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On the Movement of a Permutation Group
Journal of Algebra 214, 631᎐635 Ž1999. Article ID jabr.1998.7705, available online at http:rrwww.idealibrary.com on
On the Movement of a Permutation Group Peter M. Neumann The Queen’s College, Oxford OX1 4AW, England
and Cheryl E. Praeger Department of Mathematics, Uni¨ ersity of Western Australia, Perth, Western Australia 6907, Australia Communicated by Alexander Lubotzky Received July 23, 1997
1. THE THEOREM In w1x the second-named author defined the movement of a permutation group Ž G, ⍀ . by the formula m [ move Ž G . [ sup < ⌫ g _ ⌫ < ⌫ : ⍀ , g g G4 and proved that if G has no fixed points and m is finite then n F 5m y 2, where n is the degree of G, that is, n [ < ⍀ <. In w2x it is shown that equality holds if and only if n s 3 and G is transitive. In this note we aim to improve the bound: Ž1. If G has no fixed points and moveŽ G . s m then THEOREM. n F 12 Ž9m y 3.; Ž2. equality holds infinitely often; Ž3. moreo¨ er, if n s 12 Ž9m y 3. then either n s 3 and G s SymŽ ⍀ . or G is an elementary abelian 3-group and all its orbits ha¨ e length 3. Part Ž2. is proved by examples; Ž1. and Ž3. will be proved using a sequence of lemmas. 631 0021-8693r99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
632
NEUMANN AND PRAEGER
2. THE PROOF EXAMPLES. Let d be a positive integer, let G [ Z3d, let t [ 12 Ž3 d y 1., and let H1 , . . . , Ht be an enumeration of the subgroups of index 3 in G. Define ⍀ i to be the coset space of Hi in G and ⍀ [ ⍀ 1 j ⭈⭈⭈ j ⍀ t . If g g G _ 14 then g lies in 12 Ž3 dy 1 y 1. of the groups Hi and therefore acts on ⍀ as a permutation with 12 Ž3 d y 3. fixed points and 3 dy1 disjoint 3-cycles. Taking one point from each of these 3-cycles to form a set ⌫ we see that mŽ G . G 3 dy 1 , and it is not hard to prove that in fact mŽ G . s 3 dy 1. Thus n s 3t s 12 Ž3 dq 1 y 3. s 12 Ž9m y 3.. This proves Part Ž2. of the theorem. To prove Parts Ž1. and Ž3. we introduce the following notation: r 3 [ number of G-orbits of length 3 on which G acts as Alt Ž 3 . ; r 3X [ number of G-orbits of length 3 on which G acts as Sym Ž 3 . ; and then r 2 [ number of G-orbits of length 2; r4 [ number of G-orbits of length 4; s [ number of G-orbits of length G 5. The orbits are labelled accordingly: thus ⍀ 1 , . . . , ⍀ r 3 are those of length 3 on which G acts as AltŽ3.; ⍀ r 3q1 , . . . , ⍀ r 3qr X3 are those of length 3 on which G acts as SymŽ3.; ⍀ r 3qr X3q1 , . . . , ⍀ r 3qr X3qr 2 are those of length 2, etc. Define 0qr 4 t [ r 3 q r 3X q r 2 q r4 q s, t 0 [ r 3 q r 3X q r 2 , ⌺4 [ D tist ⍀ i , and ⌺5 [ 0 q1 t D ist 0qr 4q1 ⍀ i . For 1 F i F t 0 let K i be the kernel of the action of G on ⍀ i and for g g G let k Ž g . be the number of i in that range for which g f K i . For g g G and a G-invariant set ⌺ define fix ⌺ Ž g . [ the number of fixed points of g in ⌺, supp ⌺ Ž g . [ the size of the support of g in ⌺ Žso that fix ⌺ Ž g . q supp ⌺ Ž g . s < ⌺ <., and odd ⌺ Ž g . [ the number of non-trivial cycles of g in ⌺ that have odd length. LEMMA 1. With this notation, let ⌺ [ D tist 0q1 ⍀ i and let g g G. Then k Ž g . q 12 supp ⌺ Ž g . y 12 odd ⌺ Ž g . F m. Proof. For each i such that 1 F i F t 0 and g f K i choose a point of ⍀ i not fixed by g; then let ⌫0 be the set of chosen points. For each non-trivial cycle Ž ␣ 1 ␣ 2 ⭈⭈⭈ ␣ k . of g in ⌺ adjoin the points ␣ 1 , ␣ 3 , . . . , ␣ k ⬘ to ⌫0 , where k⬘ is odd and k y 2 F k⬘ F k y 1. Let ⌫ be the resulting set. It has been constructed so that ⌫ g l ⌫ s ⭋. Therefore < ⌫ < F m. Since < ⌫ < s k Ž g . q 12 supp ⌺ Ž g . y 12 odd ⌺ Ž g ., we have the stated inequality.
THE MOVEMENT OF A PERMUTATION GROUP
633
LEMMA 2. With the notation defined abo¨ e, Ý g g G k Ž g . s < G <Ž 23 r 3 q r q 21 r 2 ..
5 X 6 3
0 Proof. First observe that k Ž g . s Ýtis1 k i Ž g ., where k i Ž g . [ 1 if g f K i and k i Ž g . [ 0 if g g K i . Now
¡
ggG
2 3
if 1 F i F r 3 ,
5 6
if r 3 q 1 F i F r 3 q r 3X ,
1 2
if r 3 q r 3X q 1 F i F t 0 ,
and so t0
kŽ g. s
Ý ggG
Ý Ý
t0
ki Ž g . s
ggG is1
Ý Ý k i Ž g . s < G < Ž 23 r 3 q 56 r 3X q 12 r 2 . , is1 ggG
as the lemma states. n - 92 m y Ž 34 r 3X q 14 r 2 q 54 r4 q 12 Ž<Ý5 < y 3s ...
LEMMA 3.
Proof. We intend to exploit Lemma 1 by averaging over G and in order to do this we examine Ý g g G Žsupp ⌺ jŽ g . y odd ⌺ jŽ g .. for j g 4, 54 , where, recall, ⌺4 is the union of the orbits of length 4 and ⌺5 is the union of the orbits of length G 5. For any orbit ⍀ i of length 4 we find that Ý g g G odd ⍀ iŽ g . F 23 < G < Žthe sum is 0 if G ⍀ i is a 2-group, it is 23 < G < if G ⍀ i s AltŽ ⍀ i ., and it is 13 < G < if G ⍀ i s SymŽ ⍀ i ... Therefore Ý g g G odd ⌺ 4Ž g . F 23 < G < ⭈ r4 . Also,
Ý Ž < ⌺4 < y fix ⌺ Ž g . . s < G < Ž <Ý4 < y r4 . ,
supp ⌺ 4Ž g . s
Ý ggG
4
ggG
by Not Burnside’s Lemma, and therefore
Ý Ž supp ⌺ Ž g . y odd ⌺ Ž g . . G < G < Ž < ⌺4 < y r4 y 23 r4 . s < G < ⭈ 73 r4 . 4
4
ggG
Similarly, since odd ⌺ 5Ž g . F 13 supp ⌺ 5Ž g .,
Ý Ž supp ⌺ Ž g . y odd ⌺ Ž g . . 5
5
ggG
G
2 3
Ý ggG
supp ⌺ 5Ž g . s
2 3
Ý Ž < ⌺5 < y fix ⌺ Ž g . . s 23 < G < Ž < ⌺5 < y s . . 5
ggG
634
NEUMANN AND PRAEGER
Now take the inequality k Ž g . q 12 supp ⌺ Ž g . y 12 odd ⌺ Ž g . F m from Lemma 1 and sum over G. Using Lemma 2 and the above inequalities we find that m< G< )
Ý k Ž g . q Ý 12 Ž supp ⌺ Ž g . y odd ⌺ Ž g . . 4
ggG
q
4
ggG
Ý 12 Ž supp ⌺ Ž g . y odd ⌺ Ž g . . 5
5
ggG
G < G < Ž 23 r 3 q 56 r 3X q 12 r 2 q 76 r4 q 13 < ⌺5 < y 13 s . , where the strict inequality recognizes the fact that the inequality of Lemma 1 is strict for the identity element of G. Now n s 3r 3 q 3r 3X q 2 r 2 q 4 r4 q < ⌺5 < and so m ) 23 r 3 q 56 r 3X q 21 r 2 q 67 r4 q 31 < ⌺5 < y 13 s s 29 n q 16 r 3X q Multiplication by
9 2
r q
1 18 2
r q 19 < ⌺5 < y 13 s.
5 18 4
and rearrangement yields the inequality of the lemma.
Now define [ 9m y 2 n. Clearly is an integer and from Lemma 3 we have that 2 ) 3r 3X q r 2 q 5r4 q 2 Ž < ⌺5 < y 3s . G 0. Since ) 0, in fact G 1. To prove the theorem we suppose that F 3 and seek to discover what configurations may occur. Then 2 F 6 and so, since certainly < ⌺5 < G 5s there are only the following possibilities: ŽI. r 3X s r4 s 0, r 2 is 0 or 1, s s 1, and < ⌺5 < s 5; ŽII. r 3X s r 2 s 0, r4 s 1, < ⌺5 < s s s 0; ŽIII. 3r 3X q r 2 F 5, r4 s < ⌺5 < s s s 0. Cases ŽI. and ŽII. cannot arise for arithmetical reasons: in these cases we would have s 3 Žgiven the assumption that F 3., so that, since 2 n s 9m y , the degree n would be a multiple of 3, in contradiction to the fact that n is 3r 3 q 5 or 3r 3 q 7 or 3r 3 q 4 in these cases. Case ŽIII. falls to the following lemmas. LEMMA 4. If r 3 s r4 s s s 0 then n F 4 m y 1 F 12 Ž9m y 3.. The equality n s 21 Ž9m y 3. holds only when m s 1, r 2 s 0, r 3X s 1, so that G s SymŽ3..
THE MOVEMENT OF A PERMUTATION GROUP
635
Proof. Suppose that r 3 s r4 s s s 0. It follows from Lemmas 1 and 2 that then < G < m ) < G <Ž 12 r 2 q 56 r 3X .. But also n s 2 r 2 q 3r 3X and therefore m ) 41 n q 121 r 3X . Thus n - 4 m, and so n F 4 m y 1. Since 4 m y 1 s 1 1 1 Ž . Ž . Ž . 2 9m y 3 y 2 m y 1 , if n s 2 9m y 3 then m s 1, n s 3, and G s SymŽ3.. LEMMA 5.
If r 3X s r 2 s r4 s s s 0 then n F 12 Ž9m y 3..
Proof. In this case our inequalities merely say that 2 n - 9m. On the other hand, n s 3r 3 , so that n is a multiple of 3. Therefore 2 n F 9m y 3, as required. To complete the proof of the theorem we need to deal with Case ŽIII.. Suppose therefore that r4 s < ⌺5 < s s s 0. Suppose also that r 3 ) 0 and r3 r 3X q r 2 ) 0. Define ⌺1 [ D is1 ⍀ i , the union of the orbits of length 3 on t0 which G acts as AltŽ3., and ⌺ 2 [ D isr ⍀ i , the union of those orbits of 3 q1 length 3 on which G acts as SymŽ3. and those of length 2. Then define K 1 to be the kernel of the action of G on ⌺ 1 and K 2 the kernel of its action on ⌺ 2 . Clearly, K 1 l K 2 s 14 since G acts faithfully on ⍀. Now let H be the subgroup of G generated by its 2-elements. Then G ⌺ 2 s H ⌺ 2 , that is, G s HK 2 . But H F K 1 and therefore G s K 1 K 2 , that is, G s K 1 = K 2 . It follows easily that if m1 [ moveŽ G ⌺ 1 . and m 2 [ moveŽ G ⌺ 2 . then m s m1 q m 2 . Defining n1 [ < ⌺ 1 < and n 2 [ < ⌺ 2 <, we have from Lemma 5 that n1 F 12 Ž9m1 y 3., and from Lemma 4 that n 2 F Ž4 m 2 y 1., and so n s n1 q n 2 F 21 Ž9m y m 2 y 5. - 21 Ž9m y 3.. What this has shown is that in Case ŽIII., if F 3, that is, if n G X 1 Ž . 2 9m y 3 , then either r 3 s 0 or r 2 q r 3 s 0. If r 3 s 0 then we have the 1 situation of Lemma 4, so that n F 2 Ž9m y 3. with equality if and only if G s SymŽ3.. If r 2 q r 3X s 0 then of course r 2 s r 3X s 0 and we have the situation of Lemma 5. In this case n F 12 Ž9m y 3. by that lemma, and, since G is a subdirect product of copies of AltŽ3., it is an elementary abelian 3-group. Thus the proof of the theorem is complete. ACKNOWLEDGMENT We are grateful to Avinoam Mann who drew our attention to mistakes in the first version of this note.
REFERENCES 1. C. E. Praeger, On permutation groups with bounded movement, J. Algebra 144 Ž1991., 436᎐442. 2. J. R. Cho, P. S. Kim, and C. E. Praeger, The maximal number of orbits of a permutation group with bounded movement, J. Algebra, in press.
On the Movement of a Permutation Group Peter M. Neumann The Queen’s College, Oxford OX1 4AW, England
and Cheryl E. Praeger Department of Mathematics, Uni¨ ersity of Western Australia, Perth, Western Australia 6907, Australia Communicated by Alexander Lubotzky Received July 23, 1997
1. THE THEOREM In w1x the second-named author defined the movement of a permutation group Ž G, ⍀ . by the formula m [ move Ž G . [ sup < ⌫ g _ ⌫ < ⌫ : ⍀ , g g G4 and proved that if G has no fixed points and m is finite then n F 5m y 2, where n is the degree of G, that is, n [ < ⍀ <. In w2x it is shown that equality holds if and only if n s 3 and G is transitive. In this note we aim to improve the bound: Ž1. If G has no fixed points and moveŽ G . s m then THEOREM. n F 12 Ž9m y 3.; Ž2. equality holds infinitely often; Ž3. moreo¨ er, if n s 12 Ž9m y 3. then either n s 3 and G s SymŽ ⍀ . or G is an elementary abelian 3-group and all its orbits ha¨ e length 3. Part Ž2. is proved by examples; Ž1. and Ž3. will be proved using a sequence of lemmas. 631 0021-8693r99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
632
NEUMANN AND PRAEGER
2. THE PROOF EXAMPLES. Let d be a positive integer, let G [ Z3d, let t [ 12 Ž3 d y 1., and let H1 , . . . , Ht be an enumeration of the subgroups of index 3 in G. Define ⍀ i to be the coset space of Hi in G and ⍀ [ ⍀ 1 j ⭈⭈⭈ j ⍀ t . If g g G _ 14 then g lies in 12 Ž3 dy 1 y 1. of the groups Hi and therefore acts on ⍀ as a permutation with 12 Ž3 d y 3. fixed points and 3 dy1 disjoint 3-cycles. Taking one point from each of these 3-cycles to form a set ⌫ we see that mŽ G . G 3 dy 1 , and it is not hard to prove that in fact mŽ G . s 3 dy 1. Thus n s 3t s 12 Ž3 dq 1 y 3. s 12 Ž9m y 3.. This proves Part Ž2. of the theorem. To prove Parts Ž1. and Ž3. we introduce the following notation: r 3 [ number of G-orbits of length 3 on which G acts as Alt Ž 3 . ; r 3X [ number of G-orbits of length 3 on which G acts as Sym Ž 3 . ; and then r 2 [ number of G-orbits of length 2; r4 [ number of G-orbits of length 4; s [ number of G-orbits of length G 5. The orbits are labelled accordingly: thus ⍀ 1 , . . . , ⍀ r 3 are those of length 3 on which G acts as AltŽ3.; ⍀ r 3q1 , . . . , ⍀ r 3qr X3 are those of length 3 on which G acts as SymŽ3.; ⍀ r 3qr X3q1 , . . . , ⍀ r 3qr X3qr 2 are those of length 2, etc. Define 0qr 4 t [ r 3 q r 3X q r 2 q r4 q s, t 0 [ r 3 q r 3X q r 2 , ⌺4 [ D tist ⍀ i , and ⌺5 [ 0 q1 t D ist 0qr 4q1 ⍀ i . For 1 F i F t 0 let K i be the kernel of the action of G on ⍀ i and for g g G let k Ž g . be the number of i in that range for which g f K i . For g g G and a G-invariant set ⌺ define fix ⌺ Ž g . [ the number of fixed points of g in ⌺, supp ⌺ Ž g . [ the size of the support of g in ⌺ Žso that fix ⌺ Ž g . q supp ⌺ Ž g . s < ⌺ <., and odd ⌺ Ž g . [ the number of non-trivial cycles of g in ⌺ that have odd length. LEMMA 1. With this notation, let ⌺ [ D tist 0q1 ⍀ i and let g g G. Then k Ž g . q 12 supp ⌺ Ž g . y 12 odd ⌺ Ž g . F m. Proof. For each i such that 1 F i F t 0 and g f K i choose a point of ⍀ i not fixed by g; then let ⌫0 be the set of chosen points. For each non-trivial cycle Ž ␣ 1 ␣ 2 ⭈⭈⭈ ␣ k . of g in ⌺ adjoin the points ␣ 1 , ␣ 3 , . . . , ␣ k ⬘ to ⌫0 , where k⬘ is odd and k y 2 F k⬘ F k y 1. Let ⌫ be the resulting set. It has been constructed so that ⌫ g l ⌫ s ⭋. Therefore < ⌫ < F m. Since < ⌫ < s k Ž g . q 12 supp ⌺ Ž g . y 12 odd ⌺ Ž g ., we have the stated inequality.
THE MOVEMENT OF A PERMUTATION GROUP
633
LEMMA 2. With the notation defined abo¨ e, Ý g g G k Ž g . s < G <Ž 23 r 3 q r q 21 r 2 ..
5 X 6 3
0 Proof. First observe that k Ž g . s Ýtis1 k i Ž g ., where k i Ž g . [ 1 if g f K i and k i Ž g . [ 0 if g g K i . Now
¡
ggG
2 3
if 1 F i F r 3 ,
5 6
if r 3 q 1 F i F r 3 q r 3X ,
1 2
if r 3 q r 3X q 1 F i F t 0 ,
and so t0
kŽ g. s
Ý ggG
Ý Ý
t0
ki Ž g . s
ggG is1
Ý Ý k i Ž g . s < G < Ž 23 r 3 q 56 r 3X q 12 r 2 . , is1 ggG
as the lemma states. n - 92 m y Ž 34 r 3X q 14 r 2 q 54 r4 q 12 Ž<Ý5 < y 3s ...
LEMMA 3.
Proof. We intend to exploit Lemma 1 by averaging over G and in order to do this we examine Ý g g G Žsupp ⌺ jŽ g . y odd ⌺ jŽ g .. for j g 4, 54 , where, recall, ⌺4 is the union of the orbits of length 4 and ⌺5 is the union of the orbits of length G 5. For any orbit ⍀ i of length 4 we find that Ý g g G odd ⍀ iŽ g . F 23 < G < Žthe sum is 0 if G ⍀ i is a 2-group, it is 23 < G < if G ⍀ i s AltŽ ⍀ i ., and it is 13 < G < if G ⍀ i s SymŽ ⍀ i ... Therefore Ý g g G odd ⌺ 4Ž g . F 23 < G < ⭈ r4 . Also,
Ý Ž < ⌺4 < y fix ⌺ Ž g . . s < G < Ž <Ý4 < y r4 . ,
supp ⌺ 4Ž g . s
Ý ggG
4
ggG
by Not Burnside’s Lemma, and therefore
Ý Ž supp ⌺ Ž g . y odd ⌺ Ž g . . G < G < Ž < ⌺4 < y r4 y 23 r4 . s < G < ⭈ 73 r4 . 4
4
ggG
Similarly, since odd ⌺ 5Ž g . F 13 supp ⌺ 5Ž g .,
Ý Ž supp ⌺ Ž g . y odd ⌺ Ž g . . 5
5
ggG
G
2 3
Ý ggG
supp ⌺ 5Ž g . s
2 3
Ý Ž < ⌺5 < y fix ⌺ Ž g . . s 23 < G < Ž < ⌺5 < y s . . 5
ggG
634
NEUMANN AND PRAEGER
Now take the inequality k Ž g . q 12 supp ⌺ Ž g . y 12 odd ⌺ Ž g . F m from Lemma 1 and sum over G. Using Lemma 2 and the above inequalities we find that m< G< )
Ý k Ž g . q Ý 12 Ž supp ⌺ Ž g . y odd ⌺ Ž g . . 4
ggG
q
4
ggG
Ý 12 Ž supp ⌺ Ž g . y odd ⌺ Ž g . . 5
5
ggG
G < G < Ž 23 r 3 q 56 r 3X q 12 r 2 q 76 r4 q 13 < ⌺5 < y 13 s . , where the strict inequality recognizes the fact that the inequality of Lemma 1 is strict for the identity element of G. Now n s 3r 3 q 3r 3X q 2 r 2 q 4 r4 q < ⌺5 < and so m ) 23 r 3 q 56 r 3X q 21 r 2 q 67 r4 q 31 < ⌺5 < y 13 s s 29 n q 16 r 3X q Multiplication by
9 2
r q
1 18 2
r q 19 < ⌺5 < y 13 s.
5 18 4
and rearrangement yields the inequality of the lemma.
Now define [ 9m y 2 n. Clearly is an integer and from Lemma 3 we have that 2 ) 3r 3X q r 2 q 5r4 q 2 Ž < ⌺5 < y 3s . G 0. Since ) 0, in fact G 1. To prove the theorem we suppose that F 3 and seek to discover what configurations may occur. Then 2 F 6 and so, since certainly < ⌺5 < G 5s there are only the following possibilities: ŽI. r 3X s r4 s 0, r 2 is 0 or 1, s s 1, and < ⌺5 < s 5; ŽII. r 3X s r 2 s 0, r4 s 1, < ⌺5 < s s s 0; ŽIII. 3r 3X q r 2 F 5, r4 s < ⌺5 < s s s 0. Cases ŽI. and ŽII. cannot arise for arithmetical reasons: in these cases we would have s 3 Žgiven the assumption that F 3., so that, since 2 n s 9m y , the degree n would be a multiple of 3, in contradiction to the fact that n is 3r 3 q 5 or 3r 3 q 7 or 3r 3 q 4 in these cases. Case ŽIII. falls to the following lemmas. LEMMA 4. If r 3 s r4 s s s 0 then n F 4 m y 1 F 12 Ž9m y 3.. The equality n s 21 Ž9m y 3. holds only when m s 1, r 2 s 0, r 3X s 1, so that G s SymŽ3..
THE MOVEMENT OF A PERMUTATION GROUP
635
Proof. Suppose that r 3 s r4 s s s 0. It follows from Lemmas 1 and 2 that then < G < m ) < G <Ž 12 r 2 q 56 r 3X .. But also n s 2 r 2 q 3r 3X and therefore m ) 41 n q 121 r 3X . Thus n - 4 m, and so n F 4 m y 1. Since 4 m y 1 s 1 1 1 Ž . Ž . Ž . 2 9m y 3 y 2 m y 1 , if n s 2 9m y 3 then m s 1, n s 3, and G s SymŽ3.. LEMMA 5.
If r 3X s r 2 s r4 s s s 0 then n F 12 Ž9m y 3..
Proof. In this case our inequalities merely say that 2 n - 9m. On the other hand, n s 3r 3 , so that n is a multiple of 3. Therefore 2 n F 9m y 3, as required. To complete the proof of the theorem we need to deal with Case ŽIII.. Suppose therefore that r4 s < ⌺5 < s s s 0. Suppose also that r 3 ) 0 and r3 r 3X q r 2 ) 0. Define ⌺1 [ D is1 ⍀ i , the union of the orbits of length 3 on t0 which G acts as AltŽ3., and ⌺ 2 [ D isr ⍀ i , the union of those orbits of 3 q1 length 3 on which G acts as SymŽ3. and those of length 2. Then define K 1 to be the kernel of the action of G on ⌺ 1 and K 2 the kernel of its action on ⌺ 2 . Clearly, K 1 l K 2 s 14 since G acts faithfully on ⍀. Now let H be the subgroup of G generated by its 2-elements. Then G ⌺ 2 s H ⌺ 2 , that is, G s HK 2 . But H F K 1 and therefore G s K 1 K 2 , that is, G s K 1 = K 2 . It follows easily that if m1 [ moveŽ G ⌺ 1 . and m 2 [ moveŽ G ⌺ 2 . then m s m1 q m 2 . Defining n1 [ < ⌺ 1 < and n 2 [ < ⌺ 2 <, we have from Lemma 5 that n1 F 12 Ž9m1 y 3., and from Lemma 4 that n 2 F Ž4 m 2 y 1., and so n s n1 q n 2 F 21 Ž9m y m 2 y 5. - 21 Ž9m y 3.. What this has shown is that in Case ŽIII., if F 3, that is, if n G X 1 Ž . 2 9m y 3 , then either r 3 s 0 or r 2 q r 3 s 0. If r 3 s 0 then we have the 1 situation of Lemma 4, so that n F 2 Ž9m y 3. with equality if and only if G s SymŽ3.. If r 2 q r 3X s 0 then of course r 2 s r 3X s 0 and we have the situation of Lemma 5. In this case n F 12 Ž9m y 3. by that lemma, and, since G is a subdirect product of copies of AltŽ3., it is an elementary abelian 3-group. Thus the proof of the theorem is complete. ACKNOWLEDGMENT We are grateful to Avinoam Mann who drew our attention to mistakes in the first version of this note.
REFERENCES 1. C. E. Praeger, On permutation groups with bounded movement, J. Algebra 144 Ž1991., 436᎐442. 2. J. R. Cho, P. S. Kim, and C. E. Praeger, The maximal number of orbits of a permutation group with bounded movement, J. Algebra, in press.