on primitive monodromy groups of genus zero and one 1

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On primitive monodromy groups of genus zero andone 1Michael G Neubauer aa Department of Mathematics , The University of Teras at Austin , Austin, Texas,78712Published online: 27 Jun 2007.

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COMMUNICATIONS IN ALGEBRA, 2 1 (3), 7 1 1-746 (1993)

On primitive monodromy groups of genus zero and one I

Department of Mathematics, The University of TE.T(I.; n l .-lustin, Austin, Tezas, '18712

Abstract

Let G be a primitive permutation group on R wit11 F' (G) 2 C i and p an odd prime. If G is generated by the T elements of the tuple X = (xl, ..., x,) whose product is the identity, then C:=l ind(xi) = 2(\ R ) tg - 1) where g >_ 0 is the genus of the triple (G, R , X). We give a classification of all such groups when g = 0 and g = 1.

1 Introduction

The groups we are interested in arise from the fdh ing <\\~at'\~n, l e t X be a compact connected Rieman surface of nenm 4 2nd L1

be a nonconstant meromorphic function of degeree n with a set of branch points S = {sl, ..., s,). The fundamental group ro(po, P' - S) is generated by r elements 71, ...,-y, whose product is the identity, i.e. n;'=, yi = 1. For each 1 5 i 5 r the element 7; induces a permutation on the n preimages of PO. We denote this induced permutation by xi. We refer the reader to the excellent introduction of (91 for a detailed acount.

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The monodromy group Mon(X, 4) of the cover I is the subgroup of S, generated by the 2:s. Since X is connected Mon(X, 4) is a transitive subgroup of s,. Recall that if the group G acts on the finite set R we define, for

s E G, the index of x by

where orb(x) is the number of orbits of the group generated by a on 0. The generators xi, the genus g of X and the number of branch points r are related as follows

= 1, and i=l

order(xi) = di > 1.

We say G is a genus g group of type ( d l , ..., d,) if equations 2 through 5 hold. Equation 2 is the well known Riemann-Hurwitz formula. Riemann's existence theorem (see [6]) states that conversly for each group satifying the above equations there exists a cover whose monodromy group is the given

group. The primitive groups of genus zero and one in this setup play a fundamental role as was shown in [9] and (101. The composition factors of monodromy groups are contained in the composition factors of primitive monodromy groups. Aschbacher and Scott ([3] ) gave a classifaction of finite primitive groups into five classes one of them being the class for which F*(G) is abelian which is the case of interest here and which includes the class of primitive solvable groups.

In (91 it was shown that there are only finitely many primitive groups of genus zero with F*(G) abelian besides the well-known examples given by Zariski-Ritt (see theorem 1.3). In (151 the author showed that for any fixed genus g > 0 there are only finitely many groups of that genus. The precise statements are given now.

.dem.-tsry

1 . Cip- P-w-oup. then one of the jo&=..,,,ing holdS:

= 1 and e = 1 or 2,

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PRIMITIVE MONODROMY GROUPS 713

4, p = 5 and e = 2 or 3.

5 . p = 7 or 11 and e = 2.

The previous theorem is essentially Theorem A in [9], except for the improved bounds for e for the primes 2 and 3. There it was shown that for p = 2, e < 16 and for p = 3, e 5 6.

Theorem 1.2 if G is a primitive genus one group with F'(G) 2 C; an elementary abelian p-group, then one of the following holds:

4. p = 5 or 7 and e = 2 or 3.

The object here is to find all primitive groups G of genus zero and one with F8(G) r C," elementary abelian assuming p > 2 while the case p = 2 is the content of [16]. In [9] all genus zero groups for p 2 5 were already determined. ~his&ult is-included in theo~em Lb, i&(\sk'\ 111, pp , &I\\ found all groups of genus zero assumin9 6" = 1.

I

Theorem 1.3 (Zariski) Assume that G =< X I , ..., x, >, nLl xi = 1 and order(x;) = di > 1 for i E (1, ..., r ) , r 2 2. If G" = 1 and Ci=, ind(x;) = 2n - 2, then H 2 G / N has order 1,2,3,4 or 6 (with p [JHJ) and n = p if p 5 1( mod I H I). If p f 1( mod I H I), then n = p2 . Moreover either r = 4 and dl = ... = d4 = 2 o r r _< 3.

A proof of the corresponding result for groups of genus one can be found in (151, proposition 4.3.

Proposition 1.4 Assume G is as in Zariski's theorem 1.3. If GI1 = 1 and &2 ind(x;) = Bn, then G is one of the following:

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Remark: If K and H are groups K x H denotes the direct product, K . H the semidirect product of I{ with H , I{ r H the central product and H 1 I< the wreath product. Furthermore Cn denotes the cyclic group group or order n, Dn the dihedral group of order 2n, Sn ( A , ) the symmetric (alternating) group of degree n and GL,(p), SLn(p) and PSL,(p) the general, special and projectve special linear group of dimension n over the field with p elements.

Theorem 1.5 Let G be a primitive group with G'' # 1 and F8(G) Ci, p > 2. If ind(q) = 2pe - 2, then one of the following is true:

I . p = 11, e = 2 and G is oftype (2,3,8) with GIN 2 GL2(3).

2. p = 7, e = 2 and G is of type (2,4,6) with GIN 2 C3 . D4.

3. p = 5, e = 3 and G 2s of type (2,3,8) with G/i\'Z Ci S3,

4. p = 5 , e = 2 and G is o f t y p e

(a) (2,3,10) with GIN r S3.

(6) (2,3,12) with GIN E SL2(3) * C4.

(c) (2,3,20) with [(GIN) : SL1(5)/ = 2.

(d) (2,4,8) with GIN 2 2-Sylow subgroup of GL2(5).

l,) (3,4,4) with I GIN I = 96 and ((GIN) f~ SL2(5)) 2 SL2 (3).

5. p = 3, e = 2, r 5 6 and GIN is isomorphic to one of the groups in the set {GLz(3), SL2(3), Sylow 2-subgroup of GL2(3), D4}.

6. p = 3, e = 3 and G is of type

{a) (2,2,2,3) with GIN 2 SL3(3) or GIN 2 S4.

(b) (2,3,13) with G/N E SL3(3).


The Zero stems that give rise to the groups in dimension e = 2

for = 3 are too numerous to be listed The author has a list which the

=der esily reproduce with the results in section 5.

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Theorem 1.6 Let G be a primitive group with G # 1 and F*(G) " C;, p > 2. If xi=, ind(x,) = 2pe, then one of the following is true:

1. p = 11, e = 2 and G is of type (2,3,10) with GIN g Ar x C2.

2. p = 7 , e = 2 and G is of type

(a) (2 , 2 ,2 ,3) with GIN I D6. (b) (2,6,6) with G / N 2 D6 x C3.

(c) (2,3,16) with [(GIN) : SL2(7)] = 2.

3. p = 7, e = 3 and G is of type (2,3,'i) with GIN E PSL2(7 ) .

4, p = 5 , e = 2 and G is of type

(a) (2,4,10) with GIN 2 04.

(b) (2,4,12) with G/A' I Ds * C4

(c) (3,4,5) with G/N SL2(5) .

(4 (3 ,4 ,4) with G/N " GL2(5).

(e) (3 ,3 ,4 ) with GIN Z SL2(3 ) .

( f ) (2,2,2,4) with GIN g D4,

5, p = 5, e = 3 and G is of type

(a) (2 ,4 ,5) with GIN 2 S5 .

6. p = 3, e = 2, r < 6 and GIN is isomorphic to one of the groups in the set {GL2(3), SL2(3), Q,, D4, Sylow 2-subgroup of GL2(3)).

7, p = 3, e = 3, r 5 4 and G is isomorphic to one of the groups in the set {A& 5'4, S 4 x C2, SL3(3), GLs(3)).


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(e) [2 ,2 ,2 ,3) with GIN A5 x C2, GIN 2 A,, GIN 2 4 .A5.C,2 2s of order 480 or GIN is solvable of order 96, 192, or 768 and is contained in the normalizer of an extraspecial group of order 32.

The genus one systems that give rise to the groups in dimension e = 3 for p = 3 are listed in section 5. For the genus one systems in dimension e = 2 the same as was said for the genus zero systems applies.

The proof of theorems 1.5 and 1.6 consists of two parts which are quite different in flavor. One part of course is showing that the genus zero and

genus one systems given above have the properties stated there. In the second part it has to be shown that those are actually the only genus zero and genus one systems that give rise to primitive groups.

The first part of this program is very computational in nature. In most cases the structure of the group that arises from a given genus zero or genus one system can be pinned down to a large extent. However, especially for p = 3 and dimension e = 4 the computational aspects of the problems arising in determining the structure of the groups were beyond the capabilities of the author and the help of the group theory package Cay ley V3.7 was invoked to bridge those difficulties. Whenever this was done a remark was made to that extent.

The second part of the proof has a combhatorial flavor. We have to show that the above genus zero and genus one systems are the only ones occurin& Thj~ mou~t to sjjowjn~ that the only solutions of& ind(xi) = 2(n+g-1),

g = O,I are the ones given in theorems 1.5 and 1.6. The proof is split up into three parts. The case when p 2 7 is discussed

in section 3, the case p = 5 in section 4 and the case p = 3 in section 5. The case p = 2 as was mentioned earlier will be dealt with in [16]. In the next section we review the results from (91 and [15] necessary to carry out the proof.

I want to thank R. Guralnick for many helpful discussions about group cohomology.

2 Primitive permutation groups with F*(G) ab elian.

We assume that G acts faithfully and transitively on a finite set 0, with (0 1 = .. If x E G, set Fix(x) = {r E D I xw = w) and set f(x) = Fi*(x) 1. we denote by c ( x ) the *umber of orbits of < x > 0 $2 Set i d . ) = ~ - c ( x ) .

that the definition of index impkc that it is a C ~ S function on G.

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PRIMITIVE MONODROMY GROUPS 7 17

Following 121 say (G, 9, E) is a genus g system if G acts faithfully on a, E = (xi, ..., x,) E G', xi # 1, n:=, xi = 1, z;=, ind(x;) = 2(n + g - 1) and G =< xl, ..., x, >. Furthermore we say (G, Q, E ) is of type (dl, ..., d,) if order xi = d;.

If E' = (xl, x;2,x3, ,... xr), then (G, R, E') is also a genus g system. More generally let

and denote by Braid(G,r) the group generated by the maps Q;, 1 < i < r. Define two genus g systems (G, 0, E) and (H, I?, F) to be equivalent if there exists a permutation isomorphism n : (G, !I) --+ (X, I') and a b E Braid(G,r) such that C(E) = F . This actually defines an equivalence relation on the set of genus g systems (see (21 ). If 4 5 ... <_ d, we call (G,!I, E) an ordered genus g system. It is easy to that in the orbit of (G, !I, E ) under the action of Braid(G,r) there is an element (G, 51, E') which is an ordered genus g system. More generally given any property invariant under conjugation there is an element (G, 0, E') in the orbit of (G, R, E) under the action Braid(G,r) such that E' = (xi, ..., x:) is in any given order with respect to this property. Here we will always assume dl 5 ... 5 d,.

A proof of the next proposition can be found in [14],11.4.

Proposition 2.1 Suppose G =< XI, ..., x , >, nI=, xi = 1 and order($;) = di > 1 for i E {I, ..., r ) , r > 2. Then one of the following is true:

3. T = 3 and (d*,dz,d3) =

(a) (3,3,3), (2,3,6) or (2,4,4) and G" = 1 .

(b) (2,2, d ) and G is dihedral.

(c) (2,3,3) and G S A4.

(d) (2,3,4) and G % S4.

(e) (2,3,5) and G r As.

4. = 2 and G is cyclic.

Now we arsume in addition that G is a primitive group with elementary abelian normal psubgroup N 2 Ci.In this case G = HN where H = G, = {X 6 G l ~ ( u ) = u}. More precisely there is the following result of Burnside's.

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Proposition 2.2 If G, H and N are as above then

for some prime p and for a natural number e . Furthermore the map 4 : N -+ R, v +-+ vw is a bijection, and if h E H , then )(huh-') = h#(v) . In particular f ( h ) = I Ch~(h) J.

For a proof of see [Ill , Satz 3.2, p.159. If we consider N as a vector space over GF(p), we see that conjugation

by x E G induces a linear transformation on N . So we get a homomorphism $ : G ---+ GL(GF(p) ) with ker($) = Cc(N) = N. Thus $(G) % H . For x E G let det(r) denote the determinant of $(a) in the usual sense. More importantly the permutation representation of H on R is equivalent to the representation of H on N via $J. That allows us to compute the index of an element h E H regarded as an element of GL,(GF(p)) acting on 6,'. Since H is maximal in G, i t acts irreducibly on Ci.

We introduce some notation. Let L =< X I , ..., 2, > be a finite group acting on the F-vector space V. For each x 6 L we define Nz = [x, V] and e, = dimF 1,. For a field E 3 F we set N~ = I/ @F E.

In if G = HN, H =< h l , ..., h, >, then H acts on the GF(p)- vector space N and so the above definitions apply, i.e. Ni = Nh, = [hi, N ] and ei = eh, = dimNi as a GF(p)-vector space. Since fl =" $(G) C G L ~ ( P ) =

GLc(GF(p)) we let for h € H the element -h denote -$(h), and N: =

i-A;, N I and el = dim!( let ddenote the order of$ = h ~ l , let d' denote the order of h and let di denote the order of z;.

For general results on the index see 191. Here we only summarize those results relevant in our particular case. For the proofs of the results see [9] or

P51.

Lemma 2.3 1. If x = hv E G with h E H and v E N , then ind(x) 2 ind(h) with equality if and only if v E Nh. Furthermore f ( x ) > 0 ifl v E Nz.

2. Let x E G of order d and f ( x ) = 0 . Then ind(x) 2 pn. In particular

if d = p, fhen ind(x) = y n .

3. Suppose x E G of order d and f ( x ) > 0. Then ind(z) 2 y y n

where c = min{dim[xj, N ] I 1 < j < d ) .

4. Let z = hv E G = HN o j o ~ d e r d > 1 with h E H , v E N o n d p l d . d ir a power) then indjz) > ~ y n where b is the smallwt

.positive integer such that pb 5 1mod '9

d-1 Gn, g det(z) = 1, then ind(x) >- T P

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The above results indicate a relation between e, and ind (I) in the sense that ind (3) increases as e, increases. The next result which is due to L.L. Scott, 1181, and its corollaries relate the values e; to the dimension e of N.

Theorem 2.4 Let L =< X I , ..., x , >, with n:=, xi = 1, be a finite group acting on the (finite dimensional) F-vector space V . Then C:=l e; 2 e t dim[L,V] - dimCv(L).

Corollary 2.6 Assume L =< X I , ..., x, >, nI=, xi = 1, acts absolutely irreducibly on the F-vectorspace V . Let el, ..., a, be elements of an algebraically dosed field E containing F such that n:=l € i = 1. Set X: = Xiti, V' = vE and el = dirns[ri, v']. Then

In particular if J c (1, ..., r ) with I J I even. Then xi, J e: t Ci, J c e; L 2e where e: = dim[-xi, N ] .

By considering the induced action of L for example on V 8 V or V A V we can obtain further information about the elements xi. The proof of proposition 2.7 is due to R. Guralnick (see [15]).

Proposition 2J Suppose [I =( $1, ,,,,s, ) , ri, $ = 1, ads ahso\u\e\gl irreducibly on V . Let mi = dim[xi, W] .

1. I f W = V @ V andC{=,m;<2dimW, then V Z V *

2. If W = V @ V*, then C[=, mi 2 2 dim W - 2.

3. If W = A* V and '& mi < 2 dim W , then L _< Sp(V).

4. If char(F) # 2 and W = Sym2(V) and C;'=, m; < 2dim W , then L < Of ( V ) .

In the situation of G = HN we will apply the above results by setting L = H, V = N and F = GF(p) without stating them explicitly here as corollaries.

The first step in classifying primitive genus zero and one groups consists of finding conjugacy classes CI, ..., C, of G such that for yi E Ci we have E., ind yi = 2(n + g - l), g = 0, l . Once we have found such a set of conjugacy dasses we then have to find all tuples ( x l , . .. , x,) E C, x . . . x C,

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17Ll xi = 1 and < X I , ..., x, > is primitive. As it is easier to compute in GIN 2 H < GL,(p) we will first find elements hi E H which are conjugate to such that n:=, h; = 1. In a second step we will decide if such a tuple ( h l , ..., h,) can be lifted, i.e. if there exist vi E N such that hiai is conjugate to yi, flT hiv; = 1. Finally we have to deteremine all groups C; generate by those tuples The following results give an answer to some of those questions.

Let Der(H, N) = {6 : H 4 N 1 6 is a derivation} and Inn(H, N) the inner derivations. Then H 1 ( H , AT) = Der(H, N)/Inn(H. .\') acts faithfully on the conjugacy classes of complements of N in G (see 121.1 7.7). In the case of H =< h l , ..., h, >, n:=, hi = 1, set

and set S = S,. For any 6 E Der(H,N), order(h) = ordcr(hfi(h)). We set M h = {v E N ( order(h) = order(hv)) and Mi = hfh,. Sct c , = dim Mi.

Theorem 2.8 Let H =< h l , ..., h, >, HI=, hi = 1: act irreducibly on N . Assume M j + N j for j E { t + 1, ..., r ) , t 5 r . The following are equivalent:

1. 3vi E Ni for 1 5 i 5 t and 3vj E Mj - Ni for t < j 5 r such that G =< h lvr , ..., h,v, > and flL1 hiv; = 1.

Since order(h) = order(h&(h)) for any derivathn 6 we have Corollary 2.9 Let H =< hl , ..., hT > with ni=, hi = 1 and order(hi) = d:. If 3vi E N for all i E {I, ..., r ) such that n:=, h;v; = 1 and orderjhivi) > di for some i E (1, ..., r ) , then H N =< hlvl , ..., h,v, >.

The next two results explain the situation when f ( x i ) > 0 for all i E { 1 , ..., r } .

Corollary 2.10 Let H be as in theorem 2.8. The following are equivalent:

1. 3vi E N such that H N =< hlv1, ..., h , ~ . >, n:=l hiv; = 1 and f (hivi) > 0-

2. Ci='=l ei > e + dim S.

Corollary 2.11 Let G = < X I , ..., x, > with xi = hivi, hi E H , vi E N and

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In case the number of generators r = 3, a case which will take a lot of attention later, the results above take on a somewhat nicer form and are stated here. For the proofs see 191 or [15].

Corollary 2.12 Let H =< h l r h2, h3 >, hlh2h3 = I . The following are equivalent:

1. 3vi E N; such that xi = hivi i s conjugate to hi with ~ 1 x 2 ~ 3 = 1 and HAT =<xl, $2, x3>.

Corollary 2.13 Let G =< X I , X ~ , X ~ > , x1x2x3 = I and !(xi) > 0 for i E { 1 2 3 Then el + e2 > e, e2 + e3 > e and el t es > e . In particular if el = 1, then e2 = e3 = e and dims = 0.

Furthermore e\ + e: t e3 2 2e, el t e2 t e j 2 2e and el t e; t ei > 2e.

Corollary 2.14 Let H =< hl, h2, hS >, hl h2h3 = 1, act irreducibly on N. Assumep>2. Ifdl = 2 , then e-e; < e l 5 e ; fo r i E {2,3).

In many instances it is possible to determine the structure of GIN from the data given by the conjugacy classes (Cl, ..., C,) under the assumption that such group exists. We illustrate this with the example of the genus one group of type (2,6,6) for p = 7 and e = 2. (For the details see the next section.)

For such a tuple it ~0110~s that h! and hi generate a central subgroup of GIN of order 6 with quotient of type (2,2,3). Now < hl, h;, h;'h, > is group of type (2,2,6) contained in GIN. Furthermore this subgroup which is isomorphic to Dg together with the central element h; generates GIN. Hence we concluce GIN EZ Dg x C3. It remains to be shown that this group is generated inside GL2(3) by elements of the right type, i.e. there are elements hi E Ci such that GIN =< hl, h2, h3 > and hlh2h3 = 1. In our example the condition of irreducibility lets us assume that

in GL2(?), a # 0 # 6. It is not hard to see now that for any a = b-I E GF(7) we get the elements of the right type such that < hl, h2, h3 > acts irreducibly. Finally theorem 2.8 implies that such a tuple can be lifted as H1(G/N, N) = 0.

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The analogous computations involved in determining the matrix representation of GIN is omitted in those cases where the computations are as straightfrorward as in the above example which is the case especially for e = 2.

3 The groups of genus zero and one for p 2 7.

In this section we give a proof of those parts of theorem 1.5 and theorem 1.6 that relate to the primes p > 7. In particular we will assume that GI1 # {I). (For the case G" = 11) see theorem 1.3 and proposition 1.4.)

We restate the relevant parts of theorems 1.1 and 1.2 here: If G is primitive genus zero or genus one group with FW(G) " C,', p > 7, then

2. p = 7 and e _< 3.

In particular there are no genus zero or genus one groups for p > 11. In view of theorem 1.3 and proposition 1.4 we restrict our attention to the case G" # (1) and in particular e > 1.

3.1 The groups of genus zero and genus one for p = 11.

1. If d = 2 and e, = 1, then i n d ( x ) = 55; if x is a scalar, then i n d ( x ) = 60.

2. If d = 3, then i n d ( x ) = 80. I

3. If d = 4 , then i n d ( x ) = 90.

4. If d = 5, then i n d ( x ) = 88 or i n d ( x ) = 96. In the first case x is a reflection.

5. If d = 6 , then ind(x) = 100.

6. If d = 8, then i n d ( x ) = 105.

7. If d > 8 and e, = 2, then i n d ( x ) > 108 unless d = 10 and de t (x ) = -1 in which case ind(x) = 107 or d = 10 and d e t ( x ) # +1 in which case ind(z) = 103 or ind(x) = 104.

8. If f (s) = 0, then ind(x) = 110.

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It follows now immediately that if G is a genus zero or genus one group, i.e. ind(x i ) = 240 respectively 242, then the number of branch points

r = 3. Furthermore we conclude that f ( x i ) > 0 for all i E ( 1 , ..., r ) . Hence proposition 2.5 implies that el + e2 + e3 > 4. We assume that G is of type (dl ,d2,d3) where dl 5 d2 5 d3. Then dl = 2 and $ 5 4. Since GI' # 1 we may assume xl is not a scalar. Hence we have i n d ( x l ) + ind(x2) = 145 if dz = 4 and i n d ( x l ) + i n d ( x 2 ) = 135 if d2 = 3. In addition e3 = 2 and det(x3) = -1. If G is of genus zero, then we see that G is necessarily generated by a triple of type (2,3,8) and if G is of genus one, then G is generated by a triple of type (2,3,10). We discuss those two groups now.

The genus zero group of type (2,3,8). Note that h: = -E implies that ( G I N ) / < h: > is a group of type

(2 ,3 ,4) . It follows that I G I N I= 48 and G I N has a quotient isomorphic to S4. Hence G I N r GL2(3) which indeed imbeds into GL2(11) acting irreducibly. As G I N and N have coprime orders we have H 1 ( G / N , N ) = 0, Hence such a triple lifts by corollary 2.10.

The genus one group of type (2,3,10). Note that hi; is scalar and < hl h!, h2, h! > is a (2 ,3 ,5) group contained

in G I N . Hence G I N E As x Cz. As H 1 ( G / N , N ) = 0 such a triple can be lifted by corollary 2.10. 0

This shows that the two groups of theorem 1.5 and 1.6 for p = 11 are the only genus zero or genus one groups occuring for p = 11 and that they have the properties stated there.

3.2 The genus zero and genus one groups for p=7.

As was mentioned earlier the only genus zero groups for p = 7 occur for e 5 2 and the genus one groups occur for e 5 3. For the groups occuring in dimension e = 1 see theorem 1.3 and proposition 1.4.

Case 1: e = 2 We first list the values of indices of elements.

Lemma 3.1 1. If d = 2, then ind(x) = 21 for e, = 1 and ind(x) = 24 for e, = 2. In the second case x is a scalar.

2. If d = 3, then ind(x) = 28 or ind(x) = 32.

3. If d = 4, then ind(x) = 36.

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4. If d = 6 , then ind(x {37,38,39,40).

5. If d = 7 and f (x) > 0 ,

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) = 35 unless e, = 2 in which case ind(x) E

then ind(x) = 36. I f f (x) = 0, then ind(z) = 42.

6. If d J 48 and d 2 8, then ind(x) = 48y

7. If d = 14, then ind(z) = 42.

It follows that if we assume G is group of genus zero or genus one, i.e.

C;'=, ind(x;) = 96 or 98, then r 5 4. If the number of branch points r = 4, then f (xl), ..., f (x4) > 0. Thus

el $ ... t e4 > 4 by proposition 2.5. This implies that there are no genus zero - group with four branch points and if G is of genus one, then G is of type

(2,2,2,3) with el = e2 = 1 and e3 = e4 = 2. Note that if such a group exists, then GIN =< hl,h2,h3,h4 >= < h l , h 2 , h 3 h 4 > S D6. It is easy to see that D6 has indeed a faithful irreducible representation on N with the right type of elements. Thus GIN 2 D6. AS GIN and N have coprime orders we have H1(G/N, N) = 0. By corollary 2.10 such a tuple can be lifted.

Assume now that the number of branch points r = 3. As usual we assume

dl < d2 5 d3. If G is a genus zero or genus one group, then dl < 3. If dl = 3 and G is a genus zero group, then G is of type (3,3,4) with ind(31) = 28,

ind(,,J = 32 and ink] = 36, As det(iIjdet[~z) = 1 that no such triple exists. If dl = 3, then we see that no combination of indices whose associated elements satisfy also proposition 2.5 will sum up to 98.

If dl = 2, then ind(xl) = 21 as we assume G" # (1). Thus if G is a genus zero group, then ind(x2) + ind(x3) = 75 and if G is genus one group, then ind(x2) + ind(x3) = 77. Furthermore by proposition 2.5 e, = 2 for i = 2,3 unless f (xi) = 0. This implies that dz f (3,4,6).

First we consider the genus zero case: If dz = 6, then d3 = 6 and ind(x2) = 37 and ind(x3) = 38. However in this case det(xlx2xs) # 1. If d2 = 3, then ind(xs) = 43. However there is no such element x3. If dz = 4, then ind(zs) = 39 which implies that d3 = 6. We will discuss this group now. See

The genus zero group of type (2,4,6). Let fl be an eigenvector for the eigenvalue 1 of hl and let f2 be an

eigenvector for the eigenvalue of order 3 of ha. Since GIN acts irreducibly B = { fl, f2} is a basis for N. With respect to B we have:

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= ( ) and 0 -1

Since trace(h2) = trace(h3hl) = 0, a = b = -1. In both cases < hi > dG/N and h:h2ht = hi' Hence G / N E C3 . DI. Thus H1(G/:\' ;Y) = 0, Since

e l $ el $ e3 > 4 we can apply corollary 2.10 to conclude t h a t such a tuple can be lifted. 0

Now we turn to the genus one case: If d2 = 4, then ind(r3) = 11. However no such element 53 exists. If d2 = 3, then ind(x3) = 45 which implies d3 = 16 and if d2 = 6, then d3 = 6, ind(x2) = 39 and ind(x3) = 3s. \\'c shall discuss

those groups now.

The genus one group of type (2,3,16). Let fl be an eigenvector for the eigenvalue -1 of hl and let f2 be an

eigenvector for the eigenvalue 4 of h2. Since GIN acts irreducibly B = {fl, f2) is a basis for N. With respect to B we have:

hl = (7 :) and h2 = (: i ) . A computation shows that H acts irreducibly and h3 has order 16 if and

only if a = 1,2,4 or 6. If a = 1, then h;h2hi2h;' is a transvection . If a = 2, the same computation shows that h$$$ hi1 is a transvect,jon, If a = 4 or 6, then hih2 is a transvection. Hence [(GIN) : SL2(7)] = 2. Thus H1(G/N, N) = 0 and by corollary 2.10 such a tuple can be lifted.

The genus one group of type (2,6,6). We may assume ind(x2) = 39 and ind(x3) = 38. See section 2 for a

discussion of this group. This shows that the groups of theorem 1.5 and 1.6 for p = 7 and e = 2

are the only such genus zero or genus one groups occuring and that they have the properties stated there.

The Case e = 3: We already know that in dimension e = 3 there do not exist genus zero

groups (see [9] ).

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First we discuss the genus one group of type (2,3, 7), where el = ea = e3 = 2 and f (xg) = 0. Computing the indices we see that this is in fact a genus one system.

Let { fi, f 2 } be a basis for the -1 ejgenspace of hl and let f3 be an eigenvector for the eigenvalue 1 of ha. Since GIN acts irreducibly 3 = { f i , f i , f3) is a basis for N . With respect to B we have:

-1 0 a 2 0 0 ( 0 0 -1 0 1 ) a n d h 2 = ( 0 c d l 4 0 ) .

A computation shows that (hlh2)-l = ha satisfies ( x - 1)3 but not ( x - if and only if ac = 2 and bd = -1. After conjugating we may assume a = 1, c = 2, b = 1 and d = -1. Thus the isomorphism type of GIN is unique. This shows also that GIN acts irreducibly.

Note that PSL2(7) is a group of type (2 ,3 ,7) . For example we can take

as generators for SL2(7). Let V = GF(7)2 with basis i n l , n2). We have an induced action of PSLz(7) on V @ V . The subspace S = span{n; €3 nj t . . nj @ n; I Z , J = 1,2) c V @ V is invariant under this action. Note that

By [12], H1(G/N, N) = 0 which implies that such a triple can be lifted by corollary 2.10. 0

It remains to be shown that this is actually the only genus one group and that there are no groups of genus zero ocuuring for p = 7 in dimension e = 3.

We need some facts about the values of indices take in dimension e = 3.

Lemma 3.2 1. If d = 2, then ind(x) = 147,168 or 171 depending on whether e, = 1,2 or 3.

2. If d = 3, then ind(x) = 196,224 or 228 depending on whether e, = 1,2 or 3.

3. If d = 4, then ind(x) = 252 or 255.

4. If d = 6 and e, = 1 , then ind(x) = 245. If e. = 2, then ind(x) = 259, 266,273 or 280. If e, = 3, then ind(x) = 273, 274,275,277,282,286 or 285.

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5. If d = 7 and f ( x ) > 0, then ind(x) = 252 or i nd (x ) = 288.

6. If d = 8, then i n d ( x ) = 294 or i n d ( x ) = depending on whether e, = 2 or ex = 3.

7. If d > 8, then i n d ( x ) 2 304,

8. I f f ( x ) = 0, then i n d ( x ) > 294.

Note that by proposition 2.5 el + ... e, > 6 unless f (x i ) = 0 for some i 1 , , r Hence for a genus one group we have r = 3 and dl = 2. Since G" # {I} we may assume that xl is not a scalar. If f ( x i ) = 0 for some i E {I, 2,3) , then G is necessarily of type (2 ,3 ,7 ) which we discussed above. So we may assume that f ( x i ) > 0 for all i E {1,2 ,3} . If el = 2, then i n d ( x 2 ) + i nd (x3) = 518. Furthermore e2 + e3 > 4 and det(x2)det(x3) = 1. It follows that d2 = 3 or 4. If d2 = 3, then i n d ( x 3 ) = 294 and es = 3 or i nd (x3) = 290 and 3 1 d3. However no such elements x3 exist.

If el = 1, then i n d ( x 2 ) t i nd (x3) = 539. Furthermore e2 = e3 = 3 and no eigenspace of h2 or It3 is more than one dimensional. Also -1 cannot be an eigenvalue for both h2 and h3 and det(x2)det(x3) = -1 . In particular d2 $1 3 and if dz = 4, then d3 @ (6,s). Hence no such triple exists.

This shows that theorems 1.5 and 1.6 hold for p > 7.

4 The groups of genus zero and one for p = 5.

In this section we give a prod of thm parts d theorem 1.5 and &emern 1.6 that relate to the prime p = 5. Note that we assume that G" # (1). (For the case G" = (1) see theorem 1.3 and proposition 1.4.)

In 1151 it was shown that if G is primitive genus zero or genus one group with F*(G) E Ci, then e 5 3. In view of theorem 1.3 and proposition 1.4 we restrict our attention here to the cases e = 2 and e = 3.

For e = 2 all the necessary information about the elements is given in the following table. Recall that if x = hv, then d' denotes the order of h which we will only list if it differs from the order of x .

d d' f(x> ex de t (x) ind(x) 2 > 0 1 -1 10

(continued)

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Using this table we will discuss all tuples such that i n d ( x l ) + ... ind(x,) = 48 or 50 and which satisfy in addition el + ... + e, 2 4 and more specifically el t ... t e, > 4 if f ( x i ) > 0 for all 1 5 i 5 r. Furthermore we will use the condition d e t ( x l ) . . . det(x,) = 1. This implies r < 4.

We first eliminate two tuples that satisfy the above conditions but which nevertheless cannot be realized in GL2(5). A tuple of type (2 ,2 ,2 ,5 ) of genus zero with ind(x , ) = i n d ( s 2 ) = 10, ind(xs ) = 12 and i n d ( x 4 ) = 16 generates a group G such that GIN Z Dl& However this group has no irrducjble representation on N . Similarly we check that a tuple of type

(dl, d2, d3) = (3 ,3 ,5) of genus zero does not exist ;n c L 2 (5). The tuples that do correspond to groups are listed below.

The genus zero group of type(2,2,2,4). As h4 is a scalar < h4 > is normal in H and H / < h4 > is of type (2,2,2), i.e. H / c. h4 >S C2 x C2. Thus H is a subgroup of order 16 in the subgroup K of GL2(5) which consists of all elements of determinant f 1, i.e. H is a Sylow 2-subgroup of K. This group does act irreducibly on N. As G / N and N have coprime orders, H1(G/N, N) = 0. Hence such a tuple can be lifted by corollary 2.10.

The genus zero group of type (2,2,2,3). Since h3 is a scalar we have G / N =< hl, h2, hs, h4 >= < h2, ha, h1h4 > is a group of type (2,2,6) and which is isomorphic to Ds. As G I N and N have coprime orders we have H 1 ( G / N , N) = 0. Hence such tuple can be lifted by corollary 2.10. 0

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The genus zero group of type ( 2 , 3 , 1 0 ) . Since x i E N, GIN is a group of type ( 2 , 2 , 3 ) and which is isomorphic to &.There exists a faithful irreducible representation of S3 on N As G I N and A' have coprime order we have H1(G/h ' , N ) = 0 . Hence such a tuple can be lifted by corollary 2.9. 0

The genus zero group of type (2,3,12). Since h: is a scalar ( G I N ) / < ha > is of type ( 2 , 3 , 3 ) . Thus G I N 2

S L 2 ( 3 ) *C4. This group has a faithful irreducible representation on N . Since H 1 ( G / N , N ) = 0 corollary 2.10 implies that such a tuple can be lifted. 0

The genus zero group of type ( 2 , 3 , 2 0 ) . Since h: is a scalar of order 4 , ( G I N ) / < h: > is a group of type ( 2 , 3 , 5 ) .

Hence G I N is of order 240 which is the order of the subgroup of G L 2 ( 5 ) consisting of all elements of determinant f 1. This group is indeed of type ( 2 , 3 , 2 0 ) . As H 1 ( G / N , N ) = 0 corollary 2.10 implies that such a triple can be lifted. o

The genus zero group of type ( 2 , 4 , 8 ) . Since hi = 2 E is a scalar, ( G I N ) / < h: > is of type ( 2 , 2 , 4 ) , i.e it

is isomorphic to D4. In particular I G I N I= 32, which implies that G I N is isomorphic to a Sylow 2-subgroup of G L 2 ( 5 ) . This group does act irreducibly on N and is of type ( 2 , 4 , 8 ) . As G I N and N have coprime orders we have H 1 ( G / N , N ) = 0 . Hence such a tuple can be lifted by corollary 2.10. 0

The genus zero group of type ( 3 , 4 , 4 ) .

For this tuple we have to compute with the eirnenks in GL$j), We assum z'nd(r2) = 15 and ind(x3) = 17. Let f l be an eigenvector for the eigenvalue 2 of h2 and let f2 be an eigenvector for the eigenvalue 2 of h3. Since G I N acts irreducibly, B = {fi, f2) is a basis for N. With respect to B we may assume

A computation shows that (GIN1 = 96 and ( G I N ) n S L 2 ( 5 ) 2 S L 2 ( 3 ) . As H1(G/N, N ) = 0 such a tuple can be lifted by corollary 2.10. o

Next we list the groups of genus one:

The genus one group of type ( 2 , 2 , 2 , 4 ) . If we assume h3 is a scalar, then (h3h4)2 = h3 and so < h l , ..., h4 >= < h l , h2, h3h4 > which is a group

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of type (2,2,4). Hence H Dq. As GIN and N have coprime orders, H1(G/N, N) = 0. Hence such tuple can be lifted by corollary 2.10. o

The genus one group of type (3 ,4 ,5) . Since hi is a scalar, GjiV has a homorphic image which is isomorphic to

As. Hence GIN 2 SL2(5). As H 1 ( G / N , N ) = 0 corollary 2.10 implies that such a tuple can be lifted. o

The genus one group of type (3,4,4). For this tuple we have to compute with elements in GL2(5).

Let fi be an eigenvector for the eigenvalue 2 of h2 and let f2 be an eigenvector for the eigenvalue -1 of hS. Since GIN acts irreducibly, B =

ifl, f2) is a basis for A'. With respect to B we may assume

A computation shows that h:;'h? is a transvection. This implies GIN %

GL2(5). As H1(G/N, N ) = 0 such a tuple can be lifted by corollary 2.10. 0

The genus one group of type ( 3 , 3 , 4 ) . Since hi is a scalar of order 2 the group (GIN)/ < hi > is of type ( 2 , 3 , 3 ) and so G I N 2 Ad. As S4 and Ag K CZ ale not groups of type (3 ,3 ,4) we conclude that G I N SL2(3).

AS G I N and N have copr;me orders we have U'(C/N, NJ = O, Hence such tuple can be lifted by corollary 2.10. 0

The genus one group of type ( 2 , 4 , 1 0 ) . Since x$ E N , G I N is a group of type ( 2 , 2 , 4 ) and hence G I N r D4. There exists a faithful irreducible representation of D4 on N. By corollary 2.9 such a tuple can always be lifted. 0

The genus one group of type ( 2 , 4 , 1 2 ) . As hg generates a central subgroup of order 4. Hence ( G I N ) / < hz > is

of type ( 2 , 2 , 3 ) , i.e it is isomorphic to S3. Moreover < hl , hzh l , h ~ ' > 2 Ds. Since those elements together with hi generate G I N we see that G I N D6 * C4. As H 1 ( G / N , N ) = 0 such a tuple can be lifted by corollary 2.10.

The case e = 3 The necessary information about the indices for e = 3 is listed in the table below.

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We are looking for all tuples ( x l , . .. , x , ) such that i n d ( x 1 ) + . .. t i n d ( x , ) = 248 or 250 which further satisfy the conditions el t ... + e , 2 6 and more specifically el + ... $ e, > 6 if f ( x i ) > 0 for all 1 5 i 5 r. -

This implies immediately that r = 3. In that case we can furthermore conclude that dl = 2 using corollary 2.6 and the fact that GIN acts irreducibly. We will carry out a detailed discussion for the case el = 2 and leave it to the reader to complete the argument when el = 1 in which case there will be no groups occuring. Note that el # 3 since in that case hl is a scalar and hence G" = 1.

If el = 2, then ind(x2) + ind ( x3 ) = 198 or 200. Furthermore e2 + e3 2 5 and more specifically e2 + e3 > 6 if f ( x i ) > 0 for 1 _< i _< 3. The information supplied in the table above implies that d2 = 4 or dz = 5. If d2 = 4 , then the only possibilty for a genus zero or genus one group arises from a tuple of type ( 2 , 4 , 8 ) where i n d ( x 2 ) = 90 and ind(x3) = 108. However by corollary 2.6 such a tuple does not act irreducibly. If d2 = 5 , then e3 = 2 and hence f ( x 2 ) = 0 by proposition 2.8. Thus ind (x3 ) = 98 or ind (x3 ) = 100, det (x3) = 1 and e3 = 3. The only element satifying those conditions is an element of order 6 with ind (x3 ) = 98.

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For el = 2 we get the tuples (2,3,8) and (2,5,6) for genus zero and the tuples (2,4,6) and (2,4,5) for genus one. A discussion of the groups arising from those tuples is given now.

The genus zero system of type (2,5,6). We assume ind(xl) = 50, ind(x2) = 100 and ind(x3) = 98. Let {fl, f2)

be a basis for the 1 eigenspace of hl and let f3 be an eigenvector for the eigenvalue 1 of h2. Since H acts irreducibly, B = {fl, f2, f3) is a basis for N. With respect to B we may assume that

where a # 0. It can be checked now that for no choice of a, b and c is hlha in the right conjugacy class. Thus this tuple does not give rise to group of genus zero. 0

The genus zero group of type (2,3,8). We assume ind(xl ) = 60, ind(x3) = 80 and ind(x3) = 108 Let ifi, fi) be a basis for the -1 eigenspace of hl and let f3 be an eigenvector for the eigenvalue 1 of h2. Since H acts irreducibly, B = { fl , f2, f3) is a basis for N. With respect to B we may assume that

Now a computation shows that hl hz is of order 8 if and only if ( a , b, c, d) E {(0,1,0,1), (0,1,1, O), (1,0,1,1)). In all cases it can be seen that H % C4.5'3.

As H1(G/N, N) = 0 such a tuple can be lifted corollary 2.10. 13

The genus one group of type (2,4,5). We assume that ind(xl) = 60, ind(x2) = 90 and ind(x3) = 100. Note

that f(z3) = 0. Let { fi, f2,f3) be a basis of N such that {fl, f2) span the eigenspace for the eigenvalue -1 of hl and f3 is the eigenvector for the eigenvalue -1 of h2. Thus we may assume that up to conjugation

-1 0 a 2 0 0 h l = ( ;I ;) and h2= (! ;2

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A computation shows that if h i 1 = hlh2 satisfies the polynomial ( A - 1)3 but not ( A - then ac = -1 and bd = -2. Thus all groups generated by hl and h2 for the various choices of a, b, c and d are conjugate in GL3(5) and hence are isomorphic. Now Ss is a group of type (2 ,4 ,5 ) which has a faithful irreducible representation on N such that the generators map to elements of the given form.

As all elements of H 1 ( G / N , N) are inner on h3 theorem 2.8 implies that such a triple can be lifted. 0

The genus one group of type (2 ,4 ,6) . We assume i n d ( x l ) = 60, ind(x2) = 92 and ind(x3) = 98. Let { f i , f2, f3)

be a basis of N such that {fi, f 2 ) span the eigenspace for the eigenvalue -1 of hl and f3 is the eigenvector for the eigenvalue -1 of h2. Thus we may assume that up to conjugation

-1 0 a 2 0 0 = ( 0 -1 ) a n d ( -2 0 ) .

0 0 1 d -1

A computation shows that if hlh2 is of order 6 with no eigenvalue of order 6 then ac = 1 and bd = -2. Thus all groups generated by hl and h2 for the various choices of a, b, c and d are conjugate in GL3(5) and hence they are isomorphic. Now Ss is a group of type (2 ,4 ,6) which has a faithful irreducible representation on N such that the generators map to elements of the right form. So GIN Z $5. Since dim H1(S5, N) = 1 corollary 2.10 implies that such a tuple can be lifted. 0

This shows that theorems 1.5 and 1.6 hold for p = 5 .

5 The groups of genus zero and one for p = 3

In this section we give a proof of those parts of theorem 1.5 and theorem 1.6 that relate to the prime p = 3. In particular we assume that G" # (1). (For the case G" = (1) see theorem 1.3 and proposition 1.4.)

In [15] it was shown that if G is primitive genus zero or genus one group with F*(G) Ci, then e 5 4. In view of theorem 1.3 and proposition 1.4 we restrict our attention here to the cases e = 2 , 3 and 4.

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Some of the computations, especially those necessary to determine the structue of the groups in dimension e = 4, were performed using the group theory package Cayley V3.7.

The groups of genus zero and genus one in dimension e=2. We list the necessary information about the elements in dimensions e = 2

in the following table.

d d' f(x) e x det (x) ind(s) 2 > 0 1 - 1 3

The irreducible subgroups of GL2(3) are GLz(3), SL2(3), Dd, the quater- nions Qs and the Sylow 2-subgroup of GL2(3). As GIN is isomorphic to one of those groups, we have H1(G/N , N) = 0 (see [12] ).

It it easv to see that all of those groups do indeed occur as genus one - -

~ I O U ~ Q . For genus zero all of them except Qi occurl The author a list

of all possible tuples and their corresponding groups. We just mention here that for genus zero r < 5 and for genus one r _< 6 which follows from the information listed in the table above.

T h e groups of genus zero and genus one in dimension e=3. Similar to the above we first list the information about the elements in

dimension e = 3.

d d' f(x) e, det (x) ind(x) 2 > 0 1 -1 9

(continued)

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The subgroups of GL3 acting irreducibly on the natural module are GL3(3 ) , SL3(3 ) , S4, A4, $4 x C2, A4 x C2, C13, Ci3 (73, C26 and C26 * C3

(see [4] ). In all cases dimH1(G/N, N) 5 1 (see [12] ). Since we assume G" # 1 we can assume GIN # CI3, C13 C3 ,C26 or C26' C3.

We assume ind(x l ) + ... + ind(x , ) = 52 or 54. Furthermore el + ...+ e, 2 6 and more specifically el + ... + e, > 6 if f ( x i ) > 0 for 1 5 i 5 T . Thus T 5 4 for genus zero and T < 5 for genus one.

The groups of genus zero in dimension e=3. If r = 4 the only possibility is that G is of type ( 2 , 2 , 2 , 3 ) with i n d ( x l ) =

i n d ( x 2 ) = ind (x3 ) = 12 and i n d ( x 4 ) = 16. Let { f l , f 2 ) be a basis for the -1 eigenspace of hl and let f3 be a basis

for the intersection of the -1 eigenspaces of h2 and h3, Since GIN acts irreducibly we may assume that { f l y f 2 , f 3 ) is a basis for N. With respect to this basis

-1 0 a 1 0 0

W x

and h 3 = (: z ) , f -1

where w , x , y, z are such that d3 = 2 and e3 = 2.

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Now a computation using CAYLEY shows that GIN S SL3(3), S4 or A4 x C2. Since in all cases H'(G/hT, N) 5 1 and C!=, e; = 8 > 2e t dimH1(G/N, N ) , corollary 2.10 implies that such a tuple can be lifted.

If r = 3 the only possibility occurs when G is of type (2,3,13), with ind(xl) = 12, ind(x2) = 16 and ind(x3) = 24. It can be checked that such a tuple exists by using by direct computation or by use of a character table. It follows that GIN 2 SL3(3). Since H1(G/N , N ) = 0 corollary 2.10 implies that such a tuple can be lifted. o

The groups of genus one in dimension e=3. Case 1: If r 2 5, G is of type (2,2,2,2,2), ind(zl) = ind(xz) = 9 and ind(x3) = ind(x4) = ind(x5) = 12. This leaves only the possibilities GIN 2 GL3(3), $4, S4 x C2. However S4 cannot be generated by elements of the given type inside GL3(3). Both GL3(3) and S4 x C2 can be generated by elements of this type. In both those cases the generators can be lifted by corollary 2.10.

Case 2: If r = 4, then we have the following two possibilities: G is of type (2,2,2,4) with ind(xl) = ind(x2) = ind(x3) = 12 and ind(x4) = 18.

Let fi be a common eigenvector for the eigenvalue 1 of hl and h2 and let fi, f2 form a basis for the -1 eigenspace of h3. Since GIN acts irreducibly 8 = (f i fil f3) is a basis of N. With respect to B we may assume that

1 0 0 and t i 3 = (. 2 0 ) . .

d o 2

where ( : ) is a matrix of order 2 and trace 0.

Using Cayley we find that GIN 2 SL3(3) or or GIN r S4. In both cases H 1 ( G / N , N) = 0. Since el + ... + e4 = 8 corollary 2.10 implies that such a tuple can be lifted. o

G is of t ype (2,2,2,13) with i n d ( x l ) = ind(x2) = 9 , ind(x3) = 12 and ind(x4) = 24. The existence of such a genus one system can easily be verified. As H contains an element of order 13 and an element of order 2 and determinant -1 it follows that GIN r GL3(3). By corollary 2.10 such a tuple can be lifted.

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Case 3: r=3. In addition we to proposition 2.5, corollary 2.10 and corollary 2.6 we will apply corollaries 2.12-2.14. In particular ei + e j > 3 for any two i, j E {1 ,2 ,3) such that f ( x i ) , f ( x j ) > 0 .

We will again not discuss cases that can be eliminated with one of these observations. We also remark here that there is no genus 1 system of type ( 3 , 3 , 8 ) with el = e2 = 2 and there is no genus 1 system of type ( 3 , 6 , 6 ) with el = ez = 2 and e3 = 3. This can be checked by using a character table of SL3(3) (see [41).

Of the remaining cases we will first deal with those that contain an element of order 8,13 or 26. The existence of those tuples can be checked by direct computation or the use of the character tables. In all cases GIN E SL3(3) or G I N E GL3(3) . For the other tuples it will sometimes be necessary to compute explicitly if and what group the elements generate. The fact that those can be lifted is guaranteed by theorem 2.8 and its corollaries.

1, G is of type ( 2 , 4 , 1 3 ) with i n d ( x l ) = 12, ind (x2) = 18 and ind (x3) = 24; G I N 2 SL3(3 ) .

2. G is of type (2 ,6 ,13 ) with i n d ( x l ) = 9, ind (x2) = 21, e2 = 3 and ind (x3) = 24; G I N e GL3(3) .

3 . G is of type ( 2 , 6 , 1 3 ) with i n d ( x l ) = 9, i n d ( x 2 ) = 21, f ( x 2 ) = 0 and ind (x3) = 24; G / N S GL3(3) .

4. G is of type ( 2 , 6 , 1 3 ) with i n d ( x l ) = 12, i n d ( x 2 ) = 18 and ind(x3) = 24; G I N E GL3(3) .

5. 0 is of type (2,6,26) with i n d ( x , ) ' = 12, i n d ( x z ) = 17, f ( x z ) = 0 and ind (x3 ) = 25; G I N 2 G L 3 ( 3 ) .

6 . G is of type ( 3 , 3 , 8 ) with i n d ( x l ) = ind(x2) = 16 and i n d ( x 3 ) = 22; G I N S SL3(3) .

7. G is of type ( 2 , 4 , 9 ) with i n d ( x l ) = 12, i n d ( x 2 ) = 18 and i n d ( x 3 ) = 24. Since G I N is of type ( 2 , 3 , 4 ) we have G I N r S4 which is generated a s a subgroup of G L 4 3 ) by elements of right type.

8. G is of type ( 2 , 6 , 6 ) with ind(x1) = 12, ind(x2) = ind(x3) = 21 and e2 = e3 = 3 . Since h; = hi = -E a scalar we have ( G I N ) / < h$ >r A4. Hence G I N % A4 x C 2 .

9 . G is of type ( 2 , 6 , 9 ) , i n d ( x l ) = 9 , ind(x2) = 21, e2 = 3 and i n d ( x s ) = 24. As hi = -E, it follows that ( G I N ) / < hi > is of type ( 2 , 3 , 3 ) which is isomorphic to A4. Hence G I N 2 A4 x C 2 .

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10. G is of type (3,3,6) with ind(x1) = ind(x2) = 16 and ind(x3) = 22, f ( ~ 3 ) = 0.

Let f l be an eigenvector for the eigenvalue 1 of hl and let f2 be an eigenvector for the eigenvalue 1 of h2. Let f3 E N such that B = {fl, f2, f 3 ) is a basis for A'. With respect to B we may assume that

h ( : ) a n d ( P :), O c d Y 0 %

where the variables have to be chosen such that hl, h2 are of order 3 and hl - E,h2 - E are of rank 2.

Now hs is of order 2 or of order 6. Using Cayley we find that in the first case GIN E Aq and in the second case GIN E GL3(3).

11. G is of type (3,4,4), ind(xl) = 16 and ind(x2) = ind(x3) = 19.

Let f2 be an eigenvector for the eigenvalue -1 of h2 and let f 3 be an eigenvector for the eigenvalue -1 of hJ. Choose fi E N such that B = {fl, f2, f3) is a basis for N . With respect to B we may assume that

a O b w x 0 h Z = e 2 f a n d h 3 = y z 0 , [, 0 d ) ( $ 2 )

where ( : ) and ( : ) are matrices of order 4 such that h2h3

is of order 3 and hzh3 - E is of rank 2.

Using Cayley we find that under those conditions GIN Z GL3(3).

12. G is of type (3 ,6 ,6) , ind(xl) = 16,ind(x2) = 17, ind(x3) = 21 and e3 = 2 . Clearly GIN P Cz6.C3. Since hz is not central GIN + S4 x C2. Hence GIN Z GL3(3).

13. G is of type (4 ,4 ,6) with ind(xl) = 18, ind(xz) = 19 and ind(x3) = 17. Clearly GIN # C26 .C3. Since hz is not central GIN + S4 x C2. Hence GIN 2 GL3(3).

1 4 . G is of type (4 ,6 ,6) with ind(xl) = 19, ind(z2) = 17 and ind(x3) = 18. Clearly GIN P C26 a C3, Since h; is not central GIN + $4 x C 2 Hence GIN GL3(3).

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The groups of genus zero and genus one in dimension e=4. Just the sheer number of conjugacy classes in dimension e = 4 makes it

impossible to write down a similar list as we did in dimension e < 3. Also the few number of groups occuring in dimension e = 4 suggest that proceeding as in dimension e 5 3 is inefficient,

First we show here that the groups of genus zero for p = 3 in dimension e = 4 of theorems 1.5 and 1.6 have the properties stated there. We start with the example of a tuple of genus zero which cannot be lifted.

There is no group of type (3,3,4). There exists a tuple of type ( 3 , 3 , 4 ) such that i n d ( x l ) t i n d ( r z ) t i n d ( x 3 ) =

160. We assume el = e2 = 2 , e3 = 4 and f ( x l ) = 0 while f i x : ) . f ( x 3 ) > 0. Note that H acts absolutely irreducibly. By considering the action of H on N 8 N we see that (h l - E)2 # 0 # (h2 - E ) 2 by proposition 2.7. This determines the conjugacy class of hl and h2.

Let f i , f 2 form a basis for the 1 eigenspace of h l and let f3: f4 form a basis for the 1 eigenspace of h a . Since H acts irreducibly B = { f l , ..., f 4 ) is a basis for N and with respect to B we may assume

l O a b 1 0 0 0 O l l c 1 1 0 0 h [ 0 0 1 1 ) a n d h 2 = ( w l l 0 1 . 0 0 0 1 y z o 1

An inspection using Cayely shows that if hlh2 is in the right conjugacy class and < h l , ha, h3 > acts irreducibly, then GIN G As. In this case

Since f ( x 1 ) = 0 we have to apply theorem 2.8 in its full force here. We see that el = 3, ez = 2 and e3 = 4.

The space of derivations S2 of A$ to N which are inner on h2 and hs contains an outer derivation. Thus dim S2 2 5 and thus such a tuple cannot be lifted by theorem 2.8. o

Now we discuss the tuples that gives rise to groups of genus zero.

The group of type (2,4,5). Choose f l , f 2 , f3 to form a basis for the -1 eigenspace of h l . If f4 is an

eigenvector for the eigenvalue 1 of h2, then B = { f i , ..., f 4 ) is a basis for N as H acts irreducibly. With respect to B we have D

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Since GIN acts irreducibly c # 0 # z , {a, b ) # (0) # {x, y) . An inspection using Cayley shows that such a group exists and GIN S5. AS H' (GIN, N) = 0 corollary 2.10 implies that such a tuple can be lifted. 0

The groups of type (2,4,6). Let f3 be an eigenvector for the eigenvalue -1 of h2 and let f4 be an eigen-

value for the eigenvalue 1 of h2 . Let Ul be the eigenspace for the eigenvalue 1 of hl and let U2 be the eigenspace for the eigenvalue -1 of hl . As H acts irreducibly Ul or U2 is a complement of the span of {f3, f4 ) . Assume first Ul

is a complement. Choose fi, f2 E Ul such that B = {fi, ..., f4) is a basis for N , With repsect to B we may assume

l O a b 0 1 0 0 O l c d 2 0 0 0

w x 2 0 0 0 0 2 y z O l

An inspection using Cayley shows that GIN 2 Ss, GIN Ss x C2 or GIN is solvable and the order of GIN is 384, in which case it is not hard to see that GIN Z C2 1 S4. Since H 1 ( G / N , N ) = 0 such a tuple can be lifted by corollary 2.10. 0

Next we verify that the groups of genus one of theorem 1.6 for p = 3 in

T h e group of t y p e (2,3,10). Note that h: = -E and thus ( G I N ) / < hz >S As. As< -hl ,h?, -h3 >

is a group of type ( 2 , 3 , 5 ) contained in GIN it follows that GIN A5 x C2. The permutation representation of A5 on N yields elements of the right

type. Thus such a tuple exists and any such tuple will generate A5 x C2. AS H 1 ( G / N , N ) = 0 corollary 2.10 implies that such a tuple can be lifted.

The groups of t y p e (2,4,8). Let f3 be an eigenvalue for the eigenvalue 1 of h2 and let f4 be an eigen-

value for the eigenvalue 1 of h2. Let Ul be the eigenspace for the eigenvalue 1 of hl and let U2 be the eigenspace for the eigenvalue -1 of hl . As H acts irreducibly it follows that Ul or U2 is a complement of the span of ( f i , f2) . Assume first that Ul is a complement and choose fi, f2 E Ul such that B = { f l , ..., f2) is a basis for N . With repsect to B we may assume

l O a b 0 2 0 0 O l c d 1 0 0 0

0 0 0 2 y z O l

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An inspection using Cayley shows that GIN 2 G L 4 ( 3 ) or GIN is solvable of order 1152. Since H 1 ( G / N , N ) = 0 in both cases corollary 2.10 implies that such a tuple can be lifted. 0

The groups of type (2,5,6). Let f3 be an eigenvector for the eigenvalue 1 of h3 and let f4 be an

eigenvector for the eigenvalue -1 of h3. Let UI be the eigenspace for the eigenvalue 1 of hl and let U2 be the eigenspace for the eigenvalue -1 of hl. As H acts irreducibly it follows that Ul or U2 is complement to the span of i f 3 , f 4 ) . Assume first that Ul .is a complement. Then choose fi: fi E Ul such that B = {jl, ..., j4) is a basis for N. With repsect to B we may assume

l O a b 2 0 0 0 O l c d 0 1 0 0

h ~ = ( ~ o ) a n d h 3 = [ W X ~ O 1. 0 0 0 2 y z 0 l

An inspection using Cayley shows that GIN Z 2.S5 or G I N % S p 4 ( 3 ) . As H 1 ( G I N , N ) = 0 in both cases theorem 2.8 implies that such a tuple can be lifted.

If we assume that U2 is a complement, then the analogous computation shows that G I N 2 2.Ss or GIN Z S p 4 ( 3 ) . 0

The group of type (2,6,8). Choose fi, f2, f3 to form a basis for the 1 eigenspace of hl. If f4 is an

e i g ~ y ~ ~ t v r for t Re eigwalue 1 of hll then B i; [/I fi] i~ B biiii 01 1 as H acts irreducibly. With respect to B we may assume

l O O a 2 1 0 0 ' O l O b

h ~ = [ ~ c ) a n c i h 2 = ( o 0 0 1 0 O ) ' 0 0 0 2 l d l l

where b # 0 # c. Using Cayley we find that if GIN act irreducibly, then G I N is solvable of order 2304. As H 1 ( G / N , N ) = 0 corollary 2.10 implies that such a tuple can be lifted. o

The groups of type (2,2,2,3). Here el = ez = e3 = e4 = 2 and f(x4) = 0. Note that for i E {1,2,3},

Ni = [hi, N ] is the - 1 eigenspace of hi . As H acts irreducibly, Nl + N2+ N3 = N which implies that Ni+ N, = N for some 2 , j E { 1 , 2 , 3 ) . Let U; denote the 1 eigenspace of hi for i E { 1 , 2 , 3 ) . As above we can conclude that Ui+Uj = N for some i, j E { 1 , 2 , 3 ) . After possibly braiding the tuple ( h l , h2, h3) we may assume that Ul + U2 = N .

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Let { f l , f 2 ) be a basis for Ul and let { f 3 , f4} be a basis for U2. By the above B = { f i , ..., f 4 ) is a basis of N and with respect to B we may assume

where A, B, C, D, X , Y E M2(GF(3)) , I is the identity matrix in Mz(GF(3) ) and 0 is the zero matrix in M2(GF(3)) . After conjugation by a block diagonal element of GL4(3) we may assume that

We will discuss only the case B = 0 in detail. Assume B = 0. Since H acts irreducibly it follows that X has rank 2.

As A2 = I we may assume, after conjugating by a block diagonal matrix in GL4(3), that X = I and

If A = I or A = -I, then D = -A and a direct computation shows that hlhZh3 is never of order 3. If

then D is of order 2 and D has trace 0. Thus

If we set T = YA + C, then

Furthermore we know that h;' - I has rank 2 and (h;' - I ) 3 is the zero matrix in M 4 ( G F ( 3 ) ) . Using this we compute that hlhzhs is in the right conjugacy class if and only if

In those cases a check with Cayley shows that GIN 2 As, GIN 2 A5 x C2, GIN =" Cz.A5.Ci has order 480, GIN r SL4(3) or GIN is solvable of order

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96 or 768 and is contained in the normalizer of an extra special group of order 32.

If B = ( ), then GIN E SL4(3), GIN Z As x C2, GIN I As,

GIN is solvable of order 192 and is contained in the normalizer of an extra special group of order 32.

If B = ( : ), then GIN E As x C2 or G'/N is of order I92 and is

contained in the normalizer of an extra special group of order 32. In all cases H1(G/N, N) = 0 and thus these tuples can be lifted by

theorem 2.8.

Finally we will show that the groups discussed above are indeed the only groups of genus zero or genus one in dimension e = 4.

Proposition 5.1 The above groups are the only genus zero and genus one groups occuring in dimension e = 4 for p = 3.

We first list some values of indices.

Lemma 5.2 1. If d = 2, then ind(x) = 27,36,39 or 40 depending on whether e, = 1,2,3 or 4.

Assume d = 3. If f ( x ) = 0, then ind(x) = 54. If f (x) > 0 and e, = 1, then ind(x) = 36. I f f ( x ) > 0 and e, = 2, then ind(x) = 48

I f e = 4, then ind(x) = 58 or 60.

Assume d = 6 . I f f ( x ) = 0, then ind(x) 2 63. I f f (x) > 0 and ex = 2, then ind(x) = 51 o r 54. If ex > 2, then ind(x) 2 56. If ex2 > 1 , then ind(x) 2 59.

If d = 5 or d > 6 , then ind(x) > 63.

The proof of the above facts follows easily by direct computation. We can now give a proof of proposition 5.1.

Proof As usual we assume dl 5 ... 5 d,. It is easy to see from lemma 5.2 that if r 2 5, then CT=, indx 2 2.81 + 2. If r = 4, then it follows directly from proposition 2.5 and lemma 5.2

that xi=, ind xi 2 2 , 8 1 + 2 unless dl = dz = ds = 2. Note that d4 > 2 as G" # 1. Without loss of generality assume el 5 ez 5 ea. As H acts

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irreducibly el 5 2. If el = 2, then EL1 ind xi 2 2 81 + 2 unless d4 = 3, el = ... = e4 and f ( x 4 ) = 0. In this case we determined the groups above. If e l = 1, then C!= , ind xi 2 2 . 81 + 2 unless dl = 4 and ei = i. However in this case we can apply proposition 2.7 to the action of H on N 8 N to show that H does not act irreducibly.

Now assume r = 3. Case 1 : dl > 3.

By proposition 2.5 e; > 2 for some i E {1,2,3). Hence C:=, indx; 2 2 . 81 + 2 unless d j = 6 and indxj = 51 for some j E {1,2 ,3) . In this case however ek > 2 or f(xk) = 0 for IC # j by corollary 2.10. Hence C:=, indx, 2 2 .81 + 2. Case 2: dl = 3.

Note that if d2 = 3, then d3 > 3 as G" # 1. If f ( x l ) = 0, then clearly ,& ind xi 1 2 81 $ 2 unless d2 = 3. In this case e3 = 4 by proposition 2.5. If d3 > 4, then C:=, ind xi 2 2 . 8 1 $ 2 . If d3 = 4, then ind x3 = 58 or 60. In the first case g = 0 and GIN S As. In the second case we observe that G I N acts absolutely irreducibly on N. However by considering the action of H on N @ N we see that this contradicts proposition 2.7.

If f ( x 1 ) > 0, then we may assume d2 > 3 as otherwise f ( x 2 ) = 0 which is equivalent to f ( x l ) = 0 which we disposed off above.

If el = 1, then H acts absolutely irreducibly and so no eigenspace (over the al ebraic closure of G F 3 of h2 or h3 is more than one dimensional. B ( ))

6 ' 1'1 Furthermore if f ( x i ) > 0, then ei = 4 by corollary 2.10. o m par lCU al d2 > 4 and if di = 6, then f ( x i ) = 0. Hence ind xi 2 2 - 8 1 + 2.

If el = 2, then e2 + e3 2 6 and e; + e$ > 6 by corollary 2.6. Furthermore if e; = 2, then f (xi) = 0 by corollary 2.10. It follows that xi=, ind xi 2 2.81+2.

Case 3: d l = 2 . Note that if d2 = 3, then d3 > 6 and if d2 = 4, then d3 > 4 as GI' f 1. If el = 1, then H acts absolutely irreducibly and so no eigenspace of h2

or h3 (over the algebraic closure of GF(3)) is more than one dimensional. Furthermore e2 4 e3 2 7 and in case of equality f (x2) = 0 or f ( x s ) = 0 by proposition 2.5 and corollary 2.10. In particular d2 # 4 # d3 and if d; = 6, then f ( x ; ) = 0 and indx, > 66 for i E {2,3). Hence ,J& indx; 2 2 . 8 1 + 2 unless d2 = 6, ez = 3, d e t ( x 2 ) = 1 and d3 = 8 (see above).

If el = 2, then e2 + e3 > 6, e; + e3 2 6 and e2 + e$ > 6 by proposition 2.5 and corollary 2.6. Furthermore ei > 2 or f ( x ; ) = 0 by corollary 2.10. If d2 = 3, then f ( x 2 ) = 0 and e3 = 4. Hence c;=~ ind xi 2 2 . 81 + 2 unless d3 = 8, or d3 = 10. In the first case we can choose a canonical basis for N and using Cayley we see that such a group does not exist. In the second case g = 1 and G/N cz A, x C 2 (see above). If d2 = 4, then ind x 2 = 57,58 or 60

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as ez > 2 by proposition 2.10. If ind x 2 = 60, then c:='=, ind xi 2 2 . 81 + 2 unless ind x3 = 64 or 66. Note that if H acts absolutely irreducibly, then dim[h3, N @ N] 2 14 by proposition 2.7. An inspection shows that this

eliminates all possibilities. If ind 2 2 = 58, then e3 = e i = 4 by corollary 2.6. Thus C;=, ind x; 2 2 81 + 2. If ind x2 = 57, then c:=, ind x, 2 2 . 8 1 t 2 unless ind x3 = 67 or 69. Note that by corollary 2.10 e3 = 4 or f(x3) = 0. In the first case dg = 8 and GIN was determined above. In the second case d3 = 6 and GIN was also determined above.

If d2 > 4, then c:=, ind xi 2 2 81 + 2 unless (dz , d3) E {(5,6), (6,8)). If d2 = 6, then we choose a canonical basis for N and using Cayley we see that such groups do not exist. If G is of type (2,5,6), then GIN was determined above.

If el = 3, then H acts absolutely irreducibly and so no eigenspace of h2 or h3 (over the algebraic closure of GF(3)) is more than one dimensional). By corollary 2.6 we have also e; t es 2 7 and e2 t e$ 1 7. It now follows easily that & ind x; >_ 2 a . 8 1 t 2 unless d2 = 4 and d3 = 5 in which case GIN 3 Ss was determined above.

Finally if el = 4, then hl = -E and G" = 1.

This shows that theorems 1.5 and 1.6 hold for p = 3.

References

[I] M. Aschbacher, Finite Group Theory, Cambridge University Press, Cambrid e 1986, 0 I

[2] M. Aschbacher, On conjectures of Guralnick and Thompson, J. Algebra, 135 No. 2(1990), 277-341.

(31 M. Aschbacher and L. Scott, Maximal subgroups of finite groups, J. Algebra, 92(1985), 44-80.

[4] J. Conway et al., Atlas of Finite Groups, Clarendon Press, Oxford, 1985.

[5] W. Feit, R.C. Lyndon and L.L Scott, A remark about permutations, J. Comb. Theory Ser A, 18(1975), 234-235.

[6j M. Fried, Galois groups and complex multiplication, Trans. Amer. Math. SOC. 237(1978), 141-162.

[7] M. Fried, Combinatorial computation of moduli dimension of Nielsen classes of covers, Graphs and Algorithms, R. Bruce (Editor), Contem- porary Mathematics 89(198g), AMS.

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183 M. Fried and R. Guralnick, The generic curve of genus g > 6 is not uniformized by radicals, preprint.

[9] R. M. Guralnick and J. G. Thompson, Finite groups of genus zero, J. Algebra, 131 No.2(1990), 303-341.

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Ill) B. Huppert, Endliche Gruppen, Springer Verlag, Berlin, 1967

[12] W. Jones and B. Parshall, On the 1-cohomology of finite groups of Lie type, Proceedings of the Conference on Finite Groups, 313-328, Aca- demic Press, New York 1976.

[13] W. Ledermann, Introduction to Group Characters, Cambridge Univer- sity Press, Cambridge 1987.

[14] W. Magnus, Noneuclidean Tesselations and their Groups, Academic Press, New York, 1974.

[15] M. Neubauer, On monodromy groups of fixed genus, J. Algebra, to ap- pear.

[16] M. Neubauer, Primitive permutation groups of genus zero and one 11,

preprln ' 1 .

117) T . Shih, A note on groups of genus zero, Comm. Alg., 19 # 10(1991), 2813-2826.

[18] L. L. Scott, Matrices and cohomology, Annals of Mathematics, 105(1977), 473-492.

[19] 0. Zariski, Collected Papers, vol. III, MIT Press, Cambridge, 1972.

Received: December 1991

Revised: May 1992 Dow

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