The Double Transitivity of a Class of Permutation Groups Citation Bercov, Ronald David (1962) The Double Transitivity of a Class of Permutation Groups. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/2T36-NC78. https://resolver.caltech.edu/CaltechTHESIS:07282011-085251999 Abstract In this thesis primitive finite permutation groups G with regular abelian subgroup H are studied. It is shown that if, for an odd prime p, H has a Sylow p-subgroup which is the direct product of two cyclic groups of different order, then G is doubly transitive. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Hall, Marshall Thesis Committee: Unknown, Unknown Defense Date: 1 January 1962 Record Number: CaltechTHESIS:07282011-085251999 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:07282011-085251999 DOI: 10.7907/2T36-NC78 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 6551 Collection: CaltechTHESIS Deposited By: Tony Diaz Deposited On: 24 Aug 2011 18:10 Last Modified: 27 Nov 2023 23:27 Thesis Files Preview PDF - Final Version See Usage Policy. 13MB Repository Staff Only: item control page

THE DOUBLE TRANSITIVITY OF A CLASS OF PERMUTATION GROUPS Thesis by Ronald David Bercov In Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1962 ACKNOWLEDGMENTS I wish to express my appreciation to my advisor Professor Marshall Hall Jr. for his interest and guidance. I am grateful to the General Electric Foundation for a fellowship. I would like also to thank Professor Helmut Wielandt for his help and encouragement. ABSTRACT In this thesis primitive finite permutation groups G with regular abelian subgroup H are studied. It is shown that if, for an odd prime p, H has a Sylow p-subgroup which is the direct product . of two cyclic groups of different order, then G is doubly transitive. - 1 - I Introduction The object of this thesis is to show that certain finite abelian groups cannot occur as regular subgroups of uniprimitive (primitive but not doubly transitive) permutation groups. Thus we conclude that primitive groups with such a regular abelian subgroup are necessarily doubly transitive. The first result of this nature was obtained by Burnside who showed that cyclic groups of order p cur as regular subgroups of given in [1] m (p prime, m uniprimitive groups. » 1) do not ocThe proof is , p. 343. For this reason Wielandt has chosen to call such abstract groups B-groups _ / Burnside conjectured that every abelian group which is not elementary abelian is a B-group_ This conjecture is not correct. A class of counter-examples was found by Dorothy Manning in 1936. This class of counter-examples has been generalized by Wielandt and will be given below. The first advance beyond Burnside's result was obtained by Schur [2J in 1933- He showed that every cyclic group of composite order is a B-group_ In 1935 Wielandt [3] generalized this result by showing that every abelian group of composite order which has at least one cyclic Sylow subgroup is a B-group. In 1937 Kochend~rffer [4J generalized the Burnside result in a different direction by showing that every abelian group of type (po(, pI?> ) with ~"> /3 is a B-group. This thesis is a simultaneous generalization of the results - 2 of Wielandt and Kochend8rffer. We show that for any odd prime p every abelian group of composite order which has at least one Sylow subgroup of type (po<., pf» with 0( >{!> is a B-group. We now give the Wielandt class of counter-examples to the Burnside conjecture. Let H = Hl x &2 ••• x Hd with 1Hl ' = IH2 f =. •• = J Hd ' = a > 2 and d > 1 (where IHil is the order of Hi). Then H is not a B-group. Thus for any such H there exists a uniprimitive group with a regular subgroup isomorphic to H. No as- sumption is made on the structure of the H .• ~ The proof is given in [5], an unpublished set of notes from lectures given by ~ielandt at TUbingen in 1954. We mention that two classes of non-abelian B-groups are known as well. Wielandt [6] showed that every dihedral group is a B-group, [7] has shown that every generalized dicyclic group is and Scott a B-group. This thesis is a direct generalization of ~4~ in the sense that the arguments apply whether or not the regular subgroup is a p-group or not. (i.e. The case in which the Sylow subgroup is cyclic f3 = 0) requires a slightly different argument, however. Thus we mention that the arguments given in this thesis can be adapted to give a somewhat different proof of 11.'ielandt' s result in the case ~ = 0, but for clarity of the presentation, we assume that the regular subgroup has a non-cyclic Sylow subgroup of type (p 0<., P p) with 0<. "> f3 (i.e. we assume that 13 1=- ° holds). - 3 II Notation, Definitions, and Theorems from the Theory of Schur Rings Let G be a given permutation group on the letters a ,···, an l with regular subgroup H. We denote the image of the letter a. under the permutation ~ g G by a . • g ~ We regard G as a permutation group on H in the following way. We distinguish the letter a lo Since H is regular there is a unique h h , taking a j l into a Clearly hI j H, which we call = 1, ••• , n. = 1, the identity element of H. one-to-o~e The for j E: / mapping j ~ h. enables us to replace the J letters a , ••• , an by the elements hl, ••• ,h of H. l n To the permutation g permutation (~g) f G (on (on H) where h g {al, ••• ,a } ) corresponds the n is the element of H uniquely determined by the formula al h g = a l hg Let R(H) be the group ring of H over the ring of rational integerso For n (j) - 'L - £ '1. = h t H l(h) hER (H) and any integer j we put h-T-H ~ (h) h j, and I '1.1 = li.'l( (h) h I = ~ i (h) • ~ With K C H we associate K E R(H) defined by K = h ~ ?f (h) h where ¥ (h) = [ Thus 1 0 if h l:. K if h ~ K I!' is the number of elements in K. Let G be the subgroup of G (considered as a permutation group l - 4 on H) consisting of those elements of G fixing 1, the identity element of R (thus G corresponds to G ). l a l = T °, T 1 , ••• ,T k be the sets of transitivity of Gl , where Ti c: H for i = O, ••• ,k. k ¥ i Ti ( ~ i integers) form an i?u additive subgroup of R(R). Clearly the elements Definition 1: A Schur-module (S-modu1e) over R is an additive subgroup of R(H) which has a basis ! l ' ••• '!t where Ki c.. H for i Ki (\ K j = l,o •• ,t, = /1 for 1 ~ i .::::: j ~ t and t i ~ K> = 1-J. = H. Let R(R, G ) be the additive subgroup of R(R) spanned by the l - Ti, J.> = °, ••• , k • Then clearly R(R,G ) is an S-module. l Definition 2: A Schur-ring (S-ring) over H is an S-module over H which is in addition a subring of R(R) containing the multiplicative identity 1 and containing 'L (-1) = i. i(h)h- l whenever it contains Cl = 2:. ¥(h) h. Theorem 1: (Schur, 1933) R(H,G ) is an S-ring. l Definition 3: An S-ring ~ is called primitive if K = 1 and K = H are the only subgroups K of H for which ! ~ ~ holds. - 5 Theorem 2: G is a primitive group if and only if R(H,G ) is a primitive l S-ring. Theorem 3: Let ..i. be a primitive S-ring, 1.. e.g, 'L ~ y.};. Then the elements h t H actually appearing in ~ (i.e. with non-zero coefficient) generate H. Theorem 4: Let -i be an S-ring over the abelian group H of order n. Let j be an integer. Let 'l. t J. Then: =1 (a) (j, n) (b) If j =/p is a prime divisor of n and if ~ is primitive, then Q(p) = S'};(modP) holds for an appropriate integer <f . (The congruence is understood, of course, to hold for the coefficients.) Proofs of theorems 1-4 are found in reference 2. They are given in terms of somewhat different, but equivalent, concepts. Definition 4: Let '1. ~ R(H). If (j,n) = 1, r.L(j) is said to be conjugate to ~ • Definition 5: If 'l = fl(j) for all j with (j,n) = 1, i.e. if fl is its only conjugate, (,L is said to be rational. - 6 Definition 6: Let '1. t R(H) Then the sum of all (distinct) conjugates of tL '1 , and is denoted by tr. the trace of is called (Il). Tr. (rz..) is obviously rational and by theorem 4(a) lies in the S -ring g whenever rz. lies in .£ . Definition 7: For h E:: H, the trace of ih} is called the elementary trace associated with h and is denoted by tr. (h). Clearly if k has non-zero coefficient in the elementary trace associated with h, then the elementary traces associated with hand with k are identical. It is also fairly easily seen that the conjugates of the Ti are again of this form: Theorem 5: .(j ) q (j .n) = 1 ~ T~ = T for some. q with 0 ~ q ~ k. Proof: To see this we note that by theorems 1 and 4 K . (j) S where ~ 't T $= o. ~ Ys = ° or 1 for s = O, ••• ,k since (j,n) = 1. T~ = We proceed by induction on TO = £l} holding for Unless I!..$ I <. have ). ¥ = 1 for some q and . (j) q which case T~ ,!il (the statement obviously )"S = ° for all S P q, in = T q as asserted, we have \i"1 for all S for which Ys p 0, since in any case we - 7 - We therefore assume that ¥s j O. with Now let j' satisfy j j' Then Ti = The[T SJ hypothesis. smaller T for i j q [ ! i ( j) (j' ) are T J q I! S I <: , Ti ( holds for all S = (j') 1 (n). = for appropriate q by the induction We have thus expressed! i as a linear combination of which is not possible since by definition Ti(l T q =~ q. Now tr. (!i) is a sum of distinct conjugates of Ti hence a sum q of distinct T • Thus tr. (!~) has only coefficients 0 and 1 and tr. (!,i) = gi where Si <= H is the set of elements of H with non-zero coefficient (! i ). in tr. We note first that the Si need not in general be different. If necessary by renumbering the Ti we may assume without loss of 1 r generality that S , •• O,S are distinct and that for any j is an i > r there ~ r with S i = S j. Clearly ~o = tr. (!o) = tr. (1) = 1. We now assert that for i, j ~ r, i j Suppose the contrary, say h ~ S i where (s,n) we have gi n sj = ~. n sj. Then h = x s = y t = (t,n) = 1, x ~ T i , y ~ T j • Let t' satisfy tt' Then x j st' __ 1 (n). i(st l ) . = y, thus T and TJ have the element y in - 8 common, and.T i(st ' ) Thus we have T j c. Si. Now since si is rational we conclude that S j c. Si. Since Si and sj here play symmetric roles we have si~ sj by the same argu- = sj, which is in contradiction to the way we ment, hence Si i numbered the T • The Si(i = O, ••• ,r) are therefore disjoint subsets of H, we r clearly have 2- Si = !!, and therefore the Si span an S-module i=O over H. Since SO i = 1 and S i(-l) = Si (since Si is rational) for = O, ••• ,r the Si generate on S-ring over H provided only that they generate a subring of R(H). To show this we prove the following: Theorem 6: Le t 1 ~ i, j ~ r. r Then Si sj = ~ ~t st for appropriately chosen integers t=o Proof: ~i, §.j ~ R(H,G l ) ~ Si sj = We need show that if h, k £S t, k L- ° q = <S' (h) = <S'"(k). We may then put ~t = ~(h) = ¥(k). Clearly it suffices to show that for h~Tt, k (S t, i(h) = c5(k). Clearly if k( Tt holds, we have = 1S"(k) = 6 t • 1S"(h) Now = h S for some h f.T t, where (s,n) = 1. i is the number of ordered pairs (u,v) with u ~ S , v f. s j, uv = h. - 9 - and conversely. # l:.... Si Moreover uS u (. Si, V S t sj ~ v E. sj since Si and sj are rational. Thus (u,v) the ~(u s tV s ) is a one-to-one correspondence between ~(h) pairs of solutions uv = h and the ~(k) pairs of solu- = k. Thus '6 (h) = ~ (k) and theorem 6 is proved. Since Si is in R(H,G ) for i = O, ••• ,r, it is clear that the S-ring generated by l tions uv the Si is a subring of R(H,G ). 1 We will use theorem 6 in the following weaker form: Theorem 6 t : :£ G. = h y U. (h) = Then h, k l:. sj ~ ~H (h) h. 6. (k). ~ ~ To assist in computing these coefficients we introduce the following notation: Let h ( H, R c H. = {r E. R I r Then R(h) -1 h 1 E. R • The coefficient of h in [B.J 2 is the number of solutions r l t R can occur in at most one such pair and it occurs in such a pair precisely when r Thus 'R(h) I 2 = r -1 hER holds. is the coefficient of h in [B.J 2, and the ele- ments of R(h) are precisely those elements of R which "hit" other elements of R in such a way as to produce an h. We introduce the following further notation. For any set K, let I K I be the number of elements in K. - 10 K c: H means that K is a subset of H. < K ;> is the smallest subgroup of H containing K and for h ~ H, K ~ H means that K is a subgroup of H. We now state the two theorems proved in this thesis. Theorem A: Let G be a primitive permutation group of degree n. Let p be an odd prime. Let H A = = A x B x e be a regular abelian subgroup of G, where c( a / is cyclic of order p 0(.. (p prime) < b""/ is cyclic of order p f3 tel= m where (m,p) = 1, 0("> f3 > ° holds. B = and Then G is doubly transitive. Theorem B: Let the hypotheses of theorem A hold. In addition let {I} tivity of G and let tr. (!i) l = TO, Tl, ••• ,Tk be the sets of transi- = H-l for i = l, ••• ,k. Then G is doubly transitive. It is clear that theorem A includes theorem B. They are stated separately since we will first prove theorem B by a not too difficult counting argument, and then devote the greater part of the paper to the proof that under the hypotheses of theorem A, tr. (!i) = !!.:.! necessarily holds for i = l, ••• ,k. , Throughout this thesis, k will denote the number of nontrivial (i.e. f. f I} ) sets of transitivity of Gl , and r will de- note the number of distinct non-trivial traces of these Ti. - 11 Thus G is doubly transitive if and only if k = 1. The additional hypothesis of theorem B is that r = 1. Let P = AB. (Jol, p) = 1, P is a Sylow p-subgroup of H. Since IA I = P We have <A }B I = p P = p ""+f; Since H is regular we have =I n .-t 13 • H' 0(-1 = Ip 0 , = I p I 101 = mp o(• We let u =aP Put u = <u "/. Thusl~u/I = p. • Le t K ~ H, 0 <f:. A~ f3 • We may express k f. K uniquely in the form k = aSpY btpA = (t,p) = 1, c f. o. ~(-\) be the set of all such ktK for which v= o. Let Ky()...) be the set of all such ktK for which V# 0, Let but where (s,p) c o(-v :>P-A • Let KZ ()"') be the set of all such k t oi..-V~P-A · Let ~ = K for which K(A) is then the subset of K consisting of all elements which have a power of ~(A b exactly divisible by pA • ) is the subset of K(A) consisting of elements of order - 12 divisible by p 0( • ICy ( )..) is the subset of K( A) consisting of the elements k not in KXCA) for which uC () < k "/ ~ ~ holds. K (A ) consists of the remaining elements of KC A. )' the k for Z which uC fl <k:> .. ~. Without loss of generality we may assume that u We put S 1 ~ Tl holds. = s. Let Co = [c ~ c t ac t: s} • We show that by appropriate choice of generators of P we may assume that C o fi~. < S""/ = H by theorem 3. Since G is primitive, Thus S must have an element of order divisible by P A as b tp C where (s,p) (t,p) 1, c ~ C. s / tpA Let a a b • l = Q(,. , say = = 0(. -1 a p 1 since 01..-1 2. f3 u ~ S ~ = 1. u s = a p ~-l E S l Obviously we have P since S is rational. = Thus we may replace a by a and put u l 0(-1 and alP E:: S as well. = alP ....<. -1 to get We therefore assume that a has been chosen in such a way that C o is non-empty. - 13 III Preliminary Lemmas We prove five preliminary lemmas; the first two of which to be used in the proof of theorem B and the remaining three to be used in the proof of theorem A. Because of theorem 6' we know that in [~i]2 certain coefficient equalities must hold. 1-5 will tell us that such Lemmas equalities can occur only if the si have a special structure. Re- peated application of these lemmas shows that this structure is incompatible with the existence of more than one non-trivial Si. We will therefore be able to show directly that S = H-l. Lemma 1: Let x E: P, 11. E. <. x ;:>, x ~ U, c ~ C. Then for any = l, ••• ,p-l there exists v prime to n with v j (xc)v = u xc. j Proof: . Since x ~ P, we may write x where 0 ~ V ~ ~ , 0 (s,p) ~ b tP~ .x ~ J3 ' = (t,p) = 1. ul::. <x/ ~oZ-.\t> f3-A, i • e. x c HX \J Ry, and x ~ U ~ 0( f V-I . f y1ng . Choose s',m' sat1s s's m'm (where m = I C I ). Then, == 1 (pO< ) 1 (p"<") 1 (p) such that - 14 - = c m ::: 1. since Thus, (a sp where Now v btp A c) l+mm's'jp c:o<. - V-l = 1 + mm'sljp 0(- V-l v 0<.,- Y>1=:>- A and x = (xc) v ~ U 9 ~ VI. V-J. ~ 0(-1, thus plpC:<- V-I and we have v Clearly ,n) v _ 1 (m), thus ( v =1 (p). = 1 as asserted. Lemma 2: LetK~H. Let h ~ H such that hK = K. Then in K K( -1) h has coeffi- cient/KI· Proof: I n _K _K(-l) , 1 h as coe ff"1C1en " t IK( • Thus in h K K(-l) = hK K(-l) = K K( -1) h has coefficient I KI • Lemma 3: Let R C H such that R is rational. Let R* = (~ V Then R* Ry)-UC c= R(u j ) for j = 1, ••• ,p - 1. Proof: R rational ~ ~, R* rational ~ g* Ry ' R fl UC rational = g* (-IT, thus r E:. R* '7 R* rational. ~ r -1 E. R*. by Lemma 1, r -1 ~ R* J" 3> r-1 u E . R*. Thus r (. R* ===;> r -1 (;; R* and R* S R(uh. "9 r -luj~ R* ~ r E R(uh. Moreover - 1.5 We are now in a position to prove the important. Lemma 4: Let x E ~ (i.e. let x be of order divisible by po<.). Let R ~ H such that R is rational. Then (R(X) I ~ I R(uj)1 holds for j = 1, ••• ,p-l and lR(x)J = IR(Uj)I for every j = l, ••• ,p-l only if (i) h ~H X' (ii) and (iii) u-jh ~ R(x) for j = l, ••• ,p-l Proof: Let j ~ f 1, ••• ,P-l}. To each element z ~ R(x)-R(u j ) we wish to associate in a 1-1 fashion an element of R(Uj)-R(x). z -1 x E: R, By lemma 3, j ~ R(u ) ~ Z \ R* = (RX \.J Ry) - UC z Now if z z -1 -1 Z i = u c (c '- C), u = u -i c -1 ~ u j =u j-i -1 c R (since f:. i j u j we have g is rational) and R since we can always simultaneously satiefy the congruences since (m,p) v - q (p) v - = 1. ~ R(u ). thus z j ~ R(U j ) -==::;> z \ (RX U Ry) - ujC, Le. z E. R Now z fl, ••• ,p-l} 1 (m) This would violate z Thus z q £ Z U (R n ujC) E R(x) ==:::;> Z -1 x ~ R, -1 x E~. - 16 Now by lemma 1, we may conclude that ujz -1x c. R, indeed we have ujz -1 x ~RX R* c::::::.. Now by lemma 3, we have ujz -1 j x E: R(u ). Suppose the contrary, i.e. (ujz-lx)-lx t R, thus zu- j f Since R = R R. C-l) we have z -1 u j t R contradicting j j Thus with each z f R(x) - R(U ) we have associated u j in R(u ) - R(x). This completes the proof that IR(uj)1 2 Z-~k /R(x)( holds. Now suppose IR(u i >/ = IR(X)I for i = l, ••• ,p-l. Then the j only elements of R(u ) - R(x) can be the elements u j z-lx where For such an h = u j z -1~, • we have E: HX ' (i) h (iii) u-jh = z-lx f R(x) (since (z-lx)-lx = z f R) and h-lx = u-jz (HZ V (u-jC). Now h i: HX j Thus from h £R(U ) 3>h ~ R* = '/ hE R(u i ) for = l, ... ,p-l. i R(x) for some j we conclude that i h 6 R(u ) - R(x) for every i We conclude from h -1 x ~ HZ \..l(u = l, ••• ,p-l. -i C) for i = l, ••• ,p-l and p ~2 that (ii) Lemma 4 says that if , R(x) I = J R(uj)/ for every j = 1, ••• ,p-l, then in [g] 2 only elements of HX can "hit" some u j but fail to "hit" x. Such elements h fail to "hit" x because the element h -1 x which they must "hit" belongs to H - R. Z For each such h there - 17 are p-l other elements 4n tr. (h~, the ujh (J' 1 • "hit" x. every u j } 4 h d0 = , ••• , p -1) , wh .c In particular since all elements of (RX U Ry) - UC do "hit" it follows that Ry - UC ~ R(x), and indeed if any element of R not in HX "hits" any u j it must hit x as well. Moreover, any element h of RX which "hits" an element of Hy to yield an x (i.e. an h for which h -1 x~Hy holds) must belong to R(x). Thus for such an h we may conclude that h -Ix; e..R holds. We now prove a further lemma which says essentially that every elementary trace of PX(A) has some element "hitting" an element of any elementary trace of py ( A ) \J P ( A ) in such a way as to yield Z the element a. Thus if we know that there is a whole elementary trace of p y ( A ) U P ( A.) in R belonging to R(a), we will be able to Z conclude that eve:y elementary trace of P ()\) occurs in R. X Lemma 5: Let Let Then there exist e,f with (e,n) (ab tp ~ )f = (a P")t.. b SP )-e a = (f,n) = 1 such that Proof: For any integer j, let j' be an integer satisfying j j' = t(tpV -8)' (mod po() f :::;:.s(tp" -s)' (mod po<.) Let e " J P v b SP " ) -efl a f' P b spA )-e fl = (a (a p" spA )Slt a-s'(tpv -s) = al+P"(s't-s't)btP~ b ~ = (a Then = " s A e Thus we have (a P b p )- a = (abtp ~ = ab tP~ )f as asserted. We now proceed to the proof of theorem B, making use of l( po<..) • - 18 lemmas 1 and 2. We will then make use of lemmas 3, 4 and 5 with R = S (occasionally R = Si with i >1) in order to show that S = H-l necessarily holds. - 19 IV Proof of Theorem B In this section we use lemmas 1 and 2 and a counting argument to show that r plies k = 1 (i.e. tr. (!i) = H- - 1, i ::: l, ••• ,k) im- = 1. Theorem B: Let G be a primitive group with regular abelian subgroup H ::: A x B x e where A = <. a '> is 0 f order po<.. B = <: b > is of order p 13 o(";)~ ;>0. Ie , ::: m where (m,p) = 1. o 1 k ::: T , T , ••• , T be the sets of transitivity of G • l Let tr. Then k i . (! , ::: H-l for i ::: 1, ••• , k. = 1, i.e. G is doubly transitive. Proof: Since U is a subgroup of H we have whenever (v,n) ::: 1 that hV£U ~h(:U. Since all elements appearing in Tr. (!i) = H - 1 are obtained i by taking such vth powers of elements of T , it follows that TiA U t ~ for i = 1, ••• , k. 1 Let T = T be the set of transitivity of G in which u occurs. l As we let j take on values congruent to 1 through p-l modulo p and prime to n we have that the T(j) run through sets of transitivity of G l (by theorem 5). All such sets of transitivity are obtained in this way since every element of U-l appears in some such T (j) • Now suppose T has s elements of U. Then each T i has s - 20 - = I U-l I = p-l. elements of U, and we have ks We note that theorem B holds even in the case p = 2, since from what we have just shown it follows that T 1 = S1 • Since the Ti are all conjugate to T (i=l, ••• ,k) we have that the n-l elements of H-l are divided by G into k sets of transil tivity, each with n;l elements. PX(A) consists of It is easily seen that .f (p~) ~(p&--\) elements A= 0, ••• ,13. for p y ( A) consists of (p <>(-1 p-\.l; ) ~ (p - p for ~-A ) elements A = 0, ••• , f3 • = p..( + fJ -l( p-l) Thus p<:>(+ [3-1 _ (p2/3+ l +1) ( p+l) Except for the p-l elements, x, of U-l, for any element xc with x E. PXU Py' c (;; C and for any j (, \ 1, ••• ,p-l} , there exists v = l(p) with (v,n) = 1 such that u j xc = (xc)v. But since v -=.1 (p) holds T(v) and! have s elements of U in common. Since distinct Ti are disjoint it follows that T(v) = T holds; thus ujxc ~T. Now taking vth powers where (v,n) [(PXV Py) - uJ = 1 takes elements of C into other such elements. Therefore if we put (Px UPy) - ~ C (\T we have that I T*I = ~ \ (PX V Py > - U \ J C I T* = = ; [ pco<. + f3 -1 ( p-l ) + p -<+13-1 - (213+1 p + 1) (p+l) (p-~ m - 21 - Now for xc £. T* we have ujxc (; T*, thus u j T* c. T*. Now since T* is a finite set it follows that j u T* = T* for j = 1, ••• , p-l. Thus by lemma 2 we have that the coefficient of u j in T* :£*(-1) is IT*I. . Thus the coefficient of u J in T T ficient of 1 in T T(-l) is 'TI (-1) is 2IT*I. The coef- ::> 'T*I. Now, since the Schur-ring R(H, G ) has the Ti as generators, l it follows that k Z i=o Now each T i Thus we have ,(i::: 0, ••• ,k) has an element of U. 'lfi :> 1T* I for i = 0, ••• , k. Thus we get the inequality, ITI2 ~ I! !<-l)\ Now ITI = n-l k u Ii; "lfi!i \ > IT-I I i~ !il ~ IT-I IHI = mp'"" + J3 k -1 Thus we have that - -'+ ~ ( p 2 ~+l + p 2)~ . mp (p+l) - ( p 2 13+1 + p (p+l) J 2)J( .<><-+13 mp -1) thus Since mp ..<. + k j3 >0 holds, we have ~ -::> 1 m 00<. p +r (.? - - (2 p ~ +1 + P 2 f3 ~ ""> r.P 0( + f3 - p2 ~J _ mL (p+l) - 22- - Thus Now 2mp2 t3 -1 ~ pmp2f3 = mp p+1 p fJ <.. 2mpo<.+ 13 _ 2mp2f3 • "'- 2mp2 fJ ~ mpo<.+f3 -1 Thus we have Thus k = 1, and theorem B is proved. - 23 V Proof of Theorem A #1 We now wish to show that S (the trace of the set of transitivity of G in which u occurs) is all of H-l. l We first show that if SX(A) I- ~ we have that SeA) consists of most of P (.A ) C (A) where C ( ~) is a subset of C, and C (f3 ) = Co = f c C J ac" s} · 6. We will then show in #2 that if / ( is the smallest )\ for Whi~h SX( ~) that C(~) I- ~ holds, we have that SX( A) I- ~ for A = ,"f , ••• , f5 , = ~ = ~ , ••• , must hold. "'=~ Co for f3 -1, that ,... , (3, that S(~) = PeA) Co for SC;j) = AC o -1, and that ..A-1.. = ° or ~ In #3 we show that. the hypothesis contradiction. In #4 we show that C conclude that S = PC -1 = H -1. o. = C. ~ = f3 leads to a We are thus able to We now prove two lemmas, the first dealing with the structure of the site») which have elements of order divisible by po(, and the i second dealing with the structure of the S (A)' \ 1\ = 0, ••• , 'P -1, which have elements of order divisible by p~. Lemma 1.1 Let 1 ~ i ~ i Let C ( F) = r. 1d (:; C , ad E S \ P ) } Let c l:: C • Then the coefficient of ac in [ Si ( i3 )] 2 is less than or equal to the coefficient of u j holds for every such j only if for j = l, ••• ,p-l and equality - 24 i (i) S«(3)= C* . -----..-;.. where C*. ~ ~ c. C. i i C ( f3)c = C ( f.3 ) (ii) Proof: i Put R =S (13)' i D = C (f3). =~ _S ~ ~) rational. S i rational Thus since ac ~Hx holds we may conclude immediately from lemma 4 that 'R(ac)' ~ IR(u j ) t holds for j = 1, ••• ,p-l. I =, R(U j ), We now assume that 'R(ac) for j = 1, ••• ,p-l. Since (m,p) = 1, for any s with (s,p) = 1 and any t with (t,m) = 1 we may find s', t' with 1 (m) t' Thus for l(po(), t' x~ P, _ t (m). dE C we have tr. (xd) :: tr. (x) tr. (d) Thus R has every element of tr. (a) D and no other element of tr. (a) C. Now tr. (a) = ~ a s • (s,p)=l Again by lemma 4, the only elements x of tr. (a) not satisfy x -1 ac (. R must satisfy x -1 ac (HZ. ~ which might Now x ~ AC 3> x -1 ac in AC and AC t1 HZ = C. Thus the only elements x of tr. (a) D which might fail to satisfy x-lac E:: R are the elements of aD. Now as we let x run through the elements al+Pwd of tr. (a) L (w F 0), the elements x -1 ac that we get are the elements a -pw d ~ -1 c - 25 which must all lie in R. Now since D is inverse closed, we have that Now, were R to have a further element, y, of AyC it would satisfy y -1 u~R by lemma 3, thus y -lac (. R by lemma 4. This is clearly not possible by the definition of D. Hence the only elements of AyC which can occur in R belong to AyDc and all such elements do occur. Thus we have that Now, again by lemma 3 since p :>2 holds, we have 2 (a d) -1 ' u E R for d ~D, hence by lemma 4, 2 (a d)-lac = a- l d- l c ~R. But the only elements of ~C in R lie in AXD. Thus d-lc ~D holds for dE: D. we have Dc Again since D is inverse closed = D as asserted. Thus we have that i S (13) = r AX + ~J~ + C!l. We now assume that A < p , and prove a lemma similar to lemma 1.1. Lemma 102 Let R = S~A) where i E: Let ~ . fl , ••• ,r}, A< f3 • J ~. Let cl.(A) = ~d{C lab sP ~ d(R for some s with (s,p) = 1 = {d l ,···, d v } ' } - 26 Let c~ C. Let Zj (A) be the subset of PZ(A) such that -L Zj (,A) d j c = R Z t1 P d -1j c for j = 1, ••• , v. Let Z**(~)£:HZ such that R = Z v -1 z.. Z":(~) d .c i=l J J + Z**(.A) 12 is less than or equal to the co- Then the coefficient of ac in [ .E efficient of u j in [g1 2 for j = 1, ••• , p-l. If equality holds for all such j then (i) i Rx=PX(A)C(A). (ii) Ry :: (iii) The coefficient of u j in ( R Z ] 2 is ~ I Zj( ~ )' ... ) PZ ( A) Ci ( A) J 2 (iv) Py(~) CiC"'). Ci(,\)c = Ci(,A) Proof: i Again by Lemma 4 since R = S (A) is rational, we have , R(ac)J ~ tR(U j ), for j=l, ••• ,p-l. j We now assume that 'R(ac)1 = tR(U ) Put D I for all j. = Ci(~ ). Since RX ~ ¢ holds there is some elementary trace say tr. (ab wp~ Now d) in RX. Then a l-p w(l-p)p~ b d (R X holds. (al-Pbw(l-P)P~ d)-l ac = aPb-w(l-P)pA d- l c lies in Hy unless we have the moment. ,;\ = 0 and 0( = f3 + 1. We exclude this case for Then by Lemma 4 we may conclude that al-Pbw(l-p)p is in R(ac) since the only elements hE RX such that h fR(ac) -1 satisfy h aC'H • Thus we conclude that Z tr. (aPb -w(l-p)p ~) d- l c lies in R. '" d - 27 Again by Lemma 4 we have that y c: R(ac) for }.. -1 every y occurring in tr. (aPb sP ) d c. Put s = w(l-p). By lemma 5 we have that as y runs through tr. (aPb sP elements y -1 a occur in every elementary trace of P ( A ) • X A ) the It thus follows that P (\.) D ~R. X i By the definition of D = C ( A ), no further elements of ~ can be in R. Thus we have that We now turn to the case A = 0, 0.( = f3 + 1. Again from lemma 4 we have that p-l of every P elements in every trace of P ( A) occurring with d ~D in R must lie in R(ac). j X From lemma 5 we have that as we let x run through such an elementary trace of P ( A) the elements x-lac lie in distinct traces of P (A)' X z indeed one in each trace of PZ(A). It follows that we must have J ZjL\) J >12IPZ(A)j for j=l, ••• ,v where D = f d1 t ••• ,d v 1- • Now for some z f Z~ <.A) we must have uz l: Z~ (A) as well. J J (uz)z-l = u ~uzE.R(u). (uz)6 t Now Z~(;\) hold if (s,p) = 1. J Z -s f Thus (uz)StR(u s ). S By lemma 4 since by hypothesis IR(ac)/ = JR(U ) Ifor all such -1 s we have that every element in tr. (uz) d . c belongs to R(ac). J -1 As we let x run through the elements of tr. (uz) d. c we get J -1 that the elements x ac belong to every trace of P (A ) d x x (A ) D s: ~ as before. It follows that P nition of D we have r Again from the defi- - 28 RX = PX("') D. ( i) A :: 0, We now show that in either case ( 0(:::: f3 + 1 or not) (ii) and (iii) hold. Le:t y ~ PYLA) Dc. Then we have y -1 ac E. P (~) D = RX. x By lemma 4 since y\H we have y-lac ER(ac), thus yf:.. R. z We therefore have that Ry ~ p y ( A) Dc. If Ry had any further element, it would necessarily be of the form xd, x ~ P, d 4DC, and such an element cannot belong to R(ac) since ~ = PX ( A ) D. Thus we conclude that ( ii ) Ry:: PX (A ) Dc. It follows exactly as above that I Zj(A)I = lpZ(A )d-~CnR' >~JpZ(~») for j=l, ••• ,v. Now it is not possible that z E R R (ae) • Thus the coefficient of u in \R(u)fl RZ' r belongs to R(u) but not to Z R ] 2 is Z = IR(u)I1R(ac)nR z I. (Elements of R Z must be multiplied by other elements of RZ to yield u since R ~H( A) where ,\ p f3.) Only elements of PDc can belong to R(ac). It is not possible that both z and uz be in PZ(~)DC - RZ since then we would have z-lac and (uz)-lae both belonging to R(u) - R(ac) which is impossible by lemma 4. Thus z c P Z ( A)Dc - R z ~ uz ~ R(ac) - R(u). can only hold if 12 [ RZ is z-l~R holds. uz t.RZ - R(u) Thus the coefficient of u in - 29 - I PZ (~ ) Dc I - I PZ ( ~ ) Dc - RZ' - 'RZ - R (u) I = J PZ(,~ )DC I -2I pZ ( ~ )DC - RZ I = 2JR Z IlPZ ( ~ )Dcl - ,PZ(A )D I =2~IZj(")/-IPZ(A)D' 2 ).. In addition since p J 2, we have a b P d. t: R( ac) for j=l, ••• ,v J 2 thus (a b P X. dj)-l aciR, and d-~C ~D for j=l, ••• ,v Since D is inverse closed it follows that D = Dc. This completes the proof of lemma 1.2. We now put i=l in lemmas 1.1 and 1.2 thus S Since t3 i = S 1 = S. J (, A ~ H( 5' )H(.t\ ) S H( y) it follows that = L/s(A) (h)Ifor hE:AC. IS(h)/ A=O Nowls( A ) (ac,)1 ~ Is( " )(uj)J holds for A = 0, ••• , /3 and i..Js( A )(ac)1 = IS(ac)1 = ~=O :t Is( A )(uj)J =IS(uj)I for C EC o and " =0 j=l, ... ,p-l by theorem 6 since u<:S and C y uj£;S since ~ is rational, is by definition the subset of C satisfying ac E.S. o conclude that for c l::: C Thus we o )s(A)(ac)/=ls( \)(uj)i for all j=l, ••• ,p-l. Let Co = [c l ,···, Cq1. From lemma 1.1 we conclude that S ( 13) = A - 1 Moreover, letting c run through the elements of C, we have o since p fi 2, that C C • o o Co was already known to be non-empty and inverse closed. we have Lemma 1.3 Co is a subgroup of C. By lemma 1.2 we have that SX( A) fi /J Thus - 30 SX()\l {:X( A) ~} b( DCJ+ 'i£.) A) Zj (.A )d-~ci +z" C>' ) for citCo where D = {dl, ••• ,dv(A)}' Z**(A) SHZ(~) and I Z j( A )' ::> ~ , PZ ( ~ ) f Choosing c. = 1 (since 1 t C ~ 0 holds) we get from Dc. = Dc. ~ J (i,j=l, ••• ,q) that D = Dc . for j=l, ••• ,q, thus that D consists of J full cosets of Co. A ••• ,cq(A) A cl, as the coset repre- Choosing sentatives (thus v( A) = q( ~ )q = q( A) I C ') we have that if o sx(A).i¢ s(~) =[pX(A)+pY(~)JCoE~+ •• +c~("J + q A z.. Z! .(A)cic. j=l qV\) +2 i=l ~J J z**(A ) where' Zij (~)I > ~ I PZ( ~ ) ' holds for i=l, ••• ,q j=l, ••• ,q(t\ ) It also follows from lemma 1.2 that the coefficient of u in [z**( A il 2 is zero. #2 We now let.)-'l.. be the smallest A such that SX(A ) .j ¢ holds. We will show in this section that for any A with x ~ A ~ 13 - 1 we have We will then show that ~= 0, Co = C and that S(P)=p(f3)C o -l, thus proving directly that S = H - 1. By hypothesis, SX(rt) .j ¢. in S where (t,p) = 1, c~C. 'S(abtP~c) ) = ,S(u)l. Thus there is an element ab tp~ c Now by theorem 6, we have By lemma 4 we know that this can occur only if S has a special structure. - 31 From #1 we have that for any Sx( A) = ¢ or =( Px("\) + py(~ )1 Co [c~ S ( A) +. •• +C:( ~) J q( ~ ) q z:. j=l ~ + A we have either i=l + z**(A ). Lemma 2.1 sA":; p- l. ~ Let Then (i) -J ¢ Sx ( ,\) (ii) q(,~ ) = 1 (iii) C~ E. C 1. 0 i.e. + ~ Zi(A) c i + Z"C,x) i=l - - where Zi (A ) is t~e subset of PZ ( "') which we earlier denoted by Zil ( A). Proof: We first note that if ~ =p the lemma is vacuously true. Thus we assume that ~ L.. f> holds. t~ Let ab p cis, «t,p) = 1, c(C). V Let 0 <:::. that (ab tp.A{ c) e c( ~ - J1..... Then since S is rational it follows loS holds where e is chosen to satisfy the con- gruences = l-p v' (p.,.(.,) , e e = By lemma 3, (ab = (ab tp 1 tp.N. (m) c) e E-.S(u) holds. Now (ab tp--<' c) -e (ab tp'" c) v~ .I"f )p ~ HZ. Thus we may conclude from lemma 4 that (ab long to sCu) - SCab tp-"1 c), thus tp~ c) e . cannot be- - 32 t p,.4.( e t p""'Cab c) ~8(ab c) holds, (abtP-"1.c)~S. i.e. (abtP.A1c)-e tp....... V Thus Cab )p '=- S holds for 0 .::: V <.. f3 -}1. t ~+V . i But ab P ~ 8~ S7 8 ;2 PX~i'V) VPY(MtV) by lemma 1.2. V t ,Ao{+V " . . . i Thus a P b p f. 8 1 would al.so h01d~ Now 8 "S· = ¢ or S i = 8. Thus we conclude that S i t = S, and that ab P far 0<. y<: p-J1. 8X(~) p ¢ holds by hypothesis. t'1 +V £ S holds Thus 8 (A) ~ ¢ X ~~~~p-1. for We have (ab P -.4.t )l-p J1 - K J [( abP~ 1- ) l-p B -".... c ; By the definition of C ~SX(,A() for j=l, ••• ,q(..-'1..). 4 that We conclude from lemma [abP,o-i ci ..'( cj o 1 and by lemma 1.1 it follows that c;~C.9 C;hOlds for j=l, ••• ,q(..4t). Since the c~ J were coset representatives it follows that = 1. q (-"'t ) Now y = aPb P By lemma X ( a Pb P \ c~ E 8 y holds by lemma 1.2 for 4 we conclude that y~8(ab "" c ~ ) -1 a b P p"1 '"1 c l ~ = ..A-f +1, ••• ,,-1. ) must hold, thus c ~ t:.. 8 h 0 1 ds • Thus by what we have just shown, since this is clearly an element of 8 (,.Ao()' we have X (c~ )-lfC o for A= -""1 +1, ••• , J3 -1, j=l, ••• ,q(A). Thus q ( A) c~ = 1. = 1 for A = ~ + l, ••• , 13 -1 and we may choose - 33 Thus we may choose c~ l = 1. This completes the proof of lemma 3.1. We now show that Zi(A) = PZ(,A) for i=l, ••• ,q The idea is to note that for for some j Z t..P ( A)' zC Z i t S ).:)1, ..• ,/3-1- ;> zC i E sj ~ 2. S j would have to have some element xd (x ~ P, d ~ C) of order divisible by po<.. since ~sj ;> : H. Then IS(zci)1 : ,S(xd)l would have to hold by theorem 6. We will use lemmas 4, 5 and 7 to show that such an equality cannot hold. Lemma 2.2 Proof: Assume the contrary. <a > <b P > Co. M Let K = We already know that S2KX by lemma 2.1. Thus there exists k~KyUKz - S with k.j 1- Let k( sj and let xdES j be an element of order divisible by p«', where x f.P, df.C. Since xd ~s, we have either xc P ( y) for some X V < /-( or We treat these cases separately. Case (i): Let x t:. PX ( V) where ~ , SX( f) = ¢ for V <A holds. By the minimali ty of f ~ ~ and therefore in( S( j » ] 2 xd has coefficient zero (since the a-exponent of every element occurring there is divisible by p and the a-exponent of x is prime to p). - 34 Since in general, f.c. L =9 H( r )H( -C) ~ H( f), it follows that the only contribution to IS(xd)1 comes from S(V) SX' thus from S pI ) [~t X( ") Co] . p Were z and uz both to belong to s(V) the coefficient of u in [s(v;]2 would be greater than zero. This would contradict ,s(u)1 = Is(a) I since tS(A)(a)/ -SlS(A)(u)/ holds for ~ = 0, ••• , 'P and }S ( v) ( Q) elements z I = 0. Thus at most half of the of ~(9") which satisfy z -1 xdc:.H(A) can occur in It therefore follows that for every element ofKX which belongs to S(xd) there is another element of KX which does not belong to S(xd). Now h f Sx = 1Sc ) h -1 k t KX c S Thus we have that Sx SS(k). Thus there is a 1-1 correspondence between the elements of S( \f) which belong to S(xd) and the elements of Sx which belong to S(xd) and since in this way we get at most one-half the elements of Sx it follows that IS(xd)( ~ lSxl holds. Moreover, we have just shown that S(k)~ SXo It not suffices to exhibit an element of S(k) which does not belong to SX. We claim that a P is such an element. aPt S holds by lemma 1.1, since 1'= C • o Thus since k £Ky U Kz holds, we have k = a 0( -~~ ~-r~ ~-}{~J3-1. Thus we have). >1 and a-PkEKy~S. spA btp"C c where - 35 Thus we have a P t:.S(k) - SX. This completes the proof that /S(k)1 the hypothesis x £ PX ( V) where ~'S(xd)1 holds, under V <:: ).1. Case (ii): We now suppose x ~ PX( Y') where V:> K holds ~ hence d ~C 0 ~ The contribution to IS(xd)1 now comes from + Z 2 (PX(A) Co) (z**(A » ~~".::.v Now precisely as in case (i) we conclude that the contribution from the 1st, 2nd and 4th terms is ~ f:3 -1 ~lpz(~)1 ICol 1\ =.)-1 Now only elements of the form xc. may be multiplied by an ~ element of C* to yield an xd. Thus the contribution from term 3 is at most 2 f Thus we have IS(xd)I ~ -£lpz(Xo)l q + 2q. ~=-'1 Again as in case 1 we have PX(A ) Co ~ S(k) for 8 Thus we have lS(k)f ? L fpX(~ )IICof, and A =1'1 = q ....... " B-1 - ~~-l B _ ~J ( .I.. ,-' ~ P(~ (p-l) - p ~ + q LP -l(p_l) -2-1 "',.~ B-1 q 2-.- ~=~ Col = 2q. ). =)-(, ••• , f3. - 3f> since the 1st term is ~ 0 and the second term is ~ 0 since This completes the proof that k l:. S. We conclude that for A = -""( , ••• , ~ -1. SX( A ) = P( A )C o + Z**( .A ) .t "'" :: ~" ~,-, Lemma 2.3 Let ~ ft P• y, -1 Z**( A ) + ~J?( ..\ )Co+Z**C~ )) + [A-~]Co+~*· Let S = Let K = ~a > < b P Then Z**( A ) = ¢ for ~ >C • Q A = 0, ••• , f3 -1. Proof: We have that/,K-l S S and that a (:. K holds. Sx = KX. Since the coefficient of a in C Sy1 2 + 2S y Sz + [ sz1 2 is clearly zero, the contribution to I S(a) 1 comes solely from [ S~1 2 + 2S X ~y + SZ ] · Now since K is a subgroup of H, for k E K we may conclude from kh = a that h = k-la l:. K. Thus the total contribution to ' Sea) ' comes from [ ~J 2. It is easily seen that [ K_l) 2 = q K t -2) [ K-l ) + (l K I -l) • 1 Thus since ) S(a) I = I S(k) 1 for k f K-l by theorem 6, it follows that there can be no contribution to any k l:. K-l from any [ Z**( "\ Da. Now let zd f Z**( A ), (z ~ P, d E C) where 0 ~ )\ <. f3 -1. 1 fj - A -1 Then z -p d f Z**( A) holds. Thus we have (zl-p f3 - " -1 d)-l (. Z**( ,x ). 2 1- 13 - ). -1 -1 ~ - ..\ -1 Thus in [Z**( ~ ~ , (z p d) zd = zP occurs with - 37 It is in PZ(P -1). non-zero coefficient. P Z ( ~ -1) C K holds since x:f:: f3 • Since p ~ 2 we have for zd ~ Z*"'(, -1), (z ~ P,d (C) that z2 d f.PZ~~':t) holds, z-2(d)-lzd = z-lE: pz~~.;n. Z··(' -1) We conclude that =¢ as well. Now S (0) < S"> = H requires that Shave a.n ele- f. ¢ since ment whose b exponent is not divisible by p. p-l By what we have just shown, we have S + C· unless ;A.{ ( i) z::::.. P (A) C +[A-il C -:J '" =~ 0 - =j3 • Thus there remain only two possibilities: fJ -1 ,.A-( = ~ = Z • • ( A) + AC - 1 + C· ~ =0 0 Z.- f3 ' i\ -1 (ii) ...::;;-"'1.= 0, S -- /L-..- + AC o - 1 + C· #3 In this section we assume throughout that..l"1 = fJ • We will show that this leads to a contradiction. Lemma 3.1 Let h~Si where i?: 2. Then ,S(h) ~2. Proof: Si must contain some element x of order divisible by p 0( since <.. S i r~12 =[sx >= + H. Then IS(h)1 = Is (x) , mus t hold. + [Sz12. Sy12 + Clearly x cannot occur in 2S y Sz + [sz12 since such elements are products of elements with a exponent divisible by p. Now s +' S X y 11 = [A- - - C0 0 In - 38 - r AC 0 J2 the only elements of HX which occur belong to S. Hence the only contribution to the coefficient of x (which does not belong to S) is from 2 Sx ~Sz - Co~. Suppose sl' s2~SZ - Co such that sl -1 x~S, s1' s2fSZ and x~~ it follows that we have sl Thus we have (sl -1 x) -1 (s2 -1 x) = s s2 l -1 -1 s2 -1 x, s2 xES. Since -1 <: x~SX_ACo. ~ACo. In [AC o - 1J2 the coefficient of s s2 -1 is IAc l o 1- 2 which is the coefficient of a in [~12 • t s2 -=? Is(s l s2- 1 ) Thus sl I> ls(a)1 which cannot occur. Thus there is at most one 51 ~ Sz - Co satisfying 51 -1 x fS and if there exists such an 51 we have Is(x)~= 2 = lS(h)J. Thus we have either IS(h)J = 0 or IS(h)1 = 2 and lemma 3.1 is proved. It now follows that \S(h) I > 2 =.., h {S if h -# 1. Lemma 3.2 Le t 0 ~ ~ ~ Let S( A) 13-1. v(A ) = ~ zj*( ~) d~ where Zr (A) ~ PZ ( \) for J=l j=l, ••• ,v(~). /\ Then for all 1 ~i,j ~v(>'), we have that Zj* (~) = Zrl< (A) is a single elementary trace of P (,.\ ). z Proof: Then r ~ J 2 contributes to the coefficient of a Unless x = y, d since 'S(aX-Yci i ~ l = d ~ j ~-l (d.) E. C x y = , S(a - ci)r\ ACo I for we cannot have d )I = qpo(_2 = IS(a), ~ x-y" >dl(d j l J 0 - 39 c i t: C o ' a x-y c i /: 1. Thus since 1 E:C 0 holds it is clear that Z~*(X) cannot have two distinct elementary traces. Suppose we have J a x-y A >. -1 (S i )(d ) j l for some i?'2. (d By lemma 1.1 we have that ad X ~ -1 i (d ) ~ S l j must hold. 1\ (d j" )-1' We have from lemma 3.1 that ls(ad \s(adi (dj~ )-~ = 0 or 2. = 0 contradicts 's(aX-Ydi (dj )-11 .> o. IS (ad~ (d; ) -11 = 2 can occur only if C * has an element of Cod~ (d~ )-1. a x-y d In this case it follows that in [S(;3)]2 .\ .\-1 (d ) occurs with coefficient 2 as well. j l X y 'S(a - d~ (d; ) -~ > 2 = 'S(ad~ (d: )-l~'. x y is avoided only i f a - d~ (d~ ) -l( SO = d'\ 1 Thus we get This contradiction t1 }, thus aX = a Y, = d ~. • J This completes the proof of lemma 3.2. Since S(o) I¢ (because the elements of S generate H) we have that S(o) =Zi*(O) (d l + •• +dvJ where d~ = d i and v(o) = v. Lemma 3.3 Let Zi* (0) = tr. (z), D Then S :::l = f d l ,.·.,d v1, hence S(o) = tr.(z)Q. <z > D - 1 Proof: For z s A :> 0 and d. z -s+tp~ ~ since S(o) is rational. For ~ = 0 there are p ~ -l( p-2) choices for s (excluding s-=..t(p». - 40 - f3"/ 1 we have IS (z tpA d Thus for for any t with (t,p) = 1, any i d ) k I > p ~ -l( p-2)? p;::> 2 A = 0, ••• ,f3 and any i,k between 1 and v. Thus z S tp'" d i dk~ Sunless f3 = 1. If 13 = 1 we have that = ACo-l + D-l + tr. (z) D since S(p)== ¥'l (mod p) must hold. Now from[ S(o) J2 we still A-I of at least p'" (p-2) ~ 1. In addition if d. ~ 1 we get a further contribution of at 1. t least 2 from 2 S(o) S(l)' namely from 2 (z dk)d i • Thus we have IS(ztdidk)1 :>2 unless d t This means z d d E: S unless d. i k 1. viously zt didk = 1. = 1 in which case we have ob- = Z t d k t: s. We also have z did k i tp d d i k = d i d k E: S in the case j3 = 1 unless = 1. Now with A = 0, t = 1, we get zd i dk (;.S for any i,k between 1 and v. We know sinee S(o) is rational that D is inverse closed. Thus D is a subgroup of C. A vary through all possible values, we tpA , -1 get z didkf.S, unless 1\ = 15, d = d k , thus S ~(z"/ D - 1. i Now as we let t and We now calculate Is(z)l. Since D is a subgroup we have l~D, hence - 41 Elements of the 3rd term have b exponent divisible by p and z C: S (0) does not. Thus the contribution to IS(z)I comes only from the 1st two terms. < z > D, and we have '>D only fory {<z;>D, the contribution to I S(z)lcomes Since S(o) has only elements of [ <::: z '> D - ~ 2 and this contribution is clearly only from I J' - 2 since z is of order p' (since it is in PZ(O)). Bu tis (a) I = qp -2 = 'C pO< -2. )S(a), = tS(z)1 ~ID'PP-2 = 'Co. , pO<_2. Thus I D I = Ic o. , pal..-p. JD 0( But 0 , 0<. ~ 13 ~ pllnl· This is impossible since D is a subgroup of C which has order Thus ...A-1.. = f3 prime to p. cannot occur, and we conclude that .1'1 = 0, 13- 1 S ;: ~ P ( A )C 0 + AC 0 - 1 + C· A"'o #4 To compDete the proof of theorem A we need now only show that C * = 0 and C o :: C. Lemma 4.1 C· = 0. Proof: Since ( I c \, p ) = 1 ' 0. C (p ) = C 0 • Thus C·(p) has no element of C • o 1 occurs in So, hence not in S. If 1 I c · E: C* held, (c·)p would occur with coefficient 1 in This would contradict theorem 4. - 42 We therefore conclude that C· = j) and that S = R! 0 - 1 where C0 ~C (since A = P ( P »•• Now = H since G is primitive, but it is evident that Thus we have that PC = R! = Hand S = H - 1. o (S> <S>=FC. o Since SO = 1, S follows that r ~ = Sl = H - 1 and S i (l S = ¢ for 2 ~ i ~ r, it 2 cannot occur. Thus we conclude that r = 1, and with the help of theorem B, we have a complete proof of theorem A. - 43 REFERENCES l. Burnside, W.: Theory of Groups of finite order (1911) 2nd ed. Cambridge University Press. 2. Schur, I.: Zur Theorie der einfach transitiven PermutationsgruPfen. Sitzungsberichte Preuss. Akad. Wiss., phys.-math. Kl. 1933) pp. 598-623. 3. Wielandt, H.: Zur Theorie der einfach transitiven Permutationsgruppen. Math. Zeitschrift (1935) vol. 40, pp. 582-587. 4. Kochend~rffer, R.: Untersuchungen Uber eine Vermutung von W. Burnside. Schriften math. Sem. Inst. angew. Math. Univ. Berlin (1937), vol. 3, pp. 155-1800 5. Wielandt, H.: 6. Wielandt, H.: Zur Theorie der einfach transitiven permutationsgruppen II. Math. Zeitschrift (1949), vol. 52, pp. 384-393. 7. Scott, W. R.: Solvable Factorizable groups. of Mathematics. (1957), vol. 1, pp. 389-394. Permutationsgruppen (1954). Illinois Journal