The 2-Length of a Finite Solvable Group Citation Gross, Fletcher Ivan (1964) The 2-Length of a Finite Solvable Group. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/TRHX-H715. https://resolver.caltech.edu/CaltechETD:etd-09192002-161906 Abstract NOTE: Text or symbols not renderable in plain ASCII are indicated by [...]. Abstract is included in .pdf document. In this paper the relationship between the 2-length [...] and the 2-exponent [...] of a finite solvable group G is studied. It is shown that [...] - 1 provided that [...]. The special case of groups satisfying [...] = 2, i.e., groups whose Sylow 2-groups are of exponent 4, is investigated to determine whether [...] in this case. This question is not answered but it is shown that a certain normal subgroup (which may be the whole group) satisfies [...]. In addition if all the elements of order 4 are contained in this subgroup, then [...] for the whole group as well. As an application of this last result, it is proved that [...] in a group of exponent 12 Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics and Physics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Minor Option: Physics Thesis Availability: Public (worldwide access) Research Advisor(s): Hall, Marshall Thesis Committee: Unknown, Unknown Defense Date: 1 January 1964 Record Number: CaltechETD:etd-09192002-161906 Persistent URL: https://resolver.caltech.edu/CaltechETD:etd-09192002-161906 DOI: 10.7907/TRHX-H715 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 3637 Collection: CaltechTHESIS Deposited By: Imported from ETD-db Deposited On: 20 Sep 2002 Last Modified: 19 Jan 2024 19:22 Thesis Files Preview PDF (Gross_f_1964.pdf) - Final Version See Usage Policy. 1MB Repository Staff Only: item control page

THE 2-LUNGTH OF A FINITE SOLVABLE GROUP Thesis by Fletcher Gross 'In Partial Fulfillment of the Requirements For the Degree of Dactor of Philosophy California Institute of Technology Pasadena, California 1964 ACKNOVLEDGLENTS Io wish to take this opvortunity to thank the Naticnal Science Foundation for the support afforded me by their award of a Cooperative Fellowship. My special thanks and sratitude are due my advisor Professor Marshall Hall, dr., for his interest, many kindnesses, and helpful suggestions in matters both scnolarly and personal. ABST ACT In this paper the relationship between the 2~length L(G) and the 2-exponent e,(G) of a finite solvable group G is studied. It is showm that £5(G)=2e0(G) ~ 1 provided thas 4£,(¢)e1. The special case of groups satisfying eo(%) = 2, Le@e, Groups whose Sylow 2-grouns are of exponents 4, is investigated to determine whether L nse, in this case. This question is not answered bug it is showm that a certain normal subgroup (which may be the whole zroup) satisfies & ner. Tn addition if ell the elements of order 4 are contained in this subgrous, then L 85 for the whole group as well. As an application of this last result, it is proved that D oge@n in a group of exponent 12. I. Introduction The primary objective of this thesis is to obtain an improved bound for the 2~length of a finite solvable group G in terms of its 2~exponent. In addition the special case where the Sylow 2~suberoup of G is of exponent 4 is studied. The first results on the p-length of a group were obtained by Hall and Higman in [2]- Following their paper we make the following definitions: (1) (2) (3) (4) (5) A finite group G is called a p -eroup, where p is a prime, provided p does not divide the order of G. G is p-solvable if each of its composition factor groups is either a p~sroup or a p —-sroup. The upper p-series l= PSN Pye eee <3= = G of a p-solvable group G is defined inductively by setting N/ Pi to be the greatest normal p ~subgroup of G/P,, and Pe Ak, the greatest normal p-subgroup of GAN) 0 The least integer @ such that a) = G is called the p-length of G and is denoted by 2G) or, if the group G is understood, simply by £. c, (6) for a prime p and a finite group G is defined to be the class of a Sylow p—subgroup of Ge = 2 uo (6) Por a finite group G and a prime p the p-exponent @, () of G is defined by the rule that p2 ) is the exponent of a Sylow p-subgroup of G, L.¢e, the greatest order of any p-element. It is shown in [3] that 7 ee in any p-solvable Pp group. An application ot this which we use repeatedly is that if Gis solvable and e,(G) = 1 then £.(G) = 1. This follows since a group of exponent 2 is necessarily avellan. For an odd prime p, Hall and Higman showed that t py if p is not a Fermat prime and Le 20, is a Fermat prime. They further showed that these inequal- if p ities are best-possible. No results, however, were obtained about the relationship between 4, and e., when p = 2. This gap was filled by A. H. M. Hoere w EJ who showed that L, “3c, - 2 in any 2-solvable group where Lozi. In the present paper it will be shown that under the sane conditions 4.a82e, ~ 1. It is not known whether or not this is best~possible and a better result may very well be true. Indeed, the author knows of no solvable group in which L. exceeds @o« The special case e,(G) = 2 is studied in more detail and one result is that if G is of exponent le, 12 II. Statement and Proof of the Main Theorem far the rest of this paper we adopt the convention that all groups referred to are assumed finite, and if G is such a group then |G[ is its order. If His a proper subgroup of G this is written H<G and if, in addition, H is normal in G we write H<dG. GF(p") denotes the finite field of p elenents. Before proceeding to our main result we first state some basic properties of the upper p-series l= Pen4Pjan<a eee 13,9 = G. Lemma 1. For idl, P,/N3 3 contains its centralizer in G/N;_, and N,/P, contains its centralizer in G/P;. Temma 2. If B/N 7 is the Prattini esuhbgrous of Ps /N 3 for ig@l, then Pi /Ps is its own centralizer in G/F, « Corollary. For ivi, G/Ps is faithfully represented as a group of automorphisms of the elementary abelian p-group P/E; Proofs of these results are to be found in [3]. Now an elementary abelian p-sroup can be considered as a vector space over GE(p). In particular from the corollary we have that G/P, is faithfully represented as a linear group operating on the vector space Pi /Py. G/P, has no normal p-group except for the identity and L,(G/P,) = £8) - Le a Now if g is an element of order po in G/P, then the minimal equation of g on a is (x-1)"=0 where x is some integer <p". If r<p", ieee, (g - 1)P -1 = 0, then ¢g is said to be exceptional. The following is proved in Gl: Lemma 3. If ¢ is of order p™ and not exceptional in G/P,, then e,(G)2m+l. We are now prepared to state our main result: Theorem 1. [f Gis a finite solvable group and A,(G)B1, then L,(G)S2e,(G) - 1. I first remark that, since Feit and Thompson have praved the long-stsnding conjecture that sroups of odd order are solvable, a 2-solvable sroup is in fact solvable. Solvability is quite important in the proof of theorem 1. Now if ZL, = 1 the conclusion is trivial and, since e, = 1 implies L, = 1, we see that 4. = 2 implies that e,22 so the result again follows. Now if £o>2, then L(G/P>) os L5(G) - 221, so that if we could prove that Bo (G/P5 Sen (E) - 1 then theorem 1 would follow by induction on the order of G. For this purpose we only need to concern ourselves with the exceptional elements in G/P, beecanse of lenma 4. Tt will be shown that if g is of order a™ and exceptional in G/Py then ge is in P,/P,. This will immediately prove that e,(G/P)me(%) ~ 1 ana theorem 1 will be proved. ~5- Theorem 2, Let G be a solvable linear group on a field F of characteristic 2 and assume @ has no normal 2-group other than the identity. Then if N is the largest normal 2'_subgroup of @ and if g ie an exceptional clement of order 2” in @, it follows that gor is in the largest normal 2~subgroup of G/N. Proof. G/P, satisfies the hypothesis of this theorem so that, by our previous discussion, theorem 1 follows from theorem 2. The rest of Uhis section is devoted to the proof of theorem 2. It should be pointed out that this theorem is a more general result than the special case needed to prove theorem 1. Although in the statement of theorem 2 it is assumed that G is finite, neither F nor the dimension of the space on which @ operates need be finite. New neither the hypothesis nor the conclusion of the theorem is affected by an extension of the field F. Thus, without loss of generality, we assume that F is algebraically closed. Since an element of order 2 cannot he exceptional (since otherwise it would have to be the identity), m ole 2 oil must be greater than l. let h=g¢ and then h” =¢ * In proving theorem 2 we define subgrouns H and Hy3 chur, . hn-€ Hy), ond g normalizes Hy (Hy need not be normal in G). It is then shown that if x is any element in the largest normal 2-subgroup of H,/Hj9N, then ~~ Om (he, x) = (h, x)°. From this it will follow that n@ is in the largest normal 2-subgroup of H) /H,AN, and, finally, from this the desired result. Our first step is to prove two lemmas which are of use later and which also motivate the definition of He. Here, and elsewhere, we denote the space on which @ operates by V. Lemma 4, If Q is any 2' -subgroup of © which is normalized by @, then h fixes every minimal character— istic F~Q submodule of V¥. Proof. A minimal characteristic F~Q submodule is simply the join of all those F-Q submodules operator isomorphic to a given irreducible P-Q submodule. Now since Qis a 2‘ —group VW can be written as the direct sum of the minimal characteristic F~Q submodules. V= Vi OY¥5 © oor Since g normalizes Q, g— must pormute the Vai among themselves. Now if n& does not fix every V, then &, as a permutation of the Vi. has a cycle of length m_ m nm 2 2 1 A 2 2m) But (e - 1) =f + eee t Bel since F is of characteristic 2. Thus (¢ - 1)? 72 could not be zero in this case, contrary to assumption. Lemma 5. If Q is any abelian 2' subgroup of G and x is any element of G normalizing Q and fixing every minimal characteristic P-Q submodule of V, then x centralizes Qe. Proof. V = V; @ V4 @ «ee where the V, are the ~ 2m minimal characteristic F-~Q submodules. Suppose (x, Q) #1. Thus there exists a V5» Vy say, such that (x, 9) is not the identity on V;. Now Q is abelian and F is algebraically closed so that Q operates on V, as a scalar multiplication, ise, if YEQ and VEV, then yw = X(y)w where H(y) is a acalar, Thus we have (this computation is taken from fe]> Xa y=) = (x ya = x4yGn) = x Xy) xy = Yy)v. Therefore (y, x)v = v for all y€éQ, vé Vi contrary to (Q, x) not being the identity on ¥V,- Thus the lemma is proved. . Now let H be the following set of elements of 7: x€H if, and only if, for every normal 2’-subsroup Q of G, x fixes every minimal characteristic F~Q submodule of V. It is easy to see that H is a normal subgroup of @ and n°€ H by lemma 4. Since the largest normal 2-subgroup and the largest normal 2" subgroup of H are normal in & we have at once that H has no normal 2—subgroup other than the identity and the largest normal 2" -subgroup of H is Han. Ne now proceed to construct a characteristic 2'-~subgroup K of H such that K is nilpotent of class 2, no c-element of H except for the identity centralizes K, and if K = K, =x K, K eee is the representation of K as the direct product of its Sylow subgroups, K,; the Sylow q,~subgroup of K for some prime q;, then each K, is of exponent q,. The importance of K is due to the fact that - & = a Sylow 2-subgroup of H can be represented faithfully as a group of automorphisms of K, and. the restricted nature of K then restricts the structure of the Sylow 2~subgroup. To construct K, first let Q be the largest normal nilpotent subgraup of H. Clearly QSHnN. Furthermore, since H is solvable (this is where solvability is crucial), Q must contain its centralizer in H. (This is proved in [{).) Since Q is nilpotent we can write G-G, xx. where Q, is a q,-group for an odd prime q,; and q; # q5 for i # i. Lemma G. ¢(@) = 2, i.ee., Q is nilpotent of class 2. Rroof. Since h°€ H, h° does not centralize q. Thus by lemmas 4 and 5, Q is not abelian. Thus c(Q)22. Now suppose ec = c(@)=3 and let = FL @DPrn@e... ef =1 he the lower central series of 4 and let n be the first integer @(c+1)/2. Clearly nwéc-1l since cel (c4+1)/2 for cB3. But from (2. Re 150] we have (P{@®. D = 17430) 41, and from Pe. De 15g (Th. PIS fo,@ = 1 since anBerl. Thus we have that [}(q) is abelian and, of course, normal in @ but not centralized by Q. But from the definition of H and lemma 5 we see that this is impossible. Hence c = 2. This naturally implies that each a is of class not ~ QF « exceeding 2e Now ay is an odd prime and se greater than 2. Thus a is a regular q,;~group. (For the definition and properties of regular q-groups see Ee. PP. 183~186].) Then the elements of order at most a3” form a character istic subgroup of Q, which will he denoted c"(G,). Set K, = o*(,) and K= Ki x Ky X eee We now prove some elementary results which imply that no 2-element of H except for the identity centralizes K, Lemma 7. If Q is a group, x &@ non-trivial auto~ morphism of order prime to lal. and if Mis a normal subgroup of Q admitting x, then x cannot centralize poth M and G/M. Proof. Suppose that x does centralize both M and Q/M. Since x is non~trivial we have that there must be a y€ Q such that y~ 4 y. Thus we mast have y* = yz where z is in Mand z #1. Now y~ = yan = yz" Since x centralizes M. By induction we have y = ya" for all n. This is a contradiction since tne orders of x and 2 are relatively prime. Lemga 8. If P is a regular p-group with e,(P) = nrl and x is a non-trivial automorphism of P of order prime to p then x does not centralize ca-lipy, the subgroup consisting of all elements of P of order dividing pot, Proof. Suppose x does centralize cB-lipy, Then by the previous lemma x cannot centralize promt), n=] But for any y€, yP must belong to o"™-"(P) and thus «me TO om must be centralized by x. Thus yF = (y*yP. But since P is regular this implies that (y7ty*P = l. Since n>1 we see that yy hy* = (y, x) is always in o@lepy, Hence x centralizes both oh (py and p/o®-t¢p) contrary to assumption. Therefore the lemma is proved. Corollary. If fF is & regular p-group, x a non-trivial automorphism of order prime to p then x does not centralize clip), Beoof. If ¢,(F) = 1, then clip) = P. Ie e(P)>L, then the result follows from the lemma and from gnmanlfon-a(py | - gha-lypy, Ags a pesult of this we see that no 2-clement (exceont for the identity) of H centralizes K. K and each K, is a characteristic subgroup of H and thus normal in ©, Since h© does not centralize XK (sinee n@ is a non-idenity 2-element of H), K cannot be abelian. Thus K is of class 2. We are now prepared to define the subgroup Hy. For this purpose decompose VV for each K, into the sum V = V3, 0 Vio @-e where the Vi are the minimal characteristic F~K, J submodules. Let Cis = {:|=€3 and (K,;, x) is 1 on ee Clearly each O45 is a uureal subgroup of H although it is not necessarily normal in @. Now let Hy be the intersection of all the 0 3 which contain 2, Tf he is not in any Cu then set Hy, equal to Ke Im any event H,SH and H, is normalized by g¢. As was the case with H, H, cannot have any normal 2-subgroup other than the identity and the largest normal 2'-~subsroap is CHAR AH, = HjaN. It will be shown that h“ is in the greatest normal 2-suberoup of Hy /HyaN. This will imply that ne is in the greatest normal e~-subgroup of H/HAN which, in tur, will imply that n> is in the greatest normal 2~subgroup of G/N. Let P be a 2-subgroup of H, such that PCH,NN)/E Ay is the largest normal 2~subgroup of H,/H AN. Since, modulo N, P is normalized by g we can take FP and g to belong to the same Sylow 2-subgroup of &, Now h°€H, so that if h°¢p then by lemma 2, h* does not centralize P/ $(P) where @(P) is the Frattini subgroup of P. Lema 9. If x€P, then (n®, x) = (h, x)*. Proof. A may or may not belong to Hy but h normalizes TP (since h and P generate a 2-gubgroup and h normalizes PF modulo a 2" ~2roup ) so that (h, x)€@ © and thus (Ch, x)“ € GP). Therefore once the lemma is proved we heave at once that n& centralizes P/ @(P) which implies that h° must be in P which will finish the proof of theorem 2. Let k = (he, x)¢h, x)" and suppose k # 1. Eut k is a 2-element of Ty and thus cannot centralize K. Hence (Ky kk) # 1 for some is Choose Vj; J is not the identity when restricted to Va such that (Ky » k) j* Now ke Hy so Uhal by lke definition of H, we must have (Kj, 2”) also not the identity on Vise (This last statement is the motivation for the definition of Hye) In vhat follows let V’ = Vig? Q@ the image of K, when restricted to v'~5 Q= qd; and x, the image of x when restricted to vee Let get be the first vower of g fixing T and let @, be the image of this element when restricted to Y'. From (3, p. 13] it follows that since g is exceptional, 6, must be exceptional, Seles ve (- 2)? “be. Now h* ig not the identity on V' and so hé is not exceptional on vi. thus nee and so h fixes V'. We define h, = ge, * » then both (q, h,*) and (Q, Kjds where k, = (hy, x), x)", are not equal to the identity. Since By is exceptional and (9, by*) Z1, Q cannot be abelian. But c(X) = 2 so that @ must be of class 2. Since, in addition, Q is of exponent q, we have 2( QQ = raga) (Z(Q) is the center of Q). Now v" is the sum of absolutely irreducible FQ submodules all of which are cperator isomorphic to each other. On an absolutely irreducible F~Q module 2(Q) has to be cyclic and generated by a scalar matriz. The Same Gust hold true for the representation of @ on v'. Thus Z2(Q) is cyclic of order q generated by a scalar matrix. Now Q #1 since Q is not abelian, Therefore we t must have 3(Q) = Q which moans that Q is an cxtra-speeial —-13- qegroup. (See E: Dp. 15 J.) Note also that if S is the e-~group generated by , and g, then (2(Q), 5) = 1 since 2(Q) is senerated by a scalar matrix, “ow let 7 be an irreducible F-QS submodule of ve By [5, p. 14] we have that v" is an irreducible F-Q module, V' is the sum of F-Q modules operator isomorphic to ve Thus (4, hy”) #1 on v. and of course g, must be exceptional on ve Then by B: De 2i) we have the following: (1) 2 ty is a power of gq, and (2) if Sy is faithfully and irreducibly represented on Q1/2' (such a Q, can always be found since hy* is not the identity on 9/0"), then Q can be written as the central preduct of Q, and a group Qo and g, transforms Qo/Q' trivially. Ye now need the following result: Lemma 10, 2 1-1 = q ana Ja,/a"| = a°. Proof. The following argument is essentially due to Cs]. my First we have 2 ~ l= q" for some n, But m22 so that q™ = 3 (mod 4) and n must be odd. Thus g@ + 1 is divisible by q + 1. Thus we must have q+ 1 = 28-1 for some s~@3.q-1= er where r is odd. Hence we have q° = 29r 41 and q@™~ 1 = re a og (By? 4 nr2®, Since r and n are odd we see that 2° is the highest power of 2 dividing 2-1. aut @=2) on m,+1 m,~L . qg~- -~ Ls 2 (2 — 1). It now follows that s = my +1 -~ 1 so my —~ 14h — m ai TL r 1 aivides o° - 1, the polynomial (t* - 1) Since ? divides ot ~ t). Thus the irreducible factors modulo q of 4° * ~ 1 are all of degree less than or equal to 2. However no element of order 4 can heave a faithful l-~dimensional representation over GF{q) since q = 3 (mod 4) so that -l is not @ quadratic residue of q. Thus since G, is a 2-element of order at least 4 and g, is faithfully and irreducibly represented on Q,/a' it follows that fa,/af a qe Now the representation of Q on v is isomorphic to the representation of Q on ¥ so that (S19 Qo) = l on V implies that (g,, 4.) = 1. ' Thus from the preceding lemma we see that the centralizer of G1 in the space 9/Q' has co-dimension 2 over GF{q). It eusily follows that this is also true for all powers of g) (except for the identity, of course). Now if g, is of order 4 then the equation of g) on 9,79" must be + + 1 = 0 so that 61° must have the representation G! QO = on a/Q'. If g, is not of order 4 then e1* can have no faithful 1-dimensional representation over GF(q) so that a /@ must give an irreducible representation of gy" These results can be summed up by stating that in the completely reduced representation of g° on q/9', there is only one non-trivial block unless 6, is of order 4 in which -15- case there are two non-trivial blocks. Now 9/Q" can be considered as a vector space M over GF{(q). HE can be givem the structure of a symplectic space as follows (3, Re 17]: Let ec he a generator of Q’. Then for any elements a, b of Q the commutator (a, b) must be a power of ce Define poas b) by the equation (a, b) =¢ (a, 5), f(a, b) is uniquely defined on Q/9 and is bilinear and skew symmetric. Since Q" = 2(Q) it follows that pr of maximum rank on M. Now since f is of maximum rank the dimension of M must be even, 2r say. Since (S,. Q') = 1, S preserves the symplectic structure of M. Thus the homomorphic image 5S of S obtained by the representation of S on @/Q' may be considered as a subgroup of a Sylow 2-subgroup of the sympleetic group S(2r, q)- (S(2r, q) is. the symplectic group on 4 space of dimension er over GF(q).) Let X,, 8 s and h, be the images of X,+ S,» and h,, respectively, in the homomorphism Sees. Prom (Q, h,°) #1 and (Q, k,) # 1 it follows that Hy? #1 and (H,*, %) # ys %)*. We now describe a Sylow 2-subgroup of S(2r, q). (This is based upon E. PPe 23-24] .) Mis of dimension 2r over GF(qg) so that M can be provided with the structure of a space of dimension r over GR(q"). For an element & € GF(a~) define of to be o¢2, Then if Uj, + Une soe, U, are & basis for M over GF(q* }, the expression where oa is a primitive fourth root of unity, is a skew symmetric bilinear form on M of rank er with values in GP(q). Since all such forms are equivalent we can C assume i is the fundamentai forn. qo -~ 1 is divisible iy ~L site +h by 2+ so that GF qo ) contains a primitige 2 . —th root of unity 6. Then 90 = on e* = —le Let t, be the transformation of GF(q~) ceap OX and let t, be the transformation <—p@ocd. mR t, is of order 2 ls 1 2 is of order 4 and together ney generate a generalized quaternion group of order 2 1 e Call this group T and let P, be a Sylow 2~subgroup of the symmetric group on the numbers l, 2, «+e, re All transformations y of M of the form ¥Q oGuy) = 285 OF 5 gyi ys where the fT, are taken from T and G™is a permutation from Pj, form a Sylow 2-subgroup of S(2r, q). Thus we may suppose G to be a subgroup of the group just described. We first need a lemma giving additional information ahout 83 and then we will be ready to finish the proof of lemma 9. Lemma 11. The permutation o associated with g, is the identical permutation. Proaf. Tt is show in E. Pe 23] chat or is of ~17 - order smaller then the order of 61° Pirst suppose @& is of order >2. Then the order of g, is greater than 4, Thus 3° when completely reduced has only one non-trivial block. But the permutation associated with =,° is oo which has at least 2 disjoint non-trivial cycles, Clearly shis is a contradiction. Thus ef = 1. Now suppose om # 1. Assume, say, or(1) = 2, @(2) = 1. In the complete reduction of By on M there is only one non-trivial block. Thus By must be the identity on = 2 Mi ty Now €)(e<ju; + Osun) = Toda, + Th0¢u,- Thus Bo (e<, 0, + @GU5) = ToT, Ou, + TyToe<nuy. One of Boty or Tyts must not be the identity of fT. But if either one is the identity then the other is also. Thus neither one is the identity and so &° has two non-triviel blocks which con happen only if 8, is of order 4 which implies that tT is of order 2, fhus. both T,T, and ToT, must be the transformation O< —P - Xl. (This is the only element of order 2 in 7.) Thus the centralizer of 3° in M has co-dimension 4 over G¥(q) whereas it should be 2. This proves that @7= 1. Therefore @, fixes each u; and must act trivially L on ,; for all but one value of 1, 1 = 1, say. Thus 6 Boczay) = AOgdy + Be 0e5 ty m where A is an element of order 2 i in T, Then we have m,-2 f (Seu) =a oyu, + SX and 2 By “Ceduj) = -oqu, + ai ut Assume X,(eex,u,) = ZT satay? Oase I: of(L) £1. Assume, say, that at ~*¢1) = 2e ‘A straight forward calculation yields _ te 1 2 f Ba JOtgUs 4-2 ifi,2 7+ But cane i. -I (~-I is the transformation ec-—o<) rick _. L — 2 i4l,2 Tt is easy to check that this is the same 4s ay, Xj) Case II: 9f(1) = 1. It is easily verified that in this case (5°, %,) is the identity while mn. — _ 2 Ped (By, HE oxjuy) = » 1) )" eeu, + Paria t mm. Now A is of order 2 1 in a generalized quatermion group +1 of order at are A and ant, Thus 51 Wye Wy 2 Ge 4 a,)% = (we ak +? 2 (-r)(-2) = TL Thus (Bj, %,)° is also the identity. sa that the only conjugetes of A in ? Therefore it has bean shown that ~ 12 co (hy %) = (hy ’ X4) in all cases. Inis finishes the proof of lemma 9. With this, theorems 1 and 2 are also proved. -~19. III. Groups with @5 = 2 Since, if a solvable group satisfies eo1, then it also must satisfy & ne, it is perhaps a natural question to ask whether this is also true if ey = 2, This question is even more to the point when it is realized that (te the author's knowledge, at least) there have not been found any examples of grouos in which £, exceeds E50 In an argument similar to that used in the preceding section we show that if e5(G) = 2 then a normal subgroup (which may be the whole group) satisfies L o2. If all elements of order 4 are contained in this subgroup then it will follow that L (a2. As an application we show that this must happen if Gis of exponent 12, i.e., if xi@ -1 for allxinG Unfortunately, in the situation analogous to the hypothesis of theorem 2, we need an additional assumption to prove the desired result for groups of exponent le. This condition is that @ should be irreducibly represented on the space Ve Therefore before proceeding further we prove a reduction theorem which allows us to assume this extra condition. A proposition is said ta he of type A if it has the fallowing form: If Gis a finite p-solvable group satisfying condition B, then La)2(e,(G))s where f is a monotonically increasing function defined —~ 20 — for non-negative integral arguments, £(0) - O, and condition B is either vacuous or states that Op, (2) ay for some set, possibly infinite, of primes p; and non-negative integers 4,. Note that the proposition that if G is of exponent 12 then L,(G)Se(G) is certainly of type A for a group of exponent 12 is of order aazb and thus salveble by a well-imown theorem of Burnside and the condition that Gis of exponent le is equivalent to stating that en (G)se2, e,(G)ssl, and e,,()=0 for all primes p # 2, 3. We mow state and prove our reduction theoren,. Theorem Ze Im proving a proposition # of type A it suffices to prove it for the following special case: (1) G is the normal product of V by @ where ¥ is a weetor space over F, a specified finite field of characteristic p, and @ is a p-solvable linear group on V having no normal p-subgroup other than the identity. (2) All sroups of order at most Ke satisfy B. (4) V is an irreducible F-G module. Proof. First it should be explained that F may be arbitrarily picked from among the finite fields of characteristic p, but once it is chosen it is to remain fixed for the rest of the argument. In proving theorem 3 we assume the proposition 1 is valid for the special case and then prove it is wm CL valid for the sencral case. Now suppose G is the group of smallest order satisfying the hypothesis of T but not the conclusion, and let = Fat ive dy IY = 6 be the upper p-series of G. Since f(0) = O we must have £(Q)>0. Now if F,/N, is the Frattini subgroup of P,/ then, as is shown in BJ, £.(a/7,) = £4) so that if Fy # 1 then we would have a proper factor group of G Qo? satisfying the hypothesis but not the conclusion of the proposition T,. Pherefore assume Py = 1. Thus P, is an elementary abelian p~group which we identify with a vector space Vj over GF(p). G/P, is faithfully represented as a linear group & an Vi Now by 3, Pe 4] we Pind that we may assume that G has only one minimal normal subgroup. Phis subgroup must be contained in ¥, and we denote it with M. If M= V, then @ is irreducibly represented on Vy + Now it M#¥, amd if G is faithfully represented on Vy /i then we have that L Gh) = L p’@) so that we would have a contradiction to the definition of G. Finally, suppose that V, 4 M and @ is not faithfully represented on V,/M. Then the elements of & centralizing V,/M form a normal subgroup of @ greater than the identity. Let Q@ be a minimal normal - 22. subgroup of @ centralizing ¥,/M. Clearly Q must be a p -group so that V as a Q-module is completely reducible, Thus there exists 6 Q-module M, such that Vz = U@ My. Clearly Q is the identity on M, but is not the identity on M since 9 is faithfully represented on Vi-+ Now let x" be the centralizer of Q in V,~. We have uM’ normal: in ¢ since Q is normal in G, and n> M, but uu. All this contradicts the fact that M, is the unique minimal normal subgroup of G. Thus we see from the above that we can assume that G is irreducibly represented over GF(p). One consequence of this is that if H is any normal subgroup greater than the identity in G then H can have no non—zero Pixed vector in V,. For if H did have a non-zero fixed vector then all the vectors fixed by H would form a non-trivial submodule of Vy. Now F is some finite field of characteristic p so F must be a finite extension of GF(p). Let 1 = 9,16, ..., 8, be a basis for F over GF(p) and let Vis Vor coos Vi be a basis for Vi over GF(p). Finally let V be the vector space over F with basis Vz9 eees Vie Any vector of V is 3 of the form Ss r where @,;€GF(p). @ acts on V in the obvious way. Consider the group G* = GV, i.e., the normal product of V by G. Suppose that g* is of order p™ in G*, Then - 23 - e&* is the ordered pair (Z, v) for some G€ F and v€V. Now if Eis not of order p™ then © must be of order pat and cannot be exceptronal on VY. Thus there must be a V5 such that € _ ye “ # Q, Therefore — is not excep— tional on V, and so we have that €,(4)2 (a-1) +1 = Thus in agy eveot Spl Gwe, (G*)- Since for q # Pp, € gS") = e,(G) we have that G* satisfies condition B. purthermore £,(e) = £(e*) sa that if G* satisfies 7 so does G, Now suppose E is any normal p -subgroup other than the identity in @ and o SUEPOSS is & non-zero vector fixed by H. Since v # 0 the coefficient of V5 is not zero for some j, j = 1 suy. FP is a field so there exists OR EF such that ox C51 8.) = Ll. Now H must also fix Ov which can be written in the form bi tt Ov = ¥ + V¥ where g w S Yr wv Ww + = a S 6... 6.4. i me ° oj"5" j=2 isl ij id and the Cis are the coefficients for otv. It is clear . . t 1. that for H to fix O¢fv it must also fix w. Bub v is & non-zere vector of Vue Thus H can have no fixed non-zero vector in V. —- 2h o Now if V is an irreducible F-@ module then we are already at the special case of the theorem, Therefore assume U is a proper submodule. If € is not faithfully represented on V/U then let Q be a minimal normal subgroup of & centralizing V/U. Q must be s p —group so that ¥V is completely reducible as an P+Q module. Dhus there exists an FQ module U' such that ¥ = u@u. U' contains non—zero vectors since U is & proper submodule, and Q is the identity on Ue We have seen that Q cannot have any non=-zera fixed vectors so this is a contradiction. Therefore G is faithfully represented on V/U. Thus. 2(@) = &(E*/0) and of course e,(G*) Re, (G*/U) so that if G*/U satisfies proposition T se does G* and then so does G. Now we still have that any normal non—identity p —suberoup H of @ camot have any fixcd non-zero vectors in V/U since V is completely reducible as an F-H module. Thus if G is not irreducibly represented on V/U then the same argument as hefore yields that @ is faithfully represented on a non-trivial factor module of V/U. Continuing in this way we finally arrive at the case where @ is faithfully and irreducibly represented on some vector space over the field F. This is just the special case mentioned in the theorem end so theorem 4 is proved. - 25. As was stated previously, one of the results of this section is that if G is of exponent 12, then L .(a)8e5(4). Before proceeding further it might he well to justify this work. For in a eroup of order 22zb the 2-length and the 3-length can vary at most by one. Thus if it were true Lhat the 3~length of a group of exponent l2 was 1, then it would be quite trivial to state that the 2-length was at most 2. However, in E, Pe 5} is found 2 group of exponent 12 but with 3-length 2. The group obtained by taking the direct product of this group with the symmetric group on 4 letters ia a group of exponent 12 with both the eelength and the 3~length equal to 2. Thus the stated result about groups of exponent 12 camot be obtained in @ trivial way by arguing on the 3—lengeth. For the rest of this section we make the following standing assumptions: (1) G = GV, the normal product of ¥ by G, where VY is a vector space over a finite field F of churacteristic 2, (2) @ is faithfully and irreducibly represented as a linear group over Y¥, (3) @ is finite, selvable, and has ne normal 2-group other than the identity. (4) e5(G)S2, Naw we are interested in seeing under what conditions - 2 - can 23(G) exceed ¢5(G). But if e.(@) = O then both e5(G) and £, (6) are 1 and if e5(&) = 1 then A) =i and we have £68) = €5(G) = 2, Thus we may as well assume (5) e5(G) = 2. So far we haven't specified F. Our choice is given by (6) If.Q is any normal nilpotent 2 subgroup of class 22 in © then any irreducible representation of Q over F is in fact absolutely irreducible. Since there are only finitely many subgroups of G, it follows that by taking F to be a large enough extension of GF(2) we can assume (6) holds. Any finite field of characteristic 2 satisfying (6) is satisfactory for what Pullowse later we shall add to these assumptions the further one that G is of exponent le. Actually it should be pointed out that until we restrict ourselves to groups of exponent 12, we will make no use of the fact that @ is irreducibly represented on V. The approach used to investigate the structure of G is similar to that used in the vroof of theorem 2, ve will show that if Nis the largest normal 2’ subgroup of G, then a certain 2-subgroup, to be described later, | must be contained in the greatest normal 2-subsroup of G/N. In particular it £,(¢)>2 (which is she same as saying that L6O>1), we will see that there mast exist an element of order 4 of a special type in @% —- 29 — First let H be the following normal subgroup of @: x@H if, and only if, for every normal nilpotent subsroup Q of class at most 2 in G, x fixes every minimal character~ istic P-Q submodule of V. A normal nilpotent subgroup of € must be a 2'-croup since otherwise € would have a non-trivial normal e-subgroup. Thus if Q is such 4 group then VY solits inte the sum of minimal characteristic F-Q modules, so the definition of H is intelligible. Now from (5) there are elements of order 4 in @, and from (4) all such elements must be exceptional. Thus if 2 is of order 4 in @, then a must de in H by lemma 4. Hence H is greater than the identity. H has no normal 2-group except for the identity and the larsest normal 2’ ~group in H is HaN (N being the createst normal 2'~group in @). Tet Q be the greatest normal nilpotent subgroup of H. Q = Q x Q, x oe where Q, is a qy~group amd the q,; are distinet odd primes for distinct i. Now it is still true that H centralizes any normal abelian subgroup of @, Thus the proof of lemma 6 is applicable and we obtain c(Q) = 2. Now let xX, = C1(Q,) and let K=K, xk, x oe. As in the proof of theorem 2 we find that no non~identity a~element of H centralizes EK, Now let 4, be the subgroup of @ consisting of all elements which fix every minimal characteristic R-K; module for all i. Since o(K,)S2, THH,6H. oH, has no normal 2-subgroup except for the identity and its 2 greatest normal 2 -subgreup ip HAN. - 23 = tet P be a Sylow 2~subgroup of Hj. P #1 since @,(3) = 2 and if g is of order 4 then a“€ 4H. Kow the square of any element of P must be in H. Thus P/Pal is of expoment 2 and thus abelian. Therefore —_- He. we now prove two lemmas which will ther enable us to show divectly that PN/N is normal in /x,. Lemma 12. Suppose that g¢ and h are two elements of P and V ig a minimal characteristic &-K,; submodule of V. Tet 4, Sys and h, be the restrictions of Ky» Ss and h, respectively, to vi. Then, if (Q, hy~) = 1, it follows that (Q, (@j+ hy)) = 1. Proof, Assume (Q, (€ 1» 4,)) #1. Therefore neither e) = 1 then (Qs (6y2 hy) = (as (eyb,)*) ~~ ana (@s (Shy B)) = (Qe byte, hy |yhy*) = (Qs (eyp> BY) In this case simply replace BL by B hi. Therefore, without £1 nor h, centralizes Q. Now if (Qs g1 loss of gonerality, we may ascume that (q, 6,°) #1 along 2 with (Qs h,") = 1 and (Qs (21s h,)) #1. Now exactly as in the proof of lemma 9 we obtain that Q is en extra-special q-group (q = 4 since g, is of order 4 and thus exceptional so that 4 ~- 1 must be a power of q), eq’ is a symplectic space, S, and h) preserve the symplectic structure of ofa’, and we may assume that g, and h, operate on Q/Q’ as follows By (Zerg) = Aesyu, + Bocjay By 0a) = ZT oes) ~ 29 . where o is a permutation of omer 2 (since (Q, n,*) = 1), and A and the T, are chosen from a group isomorphic to the quaternion group of order 8 (since g = 3). In addition A must be of order 4 since (4, 1°) #1. Now if @ does not fix 1 then (@,, B,) would be of order 4 bul its centralizer in eq’ would have co-dimen~ sion 4 over GF(4). Thus (gy. hy) would be of order 4 but not exceptional which is impossible, Thus o fixes 1 and since (Q, hy*) = 1 we mist have E, C2e¢;u;,) = eeu, + 2 By ti Yer sys It is now an easy matter to verify that (), By) = 1 and the Lemma is proved. ' Corollary. If g, h€P and h* = 1, then (g, h) =l. Proof. (g, h) ig in P’ and thus in H. So if (g, h) is not 1 then (Ky, (=, h)}) £# 1 for some K,. The lemma states that this cannot happen. Lemma 13. If g, h€P, then (g, h)* = 1. Proof. Suppose that (2, n)“ #1. Then for some Ky (Kys (Ss h)*) #1. Choose ¥' to be a minimal characteristic F-K, submodule of V such that (Ky (gs h)“) is not the identity on v. Define Q, S12 and h, as in the previous lemma. Then if either (q, 61°) or (Q, hy*) is the sdentity, then (éy> hy) = l. Therefore assume neither 61° nor ny centralize q. Thus 8, and h, are excepvional and of order 4. Hence Q is a 3~group and $1 and h, operate on 9/Q' as follows: 6, Si ocjuj) = Aocyu, + FO - 50 = By Zexj4y) = Becjaj + ae Now if j #1 then (€,, B,) = 1 emd if j = 1 then we have Gy» By) Seay) = , Been, + > art But A and B are elements of a quatemion sroup’ so that (A, B)° = 1 and the lemma is proved. Theorem 4, PN/NSAG/N. Proof. I shall srove that PCHAN)/H, ANH, /H,AN which is equivalent to the theorem since H;@dG. Let P, be the subgroup of P such that Py CAAA) S/S is the largest normal 2~subgroup of H)/HinN. Pig P and PSAP) (since f, must contain its centralizer in P),. Thus by the corollary to lemma 12, P, contains all elements of order 2 in P, let Pp ={x]x“=1, zer}. Po 26 an elementary abelian group which is normal, modula Hn, in Hie Let C consist of those elements of H,/HnN which centralize both P, and Py/Po-e Clearly © is a normal subsroup of H,/HAAN. But any 2* element in C would then have to centralize P, contrary to Py containing its centralizer in 3, /H,AN. Thus C is a normal 2~subgroup of H,/H,nNN. But from the corollary to lemma 12 and from lemma 13, PCH,AN)/H|ANSC. faus P = Py and the theorem 1s provede Carollary. Lo(H,) el. Now let S be a Sylow 2-subgroup of @ which contains P. Brom the theorem it follows that F is normal in §&, - 31 - Lemme L4. If P conteins all elements of order 4 in S, then ¥.,(@) = 1. Proof. If S = P we are done. Therefore assume S # P. Then if x€ S-P we must have ve = 1, Also x€S~P, y€P implies that xy€S+P so that (my)° = 1 which implies that x yx = yi, Thus x induces the autormorphism yy! or FP. This can be an automorphism only if P is abelian. If both x, and Xy are in S-P, then X,X5 centralizes FP. But e,,(@) = 2 so that P dees cantain clements of order 4, Thus x,X5 cannot be in S-P. Therefore {S/P| = 2 and P is abelian. Now if x€S-P, ye P, then (x, y) = xtythy = ye Gr) and so x centralizes P/@(P). Therefore PN/N cannot be the largest normal 2-subgroup of G/N. But P is maximal in S. Then SN/N is the largest normal 2-subgroup of G/N. Thus Jn) =1. To our original assumptions (1)--(G) we now add (7) G is of exponent 12, This implies thet K must be a 3<group. Ve prove that L 2(&) = 1 in this case by showing thet the hypothesis of lemma 14 is satisfied. Now suppose there exists an element of order 4 in SP, g* is in H so (K, g“) £1. Tet V- V] @V5@... be the decomposition of V into minimal characteristic F-K modules. Since g€S-P, ¢ does not fix some Vie a dees fix each Vy and if e“ is not the identity on a Vz then ¢ must fix that = 352 - Vs for otherwise g could not be exceptional. (It is shown an [3] that if g is exceptional then the first power of g§ fixing a ¥, must be exceptional on that V,;«) Before proceeding further, we first need the following result: Lemma 15. There exist x and y in K such that (Cts 8°), Cv, 8°)) #1. Proof. Let C = {z|=€ K, (x, e°) € 2(K)f. Clearly Cpea(k) but Cc # K since then a would centralize Z4(K) and K/2C") contrary to (Kk, a”) # il. (2° eentralizes Z(K) by lemmas 4 and 5.) K/Z(E) is an elementary abelian 3-group so that there must be a GP(3)-g module of E/2(K) complementary to C/Z2(E). Thus K/2(K) = LA/Z(K) @ C/ZCE) and g normalizes L. Now for all x€I-2(K), (x, a ¢ 2K). Now suppose x, y€IKZ(K) and (x, a" )Cy, ey le a(K)e But (xy7?, ge“) = (X, @ 7 (yt, a). But since K/Z(K) is ~ 2 -1 2 . abelian GY", 8) s(,S°)G™ 1 Ss) (mod 2(Z)), and, similarly, 1 = (yt, ao) = (Ys ayy, a . ~ -l 2 2 2.—1 Thus (xy “, 6) = (=, 3°)(y, 7 = l (mod Z(K)). Thus xy7'€ Z(K). Therefore (x, z°) = > ez) (mod 2(K)) if, and only if, x = y (mod Z(£)) for x, y€L. It immediately follows from this that for any x€ Ll, there exists a y such that x = (y, em) (mod. Z2(E}). Now L cannot be abelian since ¢ normalizes L and en does not centralize it, From all this we see that there exist « 2 2 x, y@L such that ((x, 6°), (ys 3°)) #1. Now, taking « and y to satisfy the lemma, we may assume without any loss of generality that ((x, eB), (y, e“)) is not the identity on V,. Clearly this implies that ae is not the identity on V, so g must fix V,- Since g¢ does not fix every V,, assume g does not fix V,.e Therefore BK is the identity on ¥o which then also must be the case for (x, e-) and (y; a“). Wow V is an irreducible F-~@ module so that there mast be an element taking V, into Va. (This is the only place where irreducibility is made use of.) Such an element must be of the form zh where h€S and z is from a Sylow 3-subgroup of G which necessarily must comtain K. Our timate contrediction will be that z and K generate elements of order 9 whieh is impossible in a group of exponent le. we |: ~ If h¥, = V,, then 2V¥,, = Vow Set g, = hgh “. Then we have that 1 oe, 1°) + ge, g1°)) is not the identity on Vie Tt is further claimed that 6,7 # V5. For suppose €)V5 = Vo. Then gn V, = nV. and, since gVz # Vo, this implies that h” "Vp = Vj, j #2. Then we mist have that ev, = V5- But gn te S so that (gh7+)°¢ H. Thus (gh7}) fixes V5. Therefore gh 'v, Alsa (ho+y® must fix V, so we have nV; = Vo. Ve finally conclude that V5 = any, = V5 which is a contradiction. Hence g,V, # Vo. wy - 34 — a 9 Tf we now replace Vj. &. x, amd y by Vige Sy ms) + and yo ,» respectively, we may assume thst ZW, = Vo, C(x, a), C¥s g7)) is not the identity on Vy, and BV> # Vo~0 Let x = (x, a) and y; = (y, a). x, and ¥7 must be the identity on Vo since go is. Now since G is ef exponent 12, 2 must be of order 3. Thus Vy, = Vos ZVo = Vy (n #1, 2), and 2V., = Vy. Now let V' = V, @Vo@V,- V' is fixed by 2 and ’ the restrictions of Xs Fy» and 2 to Vo are 0 Oo A\. NO O\. HY a 8) a=f{B 0 Oj, x, = 10 Ir O |s 7 =] 0 I QO |, Q © 19) 0 0 Mi Oo 0 it where I is the identity and 0 the zero matrix. Now (x, ¥,) is not the identity on V, but (x). ¥,)€ 2(K) and 2(K) is represented on Vy as a cyclic group generated by a scalar matrix. Thus (M, N) = wtI where q@ is a primitive third root of unity. From a = 1 we get C = avt+g, Now Z, Xy+ and y, all helong to the same Sylow 3—subgroup of G which must be of exponent 3. 2 a XK, sa that we must have xy x1 "x1 - l. (2x, )? = 2° x5 Direct computation yields that this implies that wi, = ATA, similarly wy = atta, Thus (ih, N,) = avtlogt, wotya. But M and N generate & eroup of exponent 43 and class 2. It follows easily that (+, wt) = (mM, N) = wie er OO 9 QO O @T It is now a simple matter to verify that 2 Zz vA (a9 Vy)” Cys YZ) ys FY) Al. Phus 2X» ¥z) is a 5-element of order screater than 3. This contradiction proves that the hypothesis of lemma 14 is satisfied and thus: Theorem 5. If G is a finite group of exponent 12, then 2, (¢)Se,(4). 1. 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