A Family of Groups Containing a Nonabelian Fully Ramified Section

A Family of Groups Containing a Nonabelian Fully Ramified Section

JOURNAL OF ALGEBRA ARTICLE NO. 186, 578]596 Ž1996. 0387 A Family of Groups Containing a Nonabelian Fully Ramified Section Mark L. Lewis Department ...

224KB Sizes 7 Downloads 91 Views

JOURNAL OF ALGEBRA ARTICLE NO.

186, 578]596 Ž1996.

0387

A Family of Groups Containing a Nonabelian Fully Ramified Section Mark L. Lewis Department of Mathematics, 400 Car¨ er Hall, Iowa State Uni¨ ersity, Ames, Iowa 50011 Communicated by Walter Feit Received April 10, 1996

1. INTRODUCTION In this paper, we construct a family of groups such that each member contains a nonabelian fully ramified section. We will present a number of the properties that the groups in this class possess. Through examples, we may better understand the properties of groups containing nonabelian fully ramified sections. Furthermore, when studying the properties of groups containing abelian fully ramified sections, the extra-special p-groups are used as test cases; we believe that the family of groups constructed in this paper provide a similar testing ground for studying groups that contain a nonabelian fully ramified section. We have studied groups containing nonabelian fully ramified sections in w8]11x. In those papers, we studied the theoretical properties of groups containing a nonabelian fully ramified section, but we did not present any examples of such groups. One of the goals of studying the family of groups presented in this paper is to have a collection of examples where we can actually test the techniques developed in those papers. Furthermore, we think that this particular family of groups deserve extra study because the groups in it demonstrate some of the differences between the situation when there is an abelian fully ramified section and when such a section is nonabelian. In order to present these differences, we need to set up the environment where these groups occur. We have developed some notation for situations that we encounter when studying groups containing a nonabelian fully *Current address: Department of Mathematics and Computer Science, Kent State University, Kent, OH 44242. E-mail: [email protected] 578 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.

GROUPS CONTAINING RAMIFIED SECTIONS

579

ramified section, and we repeat it here for the convenience of the reader. We say that Ž G, K, L. is a normal triple if L : K are normal subgroups of G. Such a normal triple is called solvable, nilpotent, or abelian if KrL is solvable, nilpotent, or abelian. We call H a complement for Ž G, K, L. if G s HK and H l K s L. Next, we define Ž G, K, L, e , w . to be a basic configuration if Ž G, K, L. is a solvable normal triple, e g IrrŽ K . is G-invariant, and w g IrrŽ L. is a constituent of e L . We will denote this basic configuration as abelian or nilpotent if the normal triple Ž G, K, L. is abelian or nilpotent. We define H to be a stabilizing complement for Ž G, K, L, e , w . when H is a complement for Ž G, K, L. and w is H-invariant. From the definition of the basic configuration, we note that e is necessarily invariant under the action of H. If Ž G, K, L. is a normal triple such that Ž< G : K <, < K : L <. s 1, then we call Ž G, K, L. a coprime normal triple. Also, we say that Ž G, K, L, e , w . is a coprime configuration if Ž G, K, L, e , w . is a basic configuration with Ž G, K, L. a coprime triple. Continuing to let Ž G, K, L. be a coprime triple, we note that if H is a complement in this situation, then it follows that HrL acting on KrL is a coprime action. We will say that the action is fixed-point-free if C K r LŽ H . s 1. Since in this case all the complements are conjugate, it is easy to see that this property is independent of the choice of complement. Therefore, we may define Ž G, K, L. to be a fixed-point free coprime normal triple in this case. Given a normal triple Ž G, K, L., assume that there exists a normal subgroup M of G containing K such that Ž M, K, L. is a fixed-point-free coprime normal triple. When this situation occurs, we say that Ž G, K, L. is a controlled normal triple and that Ž G, K, L, e , w . is a controlled configuration if it is a basic configuration with a controlled normal triple such that < K : L < is odd. If N is a normal subgroup of G and w is a G-invariant irreducible character of N, then w is said to be fully ramified with respect to GrN when
580

MARK L. LEWIS

ramified sections, one for each odd prime p, and we denote this family by X . We then prove the following theorem. MAIN THEOREM. Let Ž G, K, L, e , w . be a controlled nonabelian fully ramified configuration lying in X . Then < G : L < is odd. Furthermore, if H is a stabilizing complement, then there exists a character u g IrrŽ H . that extends w such that u G has more than one irreducible constituent of odd multiplicity. We note that this theorem outlines a crucial difference between the situation when the fully ramified section is abelian and when it is not. In w2x, Isaacs has proved when Ž G, K, L, e , w . is an abelian fully ramified configuration such that < G : L < is odd that there is a complement H with the property for every character u lying in the set IrrŽ H < w . that u G has a unique irreducible constituent of odd multiplicity. As the Main Theorem shows, the groups lying in the family constructed in this paper do not have this property, and so, this clearly demonstrates that there is something different going on when we change our view from an abelian fully ramified section to a nonabelian fully ramified section. I conclude this introduction by thanking Professor I. M. Isaacs for all of his help. The work in this paper was completed while I was a student at the University of Wisconsin under Professor Isaacs’ supervision and while I was supported by a Department of Education Fellowship. Furthermore, Professor Isaacs was on sabbatical while I worked on this paper, so all his help was especially appreciated.

2. PAIRINGS AND GROUP ACTIONS The definition of a pairing, found in w8x is repeated here. Let A and B be groups. We say that Ž?, ? . is a pairing between the groups A and B if Ž?, ? . maps into the multiplicative group of the complex numbers and Ž?, ? . is a homomorphism in each coordinate. We define the subgroups A 0 s  a g A < Ž a, B . s 14 and B0 s  b g B < Ž A, b . s 14 . If A 0 s 1 and B0 s 1, then we call Ž?, ? . nondegenerate. In Lemma 3.2 of w8x, we proved that ArA 0 and BrB0 are isomorphic abelian groups. Therefore, in order to have a nondegenerate pairing between A and B, both A and B must be abelian groups. Assume that a group G acts on the groups A and B. If there is a group isomorphism f : A ª B with the property that f Ž a g . s f Ž a. g for every pair of elements a g A and g g G, then f is said to commute with the action of G. Whenever there is an isomorphism between A and B that commutes with the action of G, A and G are called isomorphic under the action of G. On the other hand, A and B are defined to be dual under the action of G if there is a G-invariant nondegenerate pairing between A and B. We begin by proving a well-known result.

581

GROUPS CONTAINING RAMIFIED SECTIONS

LEMMA 2.1. Let G act on an abelian group A. Then IrrŽ A. is dual to A under the action of G. Furthermore, if G acts on B so that A and B are dual under the action of G, then B and IrrŽ A. are isomorphic under the action of G. Proof. We define the map ²² ? , ? ::: A = IrrŽ A. ª Cq by ²² a, a :: s a Ž a. for every element a g A and character a g IrrŽ A.. Since A is abelian, we know that all the characters in IrrŽ A. are linear, and it follows that ²² ? , ? :: is a nondegenerate pairing. To show that ²² ? , ? :: is G-invariant, we consider an element g g G, and we observe ²² a g , a g :: s a g Ž a g . s a Ž a . s ²² a, a :: . Therefore, A and IrrŽ A. are dual under the action of G. Suppose that B is dual to A under the action of G, and let Ž?, ? . be the G-invariant nondegenerate pairing between A and B. Given an element b g B, we define the function a b from A to Cq by a b Ž a. s Ž a, b . for every element a g A. It is easy to see that a b g IrrŽ A. for every element b g B, and hence, we may define a map f : B ª IrrŽ A. by f Ž b . s a b . Given elements b, c g B, observe that a b c Ž a. s Ž a, bc . s Ž a, b .Ž a, c . s a b Ž a. a c Ž a. for every element a g A which shows that f is a group homomorphism. Since Ž?, ? . is nondegenerate, then kerŽ f . s 1. By Lemma 3.2 of w8x, we know since A is dual to B that < B < s < A < s
y1

g

a b g Ž a . s Ž a, b g . s Ž a g , b . s a b Ž a g . s Ž a b . Ž a . . Thus, for every pair of elements b g B and g g G, we have a b g s Ž a b . g . This implies that f commutes with the action of G, and therefore, B and IrrŽ A. are isomorphic under the action of G. We spend the remainder of this section setting up notation for a particular family of group actions. Let p be a prime, fix n to be a positive integer such that p does not divide n, and write d s ord nŽ p .. ŽThus, d is the smallest integer so that p d ( 1 mod n.. Write E and F for the Galois fields of orders p and p d, and recall that E : F and that F has dimension d over E. Since d s ord nŽ p ., we observe that n divides p d y 1. This implies that there is a unique cyclic subgroup C of Fq having order n. For every integer e, we set Me to be a group isomorphic to the additive structure of the field F, and we define the action of C on it by x ? c s xc e for every pair of elements of the field F, and we define the action of C on it by x ? c s xc e for every pair of elements x g Me and c g C, where the multiplication in the right-hand side of this equation is the multiplication in F.

582

MARK L. LEWIS

LEMMA 2.2. The action of C on Me is faithful if and only if gcdŽ e, n. s 1. Proof. Assume that C acts faithfully on Me . Then, for every nonidentity element c g C, there is an element x g Me so that x ? c / x. In F, this implies that xc e / x, and since F is a field, we see that c e / 1. Thus, we have c e / 1 for every nonidentity element c g C, and we conclude that gcdŽ e, n. s 1. Conversely, we now assume that gcdŽ e, n. s 1. This implies that c e / 1 for every nonidentity element c g C. Given an element x g Me , we observe that x ? c s xc e / x for every nonidentity element c g C. Therefore, C acts faithfully on Me . When G acts on an abelian group A, we call this action irreducible if the only proper G-invariant subgroup of A is 1. LEMMA 2.3. Fix an integer e, and write t s gcdŽ e, n.. Then C acts irreducibly on Me if and only if ord n r t Ž p . s d. Proof. Assume first that C acts irreducibly on Me , and suppose that ord n r t Ž p . s b. This implies that p b ( 1 mod n, and so, there are integers x and y so that p b y 1 s x Ž nrt . and p d y 1 s yn. If we let a s gcdŽ b, d ., then by Theorem 21.7 of w5x, it follows that p a y 1 s gcdŽ p b y 1, p d y 1.. From the Division Algorithm, there exist integers u and V so that p a y 1 s u Ž p d y 1 . q ¨ Ž p b y 1 . s uyt Ž nrt . q ¨ x Ž nrt . ( 0

mod nrt.

Thus, we observe that nrt is a divisor of p a y 1. This implies p b y 1 divides p a y 1, and so, b divides a. Since a divides b, we conclude that a s b which implies that b divides d. Write k for the Galois field having order p b , and observe that E : k : F. We define Ne to be the subgroup of Me identified with k. We will now prove that Ne is invariant under the action of C. Consider an element c g C, and observe that Ž c t . n r t s c n s 1. This implies that c t has order dividing nrt, and c t lies in k. Since t is a divisor of e, there exists an integer a so that e s at, and we note that c e s Ž c t . a g E. Given an element y g Ne , we have y ? c s yc e. Because yc e lies in k, it follows that y ? c lies in Ne . Therefore, Ne is invariant under the action of C. Since k ) 1 and Me is irreducible under the action of C, we conclude that Ne s Me . This implies that b s d as desired. Conversely, suppose that ord n r t Ž p . s d. We note that t - n, which implies that there is some element c g C such that c e / 1. Let B s  b e < b g C 4 , and observe that B is a cyclic group having order nrt. Given an element x g Me such that x / 0, we see that x ? c s xc e / x. Thus, every orbit under the action of C in Me not containing 0 has cardinality nrt. If W is a C-invariant subgroup of Me , then W is a union of C-orbits. Since W

583

GROUPS CONTAINING RAMIFIED SECTIONS

contains 0, we see that < W < ( 1 mod nrt. On the hand, since Me is a p-group, it follows that < W < is a p-power. This, along with the fact that ord n r t Ž p . s d, implies that either < W < s 1 or < W < s p d. Therefore, Me is irreducible under the action of C. We fix « for a nontrivial complex pth root of unity. Given an element a g F, we define a function from F to the complex roots of unity by waŽ x . s « TrŽ a x . for every element x g F. LEMMA 2.4. The map a ¬ wa is an isomorphism from F to IrrŽ F .. Proof. Consider the elements a, x, y g F. It is easy to see

wa Ž x q y . s « TrŽ aŽ xqy.. s « TrŽ a x .« TrŽ a y . s wa Ž x . wa Ž y . . It follows that wa g IrrŽ F .. Now consider elements a, b g F, and observe that

waq b Ž x . s « TrŽŽ aqb. x . s « TrŽ a x .« TrŽ b x . s wa Ž x . w b Ž x . for every element x g F. Thus, the map a ¬ wa is a homomorphism from F to IrrŽ F .. On the other hand, suppose that wa s w b . This implies that « TrŽ a x . s « TrŽ b x ., and we have TrŽŽ a y b . x . s 0 for every element x g F. Hence, the map a ¬ wa is one-to-one. Since < F < s
Me is dual to Mnye under the action of C.

Proof. Using Lemma 2.1, we see that to show the conclusion of this lemma, it suffices to prove that Mny e is isomorphic to IrrŽ Me . under the action of C. Define the map f : Mny e ª IrrŽ Me . by f Ž a. s wa for every element a g Mny e . By Lemma 2.4, we know that f is an isomorphism of groups, and so, to prove our result, we need to show that f commutes with the action of C. In particular, given elements a g Mny e and c g C, we need to prove that wa?c s Ž wa . c. For every element x g Me , we have

wa?c Ž x . s wac ny e Ž x . s « TrŽ ac

ny e

x.

s wa Ž xc nye . s wa Ž x Ž cy1 .

e

.

c

s wa Ž x ? cy1 . s Ž wa . Ž x . . Therefore, we have the desired result. LEMMA 2.6. Me is isomorphic to M f under the action of C if and only if gcdŽ e, n. s gcdŽ f, n. s t and frt ( Ž ert . p r mod nrt for some integer r such that 0 F r F d y 1. Proof. Suppose that Me is isomorphic to M f under the action of C. Let t : Me ª M f be an isomorphism that commutes with the action of C. ŽBy hypothesis, we know such a map exists.. Let a s t Ž1., and define the map

584

MARK L. LEWIS

t : Me ª M f by t Ž x . s t Ž x . ay1 for every element x g Me . Observe that ˆ t is an isomorphism from Me to M f . Furthermore, we have

ˆt Ž x ? c . s t Ž x ? c . ay1 s Ž t Ž x . ? c . ay1 s t Ž x . c f ay1 s Ž t Ž x . ay1 . ? c s tˆ Ž x . ? c. Thus, we may assume without loss of generality that t Ž1. s 1. Since we know that 1 generates E as an additive group, it follows that t Ž a . s a for every field element a g E. Now, fix an element c g C so that c has order n, and let m e Ž x . g Ew x x be the minimal polynomial for c e over E. Write t and b for the integers with the property that t s gcdŽ e, n. and b s ord n r t Ž p .. Let k be the Galois field having order p b , and note that k s Ew c e x. Observe that m e Ž x . has degree b and that it splits over k. We let s denote the Frobenius automorphism of k Ži.e., the automorphism of k defined by a ¬ a p for every element a g k .. By Theorem 21.10 of w5x, we know that GalŽ krE . s ² s :, and by Lemma 18.3 of w5x, we see that GalŽ krE . acts transitively on the roots of m e . This implies that 2

ky 1

 ce, cep, cep , . . . , cep 4 is the set of roots of m e . There exist elements a i g E so that we may write m e Ž x . s Ý bis0 a i x i. Since c e is a root of m e , we have that 0 s m e Ž c e . s Ý bis0 a i Ž c e . i. Applying the isomorphism t , we get b

0st

ž

Ý ai Ž c e . i is0

b

s

Ý ai Ž c f .

i

b

/

s

b

Ý t Ž ai ? c i . s Ý t Ž ai . ? c i is0

is0

s me Ž c f . .

is0

Thus, c f is also a root of the polynomial m e Ž x .. We conclude that r c f s c e p for some integer r such that 0 F r F b y 1, and this implies that r f ( ep mod n. Since t divides both e and n, we see that t must divide f. If we let s denote gcdŽ f, n., then it follows that t is a divisor of s. On the other hand, since s divides both f and n, we know that s divides ep r , and because p and s are relatively prime, s divides e. Therefore, s divides t, r and hence, s s t. In particular, we conclude that Ž c t . f r t s Ž c t .Ž e r t . p and r frt ( Ž ert . p mod nrt. Conversely, suppose that gcdŽ e, n. s gcdŽ f, n. s t and that frt ( Ž ert . p r mod nrt for some integer r. Multiplying this congruence through r by t, we find that f ( ep r mod n. Define the map t : F ª F by a ¬ a p for every a g F, and observe that t is the r th power of the Frobenius

GROUPS CONTAINING RAMIFIED SECTIONS

585

automorphism. Thus, we may view t as an isomorphism form Me to M f . Therefore, it suffice to prove that t commutes with the action of C. Given elements a g Me and c g C, observe

t Ž a ? c . s t Ž ac e . s Ž ac e .

pr

r

r

s a p c e p s t Ž a . c f s t Ž a . ? c.

Therefore, Me and M f are isomorphic under the action of C. 3. MULTISTEP GROUPS We will begin the construction of our family of groups with a class of examples constructed in Section 4 of w4x. We state the main facts found in that paper, and we recommend that the reader refer to it for proofs of these statements. We set d to be a positive integer, and we fix F to be the Galois Field of order p d. We also set s to be the augmentation of F defined by a ¬ a p for every element a g F. We take F X 4 to be the ‘‘twisted polynomial ring’’ in the indeterminate X. In other words, the elements of F X 4 are ‘‘polynomials’’ of the form a 0 q a 1 X q ??? qa n X n such that the coefficients a i lie in F. We do not assume that X commutes with the coefficients; instead, we impose the relation X a s a s X for every element a g F. ŽIt is well known that this defines a ring.. Fix an integer n, and observe that X nq 1 F X 4 s F X 4 X nq 1. This implies that X nq 1 F X 4 is a Žtwo-sided. ideal that we will denote Ž X nq 1 .. Let R s F X 4 rŽ X nq 1 ., and let x denote the image of X in R under the natural homomorphism. Then, every element of R can be uniquely written in the form a 0 q a 1 x q ??? qa n x n and < R < s Ž p d . nq 1. Also, observe that x nq 1 s 0 and x a s a s x for every element a g F, that xR s Rx is a nilpotent ideal, and that RrxR is isomorphic to F. Therefore, xR is the Jacobsen radical of R. Writing J s xR, we note that J k s x k R s Rx k for every integer k. Furthermore, it is not difficult to see that J n / 0 but that J nq 1 s 0. Given an ideal I of R, we will use the notation 1 q I to denote the set  1 q i < i g I 4 . In Lemma 4.1 of w4x, Isaacs proved that Su s 1 q J u is a group for every positive integer u. Write S s S1 , and observe that S s S1 ) S2 ) ??? ) Sn ) Snq1 s 1. Also, we note that < S < s Ž p d . n Žin particular, S is a p-group. and that S s  1 q a 1 x q a 2 x 2 q ??? qa n x n < a i g F 4 . ŽIn our terminology, we call S an n-step group where each S irSiq1 is a ‘‘step’’ in our group.. In Corollary 4.2 of w4x, Isaacs proved that w Si , S j x : Siqj for all positive integers i and j. In particular, w Si , S x : Siq1 for every positive integer i, and so, S i is normal in S. Isaacs also defined a homomorphism c i from Si

586

MARK L. LEWIS

to the additive group of F by 1 q a X i q y ¬ a , where a g F and y g J iq1. Given s g Si and t g S j for positive integers i and j with i j c i Ž s . s a and c j Ž t . s b , he proved that c iqj Žw s, t x. s ab s y ba s . 4. DEFINING THE MEMBER OF OUR FAMILY CORRESPONDING TO p In this section, we will construct the member of our family that corresponds to a given odd prime p. We first construct a larger group that contains the group we want as a subgroup. We construct this larger group for arbitrary positive integers d and n. For the particular group that shows up as a member of our family, we will have to specialize both d and n to p. Again, we take F to be the Galois field having order p d, and since Fq is a cyclic group of order p d y 1, we can take C to be the unique subgroup of Fq having order Ž p d y 1.rŽ p y 1. s p dy 1 q p dy2 q ??? qp q 1. Because the following sum will occur many times throughout the remainder of this paper, we fix this notation: given a positive integer i, define j < < v Ž i . s Ýiy1 js0 p . It is easy to see that p does not divide C . Take S to be the n-step group over F as constructed in Section 3, and define an action of C on S by

Ž 1 q a 1 x q a 2 x 2 q ??? qa n x n . ? c s 1 q a 1 c v Ž1. x q a 2 c v Ž2. x 2 q ??? qa n c v Ž n. x n , where the elements a i g F and c g C. It is easy to see that this defines an action by automorphisms of C on S and that each group Si is invariant under this action. Let G be the semidirect product obtained by C acting on S in this fashion. Set G to be the Galois group of FrE, where E is the Galois field of order p. We will now define an action of G on G. First, we set the action of G on S by

Ž 1 q a 1 x q a 2 x 2 q ??? qa n x n . ? t s 1 q a 1t x q a 2t x 2 q ??? qa nt x n , for elements a i g F and t g G . It is clear that G acts on Fq, and since C is a characteristic subgroup, G acts on C. We claim that this provides well-defined action of G on G, but since this proof contains some rather tedious calculations and provides little insight, we will omit it. We set D to be the semidirect product of G acting on G. It is easy to see that < D < s < S < < C < < G < s p d nv Ž d . d. The group corresponding to p that we include in X is in fact a subgroup of D when d and n both equal p. We have to do some more work in order to prove that the subgroup we want exists. This next lemma shows a connection to the previous section.

GROUPS CONTAINING RAMIFIED SECTIONS

587

LEMMA 4.1. SirSiq1 is isomorphic to Mv Ž i. under the action of C for e¨ ery integer 1 F i F n y 1. Proof. Recall that c i is a homomorphism from Si to F, and observe that kerŽ c i . s Siq1. Thus, we may view c i as an isomorphism from S irSiq1 to Mv Ž i. . It suffices to show that c i commutes with the action of C. Observe that

c i Ž Ž 1 q a x i q y . ? c . s c i Ž 1 q a c v Ž i. q y ? c . s a c v Ž i. s a ? c s c i Ž 1 q a x i q y . ? c. Thus, we have the desired result. We want the members of X to be fully ramified sections that satisfy the Main Theorem. By Theorem A of w8x, we know that coprime configurations do not satisfy the Main Theorem. Thus, we need p to divide < G <. This happens exactly when d s p. Therefore, we will specialize d as p, and so, for the rest of this paper F is the Galois field having order p p. LEMMA 4.2. C acts irreducibly on S irSiq1 for e¨ ery integer i such that 1 F i F p y 1. Proof. By Lemma 4.1, we know that SirSiq1 is isomorphic to Mv Ž i. under the action of D. Thus, to prove the result, it suffices to show that Mv Ž i. is irreducible under the action of C. Let t s gcdŽ v Ž i ., v Ž p ... From Lemma 2.3, we know that Mv Ž i. is irreducible under the action of C if and only if ord v Ž p.r t Ž p . s p. Since p p y 1 s v Ž p .Ž p y 1., it follows that v Ž p .rt is a divisor of p p y 1. Therefore, p p ( 1 mod v Ž p .rt. This implies that ord v Ž p.r t Ž p . divides p. To prove the result, we must show that v Ž p .rt does not divide p y 1. Since the greatest common divisor of p p y 1 and p i y 1 is p y 1, we see that v Ž p . and v Ž i . are relatively prime. It follows that v Ž p .rt / 1 Finally, we claim that p y 1 and v Ž p . are relatively prime. Thus, v Ž p .rt does not divide p y 1, and the result follows. We are now to the point where we must specialize n. In order to have a nonabelian fully ramified section, we need to take S to be a p-step group over F. In particular, we specialize n as p, and for the remainder of this paper we assume that n s p. LEMMA 4.3. Consider elements a, b g S, such that w a, b x g S p . Obser¨ e that there are elements a i g F such that a s 1 q a 1 x q a 2 x 2 q ??? qa p x p , elements bi g F such that b s 1 q b 1 x q b 2 x 2 q ??? qb p x p , and an ele-

588

MARK L. LEWIS

ment g g F such that w a, b x s 1 q g x p. Then py1

gs

Ý is1

ža b i

si pyi

y a is

py i

b pyi ,

/

and thus, TrŽg . s 0. Proof. Observe that ay1 by1 ab s 1 q g x p , and so, ab s baŽ1 q g x p .. Thus, we have that ab y ba s bag x p s g x p. The coefficients of x p in ab i py 1 s i and a q b q Ý py 1 b a s . Comand ba are a p q b p q Ý is1 a i b pyi p p is1 i py1 paring the coefficients of x p in ab y ba and g x p , we find that py1

gs

py1 s a i b pyi y i

Ý is1

py1 s bi a py1 s i

Ý is1

Ý is1

ža b i

si pyi

y a is

py i

b pyi .

/

We observe that si pyi

ža b / i

s

py i

s a is

py i

s b pyi s a is p

py i

b pyi .

This implies that TrŽg . s 0. We note that the elements of S p have the form 1 q a x p for field elements a g F. Thus, we have an isomorphism from S p to F by 1 q a x p ¬ a . Given an element b g F, we will use Lemma 2.4 to define the character w ˆb g IrrŽ S p . by wˆb Ž1 q a x p . s wb Ž a . for every field element a g F. Our next task is to determine which of the irreducible characters of S p are fully ramified with respect to SrS p . In order to do this, we need to introduce some other definitions. We begin by stating that Ž G, N, w . is a character triple if N is a normal subgroup of G and w is a G-invariant irreducible character of N. The definition and main results regarding character triples can be found in Chapter 11 of w3x. Let Ž G, K, L. be a normal triple, and let the character w g IrrŽ L. be G-invariant. We say that an L-coset gL g GrL is good with respect to the character triple Ž K, L, w . if w has a gL-invariant extension to ² L, c : for every element c g C where CrL s C K r LŽ gL.. ŽThis definition of good elements can be found in w8x.. In w1x, Gallagher proved that
GROUPS CONTAINING RAMIFIED SECTIONS

589

Žii. If the field element a f E, then the only elements of S that are good with respect to the character triple Ž S, S p , w ˆa . lie in S p . In particular, wˆa is fully ramified with respect to SrS p . Proof. By Lemma 3.5 of w8x, we know that an element a g S is good with respect to the triple Ž S, S p , w ˆa . if and if only ²² a, b ::wˆa s 1 for every element b g S such that w a, b x g S p , where we define ²² a, b ::wˆa s w ˆa Žw a, b x. for every a, b g S such that w a, b x g S p . For the moment, we will fix elements a and b in S so that w a, b x g S p . Since S p is central in S, we may write ²² a, b ::wˆa s w ˆa Žw a, b x.. There exist elements a i g F and bi g F so that a s 1 q a 1 x q a 2 x 2 q ??? qa n x n and b s 1 q b 1 x q b 2 x 2 q ??? qbn x n. Furthermore, if we assume that w a,b x g S p , then there is an element g g F so that w a, b x s 1 q g x p. Observe that we have

w ˆa Ž w a, b x . s wa Ž g . s « TrŽ ag . . This implies that ²² a, b ::wˆa s 1 if and only if Tr Ž ag . s 0. Consider first the case that a lies in E. Using the notation of the above paragraph, we have for every pair of elements a, b g S such that w a, b x g S p the identity py1

ž žÝž

Tr a

s a i b pyi y a is i

py i

b pyi

is1

/

//

py1

s a Tr

žÝž

s a i b pyi y a is i

py i

is1

b pyi

/

/

s 0.

py 1 Ž s By Lemma 4.3, we know that g s Ý is1 a i b pyi y a is b pyi .. Thus, the Ž . above identity implies that Tr ag s 0, and hence, every element a g S is good with respect to Ž S, S p , a ˆw ., and we have completed the proof of Ži.. Now, we have the case that a f E. Given an element a g S and a f S p , we need to find an element b g S, so that w a, b x g S p and w ˆa Žw a, b x. / 1. Writing a as in the first paragraph, let i be the least integer such that a i / 0. Since a f S p , we know that i - p. By Theorem 21.7 of w5x, it follows that p y 1 is the greatest common divisor of p py i y 1 and py i p p y 1. Thus, we see that a s / a since a does not lie in E. This s py i implies that a y a / 0. We now fix an element n g F so that the trace of n is not 0. We now define the element b py i g F to be py i py i nrŽ a is Ž a s y a ... We then take the element b g S to be 1 q py i b py i x . Observe that a g Si and b g S pyi . Thus, by Corollary 4.2 of w4x, i

py i

590

MARK L. LEWIS

we have w a, b x g S p , and by Lemma 4.3, we know that w a, b x s 1 q g x p , where s g s a i b pyi y a is i

py i

b pyi .

By substituting for b py i , we get

g s ai

s

ž

si

n a is

py i

ns

Žas

py i

Ža

s

py i

ya

y a is

./

i

ya.

si

y

n

Ža

s

py i

s

py i

ya.

n

py i

a is

py i

Žas

py i

ya.

.

Observe that

ž

an s as

py i

i

ya

/

as

s

as

py i

py 1

n

ya

.

Thus, we have that



Tr Ž ag . s Tr a

s Tr

ž

žž

Žas

ns a

s

py i

as

py i

si

i

ya

y a.n

py i

ya

/ /

y

n a

s

py i

ya

/0

s Tr Ž n . / 0.

This implies that a is not good with respect to the triple Ž S, S p , w ˆa ., and Žii. is completed. The final statements of both parts are based on a theorem of Gallagher found in w1x Žalso, see Problem 11.10 of w3x.. By that result,
w ˆat Ž x . s wˆa Ž x t . s « TrŽ a x

ty1

.

s « TrŽ a

t

x.

sw ˆa t Ž x . .

GROUPS CONTAINING RAMIFIED SECTIONS

591

Thus, we have w ˆa t s Ž wˆa .t . Hence, the only irreducible characters of S p fixed by G are those that correspond to elements of E. On the other hand, we know from Theorem 4.4 that these are exactly the characters that are not fully ramified with respect to SrS p , and so, they do not interest us. Instead, we consider a G- and C-invariant subgroup of S. Consider the subgroup Y s  1 q a x p < a g kerŽTr.4 of S p . THEOREM 4.5. We ha¨ e the following: Ži.

S p l w S i , C xY s Y for e¨ ery integer i such that 1 F i F p.

Žii. w Si , C xY s w S, Siy1 x for e¨ ery integer i such that 2 F i F p. Proof. We work by reverse induction on i. We prove the result when i s p. Observe that S p is isomorphic to Mv Ž p. under the action of C. Since < C < s v Ž p ., we use Lemma 2.6 to see that Mv Ž i. is isomorphic to M0 under the action of C. Thus, it follows that C centralizes S p . We now have w S p , C x s 1 and thus, w S p , C xY s Y. This implies that w S p , C xY l S p s Y. Using Lemma 4.3, we conclude that w S py 1 , S x s Y. We now consider i - p, and we assume that the result is true for i q 1. Let W s w Si , S x s w Siq1 , C xY, and observe that S p l W s Y. Note that S p is the subgroup of S containing all points that are fixed under the action of C. Since the action of C on S is coprime, S pWrW is the set of all points of SrW that are fixed by C. Applying Corollary 4.4 of w4x to SrS p , we find that Siq1 s w Si , S x S p s WS p . It is easy to see that SirW is central in SrW, and in particular, SirW is abelian. This implies that we may use Fitting’s theorem to see that SirW s Siq1rW = w Si , C x WrW. Therefore, we conclude that Siq1 l w S i , C xW s W. Now, we write U s w Si , C xW, and note that U l Siq1 s W. Furthermore, we calculate U l S p s ŽU l Siq1 . l S p s W l S p s Y and U s w S i , C xW s w Si , C xw Siq1 , C xY s w Si , C xY. This proves the first assertion. We can further assume that i G 2, and we need to prove that U s w Siy1 , S x. Since SirW is central in SrW, it follows that U is normal in S. Applying Corollary 4.4 of w4x to the group SrS iq1 , we find that Si s w Siy1 , S x Siq1. Thus, if w Siy1 , S x : U, then the observation that Y : w Siy1 , S x l S p : U l S p s Y implies that U s w Siy1 , S x, as desired. We now assume that w Siy1 , S x ­ U, and we work to get a contradiction. Observe that w Siy1 , S xU ) U and that < Si : U < s < S p : Y < s p. Since

592

MARK L. LEWIS

U w Siy1 , S x : Si , this implies that Si s w Siy1 , S xU. Fix a character w g IrrŽ S irU . such that w / 1, and note that kerŽ w . s U. We now consider the map ²² ? , ? ::w defined in w2x. Since w Siy1 , S x : S, we can view ²² ? , ? ::w as a pairing between Siy1 and S. From the linearity of w , we have ²² a, b ::w s w Žw a, b x. for every pair of elements Ž a, b . g Siy1 = S. Because U s w S i , C xY, it follows that C centralizes SirU and hence, that w is C-invariant. Therefore, ²² ? , ? ::w is a C-invariant pairing. Since w Siy1 , S x ­ U, it follows that Ž S . 0 - S and Ž Siy1 . 0 - Siy1. It is easy to see that Ž S . 0 and Ž Siy1 . 0 are C-invariant. Because w Si , S x s W : U, we observe that S i : Ž Siy1 . 0 . By Lemma 4.2, we know that C acts irreducibly on Siy1rSi . This implies that Ž Siy1 . 0 s S i . From Lemma 3.2 of w8x, we have < S : Ž S . 0 < s < Siy1 : Si < s < F < s p p and SrŽ S . 0 is abelian. Hence, it follows that w S, S x : Ž S . 0 . Also, it is easy to see that S p : Ž S . 0 , and so, S p w S, S x : Ž S . 0 . Applying Corollary 4.4 of w4x to SrS p , we find that S2 s S p w S, S x. Since < S : Ž S . 0 < s < S : S2 <, this implies that S2 s Ž S . 0 . Therefore, ²² ? , ? ::w is a nondegenerate C-invariant pairing between S1rS2 and Siy1rSi . Since there is a nondegenerate C-invariant pairing between S1rS2 and Siy1rSi , it follows that S1rS2 and Siy1rSi are dual under the action of C. By Lemma 4.1, S1rS2 is isomorphic under the action of C to Mv Ž1. , and Siy1rSi is isomorphic under the action of C to Mv Ž iy1. . Observing that v Ž1. s 1, this implies that M1 and Mv Ž iy1. are dual under the action of C. Together Lemma 2.1 and Lemma 2.5 imply that M1 is isomorphic to Mv Ž p.y v Ž iy1. under the action of C. From Lemma 2.6, we see that v Ž p . y v Ž i y 1. ( p r mod v Ž p . for some integer r such that 0 F r F p y 1. Observe that this implies that v Ž p . divides v Ž i . q p r. Since i - p and r F p y 1, it is easy to see that 0 - v Ž i . q p r - v Ž p . which is a contradiction. Therefore, the result follows. As a result of Theorem 4.5, we have S p l w S, C x s Y and < S : w S, C x< s p Žsince Y : w S, C x.. We know that both S and C normalize w S, C x. Since G acts on both S and C, it follows that G acts on w S, C x. In order to align our notation with that used in the Main Theorem, we write L s Y and K s w S, C x. Also, we will use M to denote the semidirect product of C acting on E and G for the semidirect product of G acting on M. Fix a character w g IrrŽ Y . so that w is G-invariant and w / 1. Then there is an element a g F such that w s Ž wa . Y . Since w / 1, it follows that a f k, and by Theorem 4.4, we know that wa is fully ramified with respect to SrS p . By Lemma 10.5 of w2x, we find that w is fully ramified with respect to w S, C xrY. Let e be the unique irreducible constituent of g w S, C x. It is easy to see that Ž G, K, L, e , w . is a controlled fully ramified configuration. Thus, we will take Ž G, K, L, e , w . to be the element of X that corresponds to p.

GROUPS CONTAINING RAMIFIED SECTIONS

593

5. PROOF OF MAIN THEOREM Note that S p is isomorphic to the additive group of F under the action of G . It is easy to see that there is a unique G-invariant composition series for F. ŽThat is, the action of G on F is uniserial.. Furthermore, it is clear that Y and kerŽTr. are the unique G-invariant subgroups of S p and F Žrespectively. having index p. Let Z be the unique G-invariant subgroup of Y having index p. Since w is G-invariant, it follows that kerŽ w . s Z. LEMMA 5.1.

No nonidentity element of G is good with respect to Ž K, L, w ..

Proof. Let t be a nonidentity element of G , and observe that G s ²t :. Write DrS p s C S r S pŽt . s C S r S pŽ G .. Consider an element c g S, and so, we may write d s 1 q d1 x q d 2 x 2 q ??? qd p x p for elements d i g F. Then, d g D if and only if each d i g k for every integer i such that 1 F i F p y 1. Furthermore, observe that w d, t x s 1 q Ž dtp y d p . x p and dtp y d p s w d p , t x g w F, t x. Since w F, t x is G-invariant and has index p in F, it follows that w F, t x s kerŽTr.. This implies that w D, G x s Y. Write B for the subgroup such that BrL s C K r LŽ G .. Because w D l K, G x : w D, G x s Y, we find that D l K : B. On the other hand, B : D l K, and so, we conclude that B s D l K. Note that t is good with respect to the triple Ž K, L, g . if and only if w B, t x s w B, G x : kerŽg . s Z. Let ArZ s C D r Z Ž G ., and observe that t is good with respect to Ž K, L, g . if and only if B : A. Since w Y, G x : Z, it follows that Y : A. Fixing an element a g S, we may write a s 1 q a1 x q a2 x 2 q ??? qa p x p for elements a i g F. Then, a g A if and only if each a i g K for every integer i such that 1 F i F p y 1 and TrŽ a p . s 0. We will prove that B ­ A by producing an element b g B such that b f A. Let c be any nonidentity element of C. For integers i such that 1 F i F p y 1, we define the quantity i

mi s 1

Ł Ž c v Ž j. y 1. , js1

and we take m p to be an arbitrary element of F. We now define the element m g S by m s 1 q m1 x q m 2 x 2 q ??? qm p x p. Next, we will prove that w m, c x s 1 q x y m py 1 x p. We begin with the equation m Ž 1 q x y m py 1 x p . s 1 q Ž 1 q m1 . x q Ž m1 q m 2 . x 2 q ??? q Ž m py 1 q m p y m py1 . x p .

Ž ).

Thus, for integers i such that 2 F i F p y 1, the ith coefficient of Ž). is m iy1 q m i . We know that mw m, c x s m c, and so, we need to show that Ž).

594

MARK L. LEWIS

equals m c. Recall that the ith coefficient of m c is m i c v Ž i., and we use this to show that Ž). equals m c by proving that the coefficients are the same. We begin with the coefficients of x. We have 1 qm1 s 1 q 1r Ž c y1 . s Ž c y1 q1 . r Ž c y1 . s c Ž 1r Ž c y 1 . . s m1 c. We now consider the coefficients of x i for i such that 2 F i F p y 1. Observe that iy1

m iy1 q m i s 1

Ł Ž c v Ž j. y 1. q 1 js1

s Ž c v Ž i. y 1 q 1 .

i

Ł Ž c v Ž j. y 1. js1

i

Ł Ž c v Ž j. y 1. s m i c v Ž i. . js1

Finally, we consider the coefficients of x p , but in Ž). this coefficient is m p , and in m c, it is m p c v Ž p. which equals m p since c v Ž p. s 1. Therefore we have the desired result. We claim that the element b that we want is w m, c x. It is easy to see that w m, c x lies in D l w S, C x s B. We want to show that w m, c x does not lie in A. Thus, we need to show that TrŽym py 1 . / 0. In particular, we must py 1 Ž v Ž i. prove that TrŽ1rŁ is1 c y 1.. / 0. q Since F is a cyclic group of order p p y 1 and c lies in the subgroup of order Ž p p y 1.rŽ p y 1., it follows that c s a py 1 for some element a g F. Also, we know that F s Ew a x. Note that the roots of the minimal i polynomial f Ž X . of a over E are exactly the quantities a i s a p for py 1 Ž 0 F i - p, and thus f 9Ž a . s Ł is1 a y a i .. i Ž . Ž . Now p y 1 v i s p y 1, and thus ai y a i c v Ž i. y 1 s a p y1 y 1 s . a The quantity whose trace we are trying to compute is exactly py1

a py 1

Ł Ž a i y a . s a py1rf 9 Ž a . .

is1

Using Proposition IV.5.5 of w6x, we see that TrŽ a py 1rf 9Ž a .. s 1, and this completes the proof. We now have all the tools that we need to prove the Main Theorem. Proof of Main Theorem. Observe that < G : L < s < G : M < < M : K < < K : L < s < G < < C < < S : S p < s pv Ž p . p pŽ py1. . Since p and v Ž p . are odd, it follows that < G : L < is odd.

GROUPS CONTAINING RAMIFIED SECTIONS

595

We know that G acts on C, and we can take X to be the semidirect product of G acting on C. Note that G is the semidirect product of X acting on E. Hence, the subgroup XL will be a complement for Ž G, K, L.. By Lemma 4.3 of w9x, we know that all complements for Ž G, K, L. are conjugate, and thus, we may assume that H s XL. Since L is central, H is the direct product of X and L. Therefore, w extends to H. Let u g IrrŽ H . be an extension of w , and write g s u M l H . One can see that Ž M, K, L, e , w . is a coprime fully ramified configuration. From Theorem A of w8x, we see that g M has a unique irreducible constituent c having odd multiplicity. By Theorem A of w9x, it follows that c extends e . Since g is H-invariant, we may use the unicity of c to observe that c is G-invariant. Because GrM is a group of order p, this implies via Corollary 6.17 of w3x. that there are p extensions of c to G. Note that each of these in fact extend e . Let x g IrrŽ G . be any one of these extensions of c . We want to calculate the multiplicity of x in u G. By Frobenius reciprocity ŽLemma 5.2 in w3x., this is the multiplicity of u in x H . We claim that CL s M l H contains every element of H that is good with respect to the triple Ž K, L, w .. As stated in w8x, we can view this as a property of L-cosets. Since X is isomorphic to HrL, this says that C contains every element of X that is good with respect to Ž K, L, w .. Observe that G is a Frobenius complement in X that is good with respect to Ž K, L, w .. Observe that G is a Frobenius complement in X, and so, every element of X outside of C is conjugate to an element of G Žsee Problem 7.1 of w3x.. The fact that C contains every element of X that is good with respect to Ž K, L, w . follows from Lemma 5.1. By Lemma 2.4 in w2x, we know that x Ž h. s 0 for every element of H that is not good with respect to w . By w3x, we know that the multiplicity of u in x H is

w u , xH x s 1r< H < Ý u Ž h . x Ž h . s 1r< H < Ý g Ž h . c Ž h . hgH

hgCL

s 1r< H : CL < w g , cC L x . Since w g , cC L x is the multiplicity of g in c M l H , it follows that the multiplicity of u in x H is odd. This result is true for each of the p extensions of c to G, and so, this completes the proof of the theorem.

REFERENCES 1. P. X. Gallagher, The number of conjugacy classes in a finite group, Math. Z. 188 Ž1970., 175]179. 2. I. M. Isaacs, Characters of solvable and symplectic groups, Amer. J. Math. 95 Ž1973., 594]635.

596

MARK L. LEWIS

3. I. M. Isaacs, ‘‘Character Theory of Finite Groups,’’ Academic Press, New York, 1976. 4. I. M. Isaacs, Coprime group actions fixing all nonlinear irreducible characters, Canad. J. Math. 41 Ž1989., 68]82. 5. I. M. Isaacs, ‘‘Algebra: A Graduate Course,’’ BrookrColes, Pacific Grove, California, 1994. 6. S. Lange, ‘‘Algebra,’’ 3rd ed., Addison]Wesley, Reading, Massachusetts, 1993. 7. M. L. Lewis, ‘‘A New Character Correspondence in Solvable Groups,’’ Thesis, University of Wisconsin, 1995. 8. M. L. Lewis, Nonabelian fully-ramified sections, Canad. J. Math., in press. 9. M. L. Lewis, Character correspondences and Nilpotent fully-ramified sections, submitted. 10. M. L. Lewis, A new canonical character correspondence in solvable groups, submitted. 11. M. L. Lewis, The canonical correspondent and constituents of odd multiplicity, preprint.