sets with trivial global stabilizers for primitive permutation groups which are not almost simple

ISSN 0081-5438, Proceedings of the Steklov Institute of Mathematics, 2007, Suppl. 1, pp. S100–S117.c© Pleiades Publishing, Ltd., 2007.Original Russian Text c© A.V. Konygin, 2007,published in Trudy Instituta Matematiki i Mekhaniki UrO RAN, 2007, Vol. 13, No. 1.

Sets with Trivial Global Stabilizers for Primitive

Permutation Groups

Which Are not Almost Simple

A. V. Konygin1

Received October 19, 2006

Abstract—Let G be a primitive permutation group on a finite set X such that the globalstabilizer of any subset of the set X in the group G is nontrivial. The description of G isobtained in the case when G is not almost simple.

DOI: 10.1134/S0081543807050070

INTRODUCTION

In [3] it is proved that (up to permutation isomorphism) there exist only a finite number ofprimitive permutation groups G on finite sets X such that Alt(X) � G and the global stabilizerof any subset R of the set X in the group G is nontrivial (i.e., G{R} �= 1). In this context thequestion of description of primitive permutation groups G on finite sets X such that G{R} �= 1 forany subset R of the set X is of interest (see [3]). At the same time, the question (not consideredin [3]) of an explicit description of a subset R (if any) of the set X with property G{R} = 1 seemsto be topical.

In the present paper both questions are considered in the case when the group G is not almostsimple. Recall that a group G is called almost simple if Inn(T ) � G ≤ Aut(T ) for some simplenonabelian groups T. A special paper will be devoted to description of almost simple groups G suchthat G{R} �= 1 for any subset R of the set X.

Theorem. Let G be a primitive not almost simple permutation group on a finite set X. Theneither there exists a subset R of the set X with the property G{R} = 1, or G is permutationallyisomorphic to one of the following groups:(i) S3 in the natural permutation representation of degree 3;(ii) S4 in the natural permutation representation of degree 4;(iii) A4 in the natural permutation representation of degree 4;(iv) AGL(1, 5) in the natural permutation representation of degree 5;(v) D10 in the natural permutation representation of degree 5;(vi) AGL(1, 7) in the natural permutation representation of degree 7;(vii) AGL(2, 3) in the natural permutation representation of degree 9;

1Institute of Mathematics and Mechanics, Ural Branch of the Russian Academy of Sciences, ul. S. Kovalevskoi 16,Ekaterinburg, 620219 Russia

S100

SETS WITH TRIVIAL GLOBAL STABILIZERS FOR PRIMITIVE PERMUTATION GROUPS S101

(viii) ASL(2, 3) in the natural permutation representation of degree 9;(ix) subgroup of order 144 in group AGL(2, 3) in the natural permutation representation of degree 9;(x) AGL(3, 2) in the natural permutation representations of degree 8;(xi) subgroup of order 168 in group AGL(3, 2) in the natural permutation representations of de-gree 8.

Note that the proof of the theorem gives an explicit form of a subset R of the set X withG{R} = 1 in the case such R exists.

Let G be a permutation group on a finite set X. Define the number D(G) as the minimal naturalnumber n for which there exists a function χ : X → {1, . . . , n} such that the conditions g ∈ G

and χ(g(x)) = χ(x) for all x ∈ X imply g = 1. An analogous notion for automorphism groupsof graphs was introduced in [1]. The theorem stated above asserts that if a primitive group G isnot almost simple, then either D(G) ≤ 2, or G is permutationally isomorphic to one of the groups(i)–(xi) from the theorem.

1. DEFINITIONS AND AUXILIARY STATEMENTS

Let G be a finite permutation group on a set X and R ⊆ X. By G{R} we will denote the globalstabilizer of a subset R in the group G, i.e., the set G{R} = {g ∈ G | g(R) = R}. By GR we willdenote the pointwise stabilizer of the subset R in the group G, i.e., the set GR = {g ∈ G | g(r) = r

for any r ∈ R}.Note that if H ≤ G, then D(H) ≤ D(G). A symmetric (alternating) group of the degree k is

denoted by Sk (Ak). Denote by Inv(G) the set of involutions in an arbitrary group G. Recall thatan element g ∈ G is called strongly real if hgh = g−1 for some involution h from Inv(G).

In what follows, by a graph we mean a finite undirected graph without loops or multiple edges.If Γ is a graph, then V (Γ) is the vertex set of the graph Γ and E(Γ) is the set of its edges. ByΓ(x), for x ∈ V (Γ), we will denote the neighborhood of the vertex x in the graph Γ, i.e., the set ofvertices of the graph Γ adjacent to x in Γ.

Proof of the theorem is based on the O’Nan-Scott theorem [8]. Let G be a primitive permutationgroup on a finite set X. Suppose |X| = n, x ∈ X, B = Soc(G). Since group G is primitive, we haveB ∼= T k, where T is a simple group. O’Nan-Scott theorem states that a finite primitive not almostsimple permutation group G is permutationally isomorphic to a group of one of the following types.

I. B = Zkp, G ≤ AGL(k, p) is an affine group acting on a k-dimensional space over the field

consisting of p elements, Gx = G ∩ GL(k, p) is an irreducible subgroup of the group GL(k, p).

II. k ≥ 2, T is a nonabelian group.

II (a). Let

W ={

(a1, . . . , ak).π | ai ∈ Aut(T ), π ∈ Sk, ai ≡ aj(mod Inn(T )), i, j ∈ {1, . . . , k}}

,

where elements π ∈ Sk permute components ai ∈ Aut(T ) in a natural way. Then we have W =B(Out(T ) × Sk) ≤ (Aut(T ))kSk (where Sk permutes factors in a natural way).

Putting

Wx ={(a, . . . , a).π | a ∈ Aut(T ), π ∈ Sk

}

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS Suppl. 1 2007

S102 KONYGIN

(taking the set of left cosets of Wx in W as a X), we obtain primitive representation of group W

with degree |T |k−1. For 1 ≤ i ≤ k, let Ti be a subgroup of group B, consisting of elements, allprojections of which, different from the ith one, are units, T = {T1, . . . , Tk}. The group W actson the set T in a natural way.

Let G ≤ W and let P be a permutation group induced by the action of group G on T . We saythat G is a group of type II (a) if B ≤ G and either P is primitive on T , or k = 2 and P = 1.

II (b). Let H be a permutation group on a finite set Y, and also H is either almost simpleprimitive permutation group, or primitive permutation group of type II (a). Set W = HwrSl,

where l > 1.The group W acts on X = Y l in a natural way. Let K = Soc(H), B = K l < W. We say that

G is a group of type II (b) if B ≤ G ≤ W and G transitively permutes l factors of group K l.

II (c). Let P be a transitive permutation group on a set {1, . . . , k}, Q = P1 (stabilizer of theunit in P ). Suppose that there exists a homomorphism φ : Q → Aut(T ) such that Inn(T ) ≤ Im(φ).Set

B ={

f : P → T | f(ab) = f(a)φ(b), a ∈ P, b ∈ Q}

.

Then B is a group relative to pointwise multiplication, B ∼= T k. Let P act on B as follows:

fa(a′) = f(aa′), f ∈ B, a, a′ ∈ P.

Let G be a corresponding semidirect product of groups B and P, Gx = P (thus, X is the set of leftadjacent classes of group G on subgroup P ). The group G thus defined is a group of type II (c) if itis primitive. Note that in this case the group G is contained in the group HwrSk, where H ∼= T ×T

is a group of type II (a).The proof of the theorem in the case when the group G is a group of type I is given in Section 3;

in the case when the group G is a group of type II (a), in Section 4; in the case when the group G

is a group of type II (b) or II (c), in Section 5.The following Propositions 1–13, concerning the properties of finite simple nonabelian groups,

are required only for the proof of the theorem in the case when group G is a group of type II (a)(if k �= 2, only Proposition 1 yields the proof).

Proposition 1 is one of the main principal results in paper [9].

Proposition 1. Let T be a finite simple nonabelian group. Then either T = U3(3), or T isgenerated by two elements, one of which is an involution, the other is strongly real. In particular,for any finite simple group T there exist elements t1, t2 ∈ T such that CAut(T )(t1)∩CAut(T )(t2) = 1.

Technical Propositions 2–13 can be strengthened; however in the present form they make itpossible to prove the theorem as well.

Proposition 2. Let T = An, where n ≥ 8. Then there exist elements s1, s2 ∈ T such thatT = 〈s1, s2〉, |s1| �= |s2|, |s1| �= |s−1

1 s2|, |s2| �= |s−11 s2|, s

Aut(T )1 = (s−1

1 )Aut(T ), sAut(T )2 = (s−1

2 )Aut(T )

and

(∗) |Inv(Aut(T ))| + 3∣∣sAut(T )

1

∣∣ + 2∣∣sAut(T )

2

∣∣ < |T |.

Proof. Suppose s1 = (123) and s2 = (12)(3...n) if n is even, and s2 = (3...n) if n is odd. It isdirectly verified that |s1| �= |s2|, |s1| �= |s−1

1 s2| and |s2| �= |s−11 s2|. In addition, it is well-known that

An = 〈s1, s2〉.



Let us prove inequality (∗). It is known that if s ∈ Sn and s in its expansion to independentcycles has ci cycles of the length i, i ∈ {1, ..., n}, then

|sSn | =n!∏n

i=1 ici∏n

i=1 ci!.

Consequently,

|Inv(Sn)| =�n

2�∑

k=1

n!2k(n − 2k)!k!

.

In addition, |sSn1 | =

n!3(n − 3)!

and |sSn2 | =

n!2(n − 2)

. Since

�n2�∑

k=1

n!2k(n − 2k)!k!

+ 3n!

3(n − 3)!+ 2

n!2(n − 2)

<n!2

for n ≥ 8, the proposition is proved. �

Proposition 3. Let T = A7. Then there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉,|s1| = 2, |s2| �= |s−1

1 s2| and

(∗∗) |Inv(Aut(T ))| + 3∣∣sAut(T )

1

∣∣ + 4|CT (s1)||Out(T )| < |T |.

Proof. There exist s1 and s2 from T such that T = 〈s1, s2〉, |s1| = 2, |s2| = 7 and |s−11 s2| = 5.

For group T and so chosen s1 and s2 inequality (∗∗) is verified directly. �For finite group T set m(T ) = min

{|T : CT (x)| | 1 �= x ∈ Aut(T )

}.

Validity of the following statement follows from [4] and Proposition 1.

Proposition 4. Let T be a sporadic simple group, different from M11, M12, M22, M23, M24,

J2, HS. Then there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉, |s1| = 2, sAut(T )2 =

(s−12 )Aut(T ) and

7

√2

m(T )+ 2

|Out(T )|m(T )

+1

m(T )< 1.

Proposition 5. Let T = M11. Then there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉,|s1| = 2, |s2| �= |s−1

1 s2|, sAut(T )2 = (s−1

2 )Aut(T ) and

|Inv(Aut(T ))| + 3|sAut(T )1 | + 2|sAut(T )

2 | < |T |.

Proof. Since T = M11, we conclude that |T | = 7920 and |Inv(Aut(T ))| = 165. Let s1 and s2 bestandard generators of the group T, mentioned in [11]. We have |s1| = 2, |s2| = 4, |s−1

1 s2| = 11 and

sAut(T )2 = (s−1

2 )Aut(T ). Using [4] it is verified that |sAut(T )1 | = 165 and |sAut(T )

2 | = 1980. Consequently,chosen s1 and s2 possess the required properties. �

Proposition 6. Let T be one of groups: M12, M22, M23, M24, J2, HS. Then there existelements s1, s2 ∈ T such that T = 〈s1, s2〉, |s1| = 2, |s2| �= |s−1

1 s2| and

|Inv(Aut(T ))| + 3∣∣sAut(T )

1

∣∣ + 4∣∣sAut(T )

2

∣∣ < |T |.


S104 KONYGIN

Proof. Let s1 and s2 be standard generators of the group T mentioned in [11]. Then using [4]it is verified that for T, s1 and s2 the properties mentioned in Table 1 hold.

Table 1

T |T | |Inv(Aut(T ))| |s1| |sAut(T )1 | |s2| |sAut(T )

2 | |s−11 s2|

M12 95 040 1683 2 ≤ 495 3 ≤ 4400 11M22 443 520 2871 2 ≤ 1155 4 ≤ 41 580 11M23 10 200 960 3795 2 ≤ 3795 4 ≤ 318 780 23M24 244 823 040 43 263 2 ≤ 31 878 3 ≤ 226 688 23J2 604 800 165 2 ≤ 2520 3 ≤ 17 360 7HS 44 352 000 165 2 ≤ 5775 5 ≤ 24 200 11

Direct verification shows that chosen s1 and s2 possess the required properties. �

Proposition 7. Let T = L2(q), where q ≥ 11 or q = 8. Then there exist elements s1, s2 ∈ T

such that T = 〈s1, s2〉, |s1| = 2, |s2| �= |s−11 s2|, s

Aut(T )2 = (s−1

2 )Aut(T ) and

(∗ ∗ ∗) |Inv(Aut(T ))| + 3∣∣sAut(T )

1

∣∣ + 2∣∣sAut(T )

2

∣∣ < |T |.

Proof. Set q = pk. We need the following known facts [6, 7].(i) If k = 1, then L2(q) = 〈x, y〉, where |x| = 2, |x| = 3 and |xy| = p.

(ii) If p = 2 and k > 1, then L2(q) = 〈x, y〉, where |x| = 2, |x| = 3 and |xy| = q − 1.

(iii) If p �= 2, and k > 1, then L2(q) = 〈x, y〉, where |x| = 2, |x| = 3 and |xy| =q − 1

2.

(iv) In group L2(q) there exists only one involution class. The cardinality of this class is equal

to q2 − 1 for q ≡ 0(mod 2),q(q − 1)

2for q ≡ −1(mod 4) and

q(q + 1)2

for q ≡ 1(mod 4).

(v) If p = 3, then in group L2(q) there exist two classes of elements of order 3. These classes

are permuted by an outer automorphism and have cardinalities equal toq2 − 1

2.

(vi) If p �= 3, then in group L2(q) there exists one class of elements of order 3. If q ≡ −1(mod 3),then the cardinality of this class is equal to q(q − 1). If q ≡ 1(mod 3), then the cardinality of thisclass is equal to q(q + 1).

(vii) |L2(q)| =q(q2 − 1)(2, q − 1)

.

(viii) If p = 2 and k is odd, then |Inv(Aut(T ))| = |Inv(T )| = q2 − 1.

(ix) If p = 2 and k is even, then |Inv(Aut(T ))| = |Inv(T )| +√

q(q + 1)2

= q2 − 1 +√

q(q + 1)2

.

(x) If p �= 2 and k is odd, then |Inv(Aut(T ))| = |Inv(T )| +q(q − δ)

2= q2, where |δ| = 1 and

δ ≡ q(mod 4).

(xi) If p �= 2 and k is even, then |Inv(Aut(T ))| = |Inv(T )|+ q(q − δ)2

+√

q(q+1) = q2+√

q(q+1),

where |δ| = 1 and δ ≡ q(mod 4).We set s1 = x and s2 = y, where x and y are from items (i)–(iii). Inequality (∗ ∗ ∗) is verified

directly using the listed properties of the group L2(q). �



Proposition 8. Let T be a finite simple group of the Lie type, different from L2(q) (q ≥ 4),L3(q) (q ≤ 13), L4(q) (q ≤ 5), L5(q) (q ≤ 3), L6(2), L7(2), U3(q) (q ≤ 8), U4(q) (q ≤ 3),U5(2), S4(4), S4(5), S6(2), S8(2), O−

8 (2), O−10(2), O+

8 (2), O+8 (3), Suz(8), G2(3), G2(4), F4(2), F4(3),

2F4(2)′, 2E6(2), E7(2), E7(3). Then there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉, |s1| = 2,

sAut(T )2 = (s−1

2 )Aut(T ) and

7

√2

m(T )+ 2

|sAut(T )2 ||T | +

1m(T )

< 1.

Proof. It is known [9, 10] that if T is a finite simple group of the Lie type, different fromU3(3), then T is generated by two elements s1 and s2, where s2

1 = 1, element s2 chosen from torus,is strongly real and semisimple.

Let T be a finite simple group of the Lie type, different from groups listed in the condition ofProposition 8. From [9,10] and [4] it follows that there exist elements s1, s2 such that T = 〈s1, s2〉,s21 = 1, s

Aut(T )2 = (s−1

2 )Aut(T ) and, in addition, for T and s2, the properties mentioned in Table 2are fulfilled.

Direct verification shows that now s1 and s2 possess the required properties. �

Proposition 9. Let T be one of the groups: L3(3), L3(4), L3(5), L3(7), L3(8), L4(3), L5(2)U3(3), U3(4), U4(2), U4(3), U5(2), S4(4), S4(5), S6(2), S8(2), O−

8 (2), O−10(2), O+

8 (2), Sz(8), G2(3),G2(4), F4(2), 2F4(2)′, 2E6(2). Then there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉, |s1| = 2,|s2| �= |s−1

1 s2| and

|Inv(Aut(T ))| + 3|sAut(T )1 | + 4|CT (s1)||Out(T )| < |T |.

Proof. Let s1 and s2 be standard generators of the group T, mentioned in [11]. Then using [4]it is verified that for T, s1 and s2 the properties mentioned in Table 3 are fulfilled.

Directly verification shows that now s1 and s2 possess the required properties. �

Proposition 10. Let T = U3(5). Then there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉,|s1| �= |s2|, |s1| �= |s−1

1 s2|, |s2| �= |s−11 s2|, s

Aut(T )1 = (s−1

1 )Aut(T ) and

|Inv(Aut(T ))| + 3|sAut(T )1 | + 4|sAut(T )

2 | < |T |.

Proof. Since T = U3(5), we have |T | = 126 000, |Inv(Aut(T ))| = 1575. In addition, accordingto [11] there exist generators s1 and s2 of the group T such that |s1| = 3, |s2| = 5, |s−1

1 s2| = 7 and

sAut(T )1 = (s−1

1 )Aut(T ), and moreover (see [4])∣∣sAut(T )

1

∣∣ = 3500 and∣∣sAut(T )

2

∣∣ ≤ 15 624. �

Proposition 11. Let T = L3(9). Then there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉,|s1| = 2 and

2|Inv(Aut(T ))| + 3|Inv(T )| + 4|CT (s1)||Out(T )| + max{|CT (x)| | 1 �= x ∈ Aut(T )

}< |T |.

Proof. Since T = L3(9), according to [4] we have |T | = 42456 960, |Inv(T )| = 7371,|Inv(Aut(T ))| = 80919, |Out(T )| = 4 and max

{|CT (x)| | 1 �= x ∈ Aut(T )

}= 6048. According to

Proposition 1 there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉 and |s1| = 2, and moreover(see [4]) |CT (s1)| = 5760. It is easy to see that for T and s1 the mentioned inequality is fulfilled. �


S106 KONYGIN

Table 2

T m(T ) |sAut(T )2 ||T |

An(q), n ≥ 2 ≥ 233 ≤ 2 logp(q)(n+1,q−1)

qn−1

2An(q), n ≥ 2 ≥ 279 ≤ 2 logp(q)(n+1,q+1)(q+1)

qn+1−(−1)n+1

Bn(q), n ≥ 2, p = 2 ≥ 350 ≤ 2 logp(q)

qn−1

Cn(q), p = 2 ≥ 314 ≤ logp(q)

qn−1

Cn(q), p �= 2 ≥ 314 ≤ 4 logp(q)

qn−1

2Dn(q), n ≥ 2, (4, qn + 1) = 4 ≥ 764 ≤ 32 logp(q)

(qn−1+1)(q−1)

2Dn(q), n ≥ 2, (4, qn + 1) = 2 ≥ 764 ≤ 8 logp(q)

(qn−1+1)(q−1)

2Dn(q), n ≥ 2, (4, qn + 1) = 1 ≥ 764 ≤ 2 logp(q)

(qn−1+1)(q−1)

Dn(q), n ≥ 5, n odd, (4, qn −1) = 4

≥ 495 ≤ 16 logp(q)

qn−1

Dn(q), n ≥ 5, n odd, (4, qn −1) = 2

≥ 495 ≤ 8 logp(q)

qn−1

Dn(q), n ≥ 5, n odd, (4, qn −1) = 1

≥ 495 ≤ 2 logp(q)

qn−1

Dn(q), n �= 4, n even, (4, qn −1) = 4

≥ 495 ≤ 8 logp(q)(2,q−1)2

(qn2 +(−1)

n2 )2

Dn(q), n �= 4, n even, (4, qn −1) = 2

≥ 495 ≤ 4 logp(q)(2,q−1)2

(qn2 +(−1)

n2 )2

Dn(q), n �= 4, n even, (4, qn −1) = 1

≥ 495 ≤ logp(q)(2,q−1)2

(qn2 +(−1)

n2 )2

D4(q), (4, q4 − 1) = 4 ≥ 495 ≤ 24 logp(q)(2,q−1)2

(q2+1)2

D4(q), (4, q4 − 1) = 2 ≥ 495 ≤ 12 logp(q)(2,q−1)2

(q2+1)2

D4(q), (4, q4 − 1) = 1 ≥ 495 ≤ 3 logp(q)(2,q−1)2

(q2+1)2

2B2(q), q = 22n+1, n ≥ 2 ≥ 1000 ≤ log2(q)q+

√2q+1

3D4(q) ≥ 818 ≤ 3 logp(q)

q4−q2+1

G(q) ≥ 1000 ≤ logp(q)

q2−q+12G2(q), q = 32n+1, n > 1 ≥ 1000 ≤ log3(q)

q+√

3q+1

F4(q) ≥ 1000 ≤ logp(q)

q4−q2+1

2F4(q), q = 22n+1, n ≥ 1 ≥ 1000 ≤ logp(q)

q2+q√

2q+q+√

2q+1

E6(q) ≥ 1000 ≤ logp(q)(3,q−1)

q6+q3+1

2E6(q) ≥ 1000 ≤ logp(q)(3,q+1)

q6−q3+1

E7(q), q �= 2, 3 ≥ 1000 ≤ logp(q)(2,q−1)

(q6−q3+1)(q−1)

E8(q) ≥ 1000 ≤ logp(q)

q8+q7−q5−q4−q3+q+1



Table 3

T |Out(T )| |Inv(T )| |Inv(Aut(T ))| |s1| |s2| |s−11 s2| |CT (s1)|

L3(3) 2 117 351 2 3 13 48L3(4) 12 315 1051 2 4 7 64L3(5) 2 775 3875 2 3 31 480L3(7) 6 2793 8379 2 3 19 672L3(8) 6 4599 37 303 2 3 21 3584L4(3) 4 7371 8451 2 4 13 2880L5(2) 2 6975 20 863 2 5 21 21 504U3(3) 2 63 315 2 6 7 96U3(4) 4 195 1235 2 3 13 320U4(2) 2 315 891 2 5 9 576U4(3) 8 2835 18 243 2 6 7 1152U5(2) 2 3135 22 143 2 5 11 82 944S4(4) 4 4335 5695 2 5 17 3840S4(5) 2 10 075 16 875 2 3 13 480S6(2) 1 5103 5103 2 7 9 23 040S8(2) 1 1 371 135 1 371 135 2 5 17 8 847 360O−

8 (2) 2 69 615 112 591 2 3 17 3072O−

10(2) 2 21 999 615 58 756 863 2 5 33 1 274 019 840O+

8 (2) 6 69 615 107 535 2 5 12 3072Sz(8) 3 455 455 2 4 13 64G2(3) 2 7371 10 179 2 3 13 576G2(4) 2 69 615 90 415 2 5 13 61 440F4(2) 2 355 384 575 447 507 711 2 3 17 754 974 720

2F4(2)′ 2 13 455 13 455 2 3 13 10 2402E6(2) 6 1 323 080482815 2 932 165771263 2 3 19 24 352 464568320

Proposition 12. Let T be one of groups: L3(11), L3(13), L5(3), O+8 (3), F4(3), E7(2),

E7(3). Then there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉, |s1| = 2, (s−12 )Aut(T ) = s

Aut(T )2

and

7

√2

m(T )+ 2

|Out(T )|m(T )

+1

m(T )< 1.

Proof. According to Proposition 1 there exist elements s1, s2 ∈ T such that T = 〈s1, s2〉,

|s1| = 2 and (s−12 )Aut(T ) = s

Aut(T )2 . Using [4] it is verified that for T the properties mentioned in

Table 4 are fulfilled.


S108 KONYGIN

Table 4

T m(T ) |Out(T )|L3(11) ≥ 133 2L3(13) ≥ 183 6L5(3) ≥ 121 2O+

8 (3) ≥ 1000 24F4(3) ≥ 1000 1E7(2) ≥ 1000 1E7(3) ≥ 1000 2

Direct verification shows that the required inequality holds. �

Proposition 13. Let T be one of the groups L4(4), L4(5), L6(2), L7(2). Then there existelements s1, s2 ∈ T such that T = 〈s1, s2〉, |s1| = 2, |s2| �= |s−1

1 s2| and

5

√2

m(T )+ 4

|Out(T )|m(T )

< 1.

Proof. Let s1 and s2 be standard generators of the group T, mentioned in [11]. Then using [4]it is verified that for T, s1, and s2 the properties mentioned in Table 5 are fulfilled.

Table 5

T m(T ) |Out(T )| |s1| |s2| |s−11 s2|

L4(4) ≥ 85 4 2 4 30L4(5) ≥ 165 8 2 3 39L6(2) ≥ 63 2 2 6 63L7(2) ≥ 127 2 2 7 127

Direct verification shows that now chosen T, s1, and s2 possess the required properties. �

2. PROOF OF THE THEOREM IN THE CASE WHEN G IS A GROUP OF TYPE I

In this section, the theorem is proved in the case when G is a group of type I from the O’Nan-Scott theorem (see Section 2).

In the case under consideration, T = Zp for some prime number p and B = Zkp is a unique

minimal normal subgroup of the group G. The group B acts regularly on X, |X| = pk, G ≤AGL(k, p), moreover, B is a group of translations from AGL(k, p).

It is convenient to fix on X a certain structure of vector space V (k, p), determined by thegroup B. Let e1, e2, . . . , ek be some basis of vector space X determined in this way.

Below, subset R (if any) of set X with property G{R} = 1 is specified.Case 1. k ≥ 3 and p ≥ 3.Set R = {0, e1, e2, . . . , ek, 2e1, 2e2, . . . , 2ek, e1 + e2, e2 + e3, . . . , ek−1 + ek, e1 + 2e2}. Let us

show that AGL(k, p){R} = 1. Let g ∈ AGL(k, p) and g(R) = R. Note that 0 is a unique pointfrom X through which k lines of an affine space associated with X pass and each contains threepoints from R. Then g(0) = 0. Since these lines are lines of the form 〈ei〉, i ∈ {1, . . . , k}, we



conclude that g(〈ei〉) = 〈ej〉 for any i ∈ {1, . . . , k}, where j ∈ {1, . . . , k}. For any 1 ≤ i < j ≤ k

the plane 〈ei, ej〉 contains at least six points from R if and only if j = i + 1. In addition, only fori ∈ {1, k−1} does there exist precisely one j ∈ {1, . . . , k−1} such that |〈ei, ei+1〉∩ 〈ej , ej+1〉∩R| = 3.Therefore, g(ek) ∈ {e1, 2e1, ek, 2ek} and g(ek−1) ∈ {e2, 2e2, ek−1, 2ek−1}. Since |〈e1, e2〉 ∩ R| = 7and |〈ek, ek−1〉 ∩ R| = 6, we have g(〈ek, ek−1〉) = 〈ek, ek−1〉. Consequently, g(ek) ∈ {ek, 2ek} andg(ek−1) ∈ {ek−1, 2ek−1}. Since g(ek +ek−1) = ek +ek−1, we find that g(ek) = ek and g(ek−1) = ek−1.

However, then g(〈ek−2, ek−1〉) = 〈ek−2, ek−1〉. Continuing this process, we conclude that g(ei) = ei

for any i ∈ {1, . . . , k}, i.e., g = 1.Since AGL(k, p){R} = 1, we have G{R} = 1 for any group G ≤ AGL(k, p).Case 2. k ≥ 4 and p = 2.Set R = {0, e1, e2, . . . , ek, e1 + e2, e2 + e3, . . . , ek−1 + ek, e1 + e2 + e3}. Let us show that

AGL(k, 2){R} = 1. Let g ∈ AGL(k, 2) and g(R) = R. Since 0 is a unique point from R that iscontained in the maximal number of two-dimensional subspaces containing four points from R inthe affine space, associated with X, we have g(0) = 0.

Now, arguments analogous to those of the previous case show that g(ei) = ei for any i ∈{1, . . . , k}, i.e., g = 1.

Since AGL(k, 2){R} = 1, we have G{R} = 1 for any group G ≤ AGL(k, 2).Case 3. k = 3 and p = 2.If G = AGL(3, 2) or G is a maximal subgroup of group AGL(3, 2) with Gx

∼= Z7 � Z3, x ∈ X,

then subset R with property G{R} = 1 does not exist.There remains the possibility that Gx

∼= Z7, x ∈ X, and as a set R with property G{R} = 1 onecan take any subset of the set X consisting of three elements.

Case 4. k = 2 and p > 7.Set R = {0, e1, 3e1, 2e2, 3e2}. Let us show that AGL(2, p){R} = 1. Let g ∈ AGL(2, p) and

g(R) = R. Then a line containing points 0 and e1 and a line containing points 0 and e2 are fixedunder g. It is easy to see that g = 1.

Since AGL(2, p){R} = 1, we see that G{R} = 1 for any group G ≤ AGL(2, p).Case 5. k = 2 and p = 7.SetR = {0, e1, 3e1,−e1, e2, 2e2}. Let us show that AGL(2, 7){R} = 1. Let g ∈ AGL(2, 7) and

g(R) = R. Since g(0) = 0, a line containing points 0 and e1, and a line containing points 0 and e2

are fixed under g. Then g(e1) = e1 and g(e2) = e2, i.e., g = 1.Since AGL(2, 7){R} = 1, we conclude that G{R} = 1 for any group G ≤ AGL(2, 7).Case 6. k = 2 and p = 5.Set R = {0, e1, 3e1, e2, 2e2, e1 + e2.} Let us show that AGL(2, 5){R} = 1. Let g ∈ AGL(2, 5) and

g(R) = R. Since g(0) = 0, a line containing points 0 and e1 and a line containing points 0 and e2

are fixed under g. Then g(e1) = e1 and g(e2) = e2, i.e., g = 1.Since AGL(2, 5){R} = 1, we have G{R} = 1 for any group G < AGL(2, 5).Case 7. k = 2 and p = 3.If G = AGL(2, 3), G = ASL(2, 3) or G is a subgroup of AGL(2, 3) such that Gx is a Sylow

2-subgroup from GL(2, 3), then a subset R with property G{R} = 1 does not exist.In the case when G is a subgroup of group AGL(2, 3) different from the listed ones, it is easy

to show that G{R} = 1 for R = {0, e1, 2e1, e2}.Case 8. k = 2 and p = 2.If G ≤ AGL(2, 2), then a subset R with property G{R} = 1 does not exist.Case 9. k = 1 and p > 7.


S110 KONYGIN

Set R = {0, e1, 3e1}. It is easy to show that AGL(1, p){R} = 1.Since AGL(1, p){R} = 1, we have G{R} = 1 for any group G ≤ AGL(1, p).Case 10. k = 1 and p = 7.If G = AGL(1, 7), then a subset R with property G{R} = 1 does not exist.Let G < AGL(1, 7); then |Gx| ∈ {2, 3}. If |Gx| = 2, then group G is permutationally isomorphic

to D14 in the natural action on seven points and G{R} = 1 for R = {0, e1, 3e1}. If |Gx| = 3, thenit is easy to show that G{R} = 1 for R = {0, e1, 2e1}.

Case 11. k = 1 and p = 5.If G = AGL(1, 5), then a subset R with property G{R} = 1 does not exist.If G < AGL(1, 5), then either G is permutationally isomorphic to D10 in the natural action on

five points and a set R with property G{R} = 1 does not exist, or G ∼= Z5 and Gx = 1, x ∈ X.Case 12. k = 1 and p = 3.If G = AGL(1, 3), then a subset R with property G{R} = 1 does not exist. In the other case

the stabilizer of a point in G is trivial.Case 13. k = 1 and p = 2.In this case, G{R} = 1 for R = {0}.

The theorem is thus proved for the case when G is a group of type I.

Remark. From the given proof it follows that if G ≤ AGL(k, p) and there exists a subset ofset X, a global stabilizer of which in group G is trivial, then there exists a subset R of set X suchthat |R| ≤ 3k + 1 and G{R} = 1. At the same time, obviously, if AGL(k, p){R} = 1 for some subsetR of set X, then |R| > k.

3. PROOF OF THE THEOREM IN THE CASE WHEN G IS A GROUP

OF TYPE II (a)

In this section, the theorem is proved for the case when G is a group of type II (a) from O’Nan-Scott theorem (see Section 2).

We have

W ={

(a1, . . . , ak).π | ai ∈ Aut(T ), π ∈ Sk, ai ≡ aj(mod Inn(T )), i, j ∈ {1, . . . , k}}

,

where elements π ∈ Sk permute in a natural way components ai, while W acts on X in such a waythat

Wx ={(a, . . . , a).π | a ∈ Aut(T ), π ∈ Sk

}

for some x ∈ X.

Thus, elements of the set X are left cosets (x1, . . . , xk).1Gx, where x1, . . . , xk ∈ Inn(T ), whichfor convenience are denoted by [x1, . . . , xk] (consequently, [x1, . . . , xk] = [x′

1, . . . , x′k] if and only

if x−1i x′

i = x−1j x′

j for all i, j ∈ {1, . . . , k}). In this notation for g = (g1, . . . , gk).σ ∈ W and[x1, . . . , xk] ∈ X, we have

g([x1, . . . , xk]

)=

[g1xσ(1), . . . , gkxσ(k)

]=

[1, g2xσ(2)x

−1σ(1)g

−11 , . . . , gkxσ(k)x

−1σ(1)g

−11

].

To prove the theorem in the case when G is a group of type II (a), let us set G = W and showthat there exists a subset R of the set X such that G{R} = 1.

Case 1. k > 2.



By virtue of Proposition 1 we have CAut(T )(t1) ∩ CAut(T )(t2) = 1 for some elements t1,

t2 ∈ T . As a set R let us take a subset of the set X, consisting of elements [1, 1, . . . , 1, t1],[1, 1, . . . , t1, t1], . . . , [1, t1, . . . , t1, t1] and element [1, t2, . . . , t2, t2]. Let us show that if g =(g1, . . . , gk).σ ∈ G and g(R) = R, then g = 1.

Let K = {1, . . . , k} and i ∈ K\{1, k}. Then the ith coordinates of elements from g(R) are gig−11 ,

git1g−11 , git2g

−11 , if σ(1) = 1; gig

−11 , git1g

−11 , if 1 �= σ(1) < σ(i); gig

−11 , git

−11 g−1

1 , if 1 �= σ(1) > σ(i).Thus, if σ(1) �= 1, then the set of values of the ith coordinate of elements from R consists at mostof two elements, which contradicts the choice of R. Consequently, σ(1) = 1 and g([x1, . . . , xk])= [1, g2xσ(2)g

−11 , . . . , gkxσ(k)g

−11 ].

Let us show that σ(k) = k and g = (g1, gt21 , . . . , gt2

1 ).σ. Let σ(k) = i, where i ∈ K. Then theset of values of the ith coordinate of elements from R consists of elements git1g

−11 , git2g

−11 and,

in particular, contains at most two elements. According to the choice of R it follows that eitheri = 1, or i = k. Since σ(1) = 1, we have σ(k) = k, whence gkt1g

−11 = t1 and gkt2g

−11 = t2.

Since now g([1, t2, . . . , t2]) = [1, g2t2g−11 , . . . , gk−1t2g

−11 , t2] ∈ R, we conclude that g([1, t2, . . . , t2]) =

[1, t2, . . . , t2]. Therefore, git2g−11 = t2 for any i ∈ K \ {1}. Thus, we have g = (g1, g

t21 , . . . , gt2

1 ).σ.

Let us show that g1 = 1. We have g([1, . . . , 1, t1]) = [1, gt21 g−1

1 , . . . , gt21 g−1

1 , gt21 t1g

−11 ] ∈ R. Since

t1 �= 1, it follows that [1, gt21 g−1

1 , . . . , gt21 g−1

1 , gt21 t1g

−11 ] = [1, 1, . . . , 1, t1]. Thus, gt2

1 g−11 = 1 and

gt21 t1g

−11 = t1. Consequently, gt2

1 = g1 and t1 = gt21 t1g

−11 = g1t1g

−11 . It follows that g1 ∈ CAut(T )(t1)∩

CAut(T )(t2) = 1.

Thus, g = (g1, gt21 , . . . , gt2

1 ).σ = (1, . . . , 1).σ, which together with g(R) = R obviously impliesg = 1.

Case 2. k = 2.Let s1 ∈ T and Γ is a (undirected) graph, constructed by the orbital of the group G on

X, containing ([1, 1], [1, s1]). (Note that this orbital is self-dual, because (s1, s21).π[1, 1] = [1, s1]

and (s1, s21).π[1, s1] = [s2

1, s21] = [1, 1] for π ∈ S2 \ {1}.) In addition, if [1, x], [1, y] ∈ V (Γ), then

{[1, x], [1, y]} ∈ E(Γ) if and only if x−1y ∈ sAut(T )1 ∪ (s−1

1 )Aut(T ).

Suppose that T = A5. Set s1 = (12)(34) and s2 = (135). It is verified directly that the globalstabilizer of the set {[1, 1], [1, s2], [1, s2

2s1]} ∪ Γ([1, 1]) in the group G is trivial.Further we assume that T �= A5. Let us show that it is possible to choose elements s1, s2 and s3

from T in such a way that the global stabilizer of the set {[1, 1], [1, s1], [1, s2], [1, s3]} in the groupG is trivial.

Lemma 1. Let s1, s2, s3 ∈ T such that CAut(T )(s1)∩CAut(T )(s2) = 1, |s1| �= |s2|, |s1| �= |s−11 s2|

and the following conditions hold:(i) if g ∈ Aut(T ), g2 = 1, gs1g

−1 = s−11 and gs2g

−1 = s−12 , then s3 �= gs−1

3 g−1;

(ii) s3 /∈ sAut(T )1 ∪ (s−1

1 )Aut(T ) ∪ s1sAut(T )1 ∪ s1(s−1

1 )Aut(T ) ∪ s2sAut(T )1 ∪ s2(s−1

1 )Aut(T );(iii) if g ∈ CAut(T )(s1) and g2 = 1, then s3 �= gs2g

−1;

(iv) if g ∈ Aut(T ), g2 = 1 and gs1g−1 = s−1

1 , then s3 �= gs−12 g−1;

(v) if g ∈ Aut(T ), g2 = 1, gs1g−1 = s−1

1 and gs2g−1 = s−1

1 s2, then s3gs−13 �= s1g;

(vi) if g ∈ CAut(T )(s1), g2 = s−11 and gs2g

−1 = s−12 s1, then s3 /∈ g−1Inv(〈T, g〉);

(vii) if g ∈ Aut(T ), g2 = 1 and gs1g−1 = s−1

1 , then s3 �= s1gs2g−1;

(viii) if g ∈ CAut(T )(s1), g2 = s−11 , then s3 �= s1gs−1

2 g−1.

Then the global stabilizer of the set {[1, 1], [1, s1], [1, s2], [1, s3]} in the group G is trivial.


S112 KONYGIN

Proof. Let R = {[1, 1], [1, s1], [1, s2], [1, s3]}. Let us show that a pointwise stabilizer of the setR is trivial. Suppose that (g1, g2).π ∈ GR. Then, obviously, g1 = g2 and either π = 1 and g1 = 1,or π �= 1 and g2

1 = 1. In the latter case we obtain a contradiction with condition (i). Thus, GR = 1.Let Δ be a subgraph of graph Γ generated by the set of vertices R, and H = GΔ

{R} (i.e.,H is the restriction of the group of automorphisms G{R} of the graph Γ to the subgraph Δ).Let us show that {[1, 1], [1, s1 ]} is a unique edge in the graph Δ. If {[1, 1], [1, s2 ]} ∈ E(Γ), then

s2 ∈ sAut(T )1 ∪ (s−1

1 )Aut(T ) and hence |s1| = |s2|, which contradicts the choice of s1 and s2.

If {[1, 1], [1, s3 ]} ∈ E(Γ), then s3 ∈ sAut(T )1 ∪ (s−1

1 )Aut(T ), which contradicts the condition (ii).

If {[1, s1], [1, s2]} ∈ E(Γ), then s−11 s2 ∈ s

Aut(T )1 ∪ (s−1

1 )Aut(T ) and hence |s1| = |s−11 s2|, which

contradicts the choice of s1 and s2. If {[1, s1], [1, s3]} ∈ E(Γ), then s−11 s3 ∈ s

Aut(T )1 ∪ (s−1

1 )Aut(T ),

which contradicts the condition (ii). Finally, if {[1, s2], [1, s3]} ∈ E(Γ), then s−12 s3 ∈ s

Aut(T )1 ∪

(s−11 )Aut(T ), which contradicts the condition (ii). Thus, the sets {[1, 1], [1, s1 ]} and {[1, s2], [1, s3]}

are fixed by each element from H.Let us show that H = 1. Suppose that there exists an element (g1, g2).π ∈ G{R} such that

(g1, g2).π[1, 1] = [1, 1], (g1, g2).π[1, s1] = [1, s1], (g1, g2).π[1, s2] = [1, s3] and (g1, g2).π[1, s3] = [1, s2].Then from (g1, g2).π[1, 1] = [1, 1] it follows that (g1, g2).π = (g, g).π for some g ∈ Aut(T ).

Further, by virtue of (g1, g2).π[1, s1] = [1, s1] we have g2 ∈ CAut(T )(s1), by virtue of (g1,

g2).π[1, s2]=[1, s3] and (g1, g2).π[1, s3]=[1, s2] we have g2 ∈ CAut(T )(s2), which, in view ofCAut(T )(s1)∩CAut(T )(s2) = 1, implies g2 = 1. If π = 1, then sg

1 = s1 and s3 = sg2, which contradicts

the condition (iii). If π �= 1, then sg1 = s−1

1 and s3 = (s−12 )g, which contradicts condition (iv).

Suppose that there exists an element (g1, g2).π ∈ G{R} such that (g1, g2).π[1, 1] = [1, s1],(g1, g2).π[1, s1] = [1, 1], (g1, g2).π[1, s2] = [1, s2] and (g1, g2).π[1, s3] = [1, s3]. If π = 1, then(g1, g2).π = (g, s1g).1, where g ∈ Aut(T ) and g2 = 1. In addition, sg

1 = s−11 , sg

2 = s−11 s2 and

sg3 = s−1

1 s3, which contradicts condition (v). If π �= 1, then (g1, g2).π = (g, s1g).π, where g ∈ Aut(T )and g2 = s−1

1 . In addition, sg1 = s1, gs2g

−1 = s−12 s1 and gs3g

−1 = s−13 s1. The last equality can be

rewritten as (gs3)2 = 1, whence gs3 ∈ Inv(Aut(〈T, g〉)) and hence s3 ∈ T ∩ g−1Inv(Aut(〈T, g〉)).This contradicts condition (vi).

Suppose that there exists an element (g1, g2).π ∈ G{R} such that (g1, g2).π(1, 1) = [1, s1],(g1, g2).π(1, s1) = [1, 1], (g1, g2).π[1, s2] = [1, s3] and (g1, g2).π[1, s3] = [1, s2]. If π = 1, then(g1, g2).π = (g, s1g).1, where g ∈ Aut(T ) and g2 = 1. In addition,sg

1 = s−11 , sg

2 = s−11 s3 and

sg3 = s−1

1 s2, which contradicts condition (vii). If π �= 1, then (g1, g2).π = (g, s1g).1, where g ∈Aut(T ) and g2 = s−1

1 . In addition, sg1 = s1 and s3 = s1gs−1

2 g−1, which contradicts condition (viii).�

Lemma 2. Suppose that T = 〈s1, s2〉 and at least one of the following conditions holds:

(i) orders |s1|, |s2|, |s−11 s2| are pairwise different, (s−1

1 )Aut(T ) = sAut(T )1 , (s−1

2 )Aut(T ) = sAut(T )2

and |Inv(Aut(T ))| + 3|sAut(T )1 | + 2|sAut(T )

2 | < |T |;(ii) |s1| = 2, |s2| �= |s−1

1 s2| and |Inv(Aut(T ))| + 3|sAut(T )1 | + 4|CT (s1)||Out(T )| < |T |;

(iii) |s1| = 2, (s−12 )Aut(T ) = s

Aut(T )2 and 7

√2

m(T )+ 2

|Out(T)|m(T )

+1

m(T )< 1;

(iv) |s1| = 2, |s2| = |s−11 s2| and |Inv(Aut(T ))| + 3|sAut(T )

1 | + 4|sAut(T )2 | < |T |;

(v) |s1| = 2, |s2| = |s−11 s2|, (s−1

2 )Aut(T ) = sAut(T )2 and |Inv(Aut(T ))| + 3|sAut(T )

1 | +

2|sAut(T )2 | < |T |;



(vi) |s1| = 2, (s−12 )Aut(T ) = s

Aut(T )2 and 7

√2

m(T )+ 2

|sAut(T )2 |m(T )

+1

m(T )< 1;

(vii) orders |s1|, |s2|, |s−11 s2| are pairwise different, (s−1

1 )Aut(T ) = sAut(T )1 and |Inv(Aut(T ))| +

3|sAut(T )1 |+ 4|sAut(T )

2 | < |T |;(viii) |s1| = 2 and

2|Inv(Aut(T ))| + 3|Inv(T )| + 4|CT (s1)||Out(T )| + max{|CT (x)| | 1 �= x ∈ Aut(T )} < |T |;

(ix) |s1| = 2, |s2| �= |s−11 s2| and 5

√2

m(T )+ 4

|Out(T)|m(T )

< 1;

Then there exists an element s3 ∈ T such that T, s1, s2 and s3 satisfy the condition of Lemma 1.

Proof. For any finite group H of even order with Z(H) = 1 and some x ∈ H \ {1} we have

|Inv(H)| ≤√

2|H : CH(x)| |H| (see proof (45.3) in [2]). In particular,

|Inv(T )| ≤√

2m(T )

|T |, (1)

|Inv(〈T, g〉)| ≤ 2

√2

m(T )|T | (2)

for any g ∈ Aut(T ) with g2 ∈ T.

Further, if g ∈ Aut(T ) and g2 = 1, then∣∣{t ∈ T | gtg−1 = t−1}

∣∣ ≤ |Inv(〈T, g〉)| (3)

(since from gtgt = 1 it follows that gt ∈ Inv(〈T, g〉)). In addition, if g ∈ Aut(T ) and s ∈ T, thenobviously

∣∣{t ∈ T | tgt−1 = sg}∣∣ ≤ |CT (g)|. (4)

Let T = 〈s1, s2〉 and M1, . . . ,M8 are sets of elements s3 ∈ T such that for T, s1, s2 and s3,respectively, conditions (i)–(viii) of Lemma 1 are violated. Then it is necessary to prove thatT \ (M1 ∪ . . . ∪ M8) �= ∅. Let us show that

|M1 ∪ . . . ∪ M8| < |T |. (5)

Suppose that for T, s1 and s2 condition (i) holds. By virtue of (3) we have |M1| ≤ |Inv(Aut(T ))|.Since (s−1

1 )Aut(T ) = sAut(T )1 , we have |M2| ≤ 3|sAut(T )

1 |. Since (s−12 )Aut(T ) = s

Aut(T )2 , we see that

|M3∪M4| ≤ |sAut(T )2 | and |M7∪M8| ≤ |sAut(T )

2 |. Since |s2| �= |s−11 s2|, we derive that |M5| = |M6| = 0.

Thus (5) is true.Suppose that for T, s1 and s2 condition (ii) holds. By virtue of (3) we have |M1| ≤ |Inv(Aut(T ))|.

Since |s1| = 2, we conclude that |M2| ≤ 3|sAut(T )1 | and |M3| + |M4| + |M7| + |M8| ≤

4|CAut(T )(s1)| ≤ 4|CT (s1)||Out(T )|. Since |s2| �= |s−11 s2|, we see that |M5| = |M6| = 0. Thus,

(5) is established.Suppose that for T, s1 and s2 condition (iii) holds. By virtue of (3) and (2) we have |M1| ≤

2√

2m(T )

|T |. Since |s1| = 2, by virtue of (i) we have |M2| ≤ 3|sAut(T )1 | ≤ 3|Inv(T )| ≤ 3

√2

m(T )|T |.


S114 KONYGIN

In addition, |M3 ∪ M4| ≤ |sAut(T )2 | ≤ |Out(T )|

m(T )|T | and |M7 ∪ M8| ≤ |sAut(T )

2 | ≤ |Out(T )|m(T )

|T |. By

virtue of (4) we have |M5| ≤ 1m(T ) |T |. By virtue of (2) we obtain |M6| ≤ 2

√2

m(T )|T |. Thus, (5) is

true.Suppose that for T, s1 and s2 condition (iv) holds. Then we obtain the inequalities |M1| ≤

|Inv(Aut(T ))|, |M2| ≤ 3|sAut(T )1 | and |M3| + |M4| + |M7| + |M8| ≤ 4|sAut(T )

2 |. Since |s2| �= |s−11 s2|,

we see that |M5| = |M6| = 0. Thus, (5) is true.Suppose that for T, s1 and s2 condition (v) holds. Then we obtain the inequalities |M1| ≤

|Inv(Aut(T ))| and |M2| ≤ 3|sAut(T )1 |. Since (s−1

2 )Aut(T ) = sAut(T )2 , we have |M3 ∪ M4| ≤ |sAut(T )

2 |and |M7∪M8| ≤ |sAut(T )

2 |. Since |s2| �= |s−11 s2|, we see that |M5| = |M6| = 0. Thus, (5) is established.

Suppose that for T, s1, and s2 condition (vi) holds. By virtue of (3) and (2) we have |M1| ≤

2√

2m(T )

|T | and |M5| ≤ 1m(T )

. Since |s1| = 2, by virtue of (1) we have |M2| ≤ 3|sAut(T )1 | ≤

3|Inv(T )| ≤ 3√

2m(T )

|T |. In addition, |M3 ∪ M4| ≤ |sAut(T )2 | and |M7 ∪ M8| ≤ |sAut(T )

2 |. By virtue

of (2) we have |M6| ≤ 2√

2m(T )

. Thus, (5) is true.

Suppose that for T, s1, and s2 condition (vii) holds. Then we obtain the inequalities |M1| ≤|Inv(Aut(T ))| and |M2| ≤ 3|sAut(T )

1 |. In addition, |M3| + |M4| + |M7| + |M8| ≤ 4|sAut(T )2 |. Since

|s2| �= |s−11 s2|, we have |M5| = |M6| = 0. Thus, (5) is true.

Suppose that for T, s1 and s2 condition (viii) holds. Then we obtain the inequalities |M1| ≤|Inv(Aut(T ))|, |M2| ≤ 3|sAut(T )

1 |, and |M3| + |M4| + |M7| + |M8| ≤ 4|CT (s1)||Out(T )|. In addition,|M5| ≤ max{|CT (x)| | 1 �= x ∈ Aut(T )} and |M6| ≤ |Inv(Aut(T ))|. Thus, (5) holds.

Suppose, finally, that for T, s1, and s2 condition (ix) holds. By virtue of (3) and (2) we have

|M1| ≤ 2√

2m(T )

|T |. Since |s1| = 2, we see that |M2| ≤ 3√

2m(T )

|T | by virtue of (1). In addition,

|M3| + |M4| + |M7| + |M8| ≤ 4|sAut(T )2 | ≤ 4

|Out(T )|m(T )

|T |. Since |s2| �= |s−11 s2|, we conclude that

|M5| = |M6| = 0. Thus, (5) is true. �

Lemma 3. Let T be one of the groups A6 and L2(7). Then there exist elements s1, s2 ands3 ∈ T such that T, s1, s2, and s3 satisfy the conditions of Lemma 1.

Proof. Let T = A6 and s1, s2 are standard generators of the group T, mentioned in [11]. Wehave |s1| = 2, |s2| = 4 and |s−1

1 s2| = 5. Now, direct verification shows that there exists an elements3 ∈ T such that for T, s1, s2, and s3 the conditions of Lemma 1 hold.

Let T = L2(7) and s1, s2 are standard generators of the group T, mentioned in [11]. We have|s1| = 2, |s2| = 7 and |s−1

1 s2| = 3. Now, direct verification shows that there exists element s3 ∈ T

such that for T, s1, s2, and s3 the conditions of Lemma 1 hold. �

Proof of the theorem in the case, when G is a group of type II (a).Suppose that T = An, where n ≥ 8. Let s1, s2 be the elements of group T, chosen according to

Proposition 2. Then T = 〈s1, s2〉 and for T, s1, s2 condition (i) of Lemma 2 holds. Consequently,according to Lemmas 2 and 1 there is an element s3 ∈ T such that, setting R = {[1, 1], [1, s1],[1, s2], [1, s3]}, we have G{R} = 1.



Suppose that T = A6. Let s1, s2, s3 be the elements of group T, chosen according to Lemma 3.Then, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, by virtue of Lemma 1 we have G{R} = 1.

Suppose that T = A7. Let s1, s2 be the elements of group T, chosen according to Proposition 3.Then T = 〈s1, s2〉 and for T, s1, s2 condition (ii) of Lemma 2 holds. Consequently, by Lemmas 2and 1 there is an element s3 ∈ T such that, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, we haveG{R} = 1.

Suppose that T is a sporadic simple group, different from groups M11, M12, M22, M23, M24, J2

and HS. Let s1, s2 be elements of group T, chosen according to Proposition 4. Then T = 〈s1, s2〉and for T, s1, s2 condition (iii) of Lemma 2 holds. Consequently, by Lemmas 2 and 1 there is anelement s3 ∈ T such that, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, we have G{R} = 1.

Suppose that T = M11. Let s1, s2 be elements of group T, chosen according to Proposition 5.Then T = 〈s1, s2〉 and for T, s1, s2 condition (v) of Lemma 2 holds. Consequently, by Lemmas 2and 1 there is an element s3 ∈ T such that, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, we haveG{R} = 1.

Suppose that T is one of the following groups: M12, M22, M23, M24, J2, and HS. Let s1, s2

be elements of group T, chosen according to Proposition 6. Then T = 〈s1, s2〉 and for T, s1, s2

condition (iv) of Lemma 2 holds. Consequently, by Lemmas 2 and 1 there is an element s3 ∈ T

such that, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, we have G{R} = 1.Suppose that T = L2(q), where q ≥ 11 or q = 8. Let s1, s2 be elements of group T, chosen

according to Proposition 7. Then T = 〈s1, s2〉 and for T, s1, s2 condition (v) of Lemma 2 holds.Consequently, by Lemmas 2 and 1 there is an element s3 ∈ T such that, setting R = {[1, 1], [1, s1],[1, s2], [1, s3]}, we have G{R} = 1.

Suppose that T = L2(7). Let s1, s2, s3 be elements of group T, chosen according to Lemma 3.Then, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, by virtue of Lemma 1 we have G{R} = 1.

Suppose that T is a group of Lie type, different from the following groups: L2(q) (q ≥ 4), L3(q)(q ≤ 13), L4(q) (q ≤ 5), L5(q) (q ≤ 3), L6(2), L7(2), U3(q) (q ≤ 8), U4(q) (q ≤ 3), U5(2), S4(4),S4(5), S6(2), S8(2), O−

8 (2), O−10(2), O+

8 (2), O+8 (3), Suz(8), G2(3), G2(4), F4(2), F4(3), 2F4(2)′,

2E6(2), E7(2), E7(3). Let s1, s2 be elements of group T, chosen according to Proposition 8. ThenT = 〈s1, s2〉 and for T, s1, s2 condition (vi) of Lemma 2 holds. Consequently, by Lemmas 2 and 1there is an element s3 ∈ T such that, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, we have G{R} = 1.

Suppose that T is one of the following groups: L3(3), L3(4), L3(5), L3(7), L3(8), L4(3), L5(2)U3(3), U3(4), U4(2), U4(3), U5(2), S4(4), S4(5), S6(2), S8(2), O−

8 (2), O−10(2), O+

8 (2), Sz(8), G2(3),G2(4), F4(2), 2F4(2)′, 2E6(2). Let s1, s2 be elements of group T, chosen according to Proposition 9.Then T = 〈s1, s2〉 and for T, s1, s2 condition (ii) of Lemma 2 holds. Consequently, by Lemmas 2and 1 there is an element s3 ∈ T such that, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, we haveG{R} = 1.

Suppose that T = U3(5). Let s1, s2 be elements of group T, chosen according to Proposition 10.Then T = 〈s1, s2〉 and for T, s1, s2 condition (vii) of Lemma 2 holds. Consequently, by Lemmas 2and 1 there is an element s3 ∈ T such that, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, we haveG{R} = 1.

Suppose that T = L3(9). Let s1, s2 be elements of group T, chosen according to Proposition 11.Then T = 〈s1, s2〉 and for T, s1, s2 condition (viii) of Lemma 2 holds. Consequently, by Lemmas 2and 1 there is an element s3 ∈ T such that, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, we haveG{R} = 1.

Suppose that T is one of the following groups: L3(11), L3(13), L5(3), O+8 (3), F4(3), E7(2),


S116 KONYGIN

E7(3). Let s1, s2 be elements of group T, chosen according to Proposition 12. Then T = 〈s1, s2〉and for T, s1, s2 condition (iii) of Lemma 2 holds. Consequently, by Lemmas 2 and 1 there is anelement s3 ∈ T such that, setting R = {[1, 1], [1, s1], [1, s2], [1, s3]}, we have G{R} = 1.

Suppose that T is one of the following groups: L4(4), L4(5), L6(2), L7(2). Let s1, s2 be elementsgroup T, chosen according to Proposition 13. Then T = 〈s1, s2〉 and for T, s1, s2 condition (iv) ofLemma 2 holds. Consequently, by Lemmas 2 and 1 there is an element s3 ∈ T such that, settingR = {[1, 1], [1, s1], [1, s2], [1, s3]}, we have G{R} = 1.

The validity of the theorem in the case when G is a group of type II (a) is proved. �

Remark. From the given proof it follows that if G is a group of type II (a), then there existsa subset R of the set X such that G{R} = 1 and, in addition, |R| ≤ k for k > 2, |R| ≤ 4 for k = 2and T �= A5.

4. PROOF OF THE THEOREM IN THE CASE WHEN G IS A GROUP

OF TYPE II (b) OR II (c)

In this section, the theorem is proved for the case when G is a group of type II (b) or II (c)from the O’Nan-Scott theorem (see Section 2).

SetW = SmwrSk, where m ≥ 5 and k ≥ 2, and let M = {1, . . . ,m}. We have

W = {(h1, . . . , hk)π | hi ∈ Sm, π ∈ Sk},

and if w = (h1, . . . , hk)π ∈ W, (x1, . . . , xk) ∈ Mk, then w(x1, . . . , xk) = (h1(x1), . . . , hk(xk))π. Inthe case under consideration, G is permutationally isomorphic to a subgroup of the group W.

To prove the theorem in the case when G is a group of type II (b) or II (c), we set G = W andwill show that there exists a subset R of the set X such that G{R} = 1.

Case 1. k = 2.Set R = {(3, 2), (i − 1, i), (i, i) | i ∈ {3, . . . ,m}}. Let g = (h1, h2)π ∈ G and g(R) = R. Let us

show that g = 1. Since |{(3, j) | (3, j) ∈ R}| = 3 and |{(j, i) | (j, i) ∈ R}| < 3 for any i ∈ M, wehave π = 1. Since |{(j, 2) | (j, 2) ∈ R}| = 1 and |{(j, i) | (j, i) ∈ R}| �= 1 for any i ∈ M \ {2}, wehave h2(2) = 2. By virtue of the fact that |{(3, j) | (3, j) ∈ R}| = 3 and |{(i, j) | (i, j) ∈ R}| < 3for any i ∈ M \ {3}, we have h1(3) = 3. Consequently, g(3, 2) = (3, 2) and g(3, 3) = (3, h2(3)).Since g(3, 3) ∈ R, we see that h2(3) ∈ {3, 4}. Suppose that h2(3) = 4. Then g(3, 3) = (3, 4),which, taking into account that g(3, 4) ∈ R, implies g(3, 4) = (3, 3). Now, g(4, 4) = (h1(4), 3),which, taking into account that g(4, 4) ∈ R, implies h1(4) ∈ {2, 3}. By virtue of h1(3) = 3 wehave h1(4) = 2. Consequently, g(4, 5) = (2, h2(5)), whence, taking into account that g(4, 5) ∈ R,we obtain h2(5) = 3, which contradicts the equality g(4, 4) = (2, 3). Therefore, h2(3) = 3 andg(3, 3) = (3, 3). Since g(2, 3) ∈ R, taking into account that g(3, 3) = (3, 3), we have g(2, 3) = (2, 3).However, if g(i − 1, i) = (i − 1, i) and g(i, i) = (i, i) are true for some i ∈ {3, . . . ,m − 1}, then byvirtue of g(i, i + 1) ∈ R we have g(i, i + 1) = (i, i + 1). Consequently, taking into account thatg(i + 1, i + 1) ∈ R, we obtain the equality g(i + 1, i + 1) = (i + 1, i + 1). Thus, g = 1.

Case 2. k > 2.As the elements of subset R of set X let us take the ordered set (3, 2, 2, . . . , 2) and all ordered sets

of the form (j+1, . . . , j+1) and of the form (j, . . . , j, j+1, . . . , j+1), where j ∈ {2, . . . ,m−1}.Thus, |R| = k(m − 2) + 1. Let g = (h1, . . . , hk)π ∈ G and g(R) = R. Let us show that g = 1.



Let K = {1, . . . , k}. For i ∈ K and j ∈ M we set Pi,j = |{(x1, . . . , xk) | (x1, . . . , xk) ∈ R,xi = j}|and Pi = |{Pi,j | j ∈ M}|. Since Pi,1 = 0 and Pi,j �= 0 for all i ∈ K and j ∈ M \ {1}, we havehi(1) = 1 for any i ∈ K. Since P1 = 5 and Pi < 5 for i ∈ K \ {1}, it follows that π(1) = 1.

Let us show that hi(m) = m for any i ∈ K and h1(m−1) = m−1. Since P1,m = 1 and P1,j �= 1for m �= j ∈ M, we conclude that h1(m) = m. It follows that (m,hπ(2)(m), . . . , hπ(k)(m)) =(h1(m), hπ(2)(m), . . . , hπ(k)(m)) = g(m,m, . . . ,m) ∈ R, whence hi(m) = m for any i ∈ K. Now,g(m − 1,m, . . . ,m) = (h1(m − 1),m, . . . ,m) ∈ R, whence h1(m − 1) = m − 1.

Let us show that π = 1. It was earlier proved that π(1) = 1 and h1(m − 1) = m − 1. Supposethat, for some i ∈ {2, . . . ,m − 1}, π(j) = j and hj(m − 1) = m − 1 for any j ≤ i − 1. Letx1 = . . . = xi = m − 1 and xi+1 = . . . = xk = m. Then, since hi′(m) = m for any i′ ∈ K andg(x1, . . . , xk) ∈ R, we have hi(m − 1) = m − 1 and π(i) = i. It follows that π = 1.

Suppose that hi(j′′) = j′′ holds for some j′ ∈ {3, . . . ,m} and any j′′ ∈ {j′ + 1, . . . ,m}, i ∈ K.Let us show that hi(j′) = j′ for any i ∈ K. Since g(j′, j′ + 1, . . . , j′ + 1) = (h1(j′), j′, . . . , j′) ∈ R,

taking into account that h1(j′ + 1) = j′ + 1, we have h1(j′) = j′. Suppose that for some i ∈ K

and for all i′ ∈ {1, . . . , i − 1} the following equality holds: hi′(j′) = j′. Let x1 = . . . = xi = j′ andxi+1 = . . . = xk = j′ + 1. Then from g(x1, . . . , xk) ∈ R, taking into account that hi(j′ + 1) = j′ + 1,it follows that hi(j′) = j′. Thus, hi(j′) = j′ for any i ∈ K. It follows that hi(j) = j for anyj ∈ {3, . . . ,m}, i ∈ K.

Further, for any i ∈ K we have hi(2) ∈ M \ {hi(3), . . . , hi(m)} = {1, 2}, which, taking intoaccount that hi(1) = 1, implies hi(2) = 2. Thus, g = 1.

The validity of the theorem in the case when G is a group of type II (b) or II (c) is proved. �

The proof of the theorem is thus complete.

Remark. From the given proof it follows that if G ≤ SmwrSk, where m ≥ 5 and k ≥ 2, thenthere exists a subset R of the set X such that |R| ≤ k(m − 2) + 1 and G{R} = 1.

ACKNOWLEDGMENTS

This work was supported by the Russian Foundation for Basic Research (project no. 06-01-00378).

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4. J. H. Conway, R. T. Curtis, S. P. Norton, et al, Atlas of Finite Groups (Clarendon Press, Oxford, 1985).

5. J. D. Dixon and B. Mortimer, Permutation Groups (Springer, New York, 1996).

6. P. Fong and W. J. Wong, Nagoya Math. J. 36, 143 (1969).

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Translated by S. Tarasova


sets with trivial global stabilizers for primitive permutation groups which are not almost simple

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