CHRISTOPH HERING
TRANSITIVE LINEAR GROUPS AND LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS OF PRIME ORDER*
§l. INTRODUCTION
Let G be a group of linear transformations of a vector space V of finite dimension n over a field of finite order q. Furthermore, let ~,(q) be the nth cyclotomic polynomial evaluated at q and 1 where f=(n, ~,(q)) and f" is the largest power off dividing ~,(q).. In an earlier paper (Hering, 1968) the author proved that the structure of G is very restricted if ([G[, ~*(q))~ 1. Also, in all cases known to the author, G has the property
(*) if (IGI, ~*(q))# 1, n+l, 2n+l and (n+l) (2n+l), then G is solvable or essentially a Chevalley group (of ordinary or twisted type) operating on V in the natural way. In this paper we prove, that G must have property (*) provided that G has certain (unfortunately quite special) additional properties, as for example the property that some non-solvable composition factor of G is isomorphic to one of the groups PSL (~, ~), Sp (ft, ~) or A m. Suppose that G operates transitively on the set of non-zero vectors of the underlying vector space V. Then q"- 1 [ Ial and hence ~*(q) I IGI. Further- more, we can prove that ¢* (q) # 1,n + 1,2n + 1 and (n + I) (2n + 1) unless either n ~< 2 or n > 2 and [ V] is equal to one of a certain finite number of exceptional degrees (see § 3, Theorem 3.9). So, in general, we can assume that ([G[, ~* (q)) # 1, n+ 1, 2n+ 1 and (n+ 1)(2n+ 1). This fact enables us to use all results mentioned above to obtain information about finite transitive linear groups, i.e., linear groups which operate transitively on the non-trivial vectors of the underlying vector space (see § 5). Finally, in § 6 we summarize some corollaries concerning doubly transitive permutation groups. A prime p divides (~*(q), [GI) if and only if G contains an irreducible subgroup of orderp. This establishes the relationship between our paper and the second part of the title.
* This research was supported by a National Science Foundation grant.
Geometriae Dedieata 2 (1974) 425-460. All Rights Reserved Copyright © 1974 by D. Reidel Publishing Company, Dordrecht-Holland 426 CHRISTOPH HERING
§2. DEFINITIONS AND PRELIMINARY RESULTS
We shall in general use standard notation. If G is a group, then 3G is the center of G, F (G) the Fitting subgroup of G, ~(G) the Frattini subgroup of G and G ~ the set consisting of all elements in G which are different from the identity. If H is a subgroup of G, then ~R6H is the normalizer of H in G and ~oH the centralizer of H in G. Also, every factor group of H will be called a factor of G. For each p-group P we define fl(P) to be the subgroup gen- erated by all elements of order p in P. The alternating group of degree i will be denoted by Al. If V is a vector space over a field K, then GL(V, K) is the group consisting of all non-singular K-linear transformations of V. Both ordinary and generalized quaternion groups will simply be called quaternion groups. A quaternion group of order 2 a will be denoted by Q2-. Let m and n be integers and p a prime. Then (m, n) is the greatest common divisor ofm and n, while ~p(m) is the Euler number ofm. Also, we write m ] n if m divides n and pm [1 n ifp m divides n but pm+l does not divide n. This paper is a continuation of work started by the author in an article on doubly transitive groups published in 1968. As we shall frequently refer to this article, we shall denote it here by [TL]. Also, we shall often use some results on projective representations contained in a joint paper with M. E. Harris of 1971. This paper will be denoted by [PR].
LEMMA 2.1. Let G be a p-group such that every proper characteristic sub- group of G is cyclic and contained in 3G. Then one of the following statements holds: (a) G is cyclic. (b) G is elementary abelian. (c) G is extraspeciaI. (d) p=2 and G=T'3G, where T is an extraspeciaI 2-group of order >8, 3G a cyclic group of order 4, and T c~ 3G= 3 T. Proof. If G is abelian but not elementary abelian, then fl(G) is cyclic and hence G is cyclic. Assume now that G is not abelian. Clearly G'~< • (G)~< 3G, and ~G is cyclic. Hence [a, b]P=[a, bP]=l for all a, beG. So G' is elemen- tary abelian. In fact IG'l=p, as 3G is cyclic. Suppose now that G'<~(G). Then the preimage of I2 (G/G') in G is a proper characteristic subgroup and hence cyclic. Thus GIG' and therefore G itself is cyclic. Thus ~(G)=G'. Furthermore, G' ~< 3G implies that (xy) ~= x~y l (y- Ix- lyx) t
13G1=4. The first possibility leads to extraspecial groups. Assume that I`3GI =4, and let T/G' be a complement of 3GIG' in the elementary abelian group G/G'. Then dearly G=T.`3G, I-T, 3G'I=I and Tn`3G=G'. Also 3 T~< 3 G, as G = T.3G. Thus 3 T= G', and T is extraspecial. Suppose now that I T[ = 8. As B = G, there exist elements t, and t2 of order 4 in G such that G=
LEMMA 2.2. Let q be a power of a prime p, s a prime different from p, and let G be a subgroup of GL (n, q). Suppose that G contains a normal s-group N such that N= E. 3N, where 3N is cyclic and E an extra special group oJ order s za+l. Then n>~sL lfn=s ~, then (a) ~E=3G and (b) G/E. 3 G is faithfully represented on N[3N. Proof. As the center `3E of E has order s, it is contained in every non-trivial normal subgroup of E. This implies that each completely reducible faithful representation of E has an irreducible constituent 1:1 which is faithful. Let n~ be the degree of V1. As s#p we have n>~n1 >>.s° by Huppert (1968, p. 562, Satz 16.14). Assume now that n=s °. Then n=n~, i.e., E is irreducible. Hence by Schur's Lemma the centralizer of E in Horn(V, V) is a field L. This field contains the ground field and hence has order qb for a suitable number b, which divides n. As E is linear over L, we obtain a representation of E of degree n[b. But now by the above argument n[b>~sa=n, i.e., b= 1. Thus ~GE=3G. Denote Z=`3N and considerthe normal subgroup X of G consisting of all elements which fix elementwise the factor group N]Z. Clearly 3E<.3N, which implies that N/Z ~- E/E c~ Zis abelian and therefore E" ,3 G ~ LEMMA 2.3. Assume that i>~ 9 and that the alternating group A i of degree i has a faithful representation q~ of degree n over GF (q), where q is a power of 2. Then one of the following holds: (a) A~-4 has a faithful representation of degree n- 3 over GF (q). (b) AT contains a group of order 4 consisting of transvections with common center and common axis. 428 CHRISTOPH HERING Proof. As contains a subgroup B-~ C x D, where C ~A4 and D --~Ai_ 4. Let E= (a, z) be the elementary abelian subgroup of order 4 in C. If V= V (n, q) is the underlying vector space, then we have the series V > V, 1> V (a- 1) > 0, where V, is the kernel of a -~ and hence dim V/V,=dimV(a-1)<<.n/2. Suppose now that (a) is false. Then any faithful representation of D over GF(q) has at least degree n-2. As D centralizes E, it leaves invariant V¢ and V(~r- 1). Suppose that D acts non-trivially on 1I/1Io or V(a- 1). Then the simple group D actually is faithful on one of these spaces, so that n-2 <<.n/2, i.e., n <~4. As i/> 9, we conclude that/19 has a faithful representa- tion of degree at most 4 over GF (q). But A 9 contains a subgroup isomorphic to $7, and $7 contains a Frobenius group of order 7.6. However, this leads to a contradiction (see [PR, Lemma 1.3]). So D is trivial on It'/V¢and V (~- I). As D is no 2-group, D must operate non-trivially and hence actually faith- fully on V~/V(cr-l). So dimV~/V(~-l)>~n-2 and hence dimV/V~= =dimV(a-1)=l, i.e., a is a transvection. Also dimVo/V(cr-1)=n-2, and hence D operates irreducibly on V,/V(a-1). Now • fixes V~ because • e~a. As dim V[V,= 1, ~ centralizes V[Vo, and V (~-1)<~ V,. As D leaves invariant {V(z-1)+V(a-1))/V(a-1), we have V(~-I)=V(a-1). Furthermore V, >t V (• - 1) = V (a - 1), and D leaves invariant V,[ V (a - 1) c~ c~ Vo/V(a- 1). Again using the irreducibility of D we see that V,= Vo. So C lies in the group consisting of all transvections with center V(a-1) and axis Iio. LEMMA2.4. Let m>~2 be an integer, q a power of a prime p and K a field whose characteristic is different from p. Then the degree of any non-trivial representation of PSp(2m, q) is at least (l/d) (qm_ 1), where d= (2, q- 1). Proof. This follows from [PR, Theorem 4.2], as the group PSp(2m, q) contains a subgroup isomorphic to PSp (2, q~)~-PSL (2, q~) (see Huppert, 1967, p. 228, Satz 9.24). Only for m=2 and q=2 or q=3 we might possibly have exceptions. However, it is easy to check that PSp(4, 2) does not have any non-trivial projective representation of degree <3 and PSp(4, 3) does not have any non-trivial projective representation of degree < 4. §3. SOME ELEMENTARY NUMBER THEORY Throughout this paragraph we shall use the following notation: If n is a positive integer, then p(n) is the set of prime divisors of n, e(n) the set of subsets of p Or) of even cardinality and a (n) the set of subsets of p (n) of odd cardinality. If 3~ is a finite set of primes, then n(3~)=l-Ip~tl~p. For any two integers a>~2 and n>~2 we define LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 429 I-I ( an/n - 1) ~n (a) = ~ e e tn) FI (a 1)" ~ea(n) Furthermore, ¢1 (a) =a- 1. Finally, if ~ is any finite set, then D (fl92)=s- t, where s is the number of subsets of ~ of even cardinality and t is the number of subsets of 9J~ of odd cardinality. We have LEMMA 3.1. If 9J~ is a finite set, then D (gJ~)= 0 if ~1~~ 0 and D (~1~)= 1 if =0. Proof. Let 9X#0 and P e 9T~. For 3~___~1~ define ~°=X- {P} ifPeX and X °= 3~ u {P} ifP ~. Then 0 induces a 1 - I map of the subsets of ~92 of even cardinality onto the subsets of ~ of odd cardinality. The main goal of this paragraph is the investigation of ~n (a). Before we proceed we indicate a second lemma: LEMMA 3.2. Let a be a positive integer, r a prime divisor of a-1 and r" ]1 a-1. Then (i) if k is a positive integer and r ,~ k, then r ~ 11 ak-1; (ii) /fr~2, then r ~+1 I[ a'- l. Proof Let a- 1 =r't, where r ,~ t. Then a= 1 +r't and ak=lq-(~)r~tq-(~)r2¢t2-k...q-rkatk. Hence r ~ [ak--I and r ~+1 [ at-l, as a>~l. If r'*llak--1, then r](kl) t, and hence r lk. So (i) is proved. Ifr °÷2 I at-l, then r~+2lr~+'t+(2) r2~t2 and therefore r'+2 X (2) r2¢, which implies r = 2 and tr = 1. Let n I> 2 and let r be any prime divisor of 1-I ( a"/" (~) - 1) or 1-I ( a'/" (~) - 1). ~:ee(n) ~eo(n) 430 CHRISTOPH HERING Furthermore, let : ~_oYI,.)(a'/" (~) - 1), / z__[I.,.) (a"" (~) - 1) and n=rQn~, where r 4v nr Obviously r ,t~ a. Denote by m the multiplicative order of a modulo r. Finally, let r ~ [I a~'- 1. For each ~___p(n) the conditions r[d'/*(X)-l, m[(n/n(Y,)) and 2E c_ p (n/m) are equivalent. (Clearly, m is a divisor of n.) Furthermore, if rlan/'(~)-I and r~3E, then mr'l(nln(K)) as (m,r)=l, and hence r" II a*/n(~)-1 by Lemma 3.2. So we get (*) if Q = O, then ot - [3 = trD [-p (n/m)]. Assume now that Q>0, and let r' [I a~'°-l-1. Then, again by Lemma 3.2, r ¢ 11 d'/n(~)-I if ~p(n/m) and r~, while r ~ II if ~c_p(n/m) and re~. Hence (**) if Q ~ O, then ~ - [3 = (a - z) D [p (n/m) - {r}]. As always ~- [3 >/0, ~n (a) is an integer. Furthermore, if Q= 0 and ~- [3 > 0, then m=n=nt and ~-fl=a by Lemma3.1. If Q>0 and ~-fl>0, then p(n/m)= (r} and again m=nt, as (m, r)= 1. Thus nt is the multiplicative order of a modulo r, whenever r i~,(a). Suppose now that O>0 and ~-fl>~2. Then a-z>l by Lemma 3.1, and hence r'=2 by Lemma 3.2. As nl = m Jr- 1, we get n = 2 Q. Furthermore, 2'= 2 II :-1-1 implies that Q= 1 and a-3(mod4). We always have r~-#la ~- 1. This is clear as ~-[3<,.a and r ¢ II a' -1 I [ a~"¢-I =a*-l. So ~/i,(a) [ a~-l. Combining our results we get THEOREM 3.3. (Zsigmondy, 1892). Let a>>.2 and n >>.1 be integers. (a) ~(a) is an integer and ~,(a) ] a n- 1. (b) Let r be a prime divisor of ~n (a) and let n = r On1, where r A/nl. Then n 1 is the multiplicative order of a modulo r. lf Q>O, then r 2 X~n(a), unless r=n=2 and a-=3(mod4). Denote f= (n, ~(a)) and suppose thatf~ 1. Let r be a prime divisor off and let n=rQnl, where r )( nt. Then by Theorem 3.3 nl is the multiplicative order of a modulo r. In particular, n 1 ] r-1 so that r is the largest prime divisor of n and hence uniquely defined. Therefore f is a power of r. If r 2 [ ~(a), then n=2 by Theorem 3.3. Hence never r 2 [ f so that actually f= r. We have proved THEOREM 3.4. Let a>.2 and n>>.2 be integers. Let r be the largest prime divisor of n and n=r~nl such that r X nl. If (n, ~(a))v ~ I, then (n, ~(a))=r and nl is the multiplicative order of a modulo r. LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 431 For a>~2 and n>~ 1 we define 1 • * (a) = ~-~ ~. (a), where f= (n, ¢.(a)) andf ~ is the largest power off dividing ~.(a) if f# 1, and ~= 1 otherwise. Note that ~= 1 unlessf=n=2 and a=3(mod4). With this definition we have THEOREM 3.5. Let q>~2 be a prime power and n>>, 1 an integer. Then for any prime r the following conditions are equivulent: (i) r [#*(q). (ii) r Y q and the multiplicative order of q modulo r is n. (iii) r [ qn_ 1 but r X qk_ 1 for 0 THEOREM 3.6. Let a>~2 and n>>.2 be integers. Let n=p~l...p~" and b=a "l~w'''v') where pi is a prime and e,>>. 1 for 1 <~i<. = tta#~ (n). For s---1 (mod2) we have II (1 - b -~/"(zO) fi (1 - b-') u>~ • ~e(n) 1 _ b -s /> ~=2 l_b-S I~ (1 - b -~) nb i=S (1 - b-s) ~ ~ nb--~" Here we are using the inequality (1- b-')> O- b-~) ~ "__L_b i=1 nb- 1 which can be proved by induction. (Alternatively, one can apply the Euler- Legendre theorem on pentagonal numbers (see, e.g., Knopp, 1959, p. 85). One then actually obtains a slightly better bound.) Suppose now that s is even. Then in our representation of u the numerator contains the factor 1-b -a, while of course all remaining factors are of the form 1-b -~/"('), where 3~ee(n) and pi~X, i.e., ~ee(~/pj), for a suitable j. So certainly u>~ >1(1 -b -s) u s ... us, where 1-I (1 - b-':.(~)) uj = 1 - b -pJ = =(1-b;t) -1 17 (l-b; "d"(~))>l ae • e (nl) nl >>- 0 - bYs) -s I7 (1 - b;') = i=2 "~ njb~ = (1 - bj-S) -2 1-[ (1 - b-f') >i -- = s njbj - 1 LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 433 for 1 ~< j.< LEMMA 3.7. If n=p~' ... p~', where s>~4 and e~>~ 1 for 1 <<.i<~s, then 2*(n)> >n 3. Proof. Assume at first that s=4. Then clearly cp(n)>>.-~sn. Also n~>210 = 2" 3" 5" 7, which implies 2 ~'(n)/3 t> 2Sn/105 t> n. So the statement holds for s = 4. The general case can be proved by induction" Let p I> 7 be a prime not dividing n and e a positive integer and assume that 2 ~'¢~)/3 >n, where n~>210. Then 2 ~ (npO)/3 ~. 2 ~, (n) p~ (~ - i/p,)/3 > n6p*/7 :_. = nn(6po/7)- 1 >1 n" 210 sp°/7 >np e . Assume now that ¢i* (a) ~<(n + l) (2n + 1), where a I> 2 and n > 2. Let n =p~' ... p~" be the standard representation of n, such that p~ < Pz <"" < P,. If s >i 4, then n >1 2.3.5" 7 andf <<.p~ <~n[p~lp~2"p~ 3 <~ n/30. Thus by Theorem 3.6 2 ~(n) ~< a ~'(~) ~< 2~n(a ) = 2f~* (a) ~< 2f(n + 1) (2n + 1) ~< ~n 3. However, this is impossible by Lemma 3.7. So s~<3. Assume that s=3. Then n >>.30, f<~n/6 and by Theorem 3.6 a ,(~) < ff~n(a) = f~*(a) <~ f(n + 1) (2n + 1) < 3fn 2 <~ ½n 3 . Suppose that P3 1> 7. Then 22~/7 ~< a z~/7 < f(n + 1) (2n + 1) ~< ½n3 . This implies that n<63, p3=7, n=42 and a12<7"43"85. Hence a=2 and f= 1 by Theorem 3.4. But from that we obtain 2 ~2 <43" 85, a contradiction. So P3 = 5, f~< 5 and 24nlls <~ a 4nlls = a~,(n) < 15n z . Therefore n<60 and hence n=30 and a~<3. Here a=3 implies f= 1 and 3s<31-61, as 3 has order 2 modulo 5. Hence there only remains the case an=230. Consider now the case s=2. Suppose at first that n>30. Then f<~n/p~ 434 CHRISTOPH HERING and hence anlPlPa £3 2 ~2 3 2 ~3 2 ~, (a) <~ a ~ (a) < a s/pIp2 _ 1 J -fs"* ~ T-[t~ " This implies that a=2, as otherwise 3~/3<~3-~-snS, which is impossible for nt>30. Also p1=2 as otherwise ~p(n)>~-~n, f<<.n/3 and hence 2"/3~<4~-~n3. Suppose that p2=3. Then f~<3, d'/P~Pz>~25 and 2"/3~<(322/31.5)n 2, which implies that n<45 and therefore n=36. But p~'if-1 by Theorem3.4. So in this case f= 1, which leads to a contradiction. Hence P2/> 5, tp (n) >/~n and f# l, as otherwise 22"/s < 1-6-~-n2. Thus P2 [ 22"~-1 and e I >12. If e I =2, then P2 =f= 5 and 100 [ n. In any case f <~n/8 and 22"/s < ~-sha. Thus n < 40, which is impossible as n is divisible by 8 or 100. So n<30, and one easily checks that actually n <~24. Finally assume that s=l, and suppose that n>16. Then a"(t-~/P~)< <~fn 2 <~n a. This implies that Pt > 2 and a=2. So f= 1 and 22n/3 <~n 2, which is impossible. Hence we have proved: THEOREM 3.8. Let a>~2 and n>~l be integers and suppose that ~*(a)<~ ~(n+ 1) (2n+ 1). Then one of the following holds: (a) n = 30 and a = 2. (b) n ~<24 and n has 2 different prime divisors. (c) n~<16. THEOREM 3.9. Let q be a power of a prime p and n >i 2 an integer. (a) (Zsigmondy, 1892)/f ~*(q)= 1, then qn=p~=26 orp 2. (b) If~*(q)=n+l, then qn=p~ =24, 21°, 212, 2 is, 34, 36, 56 orp 2. (c) ff ~* (q) =2n+ l, then q=p or q=p2. In the first case q~ =p~=2 a, 2 s, 22o or p 2 while in the second case q~=p2"=42, 4 a, 46 or 92. (d) ff O*(q)=(n+ 1)2, then q~=pn=74 or qn=pan=8z. (e) /f ~b* (q) = (n + 1) (2n + 1), then q" =p" = 32 s, 176 or p2. Proof. In each of the cases (a)--(d) we have n ~<30 by Theorem 3.8. Clearly, for every particular value of n, which is different from 2, there are only finitely many possibilities for a. Actually n # 2 implies a "/2 ~l. Then ~m(P) l~(q)[3"5. Hence again we have only finitely many possibilities. Inspecting the different possibilities for n one obtains the above list. We present here an independent proof for the state- ment (a), which is of special interest: Suppose that O*(a)= 1 and that n>2. Again, let n=p~' ... p~'. be the standard representation ofn. As in the beginning of the proof of Theorem 3.8 we show that s<3. Thus 24~/lS<2n, hence n<30 and s~<2. If s=l, then LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 435 f[ a- 1 and (a n- 1)/(a n/p- 1)~< f[ a- 1, a contradiction. So s=2, f<<.n/2 and 2n/a LEMMA 3.10. d(a- 1) I a"- 1 and ~* (a) [ (a n- 1/d(a- 1)), whenever a>>.2, n>>.2 and d=(n, a- 1). Proof. By Theorem 3.3 ~*(a) ] a"- 1. Furthermore (~*(a), a- 1))= 1, as n > 1 and (~* (a), d) = 1, as every prime dividing ~* (a) is larger than n by Theorem 3.5. Also (a n- 1)/a- 1 = a n- 1 + a n- 2 +... + a + 1 = n - 0 (rood d). Thus (an- l)/d(a-1) is an integer divisible by ~*(a). §4. LINEAR GROUPS CONTAINING AN IRREDUCIBLE GROUP OF PRIME ORDER Let Kbe a finite field of order q and characteristicp, Va vector space of finite dimension n over K and G a subgroup of GL(V, K). Define ~*(q) as in the preceding paragraph and assume that (~* (q), [GI) ~ 1. By 3.5 this is equivalent to saying that G contains an irreducible subgroup of prime order. Let r be a prime dividing (~*(q), ]G[), R a Sylow r-subgroup of G and S the normal closure of R in (7. Also, in this paragraph, denote by L the centralizer of S in Horn(V, V) and by F the Fitting subgroup of G. Clearly, S is irreducible and therefore L is a field of order q" for a suitable m. Also V is a vector space over L of dimension n* = n/m. We have LEMMA 4.1. Every abelian normal subgroup of G is contained in CoS. Proof. Let A be an abelian normal subgroup of G. Then AR is irreducible by 3.5. Therefore A has only one Wedderburn component in V. Hence [9~oA:~aA ] ] n (see [TL nilfssatz 5]). But r - 1 (modn) by 3.5. So R<<.~oA and S <~~oA. THEOREM A. ff F is not eontained in ~S, then thefollowing statements hold: (a) (~* (q), ]GI)= ]R] =r=n+ 1 =23+ 1 for a suitable a>~ 1. (b) ILl =q and q= 1 (rood2). (c) G conthins a normalsubgroup N such that N = T 3N, where T is an extra- special group of order 22 o+ 1, 3N is a cyclic group of order 2 or 4 and T r~ 3N = =3 T. Also N<. S, 3N~ ff.oT=3G, F=N3G andG/FisfaithfulIyrepresented on N/3N. (d) ~o/r(RF/F)=RF/F and ~rR=3G. (e) F ( G/F) <<.RF/F. Proof. Assume that IF, S] ~ 1. Then there exists a prime s such that the 436 CHRISTOPH HERING Sylow s-subgroup F~ of F is not centralized by S. Let N be of smallest pos- sible order among the subgroups with the properties N<~Fs, N Clearly Y is nilpotent and therefore contained in F. Also F/Y<<.F (G] ]I)<~ <~RY/Y. But this implies F= Y, as r )f [F[. Now (c), (d), and (e) are obvious. LEMMA 4.2. Suppose that G= MR, where M is a proper normal subgroup of G. Then M is nilpotent. Proof. By [TL, Satz 2] G is solvable. To prove that Mactually is nilpotent, we distinguish the two cases I-F, S] = 1 and [F, S] ~ 1. Assume at first that IF, S] = 1. Then S<<,F, as G is solvable. In particular R<~F. This implies that R<~G. Let M,, be a Hall r'-subgroup of M. Then R=~RMr,x JR, Mr.]. But R is cyclic and hence either R=~RMr, or R= JR, Mr.]. Here R= =[R, Mr.] implies R<<.[R,M]<~M and M=G, which contradicts our assumptions. So R=~gMr. and M<~R. But now M is cyclic by [TL, Korollar 1]. Assume now that [F, R] ~ 1. Then by Theorem A and Lemma 4.1 the order of`3F divides q- 1, and q- 1 THEOREM B. (a) SF/F is simple. (b) G/S is isomorphic to a factor of the metacyclic group FL(1, qn). (c) ~S=`3F is a subgroup of the multiplieative group of the fieM L. (d) IflZl G. If Rt ~ S, then S is nilpotent by Lemma 4.2, hence R = S and ILl = qn. On the other hand, if R1 c~ F ~ 1, then G has a non-trivial normal rl-subgroup and therefore is isomorphic to a subgroup of FL(1, q~). This again implies R<~F and L=F. So we have (d). Finally, let X]F= ~/rSF/F, and suppose that X > F. Then X<__.G and X is not nilpotent. Thus S~< X by Lemma 4.2. Hence SF[F is abelian and SF[F= = RF[F. If F~ ~S, then f£~/rRF/F= RF/F by Theorem A. If F~< ~QS, then R<~F and ILl =q'. (f) follows immediately from Lemma 4.2. Assume now that G is not solva- ble. Then S is not solvable by (b). Hence S is perfect by (f). As G is semi- linear over L, we have ILl < qn and therefore the property (iii), which implies (g). If, in addition, F~< ~S, then certainly S c~ F~< S' ta3S. THEOREM 4.3. If SF/F_-~A~, where Ai is the alternating group of degree i and i >i 5, then (a) (~* (q), IGI)=n+ 1, 2n+ 1 or (n+ 1) (2n+ 1) and (b) if r 1 is a prime dividing (~*(q), IGI), then rl<<,i<2r I and IG/Fl l I (2rl- 1)[. Proof. Let r 1 be a prime dividing (~* (q), [G]). Then rl ] IAtl byTheorem B, and hence i ~> r 1. On the other hand A t must have cyclic Sylow rl-subgroups by [TL, Korollar 1]. So rl<~i<2r 1. In particular r~ X IAtl and r12 X(~,* (q), IGI). If R 1 is a Sylow rl-subgroup of At, then the normalizer of Rt contains a Frobenius group of order rl (r~ - 1)]2. Thus a group of automorphisms of order (r 1 - 1)/2 is induced in R1. By [TL, Korollar 1] this implies that (rx-1)/2]n. Hence n<~rl-li2n. So rt=n+l or rx=2n+l, and (q~* (q), IGI) I (n+l) (2n+ 1). As we have seen above, rl <<,i<2rl. By Theorem B the factor group G[Fis isomorphic to a subgroup of the automorphism group of At, i.e., to a subgroup of St or Aut(A6) if i=6. Thus IG/FII(2r,- 1)!. COROLLARY 4.4. Assume that ~*(q) I IGI and SF/F~-Ao where At is the alternating group of degree i >/5. Then q=p or q=p2 and in the first case q~=pn=24, 26, 2 l°, 212, 2 is, 34, 36, 56 or p2 while in the second case q~=p2n=42 or 92. If in addition i >~n + 3 and q = p, then q~ takes one of the values 24, 21 a or p 2. Proof. Clearly ~*(q)=n+ 1, 2n+ 1 or (n+ 1) (2n+ 1) because of Theorem 4.3. Thus we are restricted to the cases listed in Theorem 3.8. If r I is a prime dividing @*(q), then r I [ IAtl and hence i/> rt, which implies½r! I [GL(n, q)l. This excludes the cases q~=23, 2 a, 22°, 318 and 176. Suppose now that q=4 and n = 3. Then ~b* (q) = 7 [ [Atl, which implies that SF/F contains a subgroup isomorphic to Ss and therefore a Frobenius group X of order 5.4. Also LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 439 F<~C~S<~L by Theorem A, and obviously ILl =4. So IF[ 1 3, and by the Zassenhaus Theorem SF itself contains a subgroup isomorphic to the Frobenius group X. But this is impossible by [PR, Lemma 1.3]. Consider the case q=4 and n = 6. Here ~* (q)= 13 and therefore i/> 13. Thus SF/F contains a Frobenius group Y of order 11"10. Also ILl=4, as 11 ]lSl[ [ IGL(V, L)[. So again IFI [ 3, and we obtain a linear representation of Y, which leads to a contradiction. Assume now that i/> n + 3 and q =p. We consider at first the cases qn= 2 6 and q~ = 212. By Theorem A the Fitting subgroup F is contained in if,aS and therefore in the multiplicative group of the field L. In particular, Fn S has odd order. On the other hand Fc~ S is isomorphic to a subgroup of the Schur multiplier of FS]F by Theorem B. This implies Fn S= 1, as i >/9 (see Schur, 1911). Hence we obtain a linear representation of A~ over GF(2). By Lemma 2.3 this leads to a faithful representation of A~-4 of degree n-3 (the pos- sibility (b) in Lemma 2.3 does not arise here as our ground field has order 2). However, in both cases IA~-41 ,~ IGL(n-3, 2)1. If q"=21° or 3 4, then ]&+al X IGL(n, q)]. So there only remain the cases q"=36 or 56. Here SF/F contains a subgroup isomorphic to SL(2, 8). On the other hand n is no power of 2 so that F<<.~aS<<.L. This means that we obtain a projective representation of SF/F of degree n* over L. By [PR, Theorem 4.2] this is impossible. LEMMA 4.5. Let n>~2 be an integer, q a power of a prime p, d= (n, q-l) and t a prime dividing IFL(n, q)l. (a) lf t >~( l /d) (q*-l- 1)+1, then t= (q~- l )/d (q-1) unless n= 2 and either t=q=p or t+l =q=3. (b) In any case t <~ (2/d) (qn-1 _ 1) + 1. If t = (2/d) (qn-1 _ I) + 1, then n = 2 and t =p = q or q = 2 and t = 2 n- 1. Proof. (a) Suppose that t I n. Then n >I t >t (qn-1 _ 1)/d + 1 >t (q,-1 _ 1)/(q - 1) + 1 = = (qn-2 + qn-a +...+ q + 1) + 1 /> 2(n-- 1). Hence n=2 and 2~t>>,(q-1)/d+l, which implies q=t=2 if d=l and t=2, q=3 if d=2. Thus we may assume that t~n. Then in particular (t,d)=l. This implies that (t,q-1)=l, because t>~(qn-l-1)/d+l> >(q-1)/d. Suppose now that t I qk_ 1, where 1 <~k q~- 1 q~-i _ 1 q~-i _ 1 q~-1 _ 1 q-1 d d a contradiction. 440 CHRISTOPH HERING As IrL(n, q)l =nq ~(n-l~12 (qn_ 1) (q,-1 _ 1) < (q- 1), there remain only two cases, namely t] ~*(q) or t=p. Consider the case t I ~.*(q). Then t [ (q"- 1)[d (q- 1) by Lemma 3.10. Ift < (q*- 1)/d(q- 1), then (q*- ~ - 1)[d+ +1 <~t<<.(q"-l)/2d(q-1) and hence 0 >t qn- 2q ~-~ - 2q + 2d(q - 1) + 3 I> >i q.-l(q_ 2)- 2q + 2(q - 1) + 3 = q~-l(q _ 2) + 1 >I 1, a contradiction. Therefore actually t= (q~- 1)/d (q- 1). Finally, let t =p. Then p >~ (qn - 1 _ 1)ld+ 1 >>.(q"- ~ - 1)[(q - 1 ) + 1 > q~ - 2 >~p,-2. Thus n=2 and p>~ (q-1)/d+ 1 >/(q+ 1)/2, which implies that q=p. (b) Assume that t>~2(q~-l-1)/d+l. Then we have one of the cases listed in (a). If t=p=q and n=2, then d~<2 and 2(q "-t- 1)[d+ 1 =2(q-1)/ d+l>~q=t. Clearly the case t=n=2 and q=3 is impossible. Assume t=(q"-l)/d(q-1). Then (qn-1)/d(q-1)=t>~2(q"-l-1)/d+l, which implies 0 >i q~- 2q ~-~ - 2q + d(q - 1) + 3 i> >~q"-l(q--3)+q~-I--2q +(q- 1)+3. Thus q = 2 and t = (q'- 1)[d (q- 1) = 2 ~- 1 = 2 (q~- ~ - l )/d+ 1. THEOREM 4.6. Assume that SF/F~-PSL(h, ,7), where 4 is a power of a prime p and ~>~2. If ifv~p, then ( ~*(q), IGI)[ (n+ I) (2n+ 1). Proof. Assume thatif~p. As SF/Fis simple, we have ~>~3 or ~>4. So G is not solvable, which implies that n*>~2, r>~3 and (~* (q), IGI) IISF/FI. (1) If F<~GS, then n* >1 (l/d) (~-l_ I), unless we have one of the cases =2 and 4=4 or 9; ~=3 and 4=2 or 4. Proof. If F<~ CoS, then F is contained in the multiplicative group of the field L. In particular F is abelian so that we have a linear representation of SFand a projective representation of SF[Fover L. Hence the above statement follows immediately from [PR, Theorem 4.2]. (2) If r=if, then p>>.5, IRl=r, r=n+l or r=2n+ l, and SF[F'-PSL(2, If). Proof. Assume that r=if. Then PSL (~, 4) has cyclic Sylow if-subgroups (see TL, Korollar 1). But this is only possible, if ti = 2 and 4=if. So actually IRI =r=if, as IRI I [SF/FI by Theorem B. Clearly ,0>15, as G is not solvable. If F~S, then (~* (q), [Gl)=r=n+ 1 by Theorem A. If F~<~S, then (1) implies that n>1½(if- 1)=½(r- 1), i.e., r- 1 ~<2n. Hence in this case r=n+ 1 or r=2n+l, as r-l(modn). LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 441 (3) If r=(4"-l)/d(4-1 ), where d=Qi, 4-1 ), then [RJ=r=n+l or IRI =r=2n + 1. In the second case either 4=2 or 4~=22~=42. Proof. Assume that r= (4 ~ - 1)]d (4- I). Then r = #* (4) by Lemma 3.10. Hence r 2 ~ IGL(~, q)l, and IRI =r by Theorem B. Suppose that r #n+ 1. Then r1>2n+l as r=l(modn). If n/>(4 n- t-1)/d, then r/>2n+l~> >~2(4n-l-1)/d+l and therefore r=2n+l and 4=2 by Lemma4.5. If n < (4n- t_ 1)/d, then F :~ gaS or we have one of the exceptional cases of Theorem 4.2 in [PR]. The first case is impossible by Theorem A. If li =2 and 4=4, then r=5 and n<3, so that n=2 and r=2n+ 1. For the case ~i=2 and 4=9 we obtain r=5 and n<4. Also n I r-I so that n=2. But this implies that p = 3 =ff contradicting our assumptions. Suppose now that 1i= 3 and 4=2. Then r=7, n < 3 and therefore p---7. However, this is impossible if 7 J #* (q). Finally, we might have ~ = 3 and 4= 4, which implies r = 7 and n<5. On the other hand n Jr-1. Hence actually n~<3, which again is impossible by [PR, Theorem 4.3]. (4) If r is different from p and (4 ~- 1)]d(4-1), then IRI =r=n+ 1, and one of the following statements holds: (a) F ~ ¢~S. (b) 4= 4 and ~ = 2 or 3. Proof. By Lemma 4.5 our assumption implies that r ~< (4 ~- 1 _ 1)/d. Hence n <~ r-1 < (4~- ~ -1)]d. If F ~ gaS, then ]RI = r=n + 1 by Theorem A. So we can assume that F<<,C~S, which allows us to apply (1). If ti=2 and 4=4, then r l60, and r is different from 2 and 5. So ]Rl=r=3=n+l. For ti=2 and 4= 9 we again obtain r= 3 which leads to a contradiction, as the Sylow 3-subgroups of PSL (2, 9) are not cyclic. Suppose that ti = 3 and 4= 2. Then again r=3 and therefore n=2. But this implies thatp=7 and hence r I q- 1, a contradiction. Finally, assume that ti=3 and 4=4. Then JRI =r= 5 and clearly n = 4, as n J r- 1. We have seen that never r 2 J IGI. Also either r=n+ 1 or r=2n+ 1. As this holds for every prime dividing (~* (q), IGI), our theorem is proved. We also have a corresponding statement about symplectic groups, whose proof is considerably shorter: THEOREM 4.7. If SF/F ~-PSp (2m, 4), where 4 is a power of a prime p ~ p and m> 2, then (~* (q), IGI)=r=n+ 1 = (I/d) (4'~+ 1), where d= (2, q- 1). Proof. We know that r I ISF/ell W f-I (4 O. i=1 Here certainly r #p, as the Sylow r-subgroups of G are cyclic. So r ] ~ i_ 1 or r ] 4~+ 1 for some i<~m. Let r=an+ 1 and d=(2, q- 1), and assume at first 442 CHRISTOPH HERING that F6 ~S. Then by Lemma 2.4 a ~ ~<1 ,, 2(4~-l) Hence a<(#'+l)/(#'-l)~ THEOREM 4.8. If SF/F-~J1, the Janko group of order 175560, then n=n*, S ~ Jt, G = S x F and (0" (q), [GI)= n + 1, where n = 10 or 18. If furthermore • ~*(q) I IGI, then q--E, n=18 and G~-Jx . Proof. By Theorem B we have (~* (q), IGJ) ] ]FS/F[ = IJ~] = 23. 3 • 5-7.11" • 19. IfF:~S, then r is of the form r=2"+ 1 by Theorem A. Hence in this case we only have the possibilities r = 3 and r = 5. So a ~<2, and by Theorem A the factor group SF[F can be faithfully represented on a vector space of dimension 2a over GF(2). But this is impossible, as IJll ,~ IGL(2, 2)1 and [Jll ~" [GL(4, 2)[. So F~GS and F(S)=Fc~S<.3Fc~S' is isomorphic to a subgroup of the Schur multiplier of J1. Hence F (S) = 1 by Janko (1965, Lemma ll.1). Therefore S~-J1 and G/Fis isomorphic to a subgroup of the automorphism group of S. Thus, actually G = S x F as Janko (1965, Lemma 9.1) showed that the outer automorphism group of J~ is trivial. By Janko (1965, Lemma 13.1) the smallest possible degree of a faithful representation of -/1 over any field is 7. So n/> n*~>7. On the other hand n I r-l, where r=3, 5, 7, 11 or 19. This leaves only the possibilities r=ll and r=19. If (O*(q), IG[)=ll-19, then n I (11-1, 19-1)=2, which cer- tainly is impossible. So (¢~*(q), [GI)= 11 or 19. In the first case we clearly have n=n*= 10. Suppose that (~* (q), LG[)= 19. Then a priori it might be possible that n* = (r- 1)]2=9. However, n* =9 implies that the automizer of R is a cyclic group whose order divides 9 (see [TL, Korollar 1]), which is impossible by Janko (1965, Lemma 6.1). Hence in this case again we have only the possibility n =n* =r- 1 = 18. If ~*(q)[ IGI, then q=2 and n=10 or 18 by Theorem3.5. But here LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 443 certainly n# 10, as [Jl[ ~' [GL(10, 2)1. Also we have n*=n, i.e., JL] =2, and hence F= 1. So in this case G ~ J~. THEOREM 4.9. Assume that G is not solvable. If SF]F is an epimorphic image of a finite subgroup of GL (n, C), where C is thefieM of complex numbers, then (~*(q),lGl)=n+l, 2n+l or (n+l)(2n+l). In fact, the case (@* (q), [G[)= (2n+ 1)or(n+ 1) (2n+ 1) is only possible ifSF/F~-PSL(2, r), where r = 2n + 1. Proof. We assume that SF]F~-X]N, where X is a subgroup of GL(n, C) and N is a normal subgroup of X. Let rl be an arbitrary prime divisor of (4* (q), [G[). Then the Sylow rl-subgroups of X/O,, (X) have exponent r 1. Clearly Or,(X)<~N, as SF/F is non-abelian simple. Hence the Sylow rl- subgroups of SF/F are of exponent rl and therefore of order at most r 1 because of [TL, Korollar 1]. In view of Theorem B(d) we get that r 1 ] ISF/F[ and r• )( IGI. By Feit and Thompson (1961, Theorem 1) rl~<2n+l. So rl=n+l or 2n+ 1 and (4.* (q), IGI) [ (n+ 1) (2n+ 1). Ifrt =2n+ 1, then SF/F~-PSL(2, rl) by aresult of Winter (1964). (Note that both these statements can be deduced from earlier results of Brauer (1942, Theorem 3 and Theorem 4), ifrl X [N[.) §5. TRANSITIVE LINEAR GROUPS Letp be a prime, Ka field oforderp, V a vector space of finite dimension n > 0 over K, and G a subgroup of GL (V, K), which acts transitively on the set consisting of all non-zero vectors of V. As in the preceding paragraph we define F to be the Fitting subgroup of G. However, we denote S= G (°°), the last term of the commutator series of G. Furthermore, let L be a subset of Hom (V, V) maximal with respect to the conditions that L is normalized by G, L contains the identity and L is a field with respect to the addition and multiplication in Horn(V, V). Then there exist integers m and n* such that n=mn*, ILl =pro and n* is the dimension of the vector space (V, L). Also G<<.FL(V, L), and G is transitive on the points of the projective space PG(V, L). We define 4. (p) and 4" (p) as in § 3. The transitivity of G implies that (p"-1) ] IGI and hence 4" (p) [IG}. Let • be the subgroup of G generated by all Sylow r-subgroups of G which have the property r [ 4" (p). We have examples of the following types: I. SL(V, L)<~G<~rL(v, L). II. There exists a non-degenerate skew-symmetricscalar product on (V, L), and G contains as a normal subgroup the group consisting of all isometries of the corresponding symplectic space. 444 CHRISTOPH HERING III. n* = 6, p = 2 and G contains a normal subgroup isomorphic to G 2 (2m). IV. G contains a normal subgroup E isomorphic to an extraspecial group of order 2" + 1. Furthermore, ~E= 3G, and G]E'3G is faithfully represented on E/3E. If n=2, then n*=n=2 and ILl=3, 5, 7, 11 or 23. If n>2, then n*=n=4 and ILl=3. El. G (®) ~- SL(2, 5), n* = 2 and [L[= 9, 11, 19, 29 or 59. E2. G~A 6, n*=4 and ILl=2. E3. G --- A7, n* = 4 and JL] = 2. E4. G_~ SL(2, 13), n* = 6 and ILl = 3. E5. G ~ PSU (3, 3), n* = 6 and ILl = 2. The classical examples are groups of type I or II. All transitive linear groups of type IV have been described by Huppert (1957) and Andr6 (1955, § 6). Up to isomorphism there are only finitely many elements in this class. Note that the four examples of characteristic 3 are at the same time of type I or II. Examples for the exceptional type E1 occur as stabilizers on one point in sharply doubly transitive groups, as described by Zassenhaus (1936). The symplectic group Sp (4, 2) contains as a normal subgroup a group H ~ h 6. Also, it acts transitively on the fifteen non-zero vectors in the underlying vector space W. Thus the stabilizer on any non-zero vector contains a Sylow 2- subgroup and therefore contains properly the stabilizer of H on this vector. Hence H again must be transitive. The general linear group GL (W), which is isomorphic to As, contains a subgroup J isomorphic to h 7. As in A8 all subgroups isomorphic to A6 are conjugate, we can assume that H<<.J so that J is transitive too. The type E4 has been described by Hering (1970). If q= 2 m, then PSp (6, q) contains a subgroup isomorphic to G2 (q) (see Dickson, 1915). One easily checks that this group is transitive on the cor- responding projective space. The group G2 (2) is not simple but contains a subgroup of index 2 isomorphic to PSU (3, 3). This subgroup must be transitive by the argument which we used above for A6. LEMMA 5.1. Each reducible normal subgroup of G is cyclic of order dividing pa_ 1 for a suitable integer a such that a [ n and a I u be the element of ~ containing u, and let w be an arbitrary element in V-1 u. Then there exist subspaces lw and Iu+w in ~ such that w~Iw and u + weIu÷ w. Here Iw #1~+,~, as otherwise ueI~ c~ I~ and hence w~I~. Therefore w'=(u+w)~I,~nlu+w=O and w~=0. Now by the same argument tr must be trivial on V-I~ and in particular on lB. So tr=0. (For the last argument compare Artin, 1940. Lemma 5.1 also has been proved by H. Lfineburg.) LEMMA 5.2. L is unique unless n--2, n*=l, p=3 and G is isomorphic to a quaternion group of order 8. Proof. Let L1 and L2 be two different subfields of Horn(V, V) which are normalized by G, contain the identity, and are maximal with respect to this property. IflLil=p" for i=I, 2, then clearly miln, IG:~GLiIIm~ and ]G: ~GL1 c~ ~aL2] ] mtm2. If E6LI r~ ~GL2 is irreducible, then the centralizer in Horn (V, V) of this group is a field L3 which contains L1 and L2. This field is normalized by G, as ~oL1 c~L2~_~_G. Also, it obviously contains the identity. But now LI = L2 = L3 by maximality of LI and L 2. As this contradicts our assumptions, ~t;Lt n ~GL2 must be reducible. Clearly this group is normal in G. Thus by Lemma 5.1 it is cyclic and its order divides p*-1, where a] n and a< n. Therefore p" - 1 ]IGI ~< (p~/2 _ 1) mlm 2 <~ (p~12 _ 1) n 2 and hence p~/2+ 1 <<.nz. Clearly this leaves only finitely many possibilities. However, we don't need that here, because a much stronger statement follows immediately from Theorem 3.9: As [G: ~6L 1 c~ ~6L21 [ n2, 4" (p) [ [ 1~6L1 c~ ~GL2[ and therefore 4" (p)= 1, because ~aL1 c~ ~aL2 is reducible. Thus either n~<2, or n=6 and p=2. Assume that n=6 and p=2. Then 71 I~GL1c~E6L2I, as ml~<6 and hence 1~6LIc~6L2[=7 by Lemma 5.1. Hence a Sylow 3-subgroup X of G has order 9 and is regular on V-{0}. Thus X is cyclic by Burnside. But this implies that C6L~ and ~6L2 both contain the subgroup of order 3 of X. This is impossible as [EaL~ c~ IEoL21 = 7. The case n = 1 is trivial. Assume now that n=2. Then by the above inequality p ~<3. Ifp = 2, then Horn (V, V) contains only one field containing the identi- ty of order 2 and one of order 4, and certainly L is onique. So p= 3 and IL~I =9. In this case, the normalizer of Li in GL(V) is one of the Sylow 2- subgroups of GL (V). Any two different of these Sylow 2-subgroups intersect in the normal quaternion gioup D of GL(V). As G is transitive on V-{0}, we have 8 ][G[ and hence G=D. LEMMA 5.3. lf we exclude the case n=2, n*=l, p=3 and G~Qs, then every abelian subgroup of GL(V) which is normalized by G is contained in L. Proof. Let X be an abelian subgroup of GL (V) which is normalized by G. 446 CHRISTOPH HERING As G is transitive, X has only one homogeneous component. By [TL, Hilfssatz 5] the GF(p)-algebra generated by X is a field L1. Clearly L 1 con- tains X and therefore the identity. Also Lt is normalized by G. Thus Lt ~ LEMMA 5.4. ff ~ ¢ I, then L is the centralizer of S in Horn(V, V). Proof. Assume that S¢ 1. Then Sis irreducible, and therefore the central- izer X of S in Horn(V, V) is a field. Also [G:~GL][m I n so that (~*(p), [G:~L])= 1. Hence S centralizes L, i.e., L<<,X. But this implies L = X because of the maximality of L. THEOREM 5.5. lf F ~_L, then G is of type IV or G is of type I, n*= 1, n=2 and p + 1 = 2* for a suitable a. Proof. Assume that F~L and consider at first the case n=n* =2. Then L is the prime field, and G centralizes L. As F z~ L, there exists a prime s such that the Sylow s-subgroup Fs of F is not contained in L. Let E be a subgroup of G which is minimal with respect to the properties E~G, E<~F, and E ~L. Then every proper characteristic subgroup of E is contained in E c~ L and hence is cyclic and central. Thus by Lemma 5.3 we can assume that E is not abelian. Hence E is of type (c) or (d) in Lemma 2.1, and therefore contains an extraspecial group of index ~<2 and of order s 2°+x for a suitable a. Here s I I3F~I ] IL[-1. Hence s¢p and n=2>>.s° by Lemma 2.2. Therefore s = 2, a = 1 and p ¢ 2. Thus E cannot be of type (d); actually it is a quaternion group of order 8. Now ~E=3G by Lemma 2.2. Also G/E.3G is faithfully represented on E]¢E. Clearly 36~ COROLLARY 5.6. (Hupper0. If G is solvable, then either n* = 1 and G is of type I, or G is of type IV. Proof. If G is solvable, then C6F,,< F. In the case F~< L this implies that G n L >>.F >I CoF >>.CaL and [GIGn LI <~ IG/CoLI <~[Aut (L)[, where Aut (L) is the group of automorphisms of the field L. Thus IGI <<.m(pm_ I) and (p"- 1)/(p m- 1)~ LEMMA 5.7. If n* = 3, then G is of type I. Proof. We have S#I, because in the solvable case n*= 1, 2 or 4 by Corollary 5.6. So S/F (S) has even order by the theorem of Feit and Thomp- son (1963). (Of course this fact can easily be established without the help of the Feit-Thompson theorem.) Also F(S)<~F<~L by Theorem 5.5. Hence F (S)=S nL, and S/F (S) is isomorphic to the subgroup S* of PFL(V, L) induced by S. Clearly S * <~PSL (V, L), as S * is perfect. So every involution in S is a central involution, and by Wagner (1959) S* contains PSL(V, L). This implies that S=SL(V, L). THEOREM 5.8. If S~ 1, then the following statements hold: (a) SF/F is simple. (b) G/S is isomorphic to a factor of the metacyclic group FL (1, p"). (c) C~S=3F~L. (d) ~* (p) I [SF/F[. (e) C~/rSF/F= 1. (f) Assume that ff~*(p)~ 1. Let r be a prime dividing ~*~ (p), R a Sylow r-subgroup of G and ~ the normal closure of R in G. Then S= •= ~. (g) L is the centralizer of S in Horn (V, V). Proof. (g) By Lemma 5.1 S is irreducible. Thus by Schur's lemma the centralizer £ of S in Horn (V, V) is a field, and clearly this field contains the identity and is left invariant by G, as S~G. On the other hand S <<.FL (V, L) implies S<<.GL(V, L), as S is perfect. Hence L~<£ and therefore L=L because of the maximality of L. 448 CHRISTOPH HERING (f) Assume that ~* (p) # 1, and let r be a prime divisor of~* (p). Obviously r [ ~,* (p) [ pn_ 1 I IGI, and we can apply the results of§ 4. Let R be a Sylow r-subgroup of G and ~ the normal closure of R in G. Then ~=G(~)=S by Theorem B (b) and 7= S by Theorem B (d). (d) Clearly ILl LEMMA 5.9. lf n=6 and p=2, then G is of type I, II, III or E5. Proof. It is easy to verify this by inspection. We should point out some implications of Theorem 5.8 about transitive linear groups G with the properties F<<.L and S~ 1. Note that in view of Theorem 5.5 and Corollary 5.6 such groups can be considered to be the general type of transitive linear groups. Clearly Ln G<<.F,so actually F=GnL. The group GL*/L* ~ ~/G n L* = G/F, where L* is the multiplicative group of L, is a subgroup of PFL(V, L) which acts transitively on the points of the projective space PG(V, L). This group contains the simple normal subgroup SL*/L* ~- S/S n L* = S/S n F ~ SF/F LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 449 which, being perfect, must lie in PSL (V, L). Also because of Theorem 5.8 (e) G/F is isomorphic to a subgroup of Aut (SF/F). So we obtain (*) p--if--i- [ IG/FI I IAut(SF/F)[. Also S has a faithful linear representation of degree n* over L and SF/F ~- ~S[Fc~S=S/F(S) has a faithful projective representation of degree n* over L. LEMMA 5.10. lf n* = 1, then G is of type I. lf n*=2, then G has one of the types I, IV or E 1. Proof. If n*= 1, then SL(V, L)= 1, and trivially G is of type I. Consider the case n* =2. Because of Theorem 5.5 and Corollary 5.6 we can assume that F<~L and that G is not solvable. Hence SL*/L* is a simple normal subgroup of GL*/L* contained in the two-dimensional group PSL (V, L). By the result of Dickson (1901) SL*/L* is isomorphic to PSL(2, pS), where s [ m, or SL*/L* ~-A5 and p'= _ 1 (mod 10). If SL*/L* ~- PSL(2, pS), then the number of Sylow p-subgroups of SL*[L* is pS+ 1. Each of these Sylow p-subgroups leaves invariant exactly one point of the projective line PG (V, L). On the other hand, GL*/L* is transitive on PG(V,L). This implies pS=pm=lL[ and S/F(S)~-SL*/L*~-PSL(V,L). Obviously S<~SL(V,L) and hence S=SL(V, L), which implies that G is of type I. Assume now that SL*/L* ~-As. Then by (*) pro+ 1 [ [Aut (As)l = ISsl = 120, and furthermore pro= -t- 1 (mod 10). So G is of the exceptional type El. LEMMA 5.11. lf n*> 2 and n-O(mod2), then ~(p) I IG/FI. Proof. Clearly 4,*/2 (p) lp "/2-1 I p"-1 I Ial. Suppose that there exists a prime r dividing (~*/2 (P), IFI). Then r [ 13FI. But by Lemma 5.3 always 3F<<.L. Thus r lL and hence m>~n/2 so that n* ~<2, which contradicts our hypothesis. THEOREM 5.12. If SF/F~-Ai, where A i is the alternating group of degree i>>.5, then n*=2, p"=24, or p"= 34. Proof. Certainly p"-l[ 161 because of the transitivity of G. If • *(p)> 1, then by Theorem 5.8 (f) and Corollary 4.4 we are immediately limited to finitely many cases. Clearly we can assume that n*~>3, and therefore ~,~2 (P) [ [G/F[ whenever n=0(mod2) (see Lemma 5.11). As G/Fis isomorphic to a subgroup of the automorphism group of Al, which contains the group of inner automorphisms as a subgroup of index 2 or 4, we actually have ~*/2 (P) [ IA~I. Thus no prime 450 CHRISTOPH HERING divisor of O*t2(P) is larger than i. On the other hand i<2r for any prime r dividing ~* (p), as we have seen in the proof of Theorem 4.3. This excludes the orders 2 ~°, 2 is and 56. Suppose now that n=6 and p=3. Then 13=~(3) I IA,I and hence ½13! I IGL(6, 3)1, which again is impossible. So we are left with the cases 24, 24, 212 and 34. If p"=26, then i=7 or t'=8 by Corollary 4.4. Also, by Lemma 5.9 we have five possibilities for SF/F. But none of these is isomorphic to either A 7 or As. Consider the casep" = 2 t 2. Here i = 13 or i= 14, again because of Corollary 4.4. By Theorem 5.5 the Fitting group Fis contained in L. Hence by Theorem 5.8 F=3F=~GS. So FnS is isomorphic to a subgroup of the Schur multiplier of S]FnS_~A~. On the other hand Fc~S has odd order, as FnS<~L. Therefore Fc~S=I by Schur (1911). So S~-At and G~_Si or G~_A~ by Theorem 5.8 (e). However, in the case considered here p"- 1 is odd. So the fact that G is transitive on V- {0} implies that S itself is transitive on V- {0}. But this is impossible, as neither A13 nor A14 has a subgroup of index 2 ~2-1 (see Lemma 2.5). THEOREM 5.13. If SF/F~PSL(~, 4), where 4 is a power of a prime p, 4=p m andS>t2, then G is of type I, unless we have done one of the following cases: (a) n*=2, G is of type El. (b) n*=n=4, p=2, S~-PSL(2, 9)~A6, G is of type II or E2. (c) n* = n = 4, p = 3, SF/F ~- PSL (2, 5), G is of type IV. (d) G is of type E4. (e) 4n=p" but ILl#4, t~>2. Proof. If n*~< 3, then G must be of type I or E 1 by Lemma 5.7 and Lemma 5.10 (note that S# 1, i.e., G is not solvable by assumption, and therefore not of type IV if n* = 2). So we can assume that n* > 3. In the following we have to distinguish the two cases p =p and p #p, as they are totally different. Assume at first that p #p. If ~* (p)= 1, then n = 6 and p=2, as certainly n>~n*>2. Using Lemma5.11 we find that 7= =~(2) IIG/FI. As SF/F is isomorphic to a factor of GL(6, 2), we have r~a(ti- 1)/2~<4 and hence 7 A' IG/F:SF/F[ by Theorem 5.8 (e). So actually 7 I ISF/F[. This together with the condition ISF/F[ I IGL(6, 2)1 leaves only the possibility 4=7. Here n*=6 as n*>3, hence ILl=2 and F=I. So IGI <<.IAut(PSL(2, 7))1 =6.7.8. But this number is not divisible by 2 6- 1. Hence ~* (p) # 1, and we can apply the results of § 4. In particular, we have only finitely many possibilities by Theorems 5.8 (f), 4.6 and 3.9. Ifa =2 and 4=4, then ~* (p) I IPSL(2, 4)1 =60, and ~* (p)=5. Note that every prime divisor of O* (p) is greater than 3, because n* > 2. Thus n =4 and p =2 or 3. As we assume that p #p, there remains only the case n=4 and LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 451 p=3. Since n*>3, we have n*=4 and ILl=3. Suppose now that F~L. Then IFI[2 and IG/FlllAut(PSL(2, 4))1 = 120 so that IGI ] 2"3"5. There- fore the Sylow 2-subgroups are semi-regular on V-{0}. Hence they are quaternion groups, as otherwise SF/F would have cyclic Sylow 2-subgroups. Let X be a Sylow 5-subgroup of S. As the number of Sylow 5-subgroups of S is 6, ~RoXcontains a quaternion group Q of order 8. Also Q n ~oXis a cyclic group of order 4. However, the subalgebra of Horn (V, V) generated by X is a field of order 34, and ffoXis the multiplicative group M * of M. So Q c~ ~X is the unique cyclic subgroup of order 4 in M*. Also, all elements of Q induce automorphisms of order 1 or 2 of M and hence centralize a subfield of order 9 of M. But this implies that Q is abelian, a contradiction. So F~L, and G is of type IV by Theorem 5.4. If a = 2 and 4= 9, then SF/F ~-PSL (2, 9)-~ A6. Again ~* (p)= 5 and hence n = 4 and p = 2 or 3. Because of the assumption p #if, there only remains the case n=4 andp=2. Again, n*=4 so that fL[ =2, F= 1 and S~-PSL(2, 9)~ ~A6. Now GL(4, 2)-~A8, and all subgroups isomorphic to A6 of A8 are conjugate. Therefore S is the commutator subgroup of the stabilizer Sp (f) of a suitable symplectic form f. Let X be the normalizer of S in GL (4, 2). Then X]S is isomorphic to a factor of FL(1, 24) and at the same time isomorphic to a subgroup of the outer automorphism group of S (see Theorem 5.8 (b) and (e)). The first group has cyclic Sylow 2-subgroups, while the second is an elementary abelian group of order 4. So IX]SI = 2 and X=Sp(f), which implies that G is of type II or E2. If ~ = 3 and q = 2 or 4, then ~* (p)= 7. Hence n 16, and actually n = 6, as n~>n*>3. Also p=3 or 5 by Theorem3.9. On the other hand ~*(P) I PFL(~, 8) as n*>2 (see Lemma 5.11). This leads to a contradic- tion. In the following we assume that 4#4, 9 if ~=2 and 452, 4 if h=3. If F zgL, then SF]F is isomorphic to a factor of the automorphism group of an extraspecial group of order 25 by Theorem 5.5. This means that SF/F~ '~ PSL (2, 4) ~ PSL (2, 5), and we have (c). So we can assume in addition that F<~L. Then by [PR, Theorem 4.2] n>~n*>I (4~- ~- 1)/d, where d= (~, q- 1). Together with Lemma 4.5 this implies that no prime divisor of the order of IG[F[ is larger than 2n + 1. This excludes the cases p"= 176, 3 a 8, 56, 218, and 21°, as ~*t2(p)[IG[F[, whenever 2In. So qi* (p) is equal to a prime r, and by Lemma 4.5 we have r=p or r= (~-1)ld (8-1), as r>/n+ l >/(4n-1)/d + l. Furthermore, if r=2n+l, then r=p or 8=2 and r+ I =2 n. This excludes p"=28 and p"=22°. Furthermore, p"=23 is excluded because of our con- dition n* > 3. 452 CHRISTOPH HERING By Theorem 4.6 we have ~* (p) = (n + 1), (2n + 1) or (n + 1) (2n + 1). Hence because of Theorem 3.9 and the above we are only left with the cases pn=24, 212, 34 and 36. In particular r=5, 7 or 13. To finish our proof, we have to determine, in how many ways r can be represented as a generalized Mersenne prime, i.e., in the form r= (qn-1)/ d (4-1), where # is a power of a prime and d= (g, 4-1). We see that r ~< 13 implies that a <~ 3: Suppose that ~/> 4. Then 13 I> r t> (4 4_ 1)] d (4-1)> (44-1)](4-1)=42+ 1 so that 4<-,.3. However 4=2 implies d= 1 and 4= 3 implies d~<2. Hence we obtain a contradiction in both cases. Using this it is easy to determine all possibilities: r=5, ~=2, q=4 or 9; r=7, a=2, 4=13; r=7, ~=3, 4=2 or 4; r=13, h=2, 4=25; r=13, h=4=3. Most of these cases are not allowed by the above assumption. Consider the case r=7, ~=2, 4=13: Here n=6 and p=3. Also n*16 and hence n*=6, as we are assuming that n*>3. Suppose that S~PSL(2, 13). Then S con- tains a dihedral group of order 2.7. If • is an involution in this group, then I~v~l-- 33 by [TL, Hilfssatz 5]. Also, the number of involutions in S is 7" 13, and all involutions in S are conjugate to ~. Counting the pairs consisting of a non-zero vector and an involution in S fixing this vector we obtain the equality (36-1)bl =7.13.26, where bl is the number of involutions in Sv for ve V- (0}. Note that bl is independent of the choice of v, as G is transi- tive on V- {0}. Clearly the above equality is impossible, which proves that S~PSL(2, 13), i.e., S c~F~l. On the other hand F<<,Lby Theorem 5.5. So actually F=Sr~F=L-{0}. Now it follows from Theorem 5.8 (g) and the work of Schur (1907) that S "~SL(2, 13). Also, by Theorem 5.8 (e) the factor group G]F is isomorphic to a subgroup of the automorphism group of PSL(2, 13), which is PGL(2, 13). Suppose that G>S. Then G[F ~- ~-PGL (2, 13), and G contains an involution fl which fixes a non-zero vector. The unique involution in S lies in L and hence acts as the - 1 on V. There- fore the Sylow 2-subgroups of S, which are quaternion groups, are fixed- point-free on V- {0}. In particular fl~S and hence flF does not belong to the commutator subgroup of G/F. This implies that ~/r (flF) is a dihedral group of order 4.7 and therefore 7 [ Clearly this is impossible, so that G=S~SL(2, 13), i.e., G is of type E4. Suppose that r=13, a=2 and #=25. Then n>~n*~>12, hence n=12, ILl=2, IFI=I and 212-1 [IGI] IIPFL(2, 25)1, which is impossible. If r=13, ~=3 and #=3, then n*>~8 so that again n = 12, ILl =2 and IF[ = 1. But this is impossible as 5 X [PFL(3, 3)[. There still remains the case r=p. Here ~=2 and #=p=r. If r=5, then n=4, p=2 or 3 and SF/F-~PSL(2, 5)~-PSL(2, 4). For p=3 we obtain groups of type IV, as we have seen above, while p = 2 implies that G is of type I (see the proof for the case p =.5 below), r =/~ = 7 implies n = 6, p = 3 and 13 = ~* (3) I IG/FI ] lerL (2, 7)1, a contradiction, r =/~ = 13 implies n = 12, LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 453 p=2 and n*>~6 so that 5 X IFI and hence 5 I IG/FIIIPrL(2, 13)1. So again we do not obtain any transitive linear groups. Assume now that p=p. As a consequence of the results of § 3 we obtain (**) lf p =p, then r~ = n. This holds for any dimension n*, as we can see in the following way: By Theorem 5.8 (d) we have ~* (P) ll SF/F[I GL(~, p~). Hence every prime dividing ~* (p) must divide a number of the form p~k_ 1, where k~<~. Thus nqti~>n, whenever ~*(p)~l. On the other hand ~gn(p) llPSL(a,p~)l[ IGL (n, P)I, which implies that every prime divisor of~*s (p) divides a number of the formp k- 1, where k.< U St= v. Furthermore, any two Sylowp-subgroups of S generate S. Hence ~ n ~j = 0 for i~j. As IVl=pn=4~=42, the set {~1, ~2,..., ~+1} is a congruence. This congruence determines a translation plane 9.I, and all p-elements of S are shears of 9~. It follows that S _~ SL(2, q) and that Horn(V, V) contains a field of order 4 centralizing S (see Hering, 1972, Theorems 2 and 3 and Lemma 7). So L = 4. THEOREM 5.14. SF]F cannot be isomorphic to Janko's group of order 175 560. Proof. Suppose that SF/F_~J1, where J1 is the simple group of order 175560 described by Janko (1966). Thenp=2, n= 18 and G-~Jt by Theorem 4.8. But this is impossible, as it implies that 218-11 IJ, I. THEOREM 5.15. If the Sylow 2-subgroups of G are cyclic, then n* = 1, and G is of type I. If the Sylow 2-subgroups of S are ordinary or generalized quaternion groups, then n*~<2 and G is of type I, IV, E1 or E4. Proof. Assume that the Sylow 2-subgroups of G are cyclic or the Sylow 2- subgroups of S are (generalized) quaternion groups. If G is solvable, then by Corollary 5.6 either n* = 1 and G is of type I, or G is of type IV. Suppose that G is of type IV. Then clearly the Sylow 2-subgroups of S are generalized quaternion groups. Also the case n*= 4, ILl = 3 is not possible, as in this case F contains more than one involution. Hence we can assume that G is not solvable. As a Sylow 2-subgroup Y of SF[F is isomorphic to a factor of a Sylow 2-subgroup of S, Y is cyclic, a (generalized) quaternion group or a dihedral group. By Theorem 5.8 the factor SF/F is a non-abelian simple group. If Y is cyclic, then Y= 1 and SF/F is solvable by the theorem of Feit and Thompson, a contradiction. Also Ycannot be a quaternion group by the theorem of Brauer and Suzuki (see Brauer, 1959). Thus Yis a dihedral group. By a result of Gorenstein and Walter (1962) this implies that SF[F is iso- morphic to some PSL(2, q) or to A 7. The fact that Yis a dihedral group implies that2 1F, hence 2 113FI I ILl- 1, and therefore p~2. The theorem now follows from Theorems 5.12 and 5.13. LINEAR GROUPS WHICH CONTAIN IRRDEUCIBLE SUBGROUPS 455 §6. ON COMPOSITION FACTORS OF FINITE DOUBLY TRANSITIVE GROUPS Let G be a finite doubly transitive permutation group. Denote by .4 the maximal solvable normal subgroup of G and by E (G) the group generated by all minimal normal subgroups of G/A. Then E (G) is a simple group. This follows from a result of Burnside (1911), if G does not contain a sharply transitive normal subgroup. If G does contain a sharply transitive normal subgroup, then actually G has at most one non-solvable composition factor (see [TL, pp. 159-161]). The structure of solvable doubly transitive groups is known completely by Huppert (1957). Therefore, we can assume that E (G) # 1. It is easy to see that in this case the degree of G is not larger than the square of the order of the automorphism group of E (G). So for each non-abelian finite simple group E there exist only finitely many doubly transitive permutation groups G with the property E (G)~ E. In this paragraph we report on the problem to determine all finite doubly transitive groups belonging to a given simple group E. We find that the cases E (G)~-PSL (2, q) and E (G)~`4~, the alternating group of degree i, can be handled completely. Also we prove, that E (G) is never isomorphic to ./1, Janko's group of order 175 560. Hence the map G~-,E (G), which maps each finite doubly transitive permutation group into the collection of isomor- phism types of finite simple groups, is not onto. This strengthens a result of Parker (1954) to the effect that not every non-abelian simple group has a doubly transitive permutation representation. Finally, as a consequence of our results, we obtain a classification of all finite doubly transitive groups which have a sharply transitive normal subgroup and the property that the stabilizer on two points has odd order. We present here only those cases, in which a complete answer is known. Therefore, it should be mentioned, that for many other cases, partial results exist (see, e.g., Bannai (to appear), Seitz (to appear) and § 5 of this paper). We shall use the following notation: DEFINITION. Let G be any group and A the maximal solvable normal subgroup of G. Then E (G) is the smallest subgroup of G/A containing all minimal normal subgroups of G].4. As mentioned above, we have THEOREM 6.1. If G is a finite doubly transitive permutation group, then E (G) is simple. Let G be a finite doubly transitive permutation group and let .4 be the maximal solvable normal subgroup of G. By Burnside (1911, §§ 151-154), 456 CHRISTOPH HERING G contains exactly one minimal normal subgroup H. Furthermore, A = 1 if and only if H is a non-abelian simple group, and A # 1 if and only if H is elementary abelian and sharply transitive. If A = 1, then ~aH= 1. Suppose that A# 1, and let V be the set of permuted symbols. Choose an arbitrary element O e V and define a binary operation on V by 0 ~' + 0 ~' = O~r for x, y~H. Then V together with this operation is a group isomorphic to the minimal normal subgroup H, hence an elementary abelian p-group. Also, the stabilizer Go of G on O is a group of automorphisms of V operating transitively on the set of elements different from the identity. Let L be a subring of Horn (V, V) maximal with respect to the property that it is a field containing the identity and that it is normalized by Go. Then actually Go is a group of semilinear transformations of the L-vector space (V, L) deter- mined by V and L in the natural way. Clearly G is the set of all maps of the form x i,-~ xe + p for xe V, where ve V and g belongs to the semilinear group Go. In our proofs we shall frequently use the notation introduced in this paragraph. THEOREM 6.2. Let G be a finite doubly transitive permutation group of degree n, and let Aut (E (G)) be the automorphism group of E (G). If E (G) # 1, then n< IAut (E (G))I 2. Proof. Assume that E(G)#I. If A=I, then E(G)~_H and n IIE(G)I, as H is transitive. Suppose that A ~ 1. If the Fitting subgroup of Go is not contained in L, then by Theorem 5.5 n = 3 + < 1202 < IAut (E (G))I 2 . If the Fitting subgroup of Go is contained in L, then by Theorem 5.8, qn*- 1 q.*- ~ + qn*- 2 W "" @ q + 1 - -- <~ [G/FI <~ IAut(E (G))I, q-1 where q= ILl and n* is the dimension of the vector space (V, L). But this implies that n = q"*~< q2~,-~ < (q,*-1 +...+ q + i)2 ~< [Aut(E(G))[2. COROLLARY 6.3. If E is a finite simple group, then there exist only finitely many finite doubly transitive groups G with the property E (G) ~- E. LINEAR GROUPS WHICH CONTAIN IRREDUCIBLE SUBGROUPS 457 THEOREM 6.4. Let G be a finite doubly transitive permutation group of degree n and assume that E (G)~PSL(2, q), where q>>,4. If q=4 or 5, then n=5, 6, 52, 11 z, 19 z, 292, 59 z, 24 or 3 ~. If q=7, then n= 7, 8 or 49. If q=8, then n=9, 28 or 82. If q=9, then n=6, 10, 16 or 9 z. Ifq=ll, then n=ll, 12 or 112 . If q=13, then n=14, 132 or 36. If q>13, then n=q+l or q2. Proof. Assume at first that A = 1. Then G is isomorphic to a subgroup n of PFL (2, q), as PFL (2, q) is isomorphic to the automorphism group of PSL (2, q). The group B contains a normal subgroup/~ which is isomorphic to PSL (2, q). In particular/" is simple and hence contained in the largest perfect normal subgroup of PFL(2, q), which is PSL(2, q). So actually PSL(2, q)=/~ THEOREM 6.5. Let G be a finite doubly transitive permutation group of degree n and assume that E (G)~-Ai, where i>>,5. If i=5, then n=5, 6, 52, 112, 192, 292, 592, 24 or 3 ~. lf i=6, then n=6, 10, 16 or 81. 1f i= 7, then n= 7, 15 or 16. If i=8, then n=8, 15 or 16. 1f i > 8, then n= L Proof. As As~-PSL(2, 4) and A6"~PSL(2,9), we have settled the cases i= 5 and i= 6 already in Theorem 6.4. So we can assume in the following that i~> 7. Let us consider at first the case A # 1. By Theorem 5.12, we have either n*= 2, where n* is the dimension of V over L, or n ~ {24, 34}. If n4= 2, then E is involved in PSL(2, q), where again q=lL]. Hence by Dickson (1901, Chapter XII) E (G) is isomorphic to a two-dimensional linear frac- tional group PSL(2, q) for a suitable prime power 4. But by Artin (1955) this implies that i= 5 or 6, which contradicts our assumptions. So n =24 or 34. Ifn = 34, then Visa vector space of dimension 4 over GF (3) and hence E (G) a factor of GL(4, 3). This again implies i~<6, as 7 ,~ [GL(4, 3)1. So there 458 CHRISTOPH HERING only remains the possibility n = 24. In this case E (G) is a factor of GL (4, 2) and hence i= 7 or 8. Assume now that A=I. As i>~7, Aut(E)~S~, (see Passman, 1968, Theorem 5.7). So either G~ At or G"-St. The doubly transitive permutation representations of At and St have been determined by Maillet (1895). THEOREM 6.6. Let G be a finite doubly transitive permutation group. Then E (G) cannot be isomorphic to Janko's group of order 175 560. Proof. In the case A~ 1, this follows from Theorem 5.14. If A= 1, then G~-J1, as the outer automorphism group of J1 is trivial by Janko (1966, Lemma 9.1). But, again by Janko (1966), ,/1 does not have any doubly transi- tive permutation representation. THEOREM 6.7. Let G be a doubly transitive group of permutations of a finite set V. Suppose that G contains a sharply transitive normal subgroup and that no involution in G fixes more than one point. Then there exists afield L such that (V, L) together with suitable binary operations is a vector space over L, and G is a subgroup of the group of permutations of V of the form Z F--~ Xh -]- O , where v~ V and h is a semilinear transformation of (V, L). Furthermore, we can choose L in such a way that one of the following statements holds: (a) (V, L) has dimension 1. (b) (V, L) has dimension 2 and G contains all permutations of V of the form X l--~ Xh q- V, where v e V and h is a linear transformation of (V, L) with determinant 1. (c) (V, L) has dimension 2, ILl =9, 11, 19, 29 or 59, and G~o®>=SL(Z,5), where GO is the stabilizer of G on some element of V. (d) (V, L) has dimension 2, ILl=5, 7, 11 or 23, and Go contains a normal quaternion group of order 8. (e) (V, L) has dimension 6, ILl =3, and Go~SL(2, 13). Proof. We define O, Go and L as before. Suppose there exists an involu- tion i in Go. Then I VI =0(mod2), and i is an automorphism of the abelian group V which leaves invariant no element different from O. Hence x t= -x for all elements x in V (see Zassenhaus, 1936, p. 189). In particular, i is uniquely determined. This implies that the Sylow 2-subgroups of Go always are cyclic or generalized quaternion groups. 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