb (C), and hence [x + xc + xc2 , y + yc + yc2 ] =

2 X

[xci , ycj ] = 0.

(3)

i,j=0 2i

2j

2i

2j

By (2), [xci , ycj ], i, j = 0, 1, 2, is an eigenvector for b with eigenvalue αr β r , so αr β r 6= 1, and 2i 2j by (3), there exists a pair (i, j) 6= (0, 0) such that αβ = αr β r . It is obvious that if i = 0 then j = 0, 2i 2j 2 and vice versa. Thus 1 6= α1−r = β r −1 , and since αβ 6= 1, either i = 1, j = 2, and β = α−1−r = αr , 2 2 or i = 2, j = 1, and β = α−1−r = αr . It follows that [x, y]b = αβ[x, y] = α1+r [x, y] = α−r [x, y] or 2 [x, y]b = α1+r [x, y] = α−r [x, y], proving (7.1). b contains a base consisting of eigenvectors for b; so, if the nilpotency class As a vector space over F , L b is greater than two then there exist eigenvectors u, v, w for b such that [u, v, w] 6= 0. Let ub = αu. of L 2 2 Since [u, v] 6= 0, by (7.1), vb = αr v and [u, v]b = α−r [u, v], or vb = αr v and [u, v]b = α−r [u, v]. Putting x = [u, v] and y = w in (7.1) yields wb = αs w, where s = −1, −r, or − r2 .

(4)

On the other hand, [u, v, w] = −[v, w, u] − [w, u, v]; so, either [v, w] 6= 0 or [w, u] 6= 0. Again by (7.1), b and hence of A, wb = αs w, s = 1, r, or r2 , which contradicts (4). It follows that the nilpotency class of L, is at most two. The lemma is proved. LEMMA 8. Let C be a finite non-trivial group such that its subgroup C0 , generated by all elements of prime orders, is isomorphic to Z × C1 , where (|Z|, |C1 |) = 1, Z is a cyclic group, and C1 either is trivial or isomorphic to one of the groups SL2 (3) or SL2 (5). If m > 1 is a natural number coprime to |C| then there exists a finite Frobenius group with an Abelian core containing an element of order m and with a complement isomorphic to C. Proof. It suffices to consider the case where m = pn , p is a prime. There exists a natural s such that ps − 1 is divisible by |C0 |. Then SL2 (ps ) contains a subgroup A isomorphic to C1 and the center of GL2 (ps ) contains a subgroup B isomorphic to Z such that every non-trivial element of hA, Bi = A × B has no non-trivial fixed points under the action on a natural GL2 (ps )-module V over a field of order ps . In other words, V is a C0 -module, and the natural semidirect product V C0 is a Frobenius group with core V and complement C0 . Let W be a C-module induced by the C0 -module V . Then W is a direct sum of C0 -submodules conjugated with V ; hence, W C0 is also a Frobenius group. Since C0 contains all elements of prime orders in C, W C too is a Frobenius group. Let |W | = pr and U be a direct product of r cyclic subgroups of order m. Then U/U p ' W and Aut(U )/I ' Aut(W ), where I consists of automorphisms of U acting trivially on U/U p . Since I is a p-group and p does not divide |C|, Aut(U ) contains a subgroup H which is isomorphic to C and acts on U/U p as does C on W . It follows that the semidirect product U H is the required Frobenius group. The lemma is proved. 3. REPRESENTATIONS OF SOME GROUPS LEMMA 9. Let H be an extension of a finite non-trivial 3-group V by the symmetric group S6 of degree 6. Then H contains an element of order 15 or 18. Proof. It is sufficient to consider the case where V is an absolutely irreducible S6 -module. If χ is the corresponding Brauer character, then, by [9], χ coincides with one of the characters Xi in the following table: 98

Brauer Characters of S6 (mod 3) X1 X2 X3 X4 X5 X6 X7

1A 1 1 6 4 4 9 9

2A 1 1 −2 . . 1 1

4A 1 1 2 −2 −2 1 1

5A 1 1 1 −1 −1 −1 −1

2B 1 −1 . 2 −2 3 −3

2C 1 −1 . −2 2 3 −3

4B 1 −1 . . . −1 1

Suppose that the conclusion is false. Then, clearly, dim(CV (x)) = 0 for every element of order 5 in S6 . It is easy to calculate that dim(CV (x)) = (χ(1A) + 4χ(5A))/5 = 0 for x ∈ 5A only if χ(1A) = 4. For some involution i ∈ S6 \ A6 and for an elementary Abelian subgroup U of order 4 in CA6 (i), in this instance, 1 = dim(CV (U )) < dim(CV (i)), and hence a subgroup of CA6 (i) which is isomorphic to A4 acts on CV (i) faithfully. By Lemma 6, CH (i) contains an element of order 9, and so H contains an element of order 18. LEMMA 10. Let F be a finite field of order q = pn > 3, where p is a prime, and Wi = Wi (q), i = 0, 1, . . . , p − 1, be a space of homogeneous polynomials of degree i in variables x1 , x2 over F . Let α be an automorphism of F mapping every element of F into its pth power. For j = 0, . . . , n − 1, we turn Wi into j j αj αj αj αj an SL2 (q)-module W ! i = Wi (q) by setting f (x1 , x2 )a = f (a11 x1 + a12 x2 , a21 x1 + a22 x2 ) for f (x1 , x2 ) ∈ Wi a11 a12 and a = ∈ SL2 (q). In particular, W0j is a trivial one-dimensional SL2 (q)-module. a21 a22 n−1 N j (1) Modules W (i0 , . . . , in−1 ) = Wij form a complete set of pairwise non-equivalent absolutely irrej=0

ducible SL2 (q)-modules over a field of characteristic p. If q is odd then the center of SL2 (q) acts trivially on W (i0 , . . . , in−1 ), and hence W (i0 , . . . , in−1 ) is an L2 (q)-module if and only if i0 + . . . + in−1 is even. (2) If λ, λ−1 are characteristic roots of an element a ∈ SL2 (q) then the characteristic roots of a in W1j , j j j j W2j , and W (i0 , . . . , in−1 ) are, respectively, λp , λ−p ; 1, λ2p , λ−2p ; and λi0 · · · λin−1 , where λij run over all characteristic roots of a in Wij . Proof. Part (1) is a well-known result of Brauer and Nesbitt (cf. [10]). As a base of the vector space j i Wi we can choose xi1 , xi−1 1 x2 , . . . , x2 , and this base does not depend on q. Therefore Wi (q) can be extended to Wij (q s ) for every s, and in proving (2), we may assume that a is triangular. For such a, (2) is obvious. 4. GROUPS S4 (q), q 6= 3n Let S = S4 (q), q 6= 3n . In this section we prove that h(S) = ∞. The case of even q is settled in [1], and so in proving the Main theorem, we assume that q = pn , where p is an odd prime. Throughout this section, p 6= 3. LEMMA 11. The set ω(S) consists of all divisors of numbers (q 2 + 1)/2, (q 2 − 1)/2, p(q + 1), and p(q − 1). Proof. See [11]. Proposition 1. Let L = L2 (q 2 ), where q = pn , p > 3 is a prime. Let V = W10 ⊗ W1n be an L-module, where Wij is defined as in Lemma 10. Then V can be regarded as an L-module, where L = Lhσi and σ is a field automorphism of L of order 2, and ω(L) = ω(S4 (q)) for the natural semidirect product E = V L. In particular, h(S4 (q)) = ∞. 99

Proof. We adopt the notation of Lemma [10]. Define the action of σ on W10 ⊗ W1n by the rule ! 2 2 P P aij (xi ⊗ xj ) σ = aij (xj ⊗ xi ). It is easy to check that this turns V into an L-module. For i,j=1

i,j=1

0 −1/λ

x ∈ SL2 (q 2 ), denote by x the image of x in L. Let L 3 a =

! λ σ, where λ is a primitive (q 2 − 1)th 0

root of identity. Then the order of a is even, 2

a =

0 −1/λ

λ 0

!

0 −1/λ

λ 0

!σ =

−λ1−q 0

! 0 , −λq−1

and the order of a is equal to q + 1. Vectors v1 = x1 ⊗ x1 , v2 = x1 ⊗ x2 , v3 = x2 ⊗ x1 , and v4 = x2 ⊗ x2 compose a base for V . Since v1 a = λ1+q v4 and v4 a = λ−1−q v1 , a fixes a non-trivial vector in the subspace spanned by v1 and v4 ; so, E contains an element of order p(q + 1). ! ! λ 0 λ1+q 0 2 σ. Then the order of b is even, b = , and the order of b is equal Let b = 0 −1/λ 0 λ−q−1 to q − 1. Since v2 b = λ1−q v3 and v3 b = λq−1 v2 , E contains an element of order p(q − 1). For ω(L) contains (q 2 + 1)/2 and (q 2 − 1)/2, we have ω(S4 (q)) ⊆ ω(E) by Lemma 11. Let a be an element of E. First suppose that hai ∩ V is non-trivial. Then the order of a is divisible by p. Since dim(V ) = 4 and p > 5, E cannot contain an element of order p2 , and hence the order of a is equal to ps where s is coprime to p. Thus there exists an element b ∈ L of order s which centralizes a non-trivial element in V . Let c = b2 . Then c ∈ L. Put c = d for d ∈ SL2 (q 2 ). Let α and 1/α be characteristic roots of d. By Lemma 10, the characteristic roots of c in V are α±1±q . Since one of these roots is equal to 1, the order of d divides q + 1 or q − 1; hence, the order of a divides p(q + 1) or p(q − 1). Next assume that hai ∩ V is trivial. Then the order of a is equal to an order of some element in L, and we can assume that a ∈ L. If a ∈ L then the order of a divides p, (q 2 − 1)/2, or (q 2 + 1)/2 and is so contained in ω(S4 (q)). Suppose a ∈ L \ L. Then the order of a is equal to 2s, where s is the order of a2 ∈ L. If s is divisible by p then s = p and 2s ∈ ω(S4 (q)). Thus we may assume that s is coprime to p. Let a = c where c ∈ SL2 (q)σ. Put c = dσ with d ∈ SL2 (q 2 ). Then (c2 )d = (c2 )σ , and hence the characteristic roots of c2 (∈ SL2 (q 2 )) and (c2 )σ coincide. Let α and 1/α be characteristic roots of d = c2 . Then the characteristic roots of dσ are αq and 1/αq , and so αq−1 = 1 or αq+1 = 1. Thus the order of a divides (q 2 − 1)/2 ∈ ω(S4 (q)), proving the proposition. Remark. This result, in particular, straightens out the mistake in [7] where Proposition 7 asserts erroneously that S4 (7) is recognizable.∗ 5. GROUPS S4 (3n ) The goal of this section is proving the Main theorem for S4 (3n ). Let q = 3n and S = S4 (q). We know from [7] that h(S) = ∞ for n = 1, and so further it is assumed that n > 1. LEMMA 12. µ(S) = {n2 (= (q 2 + 1)/2), (q 2 − 1)/2, 3(q + 1), 3(q − 1), 9}. In particular, s(S) = 2. Proof. See [11]. Let G be a finite group with ω(G) = ω(S). ∗I

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am indebted to A. V. Zavarnitsin who pointed up the gap in the proof of Proposition 7 in [7].

LEMMA 13. Case (3) of Lemma 1 holds. Proof. By Lemma 12, G satisfies the conditions of Lemma 1. First suppose that case (1) of Lemma 1 is realized, that is, G = HC is a Frobenius group with core H and complement C. By Lemmas 5 and 12, µ(H) = {n2 (= (q 2 + 1)/2)}, µ(C) = {(q 2 − 1)/2, 3(q + 1), 3(q − 1), 9}, and the center of C contains an element of order 2. It follows that C contains an element of order 18, a contradiction. Next assume that case (2) of Lemma 1 holds. By Lemmas 7 and 12, B is a cyclic group of order 2 (q + 1)/2, and hence |C| divides |B| − 1 = (q 2 − 1)/2. Thus A contains a Sylow 3-subgroup of G. Since 9 ∈ µ(G), A is a Sylow 3-subgroup of G, and so C is a cyclic group of order (q 2 − 1)/2. By Lemma 7, G contains an element of order 3(q 2 − 1)/2, which contradicts Lemma 12. The lemma is proved. Lemmas 13, 2, and 12 show that there exists a non-Abelian simple group P (cf. Tables 1a-c) such that P ≤ G = G/K ≤ Aut(P ) for some nilpotent normal π1 (S)-subgroup K of G, and G/P is a π1 (S)-group. Moreover, there exists j, 2 6 j 6 s(P ), such that nj (P ) = (q 2 + 1)/2 and ω(P ) ⊆ ω(S). LEMMA 14. P is isomorphic to S or L2 (q 2 ). Proof. By Lemma 12, the following holds: (14.1) m = (q 2 + 1)/2 = nj (P ) is the largest order of elements in P . If t is the least common multiple of the orders of elements in P which do not divide m then t = 9(m − 1); in particular, t/m < 9. If q > 27 and u is the order of an element in P which does not divide m then either u = m − 1, (m − 1)/2, or m/u > 4. We treat the groups in Tables 1a-c singly. Assume that P = Ar and r = p, p + 1, or p + 2. Then m = p or m = p + 2, and since q > 9, we have p > 39. Hence 3(p − 3)/2 > m, but Ar contains an element of order 3(p − 3)/2, which is a product of independent cycles of length 3, (p − 3)/2, (p − 3)/2, a contradiction with (14.1). Suppose that P = Ap−1 (r), where the pair (p, r) does not coincide with (3, 2) or (3, 4). Then m = (rp − 1)/(r − 1)(p, r − 1). This equality is impossible for q = 9, and so q > 9. Since Ap−1 (r) is isomorphic to the factor of SLp (r) by the center, and SLp (r) contains a cyclic subgroup of order rp−1 − 1, Ap−1 (r) contains an element of order u = (rp−1 − 1)/(p, r − 1) which does not divide m. By (14.1), for m − u 6= 1 we have m/u > 2. This inequality holds only for r = 2, but then the equality (q 2 + 1)/2 = 2p − 1 cannot hold. Hence this case is impossible. Similar arguments deny the case where P = 2 Ap (r), r + 1 divides p + 1. Assume that P = Ap (r) and r − 1 divides p + 1. In this case m = (rp − 1)/(r − 1), but Ap (r) contains an element of order (rp+1 − 1)/(r − 1)2 > m, a contradiction with (14.1). The case where P = 2 Ap−1 (r) is similar. The case P = 2 A3 (2) is clearly impossible. If P = Bp (3) then (q 2 + 1)/2 = (3p − 1)/2, which is impossible.  d−1  Q 2i Suppose P = Cd (r), d = 2s > 2. Then π1 (P ) = π r (r − 1) and m = (rd + 1)/(2, r − 1). Assume s

i=1

first that r is odd. Then r2 = q 2 ; in particular, r is the power of 3. Thus the order of every element in P which is coprime to rm should divide m − 1 = (rd − 1)/2. If d = 2 then P ' S. Suppose d > 4. Let h be a primitive prime divisor of rd−1 − 1, which exists by Lemma 3. Since (rd−1 − 1, rd − 1) = (r − 1) and (rd−1 − 1, rd + 1) = 2, h cannot divide 9m(m − 1) = tm. For h is the order of some element in P , this case is an impossibility. Now let r be even. Then m = rd + 1, 32n − 1 = 2rd , so n = 1 and q = 3, which is the a priori excluded case. The case where P = Bd (r), d = 2m > 4, and r is odd, can be treated similarly. If P = Cp (2) then (q 2 + 1)/2 = 2p − 1, and hence q 2 + 3 = 2p+1 , which is impossible. Likewise we can argue for the cases where P = Dp (2) and P = Dp+1 (2).

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If P = Cp (3) then (q 2 + 1)/2 = (3p − 1)/2, which is again impossible. Likewise we can argue for the cases where P = Dp (3) and P = Dp+1 (3). If P = Dp (5), then (q 2 + 1)/2 = (5p − 1)/4, and so 2q 2 + 3 = 5p , which is clearly impossible. Suppose that P = 2 Dd (r), d = 2s > 4. If r is odd then m = (q 2 + 1)/2 = (rd + 1)/2 and q m/2 > q 2 , and so Lemma 3 implies that r6 − 1 is divisible by a prime h ∈ ω(P ) which does not divide q(q 4 − 1)(q 2 − 1), and nor the order of S therefore. If r is even then (q 2 + 1)/2 = rd + 1 and q 2 − 1 = 2rd , which is impossible. The case where P = 2 Dd (2), d = 2s + 1 > 5, is similar, and the case P = 2 Dp (3), 5 6 p 6= 2s + 1, is clearly impossible. Suppose that P = 2 Dd (3), d = 2s + 1 6= p, s > 2. Then q = 3(d−1)/2 , (3d + 1, q 2 + 1) = 2, and d (3 + 1, q 2 − 1) = 4. By Lemma 3, 3d + 1 is divisible by an odd prime which, clearly, divides |P |, but not |S|. Assume P = G2 (r), 2 < r ≡ ε (mod 3), and ε = ±1. Then m = r2 − εr + 1, and so q 2 = 2r2 − 2εr + 1 ≡ 1 (mod 3), which is impossible. If P = 3 D4 (r) or F4 (r), r is odd, then (q 2 + 1)/2 = r4 − r2 + 1 ≡ 1 (mod 3), which is false. The case where P = 2 F 4 (2)0 is obviously impossible. Let P = E6 (r). Then (q 2 + 1)/2 = (r6 + r3 + 1)/(3, r − 1). If (3, r − 1) = 3 then q 2 − 1 = 2(r6 + r3 − 2)/3 and (r3 + 1, q 2 − 1) divides 4. Since r3 + 1 divides |P |, by Lemma 3, there exists a prime which divides |P |, but not |S|. It follows that (3, r − 1) = 1 and q 2 − 1 = 2(r6 + r3 ), so (r3 − 1, q 2 − 1) divides 4, and we arrive at the same conclusion. The case where P = 2 E 6 (r) is similar. Since m = (q 2 + 1)/2 > 41, all other cases in Table 1a are impossible. Suppose that P = A1 (r), r is odd. If r is a prime then (q 2 + 1)/2 = r and (q 2 − 1)/2 = r − 1; so, (r + 1)/2 ∈ ω(P ) is an odd number which is coprime to (q 2 − 1). Thus (r + 1)/2 = 3 or (r + 1)/2 = 9, which readily leads us to a contradiction. Thus r is not a prime, and hence (q 2 + 1)/2 = (r + 1)/2. In this case P = L2 (q 2 ), as required. Assume P = A1 (2s ). Then (q 2 + 1)/2 = 2s + 1, which leads us to the excluded case q = 3. The case where P = 2 A5 (2) is obviously impossible; for the cases P = 2 Dp (3), p = 2s + 1 > 5, P = G2 (r), r = 3m , and P = 2 G( r), r = 32s+1 , we obtain the respective equalities q 2 + 1 = (3p + 1)/2, √ (q 2 + 1)/2 = r2 + r + 1, and (q 2 + 1)/2 = r + 3r + 1, which are impossible. Let P = F4 (r), r = 2s > 2. Then (q 2 + 1)/2 = r4 − r2 + 1 and q 2 − 1 = 2(r4 − r2 ). Since (r4 + 1, q 2 − 1) = 1 and r4 + 1 divides |P |, by Lemma 3, there exists a prime which divides |P |, but not |S|, which is again an impossibility. √ √ Suppose P = 2 F4 (r), r = 22s+1 > 2. Then (q 2 + 1)/2 = r2 + 2r3 + r + 2r + 1, and hence r 6= 8, r > 32. For such r, we have (r2 +

√

2r3 + r +

√

2r + 1)/(r2 −

√

2r3 + r −

√

2r + 1) < 2,

√ √ and so 2( 2r3 + 2r) = 1 by (14.1), which is impossible. All other cases in Table 1b are obvious. Let P be a group in Table 1c. The case P = 2 B2 (r), r = 22s+1 > 2, is similar to the case where P = 2 F4 (r), the instance with P = 2 E6 (2) is clearly impossible, and we so let P = E8 (r). Since ni /nj < 2 and ni − nj 6= 1 for all 2 6 i, j 6 s(P ), this case is impossible by (14.1). All other cases in Table 1c are refuted readily. The lemma is proved. Proposition 2. If q = 32s then there exists an extension E of a 3-group V by L = L2 (q 2 ) such that ω(E) = ω(S4 (q)). In particular, h(S4 (q)) = ∞.

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b = SL2 (q 2 ) and V = W 2s ⊗ W 2s+1 ⊗ W 0 be an L-module b Proof. Let L specified as in Lemma 10. By 1 2 1 that lemma, V is an L-module. Let E = V L be a natural semidirect product of V ! and L. 1 0 b on V contains a First, direct computations will apply to verify that the Jordan form of ∈L 1 1 block of dimension 3, and hence E contains an element of order 9. Since L does not contain an element of order 3r for r > 1, E cannot contain an element of order 9r for r > 1. b of a such that Let a ∈ L be an element of order m not divisible by 3. We can choose a preimage b a∈L b Then the multiplicative order of α is the order of b a is 2m. Let α and 1/α be characteristic roots of b a in L. 2 2 equal to 2m, and 2m divides q + 1 or q − 1. By Lemma 10, the characteristic roots of a in V are of the form αt , where t = ±32s ± 32s+1 or t = ±2 ± 32s ± 32s+1 . It follows that E contains an element of order 3m, where m is coprime to 3 if and only if m divides q + 1 or q − 1. If the order m of an element in E is not divisible by 3, then m is the order of an element in L, and so m divides (q 2 ± 1)/2. Hence µ(E) = µ(S4 (q)), and the proposition then follows from [7, Lemma 1]. LEMMA 15. Suppose that L = L2 (34s+2 ) ≤ H ≤ Aut(L) and ω(H) ⊆ ω(S4 (32s+1 )). Then H = L or H = Lhσi, where σ is a field automorphism of L of order 2. Proof. Let q = 32s+1 . It is well known that A = Aut(L) = P GL2 (q 2 )hφi where φ is a field automorphism of order 4s + 2 (see, e.g., [12]). Obviously, H/L is a π1 (S4 (q))-group, and hence no element of H \ L can centralize a non-trivial element of L whose order divides 34s+2 + 1. Since φ2 centralizes an element of order 5, and 5 divides 34s+2 + 1, H ≤ P GL2 (q 2 )hσi. If L 6= H 6= Lhσi then a Sylow 2-subgroup of H is a dihedral or semidihedral group of order 2t+1 , where 2t is the largest power of 2 which divides q 2 − 1, and hence contains an element of order 2t 6∈ ω(S4 (q)). Hereafter, we assume that q = 32s+1 > 3. LEMMA 16. If P = S then P ' G. Proof. Suppose first that K 6= 1. Without loss of generality, we may assume that K is an elementary Abelian p-group. Let H be the full preimage of P in G. Clearly, KCH (K) = K, and so P acts on K faithfully. Assume p 6= 3. Since P contains a subgroup L isomorphic to L2 (q 2 ), which in turn contains a Frobenius subgroup F of order q 2 (q 2 − 1)/2, it follows by Lemma 6 that the full preimage of F in G contains an element of order p(q 2 − 1)/2 6∈ ω(S). Thus p = 3. Since P contains S4 (3) which contains S6 , K = 1 by Lemma 9. Thus P is a normal subgroup of G. By [12, Prop. 4.3.10], P contains a subgroup L ' L2 (q 2 ).2, and G = P NG (L). In view of the fact that L contains an element of order (q 2 + 1)/2, we have CG (L) = 1. By Lemma 15, NG (L) = L and G = P . LEMMA 17. If P = L2 (q 2 ) then G/K ' P and K is a non-trivial 3-group. Proof. By Lemma 15, G/K does not contain an element of order 9, and so 3 divides the order of K. By Lemma 1, K is a π1 (G)-group, and hence a cyclic subgroup of P of order (q 2 + 1)/2 acts on K regularly. By Lemma 5, K is nilpotent. Suppose that there exists a prime r 6= 3 which divides |K|. Let K0 be a normal subgroup of G such that K0 ≤ K and V = K/K0 is a non-trivial elementary Abelian r-subgroup. Then P acts on V , and this action is faithful. Since P contains a Frobenius subgroup of order q 2 (q 2 − 1)/2 with cyclic complement of order (q 2 − 1)/2, G contains an element of order r(q 2 − 1)/2 6= ω(S) by Lemma 6. This implies that K is a non-trivial 3-group. If G/K 6= P then, by Lemma 15, G/K ' Lhσi where σ is an automorphism of the corresponding field 2s+1 F of order q 2 = 34s+2 , mapping every f into f q . If F0 is a subfield of F of order 9 then f q = f 3 = f3

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for f ∈ F0 , so σ acts on F0 non-trivially, and hence G/K contains a subgroup S6 ' L2 (9)hδi, where δ is a non-trivial field automorphism of L2 (9). By Lemma 9, ω(G) 6= ω(S). This contradiction proves the lemma. LEMMA 18. P cannot be isomorphic to L2 (q 2 ). Proof. Assume the contrary. By Lemma 17, G/K = P and K is a non-trivial 3-group. Since G, as distinct from P , contains an element of order 3(q + 1), there exists an element of order q + 1 in G which centralizes an element k ∈ K of order 3. Let 1 = K0 < K1 < . . . < Kt = K be the upper central series of K and k ∈ Ki \ Ki−1 . Then V = Ki /Ki−1 is a P -module, and there exists an element of order q + 1 in P for which 1 is a characteristic root in V . On the other hand, every subgroup of order (q 2 + 1)/2 in G acts on K regularly, and since 5 divides q 2 + 1, an element of order 5 in P cannot have 1 as a characteristic root in V . Obviously, some absolutely irreducible component W of V also enjoys these properties. Therefore the following holds: (18.1) W is an absolutely irreducible P -module over a field of characteristic 3 such that 1 is a characteristic root in W for some element x ∈ P of order q + 1, and 1 is not contained in the set of characteristic roots in W for some element y ∈ P of order 5. By Lemma 10, W = Wit11 ⊗ . . . ⊗ Witrr , where t1 , . . . , tr are pairwise distinct numbers, 0 6 t1 , . . . , tr < 4s + 2, and i1 , . . . , ir ∈ {1, 2}. For a ∈ P = L2 (q 2 ), let a be a preimage of a in L = SL2 (q 2 ) and α and 1/α be characteristic roots of a in L. If r > 4 then, by Lemma 10, there exist numbers b1 , . . . , br ∈ {0, 1} such that, for cj = 2(bj · 3tj , j = 1, . . . , r, the set ) of characteristic roots of a in W contains all powers αd , where d ∈ r P εj cj εj = ±1, j = 1, . . . , r . By Lemma 4(c), there exist some d that are divisible by 5. So if a j=1 and a are chosen to be of order 5 then αd = 1, which contradicts (18.1). Thus r < 4. By Lemma 10, there exist exactly two numbers, say, 1 and 2, such that i1 = i2 = 1. If r = 2 then the characteristic roots of a in W are {αd | d = ±3t1 ± 3t2 }. By Lemma 4(a), d can be divisible by q + 1 only if d = ±3t (q + 1) for some non-negative integer t, and hence d cannot be divisible by 2(q + 1). On the other hand, q + 1 is divisible by 4, so, for an element a of order q + 1 in P , every preimage a of a in L is of order 2(q + 1), and hence αd 6= 1 for every d. This contradicts (18.1). Thus r = 3 and W = W1t1 ⊗ W1t2 ⊗ W2t3 . By Lemma 10, the set of all characteristic roots of a in W is equal to {αd |d = ε1 3t1 + ε2 3t2 + ε3 2 · 3t3 , ε1 , ε2 = ±1, ε3 = −1, 0 or 1}. In view of Lemma 4(b), we obtain a contradiction with (18.1), as in the previous two paragraphs. The lemma is proved. The Main theorem now follows from Propositions 1, 2, Lemmas 14, 16, and 18. 6. SOME SEPARATE SIMPLE GROUPS Proposition 3. There exists an extension E of a 13-group V by 2.S4 such that ω(E) = ω(L3 (3)). In particular, h(L3 (3)) = ∞. Proof. Let H be an extension of a group of order 2 by S4 such that a Sylow 2-subgroup of H is a quaternion group. Then µ(H) = {6, 8}. By Lemma 8, there exists an extension E of an elementary 13-group by H, which is a Frobenius group. It follows that µ(E) = {6, 8, 13} = µ(L3 (3)), and then the proposition follows from [7, Lemma 1]. Proposition 4. If G is a finite group such that ω(G) = ω(U3 (9)) then G ' U3 (9). Proof. In view of [14], we have µ(G) = {30, 73, 80}, and so GK(G) is disconnected. By Lemma 1, then, one of the following cases holds:

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Case 1. G = F H is a Frobenius group with core F and complement H. The connected components of GK(G) are ω(F ) and ω(H). Case 2. Let G = ABC, where A and AB are normal subgroups of G, B is a normal subgroup of BC, B and C are cyclic subgroups, and AB and BC are Frobenius groups. The connected components of GK(G) are ω(B) and ω(AC). Case 3. There exists a non-Abelian simple group P such that P ≤ G = G/K ≤ Aut(P ) for some nilpotent normal {2, 3, 5}-subgroup K of G, GK(P ) is disconnected, and ωj (P ) = {73} for some j, 2 6 j 6 s(P ). LEMMA 19. Case 1 is impossible. Proof. Assume the contrary. Since F is nilpotent and F is a 73-group, H contains an element of order 16. By Lemma 5(b), H contains a normal 2-complement which is of order 15. But then H contains an element of order 24, which is false. LEMMA 20. Case 2 is impossible. Proof. Assume the contrary. By Lemma 6, |B| = 73. Since |C| divides |B| − 1, the order of C is 3 or 6. By Lemma 6 again, |C| = 3, and so A = T × F , where F is a 5-group, T is a 2-group of exponent 16, and CT (C) is elementary, a contradiction. Thus Case 3 holds. LEMMA 21. G and U3 (9) are isomorphic. Proof. By Lemma 2, P is one of the groups in Tables 1a-1c. Looking over these tables, we can easily verify that P ' U3 (9). By [14], P = G/K. We may assume that K is a non-trivial elementary Abelian p-group, which can be treated as a P -module. By [14] again, P contains a Frobenius group of order 73 · 3, and hence p 6= 3 by Lemma 6. Suppose p = 5. Since P contains a subgroup isomorphic to A6 which has a cyclic Sylow 5-subgroup, the minimal polynomial of an element of P of order 5 on K is equal to (x − 1)5 by [15], and so G contains an element of order 25. Thus p = 2. By [14], the centralizer in P of some involution t ∈ P contains an elementary Abelian subgroup R of order 9. Since t cannot centralize K, and K = hCK (r) | 1 6= r ∈ Ri, t does not centralize CK (r) for some element r ∈ G of order 3. This implies that CG (r) contains an element of order 4 and G contains an element of order 12, which is false. The lemma and the proposition are proved. Proposition 5. If G is a finite group such that ω(G) = ω(3 D4 (2)) then G ' 3 D4 (2). Proof. Lemma 2 and [14] can be combined to yield the following: LEMMA 22. Let P be a finite simple group with disconnected Gruenberg–Kegel graph GK(P ). If one of connected components in GK(P ) is equal to {13} then P is isomorphic to some group in Table 2. Proof. Let G be a finite group with ω(G) = ω(3 D4 (2)). Then µ(G) = {13, 7 · 3, 7 · 4, 9 · 2, 3 · 4, 8},

(5)

and G satisfies conditions (1), (2), or (3) in Lemma 1. First assume that G satisfies (1), that is, G = F C is a Frobenius group with core F and complement C. If F is a π1 (G)-group, then F should contain an element of order 9 · 8 in view of its being nilpotent, which it does not. Thus F is a 13-group. By Eq. (5), the centralizer in C of an element of order 7 contains an element of order 6. It follows that G should contain an element of order 7 · 3 · 2, which it does not. Next suppose that G satisfies (2). By Lemma 7, then, |B| = 13 and |C| divides 12. By Eq. (5), A contains elements of orders 2, 3, and 7 and G contains an element of order 7 · 3 · 2, which is false.

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TABLE 2 P A13 A14 A15 L2 (25) L2 (27) L2 (13n ) L3 (3) L4 (3) U3 (4) S4 (5) S6 (3) O7 (3) O8+ (3) G2 (3) G2 (4) F4 (2) Sz(8) 3 D4 (2) 2 F4 (2)0 2 E6 (2) Suz F i22

µ(P ) 13, 11, 7 · 5, 7 · 3, 7 · 4, 5 · 3 · 2, 5 · 4, 9 · 2, 3 · 8 13, 11 · 3, 7 · 5, 7 · 3 · 2, 7 · 4, 5 · 9, 5 · 3 · 4, 9 · 2, 3 · 8 13, 11 · 3, 11 · 2, 7 · 5 · 3, 7 · 3 · 2, 7 · 4, 5 · 9, 5 · 3 · 4, 5 · 8, 9 · 4, 3 · 8 13, 5, 3 · 4 13, 7 · 2, 3 13, (13n − 1)/2, (13n + 1)/2 13, 3 · 2, 8 13, 5 · 4, 9, 3 · 4, 8 13, 5 · 3, 5 · 2, 4 13, 5 · 3 · 2, 5 · 4, 3 · 4 13, 7 · 2, 5 · 3 · 2, 5 · 4, 9 · 4, 3 · 8 13, 7 · 2, 5 · 3, 5 · 4, 9 · 2, 3 · 4, 8 13, 7 · 2, 5 · 3, 5 · 4, 9 · 2, 3 · 4, 8 13, 7, 9, 3 · 4, 8 13, 7 · 3, 5 · 3, 5 · 2, 3 · 4, 8 17, 13, 7 · 3, 7 · 4, 5 · 3 · 2, 5 · 4, 9 · 2, 3 · 8, 16 13, 7, 5, 4 13, 7 · 3, 7 · 4, 9 · 2, 3 · 4, 8 13, 5 · 2, 3 · 4, 16 19, 17, 13, 11 · 3, 11 · 2, 7 · 5, 7 · 3, 7 · 4, 5 · 3 · 2, 5 · 4, 9 · 2, 3 · 8, 16 13, 11, 7 · 3, 7 · 2, 5 · 3, 5 · 4, 9 · 2, 3 · 8 13, 11 · 2, 7 · 3, 7 · 2, 5 · 3 · 2, 5 · 4, 9 · 2, 3 · 8, 16

Thus condition (3) holds, that is, there exists a non-Abelian simple group P such that P ≤ G = G/K ≤ Aut(P ) for some nilpotent normal π1 (G)-subgroup K of G, and G/P is a π1 (G)-group. By Lemma 22, P is isomorphic to one of L2 (13), L2 (27), L3 (3), G2 (3), or 3 D4 (2). We handle every case singly. Let P ' L2 (13). Since 49 6∈ ω(G), 7 does not divide |K| by [15], and hence K contains elements of orders 2 and 3, centralized by an element of order 7 in G. It follows that G contains an element of order 7 · 3 · 2, which contradicts Eq. (5). Suppose P ' L2 (27). Since P contains a Frobenius group of order 27 · 13, K is a 3-group. But then G cannot contain an element of order 8, contrary to Eq. (5). Assume P ' L3 (3). Since |Aut(P )| is not divisible by 7, and P contains a Frobenius group of order 9 · 8 with cyclic complement of order 8, 7 divides |K|, and by Lemma 6, G contains an element of order 7 · 8, which is false. Let P = G2 (3). Since G does not contain an element of order 49, and P contains a Frobenius group of order 8 · 7, 7 cannot divide |K|. Hence K contains elements of orders 2 and 3, centralized by a Sylow 7-subgroup of G. It follows that G contains an element of order 7 · 3 · 2, which contradicts Eq. (5). Thus P ' 3 D4 (2) and G/K = P . Suppose K 6= 1. We may assume that K is an elementary Abelian p-group for p = 2, 3, or 7. Since P contains a Frobenius group of order 8 · 7, p 6= 7 by Lemma 6. The tables of Brauer characters for P (cf. [9]) with p = 2, 3 show that if an element of order 13 in P acts on an

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absolutely irreducible module V without fixed points then p = 2, dim(V ) = 8, and an element of P of order 21 has a non-trivial fixed point in V . This means that G should contain either an element of order 13p or an element of order 2 · 21, which it does not. It follows that K = 1 and G = P . The proposition is proved. Proposition 6. If G is a finite group such that ω(G) = ω(G2 (4)) then G ' G2 (4). Proof. By [14], µ(G) = {13, 7 · 3, 5 · 3, 5 · 2, 3 · 4, 8}.

(6)

As in the proof of Proposition 5, it is easy to show that G satisfies condition (3) of Lemma 1, that is, there exists a non-Abelian simple group P such that P ≤ G = G/K ≤ Aut(P ) for some nilpotent normal π1 (G)-subgroup K of G, and G/P is a π1 (G)-group. By Lemma 22, P is isomorphic to one of L2 (13), L2 (25), L2 (27), L3 (3), Sz(8), U3 (4), or G2 (4). If P is isomorphic to L2 (25), L3 (3), or U3 (4) then |G/K| is not divisible by 7, and hence 7 divides |K|. Since a Sylow 2-subgroup of P contains a non-cyclic Abelian subgroup, G contains an element of order 14, contrary to Eq. (6). If P is isomorphic to L2 (13) or L2 (27) then 5 does not divide |Aut(P )|, and hence 5 divides |K|. Since P contains a Frobenius group of order 13 · 6 or 27 · 13, by Lemma 6, G contains an element of order 5 · 6 or 5 · 13, contrary to Eq. (6). Suppose that P ' Sz(8). Notice that 8 ∈ µ(G), 8 6∈ ω(Aut(P )), and P contains a Frobenius group of order 13 · 4. It follows by Lemma 6 that K is a non-trivial 2-group, and so G/K ' Sz(8) : 3. By [14], G/K contains a Frobenius group of order 13 · 12, and by Lemma 6, G contains an element of order 24, contrary to Eq. (6). Thus P ' G2 (4). By [14], G/K = P . Since P contains Frobenius groups of orders 3 · 2, 4 · 3, and 16 · 5 and ω(G) does not contain 7 · 2, 9, and 25, K is a 2-group. We argue that K = 1. Suppose not. We can assume that K is an elementary Abelian group which is an absolutely irreducible P -module. By the table of 2-Brauer characters for G2 (4) in [9], combined with the fact that a Sylow 7-subgroup of P acts on K regularly, we have dim(K) = 6, dim(CK (x)) = 2, and dim(CK (y)) = 0 for x ∈ 5A and y ∈ 5C. By [14], CP (x) = hxi × A where A ' A5 and a Sylow 5-subgroup of A is generated by an element of 5C. In particular, A acts on CK (x) faithfully, and hence CG (x) contains an element of order 4, which contradicts Eq. (6). The proposition is proved. Proposition 7. If G is a finite group such that ω(G) = ω(S6 (3)) then G ' S6 (3). Proof. By [14], µ(G) = {13, 7 · 2, 5 · 3 · 2, 5 · 4, 9 · 4, 3 · 8}.

(7)

As in the proof of Proposition 5, it is easy to show that G satisfies condition (3) of Lemma 1, that is, there exists a non-Abelian simple group P such that P ≤ G = G/K ≤ Aut(P ) for some nilpotent normal π1 (G)-subgroup K of G, and G/P is a π1 (G)-group. By Lemma 22, P is isomorphic to one of the groups L2 (13), L2 (25), L2 (27), U3 (4), S4 (5), O7 (3), O8+ (3), L3 (3), L4 (3), G2 (3), Sz(8), or S6 (3). If P ' Sz(8), then P contains a Frobenius group of order 56, and by Lemma 6, K is a 2-group; so, G cannot contain an element of order 9, contrary to Eq. (7). Suppose P ' L3 (3). Since L3 (3) contains a Frobenius group of order 72 with cyclic complement, K is a 3-group by Lemma 6, and the order of G is not divisible by 7, contrary to Eq. (7). The same argument excludes the case where P ' L4 (3). Assume P ' G2 (3). Since P contains subgroups isomorphic to L3 (3) and L2 (8), we see, as in previous paragraph, that K is a 3-group, and then that G contains an element of order 21, which is impossible by

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Eq. (7). Since O8+ (3) > O7 (3) > G2 (3), a similar argument applies to show that P cannot be isomorphic to O7 (3) or O8+ (3). If P is isomorphic to U3 (4) or S4 (5) then 7 divides the order of K. But a Sylow 5- or 3-subgroup of P is non-cyclic, and so G should contain an element of order 35 or 21, which it does not. If P ' L2 (13) then |K| is divisible by 15, and since P contains a Frobenius group of order 13 · 6, G should contain an element of order 15 · 6, which it does not. If P ' L2 (25) then |K| is divisible by 7, and since P contains a Frobenius group of order 25 · 12, G should contain an element of order 7 · 12, which it does not. Similarly, if P ' L2 (27) then |K| is divisible by 5 and G contains an element of order 13 · 5. TABLE 3 G An A6 A10 L2 (q) L3 (q) L3 (3) L3 (5) L3 (9) L4 (3) L5 (3) U3 (2m ) Un (q) S4 (q) S6 (2) S6 (3) O8+ (2) O7 (3) O8+ (3) Sz(22m+1 ) 3 D4 (2) G2 (3m ) G2 (4) 2 G2 (32m+1 ) F4 (2) 2 F4 (22m+1 ) 2 F4 (2)0 2 E6 (2) Sporadic group J2

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Conditions on G n = 5, 16, p, p + 1, p + 2, p > 7 is a prime

q > 3, q 6= 9 q = 7, 2m

m>2 (n, q) ∈ {(3, 9), (3, 11), (4, 3), (6, 2)} (n, q) ∈ {(3, 3), (3, 5), (3, 7), (4, 2), (5, 2)} q = 32m+1 > 3 2m m q = 3, 3 , p , p 6= 3 is a prime

m>1

G 6= J2

h(G) 1 ∞ ∞ 1 1 ∞ 2 2 1 1 1 1 ∞ 1 ∞ 2 1 2 2 2 1 1 1 1 1 1 1 1 1 1 ∞

References [1].21-[1].27, [17] [1].23 [7] [1].1-[1].5 [1], [1].9-[1].11, [1].42 [0] [7].23 [7].24 [1].11 [17] [1], [1].42 [0], [1].12-[1].14 [7] [0] [0], [7], [1], [1].42 [7].14, [7].22 [0] [7].14, [7].22 [7].14 [7].14 [1].6 [0] [1].11, [16] [0] [1].7 [0] [1].8 [1].11 [0] [1].4, [1].14, [1].16-[1].20 [1].23

Thus P ' S6 (3). By [14], G/K = P . We argue that K = 1. Since P contains a Frobenius group of order 27 · 13, K is a 3-group by Lemma 6. Suppose K 6= 1. Since P contains a subgroup isomorphic to L = L2 (13), the table of 3-Brauer characters for L in [9] shows that G contains an element of order 21, which is false. The proposition is proved. Proposition 8. Let L = F4 (2) or L = 2 E 6 (2). If G is a finite group such that ω(G) = ω(L) then G ' L. Proof. By [14], µ(F4 (2)) = {17, 13, 7 · 3, 7 · 4, 5 · 3 · 2, 5 · 4, 9 · 2, 3 · 8, 16},

(8)

µ(2 E 6 (2)) = {19, 17, 13, 11 · 3, 11 · 2, 7 · 5, 7 · 3, 7 · 4, 5 · 3 · 2, 5 · 4, 9 · 2, 3 · 8, 16}.

(9)

Therefore the number of connected components in GK(G) is at least 3, and by Lemmas 1 and 22, P ≤ G = G/K ≤ Aut(P ), where P is isomorphic to F4 (2) or 2 E 6 (2), for some nilpotent normal π1 (G)-subgroup K of G. Since 2 E6 (2) > F4 (2) > 3 D4 (2), we can show that K = 1, as in the last paragraph of the proof of Proposition 5. By [14], P = G. COROLLARY 1. Let P be a finite simple group with disconnected Gruenberg–Kegel graph GK(P ). If one of the connected components in GK(P ) is equal to {13} then either P is isomorphic to one of the groups S4 (5) or L3 (3) and h(P ) = ∞, or P is isomorphic to one of the groups O7 (3) or O8+ (3) and h(P ) = 2, or P is recognizable by the set of its element orders. Proof. By Lemma 22, P is isomorphic to one of A13 , A14 , A15 , L2 (13n ), L2 (25), L2 (27), L3 (3), L4 (3), U3 (4), S4 (5), S6 (3), O7 (3), O8+ (3), G2 (3), G2 (4), F4 (2), Sz(8), 3 D4 (2), 2 F4 (2)0 , 2 E 6 (2), Suz, or F i22 . For the groups L3 (3), S4 (5), 3 D4 (2), S6 (3), G2 (4), F4 (2), and 2 E 6 (2), the conclusion holds by Propositions 1, 3, 5-8. For all other groups, the conclusion is known from the literature (see the Appendix). 7. APPENDIX Table 3 lists finite simple groups which are currently known to be recognizable or non-recognizable by their element order sets. In the column ‘References,’ [i].j stands for the reference [j] in [i]. [0] refers to the present article. Acknowledgement. I am thankful to A. V. Zavarnitsin who read the manuscript of the article and made a number of useful remarks. REFERENCES

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