Finite Irreducible Imprimitive Nonmonomial Complex Linear Groups of Degree 4

Journal of Algebra 236, 419–470 (2001) doi:10.1006/jabr.2000.8525, available online at http://www.idealibrary.com on

Finite Irreducible Imprimitive Nonmonomial Complex Linear Groups of Degree 4 Burkhard H¨ ofling Mathematisches Institut, Friedrich Schiller-Universit¨ at, D-07740 Jena, Germany E-mail: Burkhard.Hoefl[email protected] Communicated by Walter Feit Received April 22, 1998

The article contains a classification of the groups indicated in the title, consisting of a parametrised list of groups, each relevant group being conjugate in GL
4
 to precisely one group on the list (and each group on the list being relevant). Each group on this list is given by a generating set of matrices, whose entries are explicit functions of finitely many integers satisfying certain relations. Also included is a similar classification of the finite primitive subgroups of GL
2
. © 2001 Academic Press

Key Words: finite group; complex linear group; primitive linear group; imprimitive linear group; classification.

1. INTRODUCTION The study of finite irreducible complex linear groups of low degree is a classical problem. A century ago, many prominent mathematicians worked on it, including Jordan, Klein, and Maschke. A substantial part of the 1916 book [16] of Miller, Blichfeldt, and Dickson was devoted to this topic, and Blichfeldt’s monograph [1], published in 1917, dealt exclusively with groups of degree at most 4. The issue has stayed alive ever since. In his invited address [6] at the 1970 International Congress of Mathematicians, Feit surveyed the state of the art, and found the results complete up to degree 7. By today, they extend at least to degree 9; see [7, 12, 13]. However, these results are complete only in a weak sense: they deal mainly with primitive (or quasi-primitive) linear groups consisting of matrices with determinant 1, and often are only up to the conjugacy of their images in the relevant projective linear group. It seems taken for granted 419 0021-8693/01 $35.00 Copyright © 2001 by Academic Press All rights of reproduction in any form reserved.

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that there are adequate methods for reducing a given general problem to this case, and that it is more expedient to apply those methods to one group at a time than to try to use them to describe the general picture of, say, all groups of a given degree. Moreover, in more recent work, soluble groups were usually ignored. On the other hand, with the increasing availability of fast and versatile computer algebra systems, this point of view has started to shift. Complete lists of groups which are, in a certain sense, “small,” have gained importance, both because the mass of detail involved has become more manageable and accessible, and because computer algebra systems have made it feasible to study vast ranges of concrete examples of groups. Moreover, computers also prove helpful in ensuring the correctness and the completeness of the lists themselves. Conlon may have been the first to make a move in this new direction. In [2, 3], for each prime p, he determined all irreducible subgroups of p-power order in GL
p
 (and also in several other classical groups of degree p). His description amounts to giving a parametrised list of groups such that each finite irreducible p-subgroup of GL
p
 is conjugate to precisely one group on the list. Each group on the list is given by a generating set of matrices, and the entries of the matrices are explicit functions of finitely many parameters which range over appropriate sets of integers. A similar description of the finite irreducible 2-subgroups of GL
4
 has only recently been given by Flannery [8]. It seems desirable to exploit these developments by seeking similarly explicit descriptions of all finite irreducible subgroups of GL
n
 for n ≤ 4, to complete the project of Blichfeldt’s book to the standard set in Conlon’s articles. The most difficult part is, of course, the case n = 4. In the primitive case, the task is to obtain direct extensions of the results for PSL
4
 that were already known to Blichfeldt. Based upon his earlier work, the monomial groups in question were classified by Flannery [9]. Here we offer a description of the groups that will not be covered by either of these cases: the finite irreducible imprimitive nonomonomial subgroups of GL
4
. It is planned to make the list obtained in this article, as well as the list in [9], available as shared libraries in the computer algebra system GAP [11]. Details about the status of these and related lists and their implementation in GAP can be found on the web page [10]. The article is organised as follows. In the first part of the article, we consider the following more general situation. Let V be a vector space of dimension 2n over an algebraically closed field F, where n is a positive integer, and let G be a finite subgroup of GL
V  for which V possesses a nonrefinable system of imprimitivity V1  V2 . It will turn out that these groups G fall into two classes, namely, groups which have exactly one such

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system of imprimitivity, and groups which possess three such imprimitivity systems; see Theorem 2.4 for a more general result. For the construction of the groups G in question, we require lists of representatives of the conjugacy classes of primitive subgroups of GL
n F, of their normal subgroups, and of the automorphisms of their factor groups whose squares are inner; see Theorems 2.10 and 3.5 for details. At this stage, we do not attempt to decide which other systems of imprimitivity the groups G might have; in particular, we do not yet distinguish between monomial and nonmonomial imprimitive subgroups of GL
4
. In view of results that go back to Jordan and Blichfeldt (see, e.g., Theorem 36.13 and Corollary 50.7 of [4]), there are only finitely many conjugacy classes of finite primitive subgroups of SL
n
. Thus, the aim in Section 4 is to obtain the data required for our classification from the corresponding information about this finite list. While our results are not conclusive, this approach is always feasible if the finite primitive subgroup G of GL
n
 is the product of its centre and a primitive subgroup S of SL
n
; in particular this is always the case when the finite primitive subgroup of SL
n
 “involved” in G is perfect. Specialising to the case F =
, n = 2 in Sections 5–7, we compile lists parametrised by a finite number of integers which contain the information about finite primitive subgroups of GL
2
 required for the construction of all finite irreducible subgroups of GL
4
 which have a nonrefinable system of imprimitivity V1  V2 . The classification, up to conjugacy, of all finite primitive subgroups of GL
2
 obtained in Theorem 5.1 may be of independent interest. In Section 8, we determine which of the finite imprimitive subgroups of GL
4
 in our preliminary list are monomial, thus arriving at our final result. A summary of this list of representatives of conjugacy classes of the finite irreducible imprimitive nonmonomial subgroups of GL
4
 can be found in Section 9. Our notation is standard and follows, for example, [4, 5]. 2. GROUPS WITH MULTIPLE SYSTEMS OF IMPRIMITIVITY Let V be a finite-dimensional vector space over a field F and let G be an irreducible subgroup of GL
V . A set  = V1   Vr ,
r ∈  of subspaces of V is a system of imprimitivity for G if V = V1 ⊕ · · · ⊕ Vr and Vk g ∈  for every g ∈ G and k ∈ 1  r. (Note that the irreducibility of G implies that G permutes the Vk transitively.) A system of imprimitivity  for G refines  if every W ∈  is contained in some Vk ∈  . The system of imprimitivity  is nonrefinable if  =  for every refinement  of  . An irreducible subgroup G of GL
V  is primitive if V  is the only system of

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imprimitivity for G. Otherwise, G is imprimitive. A subgroup G of GL
V  is monomial if it possesses a system of imprimitivity which consists of onedimensional subspaces of V . Imprimitivity systems arise in the following context; see, for instance, [4, Sect. 50], or [17, Theorem 15.3]. 2.1. Lemma. Let F be a field, let V be a finite-dimensional vector space over F, and let G be a finite irreducible subgroup of GL
V . Assume that H is a subgroup of G of index k and that W is a nonzero FH-submodule of V such that k dimF W ≤ dimF V . Then, k dimF W = dimF V and Wt1   Wtk  is a system of imprimitivity of V for G, where t1   tk  is a right transversal of H in G. Proof. Let g ∈ G and i ∈ 1  k, then Hti g = Htj for some j ∈ 1  k, and so Wti g = Wtj . This proves that G permutes the Wti . In
particular, S = ki=1 Wti is an FG-submodule of V . Since 0 = W is contained in S and V is irreducible as an FG-module, we have S = V . Now, dimF V = dimF S ≤ k dimF W ≤ dimF V , and the equality holds. In particular, S is the direct sum of the Wti , and hence the latter form a system of imprimitivity of V for G. Let G be a finite irreducible subgroup of GL
4
 and assume that G is neither primitive nor monomial. Then, G has a system of imprimitivity consisting of two two-dimensional
-vector spaces. It turns out that G may have one such system of imprimitivity, or else it has three such systems. We will first investigate the latter type of linear groups. The following lemma is an easy consequence of a well-known theorem of Clifford, see, e.g., [4, Theorem 51.7] or [14, V, Satz 17.5]. 2.2. Lemma. Let V be a finite-dimensional vector space over an algebraically closed field F and let G be a finite irreducible subgroup of GL
V . Further, assume that K is a normal subgroup of G such that G/K is cyclic. If V is homogeneous as an FK-module, then K acts irreducibly on V . Therefore, if V is reducible as an FK-module, the homogeneous components of V for K form a system of imprimitivity of V for G which is permuted cyclically by G. Proof. Assume that V is homogeneous as an FK-module, then by [4, Theorem 51.7] or [5, Theorem 11.20], there exist an irreducible FK-submodule U of V , a finite-dimensional vector space W , and projective representations ρ G → GL
U and π G → GL
W  such that V = U ⊗F W , g = gρ ⊗ gπ for every g ∈ G, and K ≤ Ker π. Since V is irreducible, W must be irreducible as an F
G/K-module by [17, Lemma 15.8]. By [14, I, Satz 16.10] and [5, Theorem 11.40], π may be assumed a genuine representation. Since G/K is cyclic, it follows that dimF W = 1 and U = V , so that V is irreducible as an FK-module. The second statement now follows directly from Clifford’s theorem.

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From this, we deduce the following criterion for a finite irreducible linear group to have more than one system of imprimitivity. 2.3. Lemma. Let V be a finite-dimensional vector space over an algebraically closed field F and let G be a finite irreducible subgroup of GL
V . Further, assume that K is a normal subgroup of G such that G/K is elementary abelian of order p2 for a prime p, and that V is reducible and homogeneous as an FK-module. If K < H < G, then the FH-homogeneous components of V form a system of imprimitivity of V for G which consists of p subspaces. Therefore, G has at least p + 1 such imprimitivity systems. Proof. Lemma 2.2, applied to K
H, yields that V is reducible as an FH-module. Therefore, Lemma 2.2, used with H
G, shows that the FHhomogeneous components V1   Vn of V form a system of imprimitivity of V for G. Clearly, we have H = NG
V1 , so that n = G  H = p. Now, assume that V1   Vp are also the FL-homogeneous components of V , where K < L < G, then L = NG
V1  = H, whence G has at least p + 1 systems of imprimitivity. As the following theorem shows, the situation in Lemma 2.3 is typical for the groups we intend to study. 2.4. Theorem. Let G be a finite irreducible subgroup of GL
V , where V is a finite-dimensional vector space over an algebraically closed field F. Let p be a prime and let  be the set of all systems of imprimitivity which consist of p subspaces of V which are permuted cyclically by G. If  contains more than one system of imprimitivity, and at least one of them is nonrefinable, then the following statements hold. (a) If V1   Vp  ∈  and H = NG
V1 , then H = NG
V2  = · · · = NG
Vp  is a normal subgroup of index p in G, and V = V1 ⊕ · · · ⊕ Vp is the decomposition of V into FH-homogeneous components. (b) If V1   Vp , W1   Wp  ∈  , then V1   Vp  = W1   Wp  if and only if NG
V1  = NG
W1 . (c) Let K denote the intersection of all NG
V1 , where V1   Vp  ∈  , then G/K is elementary abelian of order p2 , K acts irreducibly on every Vi , and V is homogeneous as an FK-module. (d)   = p + 1. (e) Char F = p. Proof. (a) Let H = NG
Vj , then CoreG
H is the kernel of the action of G on V1   Vp , hence has index p in G. Therefore, H = CoreG
H is a normal subgroup of G. By [5, Proposition 10.5], a proper nontrivial FH-submodule of one of the Vi would yield a proper nontrivial Gsubmodule of V , contradicting the irreducibility of V . Therefore, V1   Vp

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are irreducible FH-modules. Since V is inhomogeneous as an FH-module by Lemma 2.2, and by Clifford’s theorem, the FH-homogeneous components have equal dimension, it follows that V = V1 ⊕ · · · ⊕ Vp is the decomposition of V into FH-homogeneous components. (b) Assume that NG
V1  = NG
W1  = H. Since V has a unique decomposition into FH-homogeneous components, it follows from (a) that V1   Vp  = W1   Wp . (c) Let V1   Vp , W1   Wp  ∈  be distinct systems of imprimitivity, and assume that V1   Vp  is nonrefinable. If V1 were reducible as FK-module, where K = NG
V1  ∩ NG
W1 , then by Lemma 2.2 its homogeneous components would form a nontrivial system of imprimitivity of V1 for NG
V1 , thus giving rise to a refinement of V1   Vp . Thus, V1   Vp are irreducible as FK-modules. Since V = V1 ⊕ · · · ⊕ Vp and V = W1 ⊕ · · · ⊕ Wp are two distinct decompositions of V into irreducible FK-modules, the V1   Vp , W1   Wp must be isomorphic as FK-modules. Thus, it remains to show that NG
Ui  ≥ K for every U1   Up  ∈  . Assume that there exists U1   Up  ∈  such that K ≤ NG
U1  and let L = NG
V1  ∩ NG
U1  and N = NG
V1  ∩ NG
W1  ∩ NG
U1  = K ∩ L. Now, suppose that V1 is reducible as FN-module, then by Lemma 2.2, V1 admits a nontrivial decomposition V1 = T1 ⊕ · · · ⊕ Tp into homogeneous components for FN. Therefore, there exist k ∈ K and l ∈ L such that T1 = T2 k = T2 l. Let h = kl−1 ∈ KL = NG
V1 , then N is properly contained in H = h N because h ∈ N implies that k ∈ K ∩ L = N. Since NG
V1   H = p, it follows that H is the normaliser of T1 in NG
V1 . Therefore, T1   Tp  is a system of imprimitivity of V1 for NG
V1 , and so V1   Vp  is refinable. This contradiction shows that the subspaces V1   Vp are irreducible as FN-modules. Thus by [15, Sect. 5], there exist uniquely determined projective representations τ G → PGL
V1  and π G → PGL
p F such that the natural projective representation of G is equivalent to τ ⊗ π, N is contained in Ker π, and τ satisfies xg = xy for every x ∈ N, g ∈ G, and y ∈ GL
V1  such that y is a pre-image of gτ . By a similar argument, there exist uniquely determined projective representations τK  G → PGL
V1  and πK  G → PGL
p F such that the natural projective representation of G is equivalent to τK ⊗ πK , K ≤ Ker πK , and xg = xy for every x ∈ K, g ∈ G, and y ∈ GL
V1 , where y is a preimage of gτK . Since N is contained in K, the projective representations τK and πK satisfy the hypotheses on τ and π, and so by their uniqueness, we must have τK = τ and πK = π. Thus, K is contained in Ker π. By the same argument, we also have L ≤ Ker π and hence NG
V1  = KL ≤ Ker π. Thus, Im π is cyclic of order dividing p, hence is reducible by [14, I, Satz 16.10]

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and [5, Theorem 11.40], and by an argument similar to [17, Lemma 15.8], G is reducible, a final contradiction. (d) If V1   Vp  ∈  , then NG
V1  contains K by (c), and NG
V1  has index p in G by (a), whence   ≤ p + 1 by (b). On the other hand, we have   ≥ p + 1 by Lemma 2.3. (e) Let V1   Vp , τK  G → PGL
V1 , and πK  G → PGL
p F be as in (c), and assume that Char F = p. Since Im πK is a p-group, it follows from [5, Lemma 11.38] that the second cohomology group H 2
Im πK  F ×  is trivial, and so πK is induced by a (genuine) representation ρ G → GL
p F with K ≤ Ker ρ. Thus, the abelian group Im ρ is reducible, and so G is reducible by [17, Lemma 15.8]. This contradiction shows that Char F = p. Note that the proof of the previous theorem allows a weaker hypothesis: instead of supposing that  contains a nonrefinable system of imprimitivity, it suffices that  contains a system of imprimitivity which cannot be refined to a system of imprimitivity with p2 subspaces on which G acts as an elementary abelian group of order p2 . To avoid additional complications in what follows, from now on we only consider the case p = 2. We introduce the following notation. Let F be a field containing a primitive fourth root ω4 of one, let n be a positive integer, let L be a finite subgroup of GL
n F, and let x y ∈ NGL
n F
L, then we define        0 x y 0 l 0  G
L x y =   l ∈ L   x 0 0 ω24 y 0 l a subgroup of GL
2n F. Note that the additional conditions x2 ∈ L, y 2 ∈ L, and
ω4 xy2 ∈ L imposed ensure that the set K consisting of the matrices,   l 0  0 l where l ∈ L, form a normal subgroup of G
L x y whose factor group is elementary abelian of order 4. If G is a linear group with more than one system of imprimitivity, then by Clifford’s second theorem [4, Theorem 51.7] G has a tensor product decomposition as the one arising in the proof of Theorem 2.4. This observation will be used implicitly to show that G is linearly isomorphic with one of the groups G
L x y defined above; cf. also Remark 2.11 at the end of this section. 2.5. Proposition. Let n be a positive integer and let G be a finite irreducible subgroup of GL
V , where V is a vector space of dimension 2n over

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an algebraically closed field F. If U1  W1 , U2  W2 , and U3  W3 , are three distinct decompositions of V into at least one of which is nonrefinable, systems of imprimitivity for G, then Char F = 2, there exists a finite irreducible subgroup L of GL
n F and elements x and y in NGL
n F
L with x2 ∈ L, y 2 ∈ L, and
ω4 xy2 ∈ L, where ω4 is a primitive fourth root of one in F, such that at least one of L x, L y, and L xy is a primitive subgroup of GL
n F, and G is linearly isomorphic with the matrix group G
L x y. Proof. Put Hi = NG
Ui ,
i = 1 2 and let K = H1 ∩ H2 . By Theorem 2.4, G/K is elementary abelian of order 4 and U1 and W1 are isomorphic irreducible FK-modules. Let α U1 → W1 be such an FK-module isomorphism and choose a basis u1   un  of U1 , then  = u1   un  u1 α  un α is a basis of V . Let ρ be the map which associates to every element g ∈ G its matrix representation with respect to . If L denotes the matrix group obtained by restricting K ρ to U1 , then L is an irreducible subgroup of GL
n F, and   l 0 ρ  k = 0 l for every k ∈ K, where l ∈ L is the restriction of kρ to U1 . Let t1 ∈ G\H1 and t2 ∈ G\H2 , then Ui ti = Wi ,
i = 1 2. It follows that   0 x1 ρ t1 = x2 0 for some x1 , x2 ∈ GL
n F. Since t12 ∈ K, we have x1 x2 ∈ L. Moreover,
kt1 ρ ∈ K ρ for every k ∈ K, and so lx1 = lx2 ∈ L for every l ∈ L. This shows that x1 and x2 belong to NGL
n F
L and x1 x−1 2 ∈ CGL
n F
L. Since L is (absolutely) irreducible, it follows from Schur’s lemma that x1 x−1 2 is a scalar matrix λ · 1L , where λ ∈ F. Replacing α by α · µ, where µ ∈ F with µ2 = λ, we obtain x1 = x2 , where x1 ∈ NGL
n F
L and x21 ∈ L. Note that we may assume that t1 ∈ H2 and t2 ∈ H1 because K is properly contained in H1 and H2 , so that H1 = t2  K and H2 = t1  K. Therefore, we have   y1 0 ρ  t2 = 0 y2 where y1 , y2 ∈ GL
n F. Since t2 normalises K, it follows that y1 and y2 are contained in NGL
n F
L. Furthermore, we have ly1 = ly2 for every l ∈ L, and so y1 y2−1 ∈ CGL
n F
L is a scalar matrix. Since t22 ∈ K, it also follows that y12 = y22 . Thus, we must have y1 = νy2 , where ν ∈ F with ν 2 = 1. If y1 = y2 , then U1 and W1 are isomorphic FH1 -modules, and so G is reducible by Lemma 2.2. Thus, ν must be a primitive square root of one

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in F, which therefore cannot have characteristic 2 (this is also clear from Theorem 2.4(e)). Hence, y1 = ω24 y2 , where ω4 is a primitive fourth root of one in F. Let x = x1 , y = y1 . Since G/K is elementary abelian of exponent 2, we have  
ω4 xy2 0 =

t1 t2 2 ρ ∈ K ρ  0
ω4 xy2 and so
ω4 xy2 ∈ L, similarly, from
t12 ρ ∈ K ρ and
t22 ρ ∈ K ρ , we obtain x2 , y 2 ∈ L. If V1   Vr  is a system of imprimitivity of U1 for L y, then it is easy to see that V1   Vr  V1 t1   Vr t1  is a system of imprimtivity of V for G which refines U1  W1 . This shows that r = 1, and L y is primitive. Similar arguments show that a system of imprimitivity of L x or L xy leads to a refinement of U2  W2  and U3  W3 , respectively. Conversely, the next proposition shows that G
L x y has three nonrefinable imprimitivity systems if L, x, and y satisfy the conditions of Proposition 2.5. 2.6. Proposition. Let L be a finite irreducible subgroup of GL
n F, where n is a positive integer and where F is an algebraically closed field of characteristic = 2. Let ω4 be a primitive fourth root of one in F and x, y ∈ NGL
n F
L with x2 ∈ L, y 2 ∈ L, and
ω4 xy2 ∈ L, then the following statements hold for the group G = G
L x y. (a)

G is irreducible.

(b) If v1   v2n  is the canonical basis for the underlying vector space V of G, then 1 = v1   vn  vn+1   v2n  2 = v1 + vn+1   vn + v2n  v1 − vn+1   vn − v2n  3 = ω4 v1 + vn+1   ω4 vn + v2n  ω4 v1 − vn+1   ω4 vn − v2n  are systems of imprimitivity of V for G, whose stabilisers in G are             y 0 0 x l 0 l 0   H1 =   l ∈ L   l ∈ L  H2 = 0 ω24 y x 0 0 l 0 l  and

 H3 =

respectively.

0 ω24 xy

xy 0

  l  0

0 l

   l ∈ L  

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(c) If L y
L x L xy is primitive, then 1
2  3  is nonrefinable. In this case, the systems of imprimitivity in (b) are the only systems of imprimitivity of V for G consisting of two subspaces of V . Proof. (a) Put U = v1  vn , W = vn+1   v2n , and H = NG
U. If T is a proper nontrivial G-invariant subspace of V , then U + T =
U ⊕ W  ∩
U + T  = U ⊕
W ∩
U + T . Since L is irreducible, U and W are irreducible FH-modules, and so W ∩
U + T  = 0 or W ≤ U + T . If W ∩
U + T  = 0, then clearly U + T = U and so T ≤ U, which is impossible. Thus, W ≤ U + T and V = U ⊕ T . Similarly, we obtain V = W ⊕ T , so that U and W are isomorphic as FH-modules. Thus, there exists a matrix s ∈ GL
n F such that ls = l for every l ∈ L, and y s = ω24 y. By Schur’s lemma, s must be a scalar matrix, and we have y = ω24 y, which is impossible since Char F = 2. (b) It is straightforward to check that the stated decompositions of V are decompositions into systems of imprimitivity with the given normalisers in G
L x y. (c) Suppose that 1 has a refinement V1   V2r , and assume without loss of generality that V1   Vr are the subspaces contained in v1   vn . Then, it is easy to see that V1   Vr  is a system of imprimitivity for the restriction of H1 to v1   vn  which is linearly isomorphic with L y, and hence r = 1. Similarly, 2 and 3 are nonrefinable if L x and L xy, respectively, are primitive. Finally if one of 1 , 2 and 3 is nonrefinable, then by Theorem 2.4, G cannot have more than three systems of imprimitivity consisting of two subspaces of V each. To decide which of the groups in Proposition 2.6 are linearly isomorphic, we need the following lemmas, the first of which can be proved by direct verification. 2.7. Lemma. Let the notation be as in Proposition 2.6, and put     1L 0 1L 0 g= and h=

0 ω24 1L 1L 0 Then, j g = j h = j for j ∈ 1 2 3, where j t = Ut  U ∈ j , and moreover G
L x yg = G
L ω24 x y and G
L x yh = G
L x ω24 y. 2.8. Lemma. Let the notation be as in Proposition 2.6, and put s1 = y, s2 = x, and s3 = ω4 xy. Then, for every permutation π ∈ Sym
3, there exists tπ = GL
2n F such that the following statements hold. = L ω4 s1π  L ω4 s2π , (a) L ω4 s1  L ω4 s2  L s1 s2  L s1π s2π . (b) G
L s2  s1 tπ = G
L s2π −1  s1π −1 . (c) jπ = j tπ for j = 1 2 3, where j tπ = Utπ  U ∈ j 

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Proof. (a) It clearly suffices to verify the statement for the permutations
1 2 and
1 2 3 generating Sym
3. This, in turn, follows directly from the fact that x2 , y 2 , and
ω4 xy2 belong to L, so that Lω24 xy =
Lxy−1 = Lyx. (b)

Let t
1 2 =



1L 1L

1L ω24 1L



 and

t
2 3 =

ω4 1L 0

0 1L

 

then for any two elements a b ∈ GL
n F, conjugation by t
1 2 maps the generators used in the definition of G
L a b to those of G
L b a, so that G
L a bt
1 2 = G
L b a. Moreover,  t  t2 3   0 0 ω4 ab 0 a
2 3 b =

0 ω24 b ω4 ab a 0 0 From this, it follows easily that G
L a bt
2 3 = G
L ω4 ab b. In particular, t
 = 1GL
2n F , t
1 2 , and t
2 3 satisfy statement (b). Using these observations, one obtains that t
1 2 3 = t
2 3 t
1 2 conjugates G
L s2  s1  to G
L s1  s3 . Moreover, using the fact that Ls1 s2 = Lω4 s3 , the matrix t
1 2 t
2 3 conjugates G
L s2  s1  to G
L ω4 s1 s2  s2  = G
L ω24 s3  s2 , and consequently conjugation with t
1 2 t
2 3 t
1 2 maps G
L s2  s1  to G
L s2  ω24 s3 . Therefore, we put t
1 3 2 = t
1 2 t
2 3 g and t
1 3 = t
1 2 t
2 3 t
1 2 h, where g and h are the matrices defined in Lemma 2.7. (c) It is straightforward to verify that t
1 2 and t
2 3 satisfy the statement for π =
1 2 and π =
2 3, respectively. Since g and h fix the j by Lemma 2.7, the statement follows from the definition of the remaining tπ . Thus, we obtain the following criterion for two groups constructed as in Proposition 2.6 to be linearly isomorphic. 2.9. Proposition. Let L1 and L2 be irreducible subgroups of GL
n F, where n is a positive integer, let F be an algebraically closed field of characteristic = 2, and let ω4 be a primitive fourth root of one in F. Assume that for j = 1 2 xj  yj ∈ NGL
n F
Lj  with x2j  yj2 ∈ Lj and
ω4 xj yj 2 ∈ Lj such that at least one of Lj  xj , Lj  yj , and Lj  xj yj  is a primitive subgroup of GL
n F. Then, the groups G1 = G
L1  x1  y1  and G2 = G
L2  x2  y2  are linearly isomorphic if and only if there exists s ∈ GL
n F such that Ls1 = L2 and L1  ω4 x1  ω4 y1 s = L2  ω4 x2  ω4 y2 . Proof. Assume that Gt1 = G2 for some t ∈ GL
2n F. Let 1 , 2 , and 3 be as in Proposition 2.6, and for j = 1 2, let Hj 1 , Hj 2 , and Hj 3 denote the stabilisers in Gj of 1 , 2 , and 3 , respectively. Now, t maps the systems of imprimitivity of V for G1 bijectively to the systems of imprimitivity

¨ fling burkhard ho

430

for G2 . Since 1 , 2 , and 3 are the only imprimitivity systems of G1 and G2 by Theorem 2.4(d), t permutes 1 , 2 , and 3 , and by Lemma 2.8, there exists x ∈ GL
2n F and x3  y3 ∈ GL
n F such that tx fixes every i , Gx2 = G
L2  x3  y3 , and L2  ω4 x2  ω4 y2  = L2  ω4 x3  ω4 y3 . Replacing t by tx, and x2 and y2 by x3 and y3 , respectively, we may thus assume that j t = j for j = 1 2 3. Let h be as in Lemma 2.7, then we may also replace t by th and we may replace y2 by ω24 y2 ; hence we may assume without loss of generality that v1  vn t = v1   vn . Thus,   s1 0  t= 0 s2 t with s1  s2 ∈ GL
n F. Since H1 j = H2 j for j = 1 2 3, we have s s
H1 1 ∩ H1 2 t = H2 1 ∩ H2 2 . It follows that L11 = L12 = L2 and ls1 = ls2 for every l ∈ L1 . Since L1 is (absolutely) irreducible, by Schur’s lemma s1 and s2 only differ by a scalar, s2 = λs1 , say. Now, t   s  0 λx11 0 x1 t = ∈ H1 s 2 = H2 2  x1 0 λ−1 x11 0 s

s

and so λx11 = x2 l = λ−1 x11 for some l ∈ L2 . This yields λ = 1 or λ = ω24 . Put s = s1 , then it follows that
L1 ω4 x1 s = L2 ω4 x2 or t
L1 ω4 x1 s = L2 ω34 x2 = L2
ω4 x2 3 . Moreover, from H1 1 = H2 1 we obtain s s
L1 ω4 y1  = L2 ω4 y2 , so that we have L1  ω4 x1  ω4 y1  = L2  ω4 x2  ω4 y2 . Conversely, assume that there exists s ∈ GL
n F such that Ls1 = L2 and L1  ω4 x1  ω4 y1 s = L2  ω4 x2  ω4 y2 . In view of Lemma 2.8, we may assume without loss of generality that L1  ω4 x1 s = L2  ω4 x2  and L1  ω4 y1 s = L2  ω4 y2 . Thus, L1 x1 s = L2 x2 or
L1 x1 s = L2 ω24 x2 ; similarly
L1 y1 s = L2 y2 or
L1 y1 s = L2 ω24 y2 . Using Lemma 2.7, replacing G
L x2  y2  by a suitable conjugate, we may suppose that
L1 x1 s = L2 x2 and
L1 y1 s = L2 y2 . But then it is easy to see Gt1 = G2 , where   s 0 t= ∈ GL
2n F

0 s The results of Theorem 2.4 and Propositions 2.5, 2.6, and 2.9 can be summarised as follows. 2.10. Theorem. Let n be a positive integer, let F be an algebraically closed field of characteristic = 2, and let ω4 be a primitive fourth root of one in F. Let  be a complete set of representatives L of the conjugacy classes of finite irreducible subgroups of GL
n F and k ∈ 1 2 3. For every L ∈ , let L be a set of elements containing exactly one coset representative x of each element of order at most 2 in NGL
n F
L/L. Moreover, let L k denote the set of all
x y ∈ L × L such that exactly k of the subgroups L x, L y, and L xy are primitive.

finite nonmonomial linear groups

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Define an equivalence relation ∼ on the set L k by
x1  y1  ∼
x2  y2  if and only if L ω4 x1  ω4 y1  and L ω4 x2  ω4 y2  are conjugate in NGL
n F
L, and let L k be a set consisting of one representative of each equivalence class in L k . Then, the set, k = G
L x y  L ∈ 
x y ∈ L k  is a complete and irredundant set of conjugacy class representatives of the finite irreducible subgroups of GL
2n F which possess exactly three systems of imprimitivity consisting of two vector spaces, exactly k of which are nonrefinable. We conclude the section by observing that the classification in terms of the groups L ω4 x ω4 y in Theorem 2.10 is not so unnatural as it might seem at a first glance. 2 11. Remark. Let the notation be as in Proposition 2.6. As in the proof of Theorem 2.4, there exist projective representations π G → PGL
n F and ρ G → PGL
2 F, and Im ρ is elementary abelian of order 4. Choose a subgroup Q ≤ GL
2 F isomorphic with the quaternion group as a representation group of Im ρ and assume that Im ρ equals the projective image of Q, then π and ρ can be lifted to projective representations π ˆ G→ GL
n F and ρ ˆ G → Q. It then turns out that Im πˆ = L ω4 x ω4 y and
Ker ρπˆ = L. In fact, it seems that the results in Lemmas 2.7, 2.8, and Proposition 2.9 can be obtained from these observations, together with the fact that there exist automorphisms of Q mapping any pair of generators of Q to any other pair, and that all these automorphisms are linear because up to equivalence, Q only has one faithful two-dimensional representation over F. 3. GROUPS WITH ONE SYSTEM OF IMPRIMITIVITY For groups with just one system of imprimitivity, we choose a construction different from the one for groups with three imprimitivity systems. Let L be a subgroup of GL
n F for a positive integer n and a field F, let N be a normal subgroup of L, let α be an automorphism of L/N and l ∈ L, and define      0 l l1 0  α G
L N α l = l  l ∈ L
Nl1  = Nl2  1L 0 0 l2  1 2 ≤ GL
2n F

In the following, the parameters L, N, α, and l will always satisfy the 2 additional conditions
Nlα = Nl and
Nxα = Nxl for every x ∈ L, so

¨ fling burkhard ho

432 that the set,



l1 0

0 l2

    l1  l2 ∈ L
Nl1 α = Nl2 

is a subgroup of index 2 in G
L N α l. 3.1. Proposition. Let n be a positive integer, let V be a 2n-dimensional vector space over a field F, and let G be a finite irreducible subgroup of GL
V  for which V possesses a nonrefinable system of imprimitivity U W . Then, there exists a finite primitive subgroup L of GL
n F, a normal subgroup N of L, there exists an automorphism α of L/N, and an element l ∈ L 2 satisfying
Nlα = Nl and
Nxα = Nxl for every x ∈ L, such that G is linearly isomorphic with the group G
L N α l. Proof. Put H = NG
U = NG
W  and let t ∈ G\H (thus Ut = W and Wt = U). Let u1   un  be a basis of U, then  = u1   un , u1 t −1   un t −1  is a basis of V . Let ρ denote the map assigning to every g ∈ G its matrix representation with respect to . Define homomorphisms ρ1  ρ2  H → GL
n F by   ρ 0 h 1  hρ = 0 hρ2 and let L = H ρ1 . If U had a proper nontrivial H-invariant subspace, then by [5, Proposition 10.5], V would have a proper nontrivial G-invariant subspace, contradicting the irreducibility of V . This shows that L is irreducible, and since U W  is nonrefinable, L must be primitive. As
uj t −1 h =
uj ht t −1 for i = 1  n and every h ∈ H, it follows that hρ2 =
ht ρ1 , thus we also have L = H ρ2 . Let C = Ker ρ1 , then C t = Ker ρ2 . Since t 2 ∈ H, the subgroups CC t and N =
CC t ρ1 =
CC t ρ2 are normal in H and L, respectively. As hρ2 ∈ N if and only if hρ1 ∈ N, the map α L/N → L/N, defined by Nhρ1 → Nhρ2 is a well-defined automorphism of L/N. Furthermore, we have uj t = uj t −1 t 2 and uj t −1 t = uj for j = 1  n, so that     0 l l 0 ρ 2 ρ and
t  = t =  1L 0 0 l where l =
t 2 ρ1 . Since t 2 ∈ H, we have l ∈ L and
t 2 ρ2 =

t 2 t ρ1 = l, so that
Nlα = Nl. 2 Let x ∈ L and choose h ∈ H such that Nx = Nhρ1 . Then,
Nxα = 2
Nhρ2 α =
N
ht ρ1 α = N
ht ρ2 = N
ht ρ1 = Nxl . Constructing a group as in Proposition 3.1 does not necessarily lead to an irreducible linear group. However, we obtain the following criterion.

finite nonmonomial linear groups

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3.2. Proposition. Let L be a finite irreducible subgroup of GL
n F for some positive integer n and for an algebraically closed field F. Moreover, let N be a normal subgroup of L, and let α be an automorphism of L/N such that there exists an element l ∈ L satisfying
Nlα = Nl and 2
Nxα = Nxl for every x ∈ L. Then, G = G
L N α l ≤ GL
2n F is reducible if and only if N = 1 and α is a linear automorphism of L. Otherwise, if v1   v2n  is a basis for the canonical underlying vector space of G, then v1   vn  vn+1   v2n  is a system of imprimitivity for G, which is nonrefinable if L is primitive. Proof. Let U = v1   vn  and let W = vn+1   v2n . Denote by H the stabiliser of U in G, then H is a subgroup of G of index 2. Moreover, U and W are irreducible as FH-modules because L is irreducible. Assume first that N = 1 and that α is a linear automorphism of L, then V is homogeneous as an FH-module. Therefore, G is reducible by Lemma 2.2. Conversely, if G is reducible, then an argument as in the proof of Proposition 2.6(a) shows that U and W are isomorphic as FH-modules. Hence, N = 1 and there exists y ∈ GL
n F such that xα = xy for every x ∈ L. Thus, α is a linear automorphism of L. Now, assume that G is irreducible, then by Lemma 2.1, U W  is a system of imprimitivity for G. Moreover, the representation of H obtained by restriction to U is equivalent to the natural representation of L. Therefore, if  is a refinement of U W , then the elements of  contained in U give rise to an imprimitivity system for L. Thus, if L is primitive, then U W  is nonrefinable. If two groups in Proposition 3.1 have exactly one system of imprimitivity with two elements, the following proposition allows us to decide whether these groups are linearly isomorphic. 3.3. Proposition. Let n be a positive integer and let F be a field. For i = 1 2, let Li be primitive subgroups of GL
n F, Ni  Li , αi ∈ Aut
Li /Ni , 2 and li ∈ Li such that
Ni li αi = Ni li and
Ni xαi = Ni xli for every x ∈ Li , and put Gi = G
Li  Ni  αi  li  acting on a 2n-dimensional F-vector space V = v1   v2n . Assume that the Gi are irreducible, and that V = v1   vn  ⊕ vn+1   v2n  is the unique decomposition of V into systems of imprimitivity with two elements for G1 and G2 . Then, G1 = Gx2 for some x in GL
2n F if y y and only if there exist y ∈ GL
n F and h ∈ L1 such that L1 = L2 , N1 = N2 , −1
N2 lα2 =
N2 ly α1 hy for every l ∈ L2 , and N2 l2 =

N1 l1 
N1 hα1
N1 hy . Proof.

Assume that Gx1 = G2 for some x ∈ GL
2n F, then clearly, V = v1   vn x ⊕ vn+1   v2n x

434

¨ fling burkhard ho

is a decomposition of V into systems of imprimitivity for G2 , and so v1   vn x = v1   vn  or v1   vn x = vn+1   v2n . Since v1   vn g = vn+1   v2n  for some g ∈ G1 , we may assume without loss of generality that v1   vn x = v1   vn  and vn+1   v2n x = vn+1   v2n . Thus, x has the form,   y 0 0 hy for suitable matrices y, h ∈ GL
n F. Let a ∈ L1 and choose b ∈ L1 such that
N1 aα1 = N1 b, then we have  x a 0 ∈ G2  0 b and so
N1 ayα2 =
N2 ay α2 = N2 bhy =
N1 aα1 hy . Furthermore, the matrix,   x    −1 0 l1 0 l2 0 y l1 hy  = 0 y −1 h−1 yl2 1L 0 1L 0 belongs to G2 . Thus,
l1 hy ∈ L2 and
N2
l1 hy α2 =
N2
h−1 y l2 ; note that the first statement implies that h ∈ L1 . Now, N2 l2 = N2 hy
N2
h−1 y l2  =
N1 hy
N1 l1 hyα2 =
N1 hy
N1 l1 hα1 hy =

N1 l1 α1
N1 hα1
N1 hy =

N1 l1 
N1 hα1
N1 hy  as required. Conversely, assume that there exist y ∈ GL
n F and h ∈ L1 such that −1 y y L1 = L2 , N1 = N2 ,
N2 lα2 =
N2 ly α1 hy for every l ∈ L2 , and N2 l2 = α1 y

N1 l1 
N1 h
N1 h . Using the above arguments, it is easy to see that Gx1 ≤ G2 , where   y 0 x=  0 hy and it follows that G1 and G2 are linearly isomorphic. To apply Proposition 3.3, we have to decide which of the irreducible groups in Proposition 3.2 actually have exactly one nonrefinable system of imprimitivity with two elements. This can be achieved by comparing the description obtained in Proposition 3.2 with the one obtained in Proposition 2.5.

finite nonmonomial linear groups

435

3.4. Proposition. Let L be a finite primitive subgroup of GL
n F, where n is a positive integer and where F is an algebraically closed field. Assume that N is a normal subgroup of L and that α is an automorphism of L/N such 2 that there exists an element l ∈ L satisfying
Nlα = Nl and
Nyα = Ny l for every y ∈ L. If G = G
L N α l is irreducible, then G has more than one system of imprimitivity consisting of two subspaces if and only if one of the following holds. (1) Char F = 2, N = 1, and there exist an α-invariant subgroup K of index 2 in L and x ∈ NGL
n F
K such that y α = y x for every y ∈ K; (2) Char F = 2, N = −1L , and there exists x ∈ NGL
n F
L such that
Nyα = Ny x for all y ∈ L. Proof. Assume that (1) holds. By proposition 3.2, α cannot be a linear automorphism of L, so that K consists of all y ∈ L such that y α = y x . In particular, we have l ∈ K. Now, it is easy to check that the subgroup,    y 0  y∈K  R= 0 yx  of G is normal in G, and that G/R is elementary abelian of order 4. Since K is irreducible by Lemma 2.2, the underlying vector space V of G is homogeneous as an FR-module. Therefore, the result follows from Lemma 2.3. Similarly, if N and α satisfy (2), then    y 0  y∈L R= 0 yx  is a normal subgroup of G, G/R is elementary abelian of order 4, and V is a homogeneous FR-module. Therefore as above, by Lemma 2.3, V has more than one system of imprimitivity for G consisting of two two-dimensional subspaces. Conversely, assume that G has more than one system of imprimitivity consisting of two subspaces of V , where V = v1  v2   v2n  is the underlying vector space of G, and let H = NG
v1   vn , so that G  H = 2. By Theorem 2.4, we have Char F = 2 and G contains a normal subgroup R ≤ H with G/R elementary abelian of order 4, such that the subspaces v1   vn  and vn+1   v2n  are isomorphic irreducible FR-modules. Thus, there exists an irreducible subgroup K of L and x ∈ GL
n F such that

 k 0   k ∈ K

R= x  0 k As R is normal in G, it contains the element     x  0 l k 0 k 0 1L = 1L 0 0 l−1 0 0 kx

0 kl

 

436

¨ fling burkhard ho

which shows that x belongs to NGL
n F
K. Now, let   l1 0 ∈ H\R

h= 0 l2 Since h normalises R, it follows that kl1 x = kxl2 for every k ∈ K, and hence l1x l2−1 ∈ CGL
n F
K is a scalar by Schur’s lemma. Since h2 ∈ R, we also have l22 =
l12 x , and thus l2 = l1x or l2 = −l1x . Moreover, R = K and H = LN; consequently L  KN = H  R = 2. Thus, if L  K = 2 and N = 1, we have kα = kx for every k ∈ K, and (1) holds. Otherwise, we have K = L and N = 2. Then, l2 = −l1x because h does not belong to R. On the other hand, one has Nl1x =
Nl1 α = Nl2 . Therefore, −1L ∈ N, and (2) holds. The above results now lead to our classification theorem. We adopt the following notation. Let G be a group with subgroups L and N with N  L, then NG
L/N denotes the set of all g ∈ G such that
Nlg = Nl for every l ∈ L; thus NG
L/N = NGL
n F
L ∩ NGL
n F
N. 3.5. Theorem. Let n be a positive integer, let F be an algebraically closed field, and let  be a complete set of representatives of the conjugacy classes of finite primitive subgroups of GL
n F. For every L ∈ , let L denote a complete set of representatives of the NGL
n F
L-conjugacy classes of normal subgroups of L. Now, for every N ∈ L , and α, β ∈ Aut
L/N, define α ∼ β if and only if there exist y ∈ NGL
n F
L/N and h ∈ L such that
Nlα = −1
Nly βhy for every l ∈ L. Choose one representative from each equivalence class modulo ∼, and let L N consist of those representatives α which satisfy the following conditions. (1) α2 is an inner automorphism of L/N; (2) if N = 1, α is not a linear automorphism of L; (3) if Char F = 2 and N = 1, then the restriction of α to any α-invariant subgroup K of index 2 in L is not a linear automorphism of K; (4) if Char F = 2 and N = −1L , then α is not induced by a linear automorphism of L. If α ∈ L N , let L N α contain exactly one representative l ∈ L of every 2 coset Nl ∈ L/N which satisfies
Nlα = Nl and
Nxα = Nxl for every x ∈ L. Then, the set,  = G
L N α l  L ∈  N ∈ L  α ∈ L N  l ∈ L N α  contains exactly one representative of each conjugacy class of finite irreducible subgroups of GL
2n F which have precisely one nonrefinable system of imprimitivity consisting of n-dimensional vector spaces.

finite nonmonomial linear groups

437

Proof. By Propositions 3.2 and 3.4, clearly every group in  has the properties in question. Now, let G be an irreducible imprimitive subgroup of GL
2n F which has a unique system of imprimitivity consisting of ndimensional subspaces. We show that there exists exactly one group G∗ ∈  which is linearly isomorphic with G. By Proposition 3.1, there exist a primitive subgroup L of GL
n F, N  L, α ∈ Aut
L/N, and l ∈ L such that G is linearly isomorphic with G
L N α l; thus it suffices to show that G
L N α l is linearly isomorphic with exactly one group in  . By the definition of , there exists exactly one L∗ ∈  which is linearly isomorphic with L; let y ∈ GL
n F be such that Ly = L∗ . Since N y is a normal subgroup of L∗ , there exists a unique N ∗ ∈ L∗ such that N yn = N ∗ for some n ∈ NGL
n F
L∗ . Replacing y by yn, we may thus assume that N y = N ∗ . Now, the map β∗ defined by −1 N ∗ x →
N ∗ xy αy for every x ∈ L∗ is an automorphism of L∗ /N ∗ whose square is inner. In view of Propositions 3.2 and 3.4, β∗ satisfies conditions (1)–(4), and so there exists a uniquely determined automorphism α∗ ∈ L∗  N ∗ and y0 ∈ ∗ ∗ NGL
n F
L∗ /N ∗  and h0 ∈ L∗ such that
N ∗ xy0 α =
N ∗ xβ y0 h0 for every ∗ x ∈ L. Replacing y by yy0 , we thus obtain that
Nxyα =
Nxαhy for every y −1 x ∈ L, where h = h0 . Let l∗ ∈ L∗ be such that N ∗ l∗ =

Nl
Nhα
Nhy . Then, ∗


N ∗ l∗ α =

Nl
Nhα
Nhyα

∗

=

Nl
Nhα
Nhαhy 2

=

Nlα
Nhα
Nhα hy =

Nl
Nhl 
Nhα hy =

Nl
Nhα
Nhy = N ∗ l∗

Thus, there exists a unique l∗∗ ∈ L∗  N ∗  α∗ such that N ∗ l∗∗ = N ∗ l∗ , and we may assume that l∗ = l∗∗ . But now Proposition 3.3 shows that G is linearly isomorphic with G
L∗  N ∗  α∗  l∗ . Since L∗ ∈ , N ∗ ∈ L∗ , α∗ ∈ L∗  N ∗ , and l∗ ∈ L∗  N ∗  α∗ , are uniquely determined by G, the set  cannot obtain two linearly isomorphic groups. 3.6. Remark. Let the notation be as in Theorem 3.5, and for every x ∈ GL
n F normalising L/N, let x denote the automorphism of L/N induced by conjugation with x. Let T be a transversal of LCGL
n F
L/N in NGL
n F
L/N, then it is easy to see that the equivalence class of α ∈ Aut
L/N with respect to the relation ∼ is  
αht  t ∈ T h ∈ L = Inn
L/Nαt

t∈T

438

¨ fling burkhard ho

Note also that if l l∗ ∈ L N α , then Nl−1 l∗ belongs to the centre of L/N. Thus, if Nz1   Nzr  ⊆ L/N is the set of all fixed points of Z
L/N under the action of α, then one may choose L N α = lz1  lz2   lzr 

4. NORMAL SUBGROUPS AND AUTOMORPHISMS OF CENTRAL PRODUCTS In this section, we collect some auxiliary results which will prove useful for determining the normal subgroups and automorphisms of factor groups of primitive linear groups. Our first well-known lemma motivates the hypotheses of the following technical results. 4.1. Lemma. Let G be a finite subgroup of GL
n F, where n is a positive integer and let F be an algebraically closed field. Let Z = Z
GL
n F and let S = GZ ∩ SL
n F, then S is finite and there exists a finite subgroup C of Z such that G  SC. In particular, the imprimitivity systems of S are precisely the imprimitivity systems of G, two elements in G are conjugate if and only if they are conjugate in SC, and every normal subgroup of G is normal in SC. In view of the previous lemma, the following result will enable us to compute the normal subgroups of the finite primitive subgroups of GL
2
 if the normal subgroups of the finite primitive subgroups of SL
2
 are known. Note that the lists described in Proposition 4.2(a) and (b) need neither be irredundant nor be disjoint, and that C0 and S0 need not equal the intersections of C and S with the arising normal subgroup of G. 4.2. Proposition. Let the finite group G be the product of a cyclic central subgroup C and of a normal subgroup S. Then, the normal subgroups of G are precisely the following. (a) all the products C0 S0 , where C0 is a subgroup of C and S0 is a normal subgroup of S; (b) all the subgroups of the form cs C0  S0 , where C0 and S0 are normal subgroups of C and S, respectively, c ∈ C\C0 , s ∈ S\S0 , and sS0 /S0 is a cyclic central factor of S which has the same order as cC0 /C0 . Proof. Since C is a central subgroup of G, it is easy to see that the subgroups described above are normal subgroups of G. Conversely, let N be a normal subgroup of G, then also X = CN ∩ SN is a normal subgroup of G. By repeated application of the modular law, we obtain that X = CN ∩ SC ∩ SN =
CN ∩ SC ∩ SN =
CN ∩ S
C ∩ SN

finite nonmonomial linear groups

439

moreover, X =
CN ∩ SN =
C ∩ SNN

The factor group N/N ∩ S ∼ = NS/S is isomorphic with a section of C, hence is cyclic, and so N = n N ∩ S N ∩ C for some n ∈ N. Since N is contained in X, there exist s ∈ CN ∩ S and c ∈ C ∩ SN such that n = cs−1 . If s ∈ N, then it follows that c ∈ N and so n ∈
N ∩ S
N ∩ C, and N is of type (a). Thus, we may assume that s ∈ S ∩ N, and that c ∈ C ∩ N. By the isomorphism theorems, we have
C ∩ SN/
C ∩ N ∼ =
C ∩ SNN/N =
CN ∩ SN/N ∼ =
CN ∩ S/
N ∩ S and since the above isomorphisms are G-isomorphisms,
CN ∩ S/
N ∩ S is a central section of S. Moreover, the first isomorphism maps
C ∩ Nc to Nc = Ns, which is mapped to
N ∩ Ss by the last isomorphism. Therefore, it is easy to see that the groups are generated by
N ∩ Cc, Nc = Ns, and
N ∩ Ss, respectively. To compute the automorphism group of a central product G = Z
GS, it is sometimes convenient to replace S by a larger subgroup E of G which is characteristic in G. Indeed, such a subgroup E is not hard to find. 4.3. Lemma. Suppose that the finite group G is the product of a characteristic central subgroup Z and of a subgroup S. If Z ∩ S is characteristic in Z and k denotes the greatest common divisor of exp
Z/Z ∩ S and exp
S/S  , then the subgroup E = S z  z ∈ Z, z k ∈ S is characteristic in G. Proof. Let α ∈ Aut
G and denote β the natural isomorphism from G/S = ZS/S to Z/Z ∩ S, then δ S → Z/Z ∩ S, defined by sδ =
Ssα β for every s ∈ S, is clearly a homomorphism. Let s ∈ S, then sα = tz with t ∈ S and z ∈ Z; thus sδ =
Z ∩ Sz. Since Im δ is abelian, it follows that S  ≤ Ker δ. Put n = exp
S/S  , then sn ∈ S  ≤ Ker δ, and so z n ∈ S. Thus,
sn α =
tzn = t n z n ∈ S. Since also z m ∈ S, where m = exp
Z/Z ∩ S, it follows that z k ∈ S and so sα ∈ E. Moreover, if z ∈ Z ∩ E, then
z α k =
z k α ∈
S ∩ Zα = S ∩ Z because S ∩ Z is characteristic in Z, hence in G. Thus, z α ∈ E and E = S
E ∩ Z is characteristic in G. The automorphism group of a central product of two characteristic subgroups can easily be described in terms of the automorphism groups of the factors. We will only need the following special case. 4.4. Lemma. Let the finite group G be the product of its characteristic subgroups E and Z, and assume that Z ≤ Z
G is cyclic of order n.

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¨ fling burkhard ho

(a) Let α be an automorphism of G, then there exist a unique automorphism η of E and a unique positive integer k ≤ n with
k n = 1 such that
ezα = eη z k for every e ∈ E, z ∈ Z, and eη = ek for every e ∈ E ∩ Z. (b) Conversely, if η is an automorphism of E and k is a positive integer with
k n = 1 such that eη = ek for every e ∈ E ∩ Z, then the map α, defined by
ezα = eη z k for every e ∈ E, z ∈ Z, is an automorphism of G. (c) Assume that α, η, and k are as in (b). Then, α2 is an inner automorphism of G if and only if k2 ≡ 1
mod n and η2 is an inner automorphism of E. (d) Let α, η, and k be as in (b). Let e ∈ E, t ∈ , and assume that Z = z, then the following statements are equivalent. t 2 (1)
ez t α = ez t and gez = gα for all g ∈ G; 2 (2) eη = ez
1−kt for some integer k with k2 ≡ 1
mod n, and xe = xη for all x ∈ E. Proof. (a) Since E and Z are characteristic in G, the restrictions of α to E and Z are automorphisms of E and Z, respectively, and since Z is cyclic, we have z α = z k for some integer k with
n k = 1. Let η be the restriction of α to E, then clearly
ezα = eη z k for every e ∈ E, z ∈ Z. Moreover, if e ∈ E ∩ Z, then eη = eα = ek . (b) Since eη = ek for every e ∈ E ∩ Z, the map α is well defined; since Z is central, α is a homomorphism of G. Since α is obviously surjective, it follows that α is an isomorphism, as required. 2 2 (c) If α2 is an inner automorphism of G, then z k = z α = z for 2 every z ∈ Z; thus k ≡ 1
mod n. Moreover, there exist e1 ∈ E and z1 ∈ Z 2 2 such that ge1 z1 = gα for every g ∈ G, and so eη = ee1 z1 = ee1 for every 2 e ∈ E. Consequently, η is an inner automorphism of E. Conversely, if 2 there exists e1 ∈ E such that eη = ee1 for every e ∈ E and k2 ≡ 1
mod n, e1 e1 α2 then
ez = e z =
ez for every e ∈ E, z ∈ Z, and so α2 is an inner automorphism of G. (d) Assume (1). Then, ez t =
ez t α = eη z tk by (b), and so eη = t −kt ez z = ez
1−kt . Since α2 is inner, we have k2 ≡ 1
mod n by (c). 2 t η2 Also, x = xα = xez = xe for every x ∈ E. Conversely, if (2) holds, then
ez t α = eη z tk = ez
1−kt z tk = ez t . If g = xz r ∈ G, where x ∈ E and 2 2 2 t r ∈ , then gα = xη z rk = xe z r =
xz r e =
xz r ez , as required. In the case when a finite primitive subgroup N of GL
2
 properly contains the subgroup Z
N
N ∩ SL
2
 or when N ∩ SL
2
 is not characteristic in N, the next result can often be used to compute its automorphisms. 4.5. Proposition. Let the group G be a central product of its subgroups Z and S such that Z ≤ Z
G and every automorphism of S/S ∩ Z is induced by

finite nonmonomial linear groups

441

an automorphism of S. Assume that N is a normal subgroup of G such that G = NZ and Z ∩ N is a characteristic subgroup of N. Then, the following statements hold. (a) Let α ∈ Aut
N and ζ ∈ Aut
Z such that α and ζ coincide on N ∩ Z. Then, there exist γ ∈ Hom
S Z and σ ∈ Aut
S such that z ζ = z γ z σ for every z ∈ S ∩ Z. Moreover, the map β, defined by
zsβ = z ζ sγ sσ for all z ∈ Z, s ∈ S, is a well-defined automorphism of G whose restriction on N coincides with α. In particular, if every automorphism of Z ∩ N can be extended to an automorphism of Z, then every automorphism of N can be extended to an automorphism of G. (b) If N = G and S/S  and S ∩ Z have coprime orders, then ζ, γ, and σ in (a) are uniquely defined by α. In this case, α is an inner automorphism of G if and only if ζ is the identity automorphism of Z, sγ = 1 for every s ∈S, and σ is an inner automorphism of S. Proof. Since N/N ∩ Z ∼ = G/Z ∼ = S/S ∩ Z, the automorphism α induces an automorphism on S/S ∩ Z; let σ be an automorphism of S which induces the same automorphism on S/S ∩ Z. Thus, if z z1 ∈ Z, s s1 ∈ S with zs z1 s1 ∈ N and
zsα = z1 s1 , then
S ∩ Zsσ =
S ∩ Zs1 . It follows that
zsα s−σ ∈ Z for every z ∈ Z, s ∈ S with zs ∈ N. Let s ∈ S and choose z z1 ∈ Z such that zs z1 s ∈ N. As z −1 z1 ∈ N ∩ Z ≤ Z
G, it follows that the map γ S → Z, defined by sγ =
zsα s−σ z −ζ does not depend on the choice of z. Since Im γ and Im ζ are contained in Z
G, the map β zs → z ζ sγ sσ is, in fact, an automorphism of G, and it is clear from the definition of γ that the restriction of β to N coincides with α. This proves (a). Now, assume that G = N and that S/S  and S ∩ Z have coprime orders. Let σ1 ∈ Aut
S, γ1 ∈ Hom
S Z, and let ζ1 ∈ Aut
Z such that sσ sγ z ζ = sσ1 sγ1 z ζ1 for all s ∈ S, z ∈ Z. Taking s = 1, we obtain ζ = ζ1 , and thus s−σ1 sσ = sγ1 −γ ∈ S ∩ Z for all s ∈ S. Since S  ≤ Ker
γ1 − γ and
S  S   S ∩ Z = 1, it follows that γ = γ1 and consequently also σ = σ1 . If α is an inner automorphism, then α acts like conjugation with an element of S because G = SZ, in particular, S α = S. Thus, Im γ is contained in Z ∩ S and it follows from
S  S   S ∩ Z = 1 that γ = 1. Therefore, sα = sσ for every s ∈ S and ζ is the identity of Z. 5. FINITE PRIMITIVE SUBGROUPS OF GL
2
 AND THEIR NORMAL SUBGROUPS For the rest of article, we will fix the following notation. For every positive integer n, let ωn be a primitive nth root of one in
, such that for every positive integer m, ωm mn = ωn . Moreover, adopting the notation used in the

¨ fling burkhard ho

442

classification of the finite primitive subgroups of SL
2
 in [1, Chap. III], we define the following matrices,       1 ω4 − 1 ω4 − 1 0 ω8 0 ω4  s=   u= w1 = 0 ω−1 0 ω−1 2 ω4 +1 −ω4 − 1 4 8   1 −
ω45 + ω5  + ω4
ω35 + ω25  ω4 v=

3 4 2 −ω4 2 ω5 + ω5 + ω4
ω5 + ω5  Note that w1 , s, u, and v have determinant 1 and orders 4, 3, 8, and 4, respectively; moreover, they satisfy the following relations: w12 = v2 , w1 w1s = 2 w1s , u2 = w1 , su =
w1 s2 = s2 w1−1 , sv = s2 , and
w1 v3 = 1GL
2
 . For every positive integer n, let L1 n = ωn 1GL
2
  and if n is even, define L2 n = w1  w1s  ωn 1GL
2
  L3 n = w1  s ωn 1GL
2
  L3∗  n = w1  ω3n s L3∗∗  n = w1  w3n s2  = Lu3∗  n  L4 n = w1  s u ωn 1GL
2
  L4∗  n = w1  s ω2n u L5 n = w1  s v ωn 1GL
2
 

Using the relations among their generating matrices given above, it is straightforward to check that the groups L2 2 L3 2 , L4 2 , and L5 2 are isomorphic with the quaternion group, SL
2 3, the binary octahedral group, and SL
2 5, and that each of the groups L3 2 , L4 2 , and L5 2 possesses a unique chief series, namely, 1 = L1 1
L1 2
L2 2
L3 2  1 = L1 1
L1 2
L2 2
L3 2
L4 2  1 = L1 1
L1 2
L5 2

Thus, the central factor groups of L3 2 , L4 2 , and L5 2 , are isomorphic with Alt(4), Sym(4), and Alt(5), respectively. The group L4∗  2 is isomorphic with GL
2 3, and the central factor groups of L3∗  2 and L4∗  2 are Alt(4) and Sym(4). We will frequently refer to these facts without giving an explicit reference.

finite nonmonomial linear groups

443

We also introduce the following sets of subgroups of L3 n , L3∗  n , L4 n , L4∗  n , and L5 n , respectively. Define as follows, 3 n = L1 m  m divides n ∪ L2 m  L3 m  m divides n m even ∪ L3∗  m  3m divides n m even  3∗  n = L1 m  m divides n ∪ L2 m  m divides n m even ∪ L3∗ m  m divides n m even n/m ≡ 1
mod 3 ∪ L3∗∗  m  m divides n m even n/m ≡ 2
mod 3 4 n = L1 m  m divides n ∪ L2 m  L3 m  L4 m  m divides n m even ∪ L4∗  m  2m divides n m even 4∗  n = L1 m  m divides n ∪ L2 m  L3 m  m divides n m even

∪ L4∗  m  m divides n m even n/m ≡ 1
mod 2  5 n = L1 m  m divides n ∪ L5 m  m divides n m even

We can now state the classification theorem for finite primitive subgroups of GL
2
. 5.1. Theorem. (a) Let j j ∗ ∈ 1 2 3 3∗  4 4∗  5, n n∗ ∈ , and assume that n is even if j = 1, and that n∗ is even if j ∗ = 1. Then, Lj n and Lj ∗  n∗ are conjugate in GL
2
 if and only if j = j ∗ and n = n∗ . (b)  = L3 n  L3∗  n  L4 n  L4∗  n  L5 n  n ∈  n even is an irredundant list of representatives of the conjugacy classes of finite primitive subgroups of GL
2
. (c) NSL
2

L3 n  = NSL
2

L4 n  = NSL
2

L4∗  n  = L4 2 , NSL
2

L3∗  n  = L3 2 , and NSL
2

L5n  = L5 2 ; moreover, Z
L3 n  = Z
L3∗  n  = Z
L4 n  = Z
L4∗  n  = Z
L5n  = L1 n for every even n ∈ . (d) If n ∈  is even and j ∈ 3 3∗  4 4∗  5, then j n is a complete and irredundant list of representatives of the NGL
2

Lj n -conjugacy classes of normal subgroups of Lj n . Proof. (a) Suppose that Lj n and Lj ∗  n∗ are conjugate in GL
2
. If N is a normal subgroup of GL
2
, then also Lj n ∩ N and Lj ∗  n∗ ∩ N are conjugate in GL
2
. Thus, if Z = Z
GL
2
, we obtain that Lj n ∩ Z = L1 n and Lj ∗  n∗ ∩ Z = L1 n∗ are conjugate. As these groups have orders n and n∗ , respectively, it follows that n = n∗ . Moreover, also ZLj n ∩ SL
2
 and ZLj ∗  n∗ ∩ SL
2
 are conjugate in GL
2
. Now, ZLj n ∩ SL
2
 = Lj2 for j ∈ 1 2 3 4 5, and ZL3∗  n ∩ SL
2
 = L3 2 and ZL4∗  n ∩ SL
2
 = L4 2 . This leaves the cases j j ∗  ⊆ 3 3∗  and j j ∗  ⊆ 4 4∗ . Now, L3 n ∩ SL
2
 = L3 2

444

¨ fling burkhard ho

and L3∗  n ∩ SL
2
 = L2 2 , so that these groups cannot be conjugate in GL
2
. Similarly, L4 n ∩ SL
2
 = L4 2 and L4∗  n ∩ SL
2
 = L3 2 . This proves that also j = j ∗ . (b) Let L be a finite primitive subgroup of GL
2
, then by Lemma 4.1, L is a normal subgroup of SC, where S is a finite primitive subgroup of SL
2
 and C a finite group of scalar matrices in GL
2
. Now, by [1, Chap. III] (see also [6, Sect. 8.5]), up to conjugacy, L3 2 , L4 2 and L5 2 are the only finite primitive subgroups of SL
2
. Thus, replacing G by a suitable conjugate in GL
2
, we may assume that L is a normal subgroup of Lj n for some n ∈ , where j ∈ 3 4 5. To determine the normal subgroups of Lj n , observe that L1 2 /L1 1 and L3 2 /L2 2 are the only central chief factors of L3 2 , that L1 2 /L1 1 and L4 2 /L3 2 are those of L42 , and that L1 2 /L1 1 is the only central chief factor of L5 2 . Now, Proposition 4.2 with S = Lj2 and C = L1 n can be used to compute the normal subgroups of Lj n . If j = 3 and n is not divisible by 3, this yields the set j n . If n is divisible by 3, then the normal subgroups of L3 n consist of the groups in 3 n together with the groups L3∗∗ m = Lu3∗ m , where m is even and where 3m divides n. In the cases j = 4 and j = 5, we obtain the sets 4 n and 5 n of normal subgroups of L4 n and L5 n , respectively. Since the groups L1 n
n ∈  are reducible and the L2 n
n ∈ , n even, are monomial, every finite primitive subgroup of GL
2
 is conjugate to a group in . Conversely, since ZL3 n ∩ SL
2
 = ZL3∗  n ∩ SL
2
 = L3 2 , ZL4 n ∩ SL
2
 = ZL4∗  n ∩ SL
2
 = L4 2 , and ZL5 n ∩ SL
2
 = L5 2 , every group in  is irreducible and primitive by Lemma 4.1. The above list  is irredundant by (a). (c) Let j ∈ 3 3∗  4 4∗  5, let n be a positive even integer, L = Lj n , and S = NSL
2

L. By Schur’s lemma CSL
2

L = Z ∩ SL
2
 = L1 2 . Now, S/S ∩ Z is isomorphic with a subgroup of Aut
L which is finite since L is finite. This proves that S is finite. In the case when j ∈ 3 4 4∗ , it is easy to see that u, and hence L4 2 , normalises L. Thus, L42 ≤ S, and since L4 2 is maximal among the finite primitive subgroups of SL
2
, we have S = L4 2 . A similar argument shows that NSL
2

L5 n  = L5 2 . Since s normalises L3∗  n , it follows that L3 2 is contained in NSL
2

L3∗  n , and since every element of SL
2
 which normalises L3∗  n also normalises L3 3n = ω3n 1GL
2
  L3∗  n , we have S = L3 2 or S = L4 2 . Assume the latter, then ω3n su = ω3n s2 w1−1 ∈ L3∗  n , and so also ω3n 1GL
2
 ∈ L3∗  n . This contradiction shows that u is not contained in NGL
2

L3∗  n , whence S = L3 2 . The statement about the centres follows directly from Schur’s lemma.

finite nonmonomial linear groups

445

(d) It follows from (a) that for every even n ∈  and j ∈ 3 3∗  4 4∗  5, the lists j n contain at most one representative of each NGL
2

Lj n -conjugacy class of normal subgroups of Lj n . Therefore, by the proof of (b), for every even n ∈ , the lists 4 n and 5 n contain precisely one such representative. A similar result holds for 3 n , observing that in the case when 3m divides n, the subgroups L3∗  m and L3∗∗ m = Lu3∗ m are conjugate in NGL
2

L3 n  because u ∈ L42 ≤ NGL
2

L3 n  by (c). Obviously, every normal subgroup of L3∗  n is normal in L33n ; similarly for L4∗  n and L42n . This shows that 3∗  n and 4∗  n are the lists of normal subgroups of L3∗  n and L4∗  n , respectively. 5.2. Corollary. (a) Let P be a finite primitive subgroup of GL
2
. Then, every subgroup L of P of index 2 is likewise primitive. (b) Let G be a finite subgroup G of GL
4
 which has more than one imprimitivity system consisting of two-dimensional spaces. If at least one of these systems is nonrefinable, then all are. Proof. The first statement follows directly from statements (d) and (b) of Theorem 5.1. Therefore, using the notations of Theorem 2.10, if
x y ∈ Lk , then L itself is primitive. But then all of L x, L y, and L xy are primitive, and k = 3. Therefore, L1 and L2 are empty, which proves the second statement. In view of Corollary 5.2, the next lemma contains the remaining information required in Theorem 2.10 to determine the finite subgroups of GL
4
 having more than one imprimitivity system consisting of twodimensional subspaces. 5.3. Lemma. Assume that G = GL
2
 and let j ∈ 3 3∗  4 4∗  5, let n ∈  be an even integer, and put  1G  ω2n 1G  u ω2n u if j = 3, j n = 1G  ω2n 1G  if j = 3, then each set j n consists of exactly one representative x of every coset Lx ∈ NG
L/L with
Lx2 = L, where L = Lj n . Moreover, if x y ∈ j n , then
ω4 xy2 = ω24 1G is contained in Lj n . Define an equivalence relation ∼ on j n × j n by
x1  y1  ∼
x2  y2  if and only if L ω4 x1  ω4 y1  and L ω4 x2  ω4 y2  are conjugate in NGL
2

L, then the following sets,   
1G  1G 
ω2n 1G  ω2n 1G 
1G  u if j = 3,
ω2n u ω2n u
ω2n 1G  u j n =  
1G  1G 
ω2n 1G  ω2n 1G  if j = 3 contain exactly one representative
x y of each equivalence class in j n × j n .

¨ fling burkhard ho

446

TABLE I Equivalence Classes for L3 n , n ≡ 2
mod 4
x y ∈ 3 n × 3 n
ω2n 1G  ω2n 1G 
1G  1G 
1G  ω2n 1G 
ω2n 1G  u
u u
ω2n 1G  ω2n u,
ω2n u ω2n u
1G  u
1G  ω2n u
u ω2n u

L ω4 x ω4 y L = L3 n L ω2n 1G  = L3 2n L ω2n u = L4∗  n L u = L4 n L ω2n 1G  u = L4 2n

Proof. The transversals j n can be obtained directly from the results in Theorem 5.1(c). We determine the equivalence classes of the tuples in j n × j n . Assume first that L = L3 n . If 4 does not divide Z
L = n, then ω4 1G ∈ L. Since ω2n  ω4 L/L is cyclic and ω24 1G , ω22n 1G ∈ L, we have Lω4 1G = Lω2n 1G . The first column of Table I shows the elements of the resulting equivalence classes (but contains only one of the elements
x y and
y x), the second column contains the corresponding group L ω4 x ω4 y. Note that the groups in the second column are nonconjugate in NGL
2

L by Theorem 5.1(a). If 4 divides n, then ω4 1G ∈ L, and so L ω4 x ω4 y = L x y. Thus, we obtain the equivalence classes and corresponding subgroups of L shown in Table II. Thus, in both cases, we may choose j n = 
1G  1G 
ω2n 1G  ω2n 1G 
1G  u
ω2n u ω2n u
ω2n 1G  u

If j ∈ 3∗  4 4∗  5 and L = Lj n , the case n ≡ 2 (mod 4) yields Table III. If 4 divides n, then we obtain the results shown in Table IV. This justifies our choice of j n . 5.4. Remark. While groups Lj n with distinct indices are nonconjugate in GL
2
, they may be (abstractly) isomorphic. For example, if n is even TABLE II Equivalence Classes for L3 n , n ≡ 0
mod 4
x y ∈ 3 n × 3 n
1G  1G 
1G  ω2n 1G ,
ω2n 1G  ω2n 1G 
1G  u
u u
1G  ω2n u
ω2n u ω2n u
ω2n 1G  u,
ω2n 1G  ω2n u,
u ω2n u

L ω4 x ω4 y L = L3 n L ω2n 1G  = L3 2n L u = L4 n L ω2n u = L4∗  n L ω2n 1G  u = L4 2n

finite nonmonomial linear groups

447

TABLE III Equivalence Classes for Lj n , j = 3 n ≡ 2
mod 4
x y ∈ 3 n × 3 n
ω2n 1G  ω2n 1G 
1G  1G ,
1G  ω2n 1G 

L ω4 x ω4 y L = Lj n L ω2n 1G  = Lj 2n

and is not divisible by 3, then the map w1 → w1 , ω3n s → ωn s extends to an isomorphism from L3∗  n to L3 n . If n is divisible by 6, the groups L3 n and L3∗  n cannot be isomorphic because ω3n s has order 3n while L3 2 has exponent 12 and therefore L3 n has exponent n if n is divisible by 12, and has exponent 2n if n ≡ 6 (mod 12).

6. AUTOMORPHISMS OF QUOTIENTS OF L3 n AND L3∗  n In this section, we collect the data about automorphisms of L/N and certain fixed points required in Theorem 3.5, where L = L3 n or L3∗  n . To describe the automorphisms whose square is inner in a convenient way, we introduce the following sets of integers, where k t ∈ . Let K
t = k ∈   1 ≤ k ≤ t k2 ≡ 1
mod t and C
t k = l ∈   0 ≤ l ≤ t − 1
k − 1l ≡ 0
mod t

(Note that K
t corresponds to the set of all automorphisms of order ≤ 2 of the cyclic group /t , and that C
t k is a set of representatives in of the fixed points of endomorphism of /t defined by t + 1 → t + k. Moreover, C
t k = rt/m  0 ≤ r < m, where m is the greatest common divisor of t and k − 1.) Let n ∈  be an even integer, j ∈ 1 2 3 3∗  m t ∈  such that n = mt, and assume that m is even if j = 1 and that 3m divides n if j = 3∗ . Put TABLE IV Equivalence Classes for Lj n , j = 3 n ≡ 0
mod 4
x y ∈ 3 n × 3 n
1G  1G 
1G  ω2n 1G ,
ω2n 1G  ω2n 1G 

L ω4 x ω4 y L = Lj n L ω2n 1G  = Lj 2n

¨ fling burkhard ho

448

L = L3 n and N = Lj m , and for every k ∈ K
t, define automorphisms of L/N and sets of fixed point representatives by 1 k

α3 n j m  Nωn 1L → Nωkn 1L  Nw1 → Nw1  Ns → Ns  l ωn 1L  l ∈ C
t k if j = 1 3 or 3∗ , 1 k 3 n j m = l l l 2 ωn 1L  ωn s ωn s  l ∈ C
t k if j = 2. If j ∈ 1 2, define 2 k

α3 n j m  Nωn 1L → Nωkn 1L 

Nw1 → Nw1 

Ns → Nsu 

and 2 k

3 n j m = ωln w1  l ∈ C
t k moreover if j ∈ 1 2 and t ≡ 0
mod 3, let 3 k

α3 n j m  Nωn 1L → Nωkn 1L   t/3 Nωn su Ns → t/3 Nωn s

Nw1 → Nw1  if k ≡ 1
mod 3, if k ≡ 2
mod 3

3∗  k

α3 n j m  Nωn 1L → Nωkn 1L  Nw1 → Nw1   2t/3 Nωn su if k ≡ 1
mod 3, Ns → 2t/3 Nωn s if k ≡ 2
mod 3 and

3 k

3 n j m

 l ωn w1  l ∈ C
t k      ωln 1L  l ∈ C
t k = ωln 1L  l ∈ C
t k  l+2t/3   ωl 1  ω s   n L l+t/3n 2 ωn s  l ∈ C
t k

if j = 1 and k ≡ 1
mod 3 if j = 1 and k ≡ 2
mod 3 if j = 2 and k ≡ 1
mod 3 if j = 2 and k ≡ 2
mod 3

If j = 2 and t ≡ 0
mod 3, let 4 k

α3 n 2 m  Nωn 1L → Nωkn s 4∗  k

α3 n 2 m  Nωn 1L → Nωkn s2  and 4 k

3 n 2 m =



Ns → Ns2k  Ns → Ns2k 

ωln s2l  l ∈ C
t k if k ≡ 1
mod 3, ωln 1L  ωln s ωln s2  l ∈ C
t k if k ≡ 2
mod 3.

finite nonmonomial linear groups

449

If j = 2 and t ≡ 0
mod 9, let 5 k

k
1+t/3

5∗  k

k
1+t/3 2

6 k

k
1−t/3

s

2k Ns → Nω2t/3 n s 

6∗  k

k
1−t/3 2

2k Ns → Nωt/3 n s 

α3 n 2 m  Nωn 1L → Nωn α3 n 2 m  Nωn 1L → Nωn

α3 n 2 m  Nωn 1L → Nωn

5 k 3 n 2 m

 = 

6 k

3 n 2 m =

2k Ns → Nω2t/3 n s 

s 

α3 n 2 m  Nωn 1L → Nωn

and

2k Ns → Nωt/3 n s 

s

s 

if k ≡ 1
mod 3, ωln s2l  l ∈ C
t k l+2t/3 l+t/3 s ωn s2  l ∈ C
t k if k ≡ 2
mod 3, ωln 1L  ωn if k ≡ 1
mod 3, ωln s2l  l ∈ C
t k l+t/3 l+2t/3 2 ωln 1L  ωn s ωn s  l ∈ C
t k if k ≡ 2
mod 3.

Finally, if j = 2 and t ≡ 3
mod 9, let 7 k

α3 n 2 m  Nωn 1L → Nωnk+2kt and

2

/9 kt/3

s



Ns → Nωnkt

2

/9

1L 

   7 k  sk
l+t/3  ωl+2t/3 sk
l+2t/3  l ∈ C 3t  k

3 n 2 m = ωln skl  ωl+t/3 n n

If t is divisible by 3, let I = 1 2 3, otherwise let I = 1 2, and define  i k  α3 n 1 1  i ∈ I k ∈ K
n     k ∈ 1 n2 + 1 if i ≤ 2 if m = 1 n ≡ 0
mod 4,    i k   α3 n 1 1  i ∈ I k ∈ K
n

3 n 1 m = k = 1 if i ≤ 2 if m = 1 n ≡ 2
mod 4,  i k  n  α  i ∈ I k ∈ K
  3 n 1 2 2    k = 1 if i ≤ 2 if m = 2,    αi k  i ∈ I k ∈ K
t if m > 2, 3 n 1 m  i k α3 n 2 m  i ∈ I k ∈ K
t if t ≡ 0
mod 3,      αi3kn 2 m  i ∈ 1  4 7

3 n 2 m = k ∈ K
t if t ≡ 3 or 6
mod 9,   i k  α  i ∈ 1



 6  3 n 2 m  k ∈ K
t if t ≡ 0
mod 9, 1 k

3 n 3 m = α3 n 3 m  k ∈ K
t 1 k

3 n 3∗  m = α3 n 3∗  m  k ∈ K
t

The above data contain, indeed, the information required to apply Theorem 3.5.

450

¨ fling burkhard ho

6.1. Theorem. Let n ∈  be an even integer and assume that n = mt, where m t ∈  and let j ∈ 1 2 3 3∗ . If j = 1, assume further that m is even, and if j = 3∗ , assume that 3m divides n. Then, L N = 3 n j m is irredundant and satisfies the conditions of Theorem 3.5, where L = L3 n i k i k and N = Lj m . Moreover, if α = α3 n j m ∈ 3 n j m , then 3 n j m , contains 2 exactly one representative x ∈ L of each coset Nx such that
Nlα = Nlx for every l ∈ L and
Nxα = Nx. Proof. Let L = SZ, where S = L3 2 , and Z = Z
L = ωn 1L  is cyclic of order n. Moreover, it follows from Theorem 5.1 that NSL
2

L = L4 2 and hence NSL
2

L/N ≤ L4 2 for every normal subgroup N of L. As a consequence, we have NSL
2

L/N = L4 2 if j ∈ 1 2 3, and NSL
2

L/N = L3 2 if j = 3∗ . Thus, if j ∈ 1 2 3, by Remark 3.6 the equivalence class of α ∈ Aut
L/N modulo the equivalence relation ∼ defined in Theorem 3.5 is Inn
L/Nα ∪ Inn
L/Nαu¯ , where u¯ is the automorphism induced on L/N by conjugation with u. Otherwise, if j = 3∗ , two automorphisms of L/N are related via ∼ if and only if they coincide modulo the inner automorphism group of L/N. Now, consider the case j = 1. Since N = m, we have N ∩ S = 1 if m is odd and N ∩ S = S ∩ Z = Z
S if m is even. As far as possible, we will treat both cases simultaneously. It is easy to check that the outer automorphism groups of S/S ∩ Z ∼ = Alt
4 and S have order 2, a nontrivial element being induced by conjugation with u. It follows that every automorphism of S/S ∩ Z is induced by a unique automorphism of S. Observe that SN ∩ Z =
S ∩ ZN, so that SN/N ∩ Z/N has order at most 2. Moreover,
SN/N = S  N/N has index 3 in S/N. Thus, Proposition 4.5 with L/N in place of both G and N can be applied to determine Aut
L/N. Therefore, if α ∈ Aut
L/N, there exist uniquely defined automorphisms σ ∈ Aut
S γ ∈ Hom
S Z/N (observe that N ∩ S ≤ Z
S ≤ S  ) and a unique integer k satisfying
k t = 1 and 0 ≤ k < t such that
Nxzα = Nxσ xγ z k for every x ∈ S z ∈ Z. Since 2 1 2 1 1 α3 2 1 1  is a transversal of Inn
S in Aut
S, where α3 2 1 1 arises by conjugation with u, the set of all automorphisms of the form Nxz → Nxxγ z k or Nxz → Nxu xγ z k for all x ∈ S, z ∈ Z, where γ ∈ Hom
S Z/N and k ∈ 1  t such that
k t = 1, is a transversal of Inn
L/N in Aut
L/N. Now assume that α2 is an inner automorphism of L/N =
SN/N
Z/N, 2 then there exists y ∈ S such that N
xzy =
Nxσ xγ z k α = Nxσ xσγ
xγ z k k = 2 2 Nxσ
xσ xk γ z k for all x ∈ S z ∈ Z. In particular,
xσ xk γ ∈ SN/N ∩ Z/N ≤ Z
SN/N for every x ∈ S. Since SN/N ∼ = S or S/N ∼ = Alt
4, its centre Z
SN/N has order at most 2. Thus, the subgroup of S consisting of all elements x with
xσ xk γ = N/N has index at most 2 in S. Since 2 2 S does not have subgroups of index 2, we have N
xzy = Nxσ z k for all

finite nonmonomial linear groups

451

x ∈ S Z ∈ Z. Hence, σ 2 is an inner automorphism of S and k2 ≡ 1
mod t. Clearly, these conditions are also sufficient for α2 to be an inner automorphism of L/N. 1 k 2 k Thus, the automorphisms of type α3 n 1 m and α3 n 1 m form a complete list of representatives α of elements of Out
L/N of order ≤ 2 for which 1 k 2 k Ker γ = S. Note that the automorphisms of types α3 n 1 m and α3 n 1 m are  invariant under conjugation with u. ¯ If exp
S/S  = 3 does not divide t, then it follows that Im γ = 1 for every automorphism α, and so in fact the above list is complete. Now, assume that t = 3r for some r ∈  and that α is an automorphism such that Ker γ = S. Then, S  S   = 3 forces Ker γ = S  , and so sγ = Nωrn 1L , or sγ = Nω2r 1L . As noted above,
sσ sk γ = N/N, and so we obtain 3 k 3∗  k the automorphisms of types α3 n 1 m and α3 n 1 m ; it is easily checked that ∗ 3 k 3 k α3 n 1 m = u¯ −1 α3 n 1 m u¯ for all k. 1 k 3 k The sets 3 n 1 m   3 n 1 m can be computed easily by observing 1 k that for every k ∈ K
t
α3 n 1 m 2 is the identity automorphism and 2 k 3 k
α3 n 1 m 2 = u¯ 2 =
α3 n 1 m 2 . Now, it is straightforward to determine the automorphisms of L/N which do not satisfy one of the conditions (2)–(4) of Theorem 3.5. Observe that L only has a normal subgroup of index 2, namely, L3 n/2 , if n is divisible by 4. Next, let j = 2, then L/N is the direct product of the cyclic groups ZN/N = Nωn 1L  of order t with SN/N = Ns of order 3, where Z = Z
L and S = L3 2 . Thus, L/N does not have nontrivial inner automorphisms, and we have to compute the automorphisms of order ≤ 2 and their fixed point sets. In the case
3 t = 1, the automorphism group is just a direct product of 1 k Aut
ZN/N with Aut
SN/N and we obtain a list consisting of the α3 n 2 m 2 k and α3 n 2 m . If t = 3r with r ∈ , let α ∈ Aut
L/N. Then,
Nωn 1L α =
Nωn 1L a
Nsb with 0 ≤ a ≤ t − 1 and 0 ≤ b ≤ 2, and, observing that
Ns3 = 1, we have
Nsα =
Nωn 1L rc
Nsd with 0 ≤ c d ≤ 2. Thus, it follows that α ∈ Aut
L/N with α2 = 1 if and only if a2 + rbc ≡ 1
mod t, b
a + d ≡ c
a + d ≡ 0
mod 3, and d 2 + rbc ≡ 1
mod 3. The fixed points α are precisely the elements
Nωn 1L x
Nsy , 0 ≤ x ≤ t − 1, 0 ≤ y ≤ 2, where x and y satisfy the equations,
a − 1x + rcy ≡ 0
mod t bx +
d − 1y ≡ 0
mod 3


∗
∗∗ 1 k

If we choose b = c = 0, d = 1, we obtain the automorphisms α3 n 2 m , 2 k and b = c = 0, d = 2 leads to α3 n 2 m , where k is used in place of a.

452

¨ fling burkhard ho

In the case b = 0 and c = 0, we have a ∈ K
t, and since c
a + d ≡ 0
mod 3, it follows that d ≡ −a
mod 3. The possibilities c = 1 and c = 2 3 k 3∗  k thus lead to the automorphisms α3 n 2 m and α3 n 2 m , where k = a. Note 3∗  k 3 k that
α3 n 2 m u¯ = α3 n 2 m . 3 k To compute the fixed points of α = α3 n 2 m , observe that y ≡ 0
mod 3 if k ≡ 1
mod 3, and thus the fixed points are the elements
Nωn 1L x , where x ∈ C
t k. Otherwise, we have k ≡ 2
mod 3. Thus, Eq.
∗∗ does not impose any restriction on x and y. By a hypothesis on k, we have r ≡
k − 1r
mod t, and so Eq.
∗ is equivalent to
k − 1
x + ry ≡ 0
mod t. Hence, given y, the solutions for
∗ are the x with x ≡ l + 3 k 2ry
mod t for some l ∈ C
t k. This leads to 3 n 2 m . The choice b = 0, c = 0 yields a = k ∈ K
t and d ≡ 2k
mod 3, and 4 k 4∗  k so α = α3 n 2 m if b = 1, and α = α3 n 2 m if b = 2, respectively. Again, ∗ 4 k 4 k 4 k u¯ −1 α3 n 2 m u¯ = α3 n 2 m . Therefore, assume that α = α3 n 2 m . Then, Eq.
∗ above is satisfied if and only if x ∈ C
t k, and by
∗∗, we have 4 k x +
2k − 1y ≡ 0
mod 3. Thus justifies our choice for 3 n 2 m . If r ≡ 0
mod 3, then two further classes of automorphisms arise. If bc ≡ 1
mod 3, then a2 ≡ 1 − r ≡
1 + r2
mod t, and since 1 + r and t are coprime, the last equation is satisfied if and only if a ≡ k
1 + r
mod t for some k ∈ K
t. Thus, if b = c = 1, we have d ≡ 2k
mod 3, and we 5 k obtain the automorphisms α3 n 2 m , similarly if b = c = 2, the resulting ∗ 5 k 5 k 5∗  k automorphisms are the α3 n 2 m ; observe that u¯ −1 α3 n 2 m u¯ = α3 n 2 m for every k. 5 k In the case α = α3 n 2 m , Eqs.
∗ and
∗∗ become
k + kr − 1x + ry ≡ 0
mod t and x −
k + 1y ≡ 0
mod 3, respectively. In particular, we have
k − 1x ≡ 0
mod r. Now, assume that k ≡ 1
mod 3 then x ≡ 2y
mod 3, by
∗∗, and
∗ becomes
k − 1x ≡ 0
mod t. Therefore, the solutions are x ∈ C
t k, y ≡ 2x
mod 3. If k ≡ 2
mod 3, Eq.
∗∗ yields x ≡ 0
mod 3 but does not restrict the possibilities for y. In particular, since x ≡ 0
mod 3 and k − 1 ≡ 1
mod 3, Eq.
∗ is equivalent with
k − 1
x + ry ≡ 0
mod t. Hence, x ≡ l + 2ry
mod t for some l ∈ 5 k C
t k, and we obtain 3 n 2 m . In a similar way, the case bc ≡ 2
mod 3 yields a ≡ k
1 − r
mod t for any k ∈ K
t. If b = 1, c = 2, it follows that d ≡ −k
mod 3, and we obtain 6 k 6∗  k the α3 n 2 m , similarly if b = 2, c = 1, the α3 n 2 m arise. Again, for every ∗ 6 k 6 k 6 k k α3 n 2 m is conjugate to α3 n 2 m via u. ¯ If α = α3 n 2 m , Eqs.
∗ and
∗∗ become
k − kr − 1x + 2ry ≡ 0
mod t and x −
k + 1y ≡ 0
mod 3, respectively. Thus, if k ≡ 1
mod 3, we have 2y ≡ x
mod 3, and so
∗ becomes
k − 1x ≡ 0
mod t. Thus, the solutions are x ∈ C
t k, y ≡ 2x
mod 3. If k ≡ 2
mod 3, we have x ≡ 0
mod 3, and Eq.
∗ is equivalent with
k − 1
x + 2ry ≡ 0
mod t. This justifies our claim about

finite nonmonomial linear groups

453

6 k

3 n 2 m . Since we exhausted all possibilities for b and c, the list 3 n 2 m is complete if r ≡ 0
mod 3. If
r 3 = 1, then we must have rbc ≡ 1
mod 3, because otherwise d 2 ≡ 2
mod 3. Therefore, d ≡ a ≡ 0
mod 3 and a2 ≡ 1
mod r. To obtain a similar parametrisation as for the previous automorphisms, observe that
a + rc2 ≡ 1
mod t and so a + rc ≡ k
mod t for a unique k ∈ K
t. Conversely, if k ∈ K
t, let rc ≡ b ≡ k
mod 3 and let a ≡ k
1 + 2r 2 
mod t, then a ≡ 0
mod 3, a2 ≡ 1
mod r, and rbc ≡ 1
mod 3. We thus obtain the last type of automorphism in the set 3 n 2 m , namely, the 7 k α3 n 2 m . Using r 2 ≡ 1
mod 3, we conclude that rc ≡ r 3 c ≡ r 2 k
mod t, and so Eqs.
∗ and
∗∗ are equivalent with
k + 2kr 2 − 1x + kr 2 y ≡ 0
mod t and kx ≡ y
mod 3, respectively. Considering the first equation modulo r, we obtain x ≡ l
mod r for some l ∈ C
r k, and so we must have x ≡ l
mod t, x ≡ l + r
mod t, or x ≡ l + 2r
mod t. Conversely, if x ≡ l
mod t, x ≡ l + r
mod t, or x ≡ l + 2r
mod t for some l ∈ C
r k and y ≡ kx
mod 3, then x and y solve
∗ modulo r and modulo 3, and hence 7 k modulo t. Since the pairs
x y also solve
∗∗, we obtain 3 n 2 m . Finally, if j = 3 or j = 3∗ , then L/N is cyclic of order t and generated by Nωn 1L , and so the result follows. The desired information about automorphisms of factor groups of L3∗  n can now be obtained from our previous results using Proposition 4.5. 6.2. Theorem. Let n ∈  be even and assume that n = mt with m t ∈  and j ∈ 1 2 3∗  3∗∗ . If j ∈ 2 3∗  3∗∗ , assume in addition that m is even. Moreover, assume that t ≡ 1
mod 3 if j = 3∗ , and that t ≡ 2
mod 3 if j = 3∗∗ . Put L = L3∗  n and N = Lj m  L, and for every k ∈ K
3t, define  Nωk3n s if k ≡ 1
mod 3, 1 k α3∗  n j m  Nw1 → Nw1  Nω3n s → if k ≡ 2
mod 3, Nωk3n su and 1 k 3∗  n j m

 l   ωn 1L  l ∈ C
t k  ωln w1  l ∈ C
t k = ωl sl  l ∈ C
3t k    3n ωln 1L  l ∈ C
t k

if if if if

Moreover, define the following sets  1 k α3∗  n 1 1  k ∈ K
3n      + 1 k ∈ 1 3n  2 1 k

3∗  n 1 m = α3∗  n 1 1  k ∈ K
3n k = 1  1 k    k ∈ K
3n  k = 1 α ∗  2   31 kn 1 2 α3∗  n 1 m  k ∈ K
3t

j j j j

= 1, k ≡ 1
mod 3, = 1, k ≡ 2
mod 3, = 2, = 3∗ or j = 3∗∗ .

if if if if

m = 1 n ≡ 0
mod 4, m = 1 n ≡ 2
mod 4, m = 2, m > 2,

¨ fling burkhard ho

454 1 k

3∗  n 2 m = α3∗  n 2 m  k ∈ K
3t 1 k

3∗  n 3∗  m = α3∗  n 3∗  m  k ∈ K
t 1 k

3∗  n 3∗∗  m = α3∗  n 3∗∗  m  k ∈ K
t of automorphisms of L/N. Let L = 3∗  n , and for every N = Lj m ∈ L , let L N = 3∗  n j m ; 1 k 1 k moreover for every α = α3∗  n j m ∈ 3∗  n j m , let L N α = 3∗  n j m . Then, the sets L N and L N α satisfy the hypotheses of Theorem 3.5. Proof. Let G = L3 3n , S = L3 2 and Z = Z
G = ω3n 1G . Since NSL
2

L = S by Theorem 5.1, every automorphism of L induced by GL
2
 is inner. Thus, by Remark 3.6 it suffices to list representatives for all elements of Out
L/N of order at most 2. Assume first that j = 1. To apply Proposition 4.5 with G/N SN/N Z/N, and L/N in place of G S Z, and N, respectively, we have to show that every automorphism of SN/SN ∩ Z is induced by an automorphism of SN/N. Since SN ∩ Z =
S ∩ ZN by the modular law, we have SN/SN ∩ Z ∼ = S/
S ∩ Z
S ∩ N = S/S ∩ Z ∼ = Alt(4). Now, Out
S has 2 1 order 2, a representative of a nontrivial element being α3 2 1 1 , and so every automorphism of S/S ∩ Z is induced by an automorphism of S. In particular, every automorphism of SN/SN ∩ Z is induced by an automorphism of SN/N. Now, let α ∈ Aut
L/N such that α2 is an inner automorphism of L/N. We show that α can be extended to an automorphism β of G/N such that β2 is an inner automorphism of G/N. Since L/N ∩ Z/N is cyclic of order t, there exists an integer k with k2 ≡ 1
mod t such that
Nzα = Nz k for every z ∈ L ∩ Z. Let t = 3ν r with integers ν ≥ 0 and r with
3 r = 1. Since k2 − 1 =
k + 1
k − 1 is divisible by 3ν , we have k ≡ 1
mod 3ν  or k ≡ −1
mod 3ν . In the first case, choose an integer l with l ≡ k
mod r and l ≡ 1
mod 3ν+1 , otherwise choose l with l ≡ k
mod r and l ≡ −1
mod 3ν+1  (note that this is always possible by the Chinese remainder theorem). Then, l ≡ k
mod t and l2 ≡ 1
mod 3t, whence ζ, defined by Nz ζ = Nz l for all z ∈ Z, is an automorphism of Z/N which coincides with α on L/N ∩ Z/N. Thus, by Proposition 4.5, there exist σ ∈ Aut
S/N and γ ∈ Hom
SN/N Z/N such that the automorphism β of G/N, defined by
Nxzβ =
Nxσ
Nxγ
Nzl for every x ∈ S, z ∈ Z, coincides with α on L/N. Clearly, ζ is the restriction of β to Z/N. Let g ∈ G such that 2
Nxα = Nxg for every x ∈ L. Since every h ∈ G can be written h = xz 2 2 2 2 with x ∈ L and z ∈ Z, we have
Nhβ =
Nxzβ =
Nxα Nz ζ = Nxg Nz = N
xzg = Nhg , and so β is the required automorphism of G/N.

finite nonmonomial linear groups

455

Hence, a complete and irredundant list of representatives of elements of order at most 2 of Out
L/N can be obtained by choosing those automor1 k 2 k 3 k 3∗  k phisms among the α3 3n 1 m , α3 3n 1 m , α3 3n 1 m , α3 3n 1 m which fix L/N. 1 k 2 k It turns out that α3 3n 1 m fixes L/N if k ≡ 1
mod 3 and that α3 3n 1 m 1 k 1 k fixes L/N if k ≡ 2
mod 3. We therefore put α3∗  n 1 m = α3 3n 1 m if k ≡ 1 k 2 k 1
mod 3, and α3∗  n 1 m = α3 3n 1 m if k ≡ 2
mod 3. The relevant fixed 1 k 2 k point sets are, of course, L ∩ 3 3n 1 m if k ≡ 1
mod 3, and L ∩ 3 3n 1 m if ∗ 3 k 3 k k ≡ 2
mod 3. The automorphisms of type α3 3n 1 m and α3 3n 1 m fix L/N if and only if k ≡ 1
mod 3 and k ≡ 2
mod 3, respectively. In this case, we 3 k 1 l have α3 3n 1 m = α3∗  n 1 m , where l is an integer with l ≡ k + t
mod 3t and ∗ 3 k 1 l 0 ≤ l < 3t, and α3 3n 1 m = α3∗  n 1 m , where l satisfies l ≡ k + 2t
mod 3t and 0 ≤ l < 3t. This completes our list in the case j = 1. 1 1 1 1 Clearly, the automorphisms α3∗  n 1 1 and α3∗  n 1 2 are the only automorphisms in our list not satisfying conditions (2) and (4), respectively, of Theorem 3.5. Since S does not have subgroups of index 2, L does not have a normal subgroup of index 2 if n is not divisible by 4, and has exactly one such normal subgroup, namely, L3∗  n/2 , if n is divisible by 4. In the latter 1 3n/2+1 case, the only automorphism not satisfying Theorem 3.5 (3) is α3∗  n 1 1 . If j = 2, the result follows from the fact that L/N = Nω3n s is cyclic of order 3t and Nsu = Ns2 ; similarly, if j = 3∗ or j = 3∗∗ , the factor group L/N = Nωn 1L  is cyclic of order t. 7. AUTOMORPHISMS OF QUOTIENTS OF L4 n , L4∗  n , AND L5 n We now determine information similar to that in Section 6 for the remaining finite primitive subgroups L4 n , L4∗  n , and L5 n of GL
2
. The following lemma will be useful when computing automorphisms of L4 n and L4∗  n . 7.1. Lemma. For every even n ∈  and for every even divisor m of n, the subgroups L3 m and L4 m are characteristic subgroups of L4 n . Moreover, if 2m divides n, then also L4∗  m is characteristic in L4 n . Proof. As a first step, we show that S = L4 2 is characteristic in L4 n . By Lemma 4.3, this is the case if n is not divisible by 4, and L4 4 is characteristic in L if n is divisible by 4. Thus, it remains to show that S is characteristic in L = L4 4 . Now, L = S  = L3 2 and thus L/L is elementary abelian of order 4. Since L is soluble, the maximal normal subgroups of L are S = L4 2 , L4∗  2 , and L3 4 , whose respective centers have size 2, 2, and 4. Furthermore, S does not split over its derived subgroup L3 2 because the only element of order 2 is central, while the subgroup generated by ω4 w1s u complements the derived subgroup L3 2 of L4∗  2 . It follows

¨ fling burkhard ho

456

that the maximal normal subgroups of L are even nonisomorphic, and S must be characteristic in L. Now, put G = L4 n and observe that L4 m = L4 2
Z
G ∩ L4 m  and L3 m = L3 2
Z
G ∩ L3 m . Since L4 2 and L3 2 =
L4 2  are characteristic in G and Z
G is cyclic, it follows that L3 m and L4 m are characteristic in G. Finally, assume that 2m divides n and consider the characteristic subgroup G = L4 2m of L4 n . Clearly, L4 m , L3 2m , and L4∗  m are normal subgroups of index 2 in G, and any two of them intersect in the characteristic subgroup N = L3 m of G. Since G/N is elementary abelian of order 4 and L4 m and L3 2m are characteristic in G, the same must be true for L4∗  m . We now introduce the data required for Theorem 3.5 in the case when L = L4 n . Let n ∈  be an even integer and m t ∈  such that n = mt. Let j ∈ 1 2 3 4 4∗  and assume that m is even if j = 1, and that t is even if j = 4∗ . Let N = Lj m  L and for every k ∈ K
t, define the following automorphisms of L/N and sets of representatives of fixed points. Let 1 k

α4 n j m  Nωn 1L → Nωkn 1L  and



1 k

4 n j m =

Nw1 → Nw1 

Ns → Ns

ωln 1L  l ∈ C
t k ωln 1L  ωln u  l ∈ C
t k

Nu → Nu

if j ∈ 1 2 4 4∗ , if j = 3.

If t is even and j ∈ 1 2 3, also define 2 k

α4 n j m  Nωn 1L → Nωkn 1L  Ns → Ns and 2k 4njm =



Nw1 → Nw1 

Nu → Nuωt/2 n 

if j ∈ 1 2, ωln 1L  l ∈ C
t k l+t/2 2
k−1l/t+k−1 t l 2
k−1l/t ωn u ωn u  l ∈ C
2 k if j = 3.

Moreover, let

3 k

α4 n 3 m  Nωn 1L → Nωkn u and 3 k 4 n 3 m

Nu → Nu

 =

2l ω2l n 1L  ωn u

 l∈C



t k 2



If t ≡ 2
mod 4, put 4 k

u α4 n 3 m  Nωn 1L → Nωk+t/2 n

Nu → Nωt/2 n 

and 4 k

4 n 3 m = ωln ul  l ∈ C
t k

finite nonmonomial linear groups

457

and if t ≡ 0
mod 24 , define 5 k

α4 n 3 m  Nωn 1L → Nωk+t/4 u n and

Nu → Nωt/2 n u

  5 k 4l
k−1/t+l 4 n 3 m = ω2l  ω2l+t/2 u4l
k−1/t+l  l ∈ C 4t  k

nu n

Finally, put  i k   α4 n 1 1  i = 1 2   n  if m = 1 n ≡ 0
mod 4,  k ∈ K
n\1 2 + 1   i k   α4 n 1 1  i = 1 2     ∈ K
n k = 1 if m = 1 n ≡ 2
mod 4,  k i k

4 n 1 m = α4 n 1 2  i = 1 2   k ∈ K
n2  k = 1 if m = 2 n ≡ 0
mod 4,    1 k n  α  k ∈ K
 k =

1 if m = 2 n ≡ 2
mod 4,  4 n 1 2  2   i k  if m > 2 t ≡ 0
mod 2, α4 n 1 m  i = 1 2 k ∈ K
t     α1 k if m > 2 t ≡ 1
mod 2, 4 n 1 m  k ∈ K
t  i k α4 n 2 m  i = 1 2 k ∈ K
t if t ≡ 0
mod 2,

4 n 2 m = 1 k α4 n 2 m  k ∈ K
t if t ≡ 1
mod 2,  1 k  if t ≡ 1
mod 2, α4 n 3 m  k ∈ K
t    i k  if t ≡ 2
mod 4,  α4 n 3 m  i = 1  4 k ∈ K
t

4 n 3 m = αi k if t ≡ 4 8 4 n 3 m  i = 1 2 3 k ∈ K
t    or 12
mod 16,    i k α4 n 3 m  i = 1 2 3 5 k ∈ K
t if t ≡ 0
mod 16, 1 k

4 n 4 m = α4 n 4 m  k ∈ K
t 1 k

4 n 4∗  m = α4 n 4∗  m  k ∈ K
t

7.2. Theorem. Let n ∈  be an even integer and t m ∈  with n = mt and put L = L4 n . Let j ∈ 1 2 3 4 4∗ , and if j = 1, assume also that m is i k even and that 2m divides n if j = 4∗ . Let N = Lj m  α = α4 n j m ∈ 4 n j m i k and put L N = 4 n j m and L N α = 4 n j m , then these sets satisfy the hypotheses of Theorem 3.5. Proof. Let S = L4 2 and Z = Z
L, so that L = SZ, and observe that by Theorem 5.1, NSL
2

L = S. Therefore, by Remark 3.6, the equivalence relation ∼ defined in Theorem 3.5 is just congruence modulo the inner automorphism group of L/N. Assume first that j = 1 and that m is odd, or equivalently, that N ≤ Z and S ∩ N = 1. Since S is characteristic in L by Lemma 7.1 and Z ∩ S has

458

¨ fling burkhard ho

order 2, the automorphisms of L/N ∼ = L4 t can be determined with the 2 1 help of Lemma 4.4, observing that α4 2 1 1 is a representative of the only nontrivial element of Out
S. If j = 1 and if m is even, we have N ∩ S = Z
S and so S/S ∩ N ∼ = Sym
4; thus every automorphism of S/S ∩ N is inner. If t is odd, then SN/N is characteristic in L/N by Lemma 4.3, and so by Lemma 4.4, the 1 k automorphisms of type α4 n 1 m form a complete list of representatives of elements of Out
L/N. Finally, if t is even, then the automorphisms of L/N = SN/N × Z/N can be determined by Proposition 4.5, noting that Proposition 4.5(b) applies. To find the automorphisms not satisfying conditions (2)–(4) of Theorem 3.5, note first that L3 n is the only normal subgroup of L of index 2 if n is not divisible by 4, and that L has precisely three subgroups of index 2 if n is a multiple of 4, namely, L3 n  L4 n/2 , and L4∗  n/2 , which 2 1 are all characteristic in L by Lemma 7.1. Thus, α4 n 1 1 is the only automorphism not satisfying Theorem 3.5 (3) for K = L3 n ; similarly if n is 1 n/2+1 2n/2+1 divisible by 4, then α4 n 1 1 and α4 n 1 1 correspond to the cases K = L4 n/2 1 1 and K = L4∗  n/2 . Since every linear automorphism of L4 n is inner, α4 n 1 1 1 1 and α4 n 1 2 are the only automorphisms not satisfying conditions (2) and (4) of Theorem 3.5. If j = 2, we have S/S ∩ N ∼ = Sym
3 all of whose automorphisms are inner. Moreover, its centre is trivial, so that L/N = ZN/N × SN/N, and the set of representatives of automorphisms of L/N can be computed as above, using Proposition 4.5. In the case when j = 3, the factor group L/N is the direct product of the cyclic groups generated by Nωn 1L and Nu of orders t and 2, respectively. If t is odd, L/N is, in fact, cyclic, and so the automorphisms of order ≤ 2 1 k are the α4 n 3 m , where k ∈ K
t. Thus, assume that t is even. Now, for every automorphism α of L/N, there exist integers a b c, and ct/2 d such that
Nωn 1L α = Nωan ub and
Nuα = Nωn ud (note that u2 ∈ N), where 0 ≤ a < t and 0 ≤ b c d ≤ 1. An element Nωxn uy is a fixed point of an automorphism α if x and y satisfy t
a − 1x + cy ≡ 0
mod t 2


∗

bx +
d − 1y ≡ 0
mod 2


∗∗

If α2 = 1, we must also have a2 +
t/2bc ≡ 1
mod t b
a + d ≡ c
a + d ≡ 0
mod 2, and d 2 + bc
t/2 ≡ 1
mod 2. Taking k = a, the cases b = c = 0; b = 0, c = 1; and b = 1, c = 0, lead to the automorphisms 1 k 2 k 3 k α4 n3 m , α4 n3 m , and α4 n 3 m .

finite nonmonomial linear groups 1 k

459

3 k

For the automorphisms α4 n 3 m and α4 n 3 m , the fixed points are easy to determine, for if c = 0, Eq. (∗) is satisfied if and only if x ≡ l
mod t for some l ∈ C
t k. If b = 0
∗∗ does not impose any restrictions on x and y we obtain 1 k 4 n 3 m . If b = 1
∗∗ is equivalent with x ≡ 0
mod 2 and our result follows by observing that   l  l ∈ C
t k l ≡ 0
mod 2 = 2l  l ∈ C 21  k

If c = 1 and b = 0, again
∗∗ holds for all x y. Solving
∗ modulo t/2, we obtain x ≡ l
mod t or x ≡ l + t/2
mod t for some l ∈ C
t/2 k. In particular,
k − 1x is divisible by t/2. Putting y=

2
k − 1x  t 2 k

we obtain all solutions for
∗, and thus the set 4 n 3 m . In the case when b = c = 1 a has to satisfy the equation a2 + t/2 ≡ 1
mod t, and d ≡ a
mod 2. If t ≡ 2
mod 4, it follows that a2 ≡ 1
mod t/2 and a2 ≡ 0
mod 2. Therefore,
a − t/22 ≡ 1
mod t/2 and
a − t/22 ≡ 1
mod 2, so that a − t/2 ≡ k
mod t for some k ∈ K
t. Conversely, if k ∈ K
t, then k is odd, hence
k + t/22 ≡ 0
mod 2, and evidently
k + t/22 ≡ k2 ≡ 1
mod t/2. Thus, the resulting automor4 k 4 k phisms are the α4 n 3 m . To determine the fixed points of α4 n 3 m , note that Eq.
∗∗ is equivalent with x ≡ y
mod 2, so that
∗ is equivalent with
k − 1x ≡ 0
mod t, and thus with x ≡ l
mod t for some l ∈ C
t k. 4 k Thus, the set 4 n 3 m defined above contains suitable representatives of the fixed points. For t ≡ 4
mod 8 or t ≡ 8
mod 16, the equation a2 + t/2 ≡ 1
mod t does not have integral solutions, for such a solution a would also satisfy a2 ≡ 3
mod 4 or a2 ≡ 5
mod 8, respectively. If t ≡ 0
mod 16, then d ≡ a ≡ 1
mod 2, and we have
a − t/42 ≡ 1
mod t. Thus, there exist k ∈ K
t such that a ≡ k + t/4
mod t. Conversely, if k ∈ K
t, then 5 k a = k + t/4 satisfies a2 + t/2 ≡ 1
mod t. Thus, we obtain the α4 n 3 m , and our list of automorphisms is complete for j = 3. 5 k To compute the fixed points of α4 n 3 m , we consider the above Eqs.
∗ and
∗∗. The latter clearly holds if and only if x ≡ 0
mod 2. Considering
∗ modulo t/2 and using x ≡ 0
mod 2, we obtain x ≡ 2l
mod t/2 for some l ∈ C
t/4 a = C
t/4 k. Thus, x ≡ 2l
mod t or x ≡ 2l + t/2
mod t. In the first case, solving
∗ for y, we obtain 4l
k − 1 4l
k + t/4 − 1 ≡ + l
mod 2

t t Observing that k + t/4 − 1 is even, we find the same solutions for y if 5 k x ≡ 2l + t/2
mod t. Thus, we obtain the fixed point set 4 n 3 m . y≡

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Finally, if j ∈ 4 4∗ , the factor group L/N = Nωn 1L  is cyclic of order t and the result is clear. Next, we determine the required automorphisms and their fixed points for the groups L4∗  n . 7.3. Theorem. Let n ∈  be an even integer, j ∈ 1 2 3 4∗ , and assume that n = mt with m t ∈ . Assume that m is even if j = 1 and that t is odd if j = 4∗ . Put L = L4∗ n and N = Lj m . For j ∈ 1 2 3, define automorphisms of L/N by 1 k

α4∗  n j m  Nw1 → Nw1 

Ns → Ns

Nuω2n → Nuωk2n

for every k ∈ K
2t, and for k ∈ K
t, let 1 k

α4∗  n 4∗  m  Nuω2n → N
uω2n k

Moreover, put 1 k 4∗  n j m

and

4∗  n 1 m

 =

if j ∈ 1 2 4∗ , ωln 1L  l ∈ C
t k l 
ω2n u  l ∈ C
2t k if j = 3,

 1 k   α4∗  n 1 1  k ∈ K
2n k ∈ 1 n + 1 if m = 1, k = α1 if m = 2, 4∗  n 1 2  k ∈ K
n k = 1   1 k if m > 2, α4∗  n 1 m  k ∈ K
2t 1 k

4∗  n 2 m = α4∗  n 2 m  k ∈ K
2t 1 k

4∗  n 3 m = α4∗  n 3 m  k ∈ K
2t 1 k

4∗  n 4∗  m = α4∗  n 4∗  m  k ∈ K
t

1 k

1 k

Then, for every α = α4∗  n j m ∈ 4∗  n j m , the sets 4∗  n j m and 4∗  n j m satisfy the conditions of Theorem 3.5. Proof. Put G = L4 2n  S = L4 2 , and Z = Z
G, and observe that by Theorem 5.1, NSL
2

L = S, so that the automorphisms of L/N induced by GL
2
 are inner. Therefore, by Remark 3.6, to find the required equivalence class representatives modulo the relation ∼ in Theorem 3.5, it suffices to find all automorphisms α with α2 ∈ Inn
L/N in a transversal of Inn
L/N in Aut
L/N. Assume first that j = 1 or 2. To use Proposition 4.5, we show that the automorphisms of SN/SN ∩ ZN are induced by automorphisms of SN/N. If j = 2, this is trivially true because SN/N ∼ = Sym
3 and hence SN ∩ ZN = N. If j = 1, we have N ≤ Z and so SN ∩ ZN =
S ∩ ZN. Thus, the statement is also trivial if m is even. Otherwise, S ∩ N = 1 and SN ∩ ZN = Z
SN. Therefore, SN/
S ∩ ZN ∼ = S/Z
S ∼ = Sym
4 which has trivial outer automorphism group, and the statement is clear.

finite nonmonomial linear groups

461

Thus, by Proposition 4.5, Aut
L/N consists of the restrictions of the automorphisms α of G/N which satisfy
L/Nα = L/N, and for each such α, there exist σ ∈ Aut
SN/N γ ∈ Hom
SN/N ZN/N, and k ∈  with
k 2t = 1, such that
Nxzα =
Nxσ
Nxγ Nz k for every x ∈ S z ∈ Z. 2 2 2 Then,
Nxzα =
Nxσ

Nxσ Nxk γ Nz k . Since SN/S  N has order 2 k is odd, and S  N/N ≤ Ker γ, it follows that

Nxσ Nxk γ = N/N for every x ∈ S. In particular, α2 fixes SN/N and ZN/N, and α2 and σ 2 coincide on SN/N. Now, assume that α2 induces an inner automorphism on L/N, and let 2 g ∈ S such that
Nxα = Nxg for every x ∈ L. We will prove that α2 is an inner automorphism of G/N. Since SN/N is isomorphic with Sym(3), Sym(4), or L4 2 , it follows that Out
SN/N has order at most 2. Therefore, there exists h ∈ S 2 2 such that
Nxα =
Nxσ = Nxh for every x ∈ S. Let l ∈ L and write 2 2 2 l = xz, with x ∈ S and z ∈ Z. Then,
Nlg =
Nlα =
Nxα
Nzα = 2 2
Nxh
Nzα =
Nlh
Nz −1 h
Nzα , which shows that gh−1 centralises ∼ LZN/ZN = G/ZN = SZN/ZN = S/S ∩ ZN. Since the latter is isomorphic with Sym(3) or Sym(4), and hence has trivial centre, it follows that Ngh−1 belongs to ZN/N ≤ Z
G/N. Therefore, we may assume that g = h. This shows that α2 acts in the same way as g on SN/N and L/N, and hence on G/N = SN/N L/N, as required. Thus, a complete and irredundant list of representatives of elements of order at most 2 of the outer automorphism group of L/N can be obtained from the list in Theorem 7.2 by choosing the restrictions to L/N of those i k automorphisms among the α4 2n j m which fix L/N as a set. 1 k Now, let k ∈ K
2t, then the α4 2n j m fix L/N and their restrictions 2 k to L/N are distinct. Put β = α4 2n j m , then
Nuω2n β = Nuωt+k 2n which is contained in L4 n N/N if t is odd. Therefore, in this case we have
L/Nβ = 2 k L/N. If t is even, then it is easy to see that the restrictions of β = α4 2n j m 1 l and α4 2n j m coincide on L/N, where l is an integer with l ≡ k + t
mod 2t and 1 ≤ l ≤ 2t. Since every linear automorphism of L is inner and L3 n is the only normal subgroup of index 2 in L, the automorphisms of L/N 1 1 1 n+1 not satisfying conditions (2)–(4) of Theorem 3.5 are α4∗  n 1 1 , α4∗  n 1 1 , and 1 1 1k 1 k α4∗  n 1 2 , respectively. Since we have 4∗  n j m = L ∩ 4 2n j m for every k, this completes the proof of the theorem if j = 1 or j = 2. In the cases j = 3 and j = 4∗ , the stated results follow easily from the fact that in these cases, L/N = Nω2n u is cyclic of order 2t and t, respectively. Finally, we determine the automorphisms of the quotients of L5 n whose square is inner.

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7.4. Theorem. Let n ∈  be even and put L = L5 n . If n = mt with m t ∈ , j ∈ 1 5, and N = Lj m is a normal subgroup of L = L5 n , define 1 k

α5 n j m  Nw1 → Nw1  2 k

Ns → Ns

α5 n 1 m  Nw1 → N
w1v 3  1 k

Nv → Nv

Ns → Ns

Nωn 1L → Nωkn 1L 

Nv → Nv3 

Nωn 1L → Nωkn 1L 

2 k

5 n j m = 5 n 1 m = ωln 1L  l ∈ C
t k

Let I = 1 2 and put  i k  α5 n 1 1  i ∈ I k ∈ K
n    

∈ 1 n2 + 1 if i = 1   k  i k   α5 n 1 1  i ∈ I k ∈ K
n2 

5 n 1 m = k = 1 if i = 1  i k  n  α  5 n 1 2  i ∈ I k ∈ K
2     k = 1 if i = 1    αi k 5 n 1 m  i ∈ I k ∈ K
t

if m = 1 n ≡ 0
mod 4, if m = 1 n ≡ 2
mod 4, if m = 2, if m > 2,

1 k

5 n 5 m = α5 n 5 m  k ∈ K
t i k

then the sets 5 n j m and 5 n j m satisfy the conditions of Theorem 3.5. Proof. Again, note that by Theorem 5.1, NSL
2

L = L5 2 , so that every automorphism of L/N induced by conjugation with an element of GL
2
 is inner. Therefore by Remark 3.6, two automorphisms of L/N are congruent modulo the equivalence relation ∼ defined in Theorem 3.5 if and only if they differ only by an inner automorphism of L/N. Assume that j = 1. Since L5 2 N/N is perfect, Lemma 4.3 yields that L5 2 N/N is characteristic in L5 n /N. Moreover, N ∩ L5 2 = 1 or Z
L5 2 , depending on whether m is odd or even. Now, the outer automorphism groups of L5 2 and L5 2 /Z
L5 2  ∼ = Alt
5 each have only one nontrivial 2 1 2 1 element, representatives being α5 2 1 1 and α5 2 1 2 respectively. Thus, the required automorphisms and fixed point sets can be computed using Lemma 4.4 with E = L5 2 N/N and Z = Z
L5 n /N, observing that 2 1 2 1
α5 2 1 1 2 and
α5 2 1 2 2 are, respectively, the identity automorphisms of L5 2 and L5 2 /L1 2 . 1 1 1 1 Clearly, α5 n 1 1 and α5 n 2 1 are the only automorphism in our list which do not satisfy conditions (2) and (3), respectively, of Theorem 3.5. Moreover, L possesses a normal subgroup of index 2 if and only if n is divisible by 4, and in that case L5 n/2 is the only such subgroup. Therefore, 1 n/2+1 α5 n 1 1 is the only remaining automorphism in our list which does not satisfy Theorem 3.5(4). If j = 5, then L/N is cyclic of order t and the result is clear.

finite nonmonomial linear groups

463

8. IRREDUCIBLE MONOMIAL SUBGROUPS OF GL
4
 Finally, we determine which of the representatives in the lists of irreducible subgroups of GL
4
 obtained so far are monomial groups. It seems to be difficult in general to decide whether a subgroup of GL
2n F which possesses a system of imprimitivity consisting of two subspaces of F 2n also has systems of imprimitivity consisting of more than two vector spaces. 8.1. Theorem. Let V be a four-dimensional vector space over
and suppose that G is a finite irreducible subgroup of GL
V  such that V has a nonrefinable system of imprimitivity U W  for G. Let n = Z
G, then G is monomial if and only if G is linearly isomorphic with one of the following groups. 3 1

where l ∈ 3 n 1 1 and n ≡ 0
mod 3

1 n+1

where l ∈ 3∗  n 1 1 and n ≡ 1
mod 3

1 2n+1

where l ∈ 3∗  n 1 1 and n ≡ 2
mod 3

G
L3 n  1 α3 n 1 1  l G
L3∗  n  1 α3∗  n 1 1  l G
L3∗  n  1 α3∗  n 1 1  l

3 1

1 n+1

1 2n+1

Proof. Assume first that G is monomial, then V possesses an imprimitivity system V1   V4  for G. Let H = NG
U and let D = NG
V1  ∩ · · · ∩ NG
V4 . Since G acts transitively on the subspace of V in any system of imprimitivity, it follows that G/H is cyclic of order 2, and that G/D is isomorphic with a transitive permutation group of degree 4. Hence G/D is cyclic or elementary abelian of order 4, dihedral of order 8, or isomorphic with Sym(4) or Alt(4). Moreover, if N1 = CG
U and N2 = CG
W , then H/N1 ∼ = H/N2 is isomorphic with a primitive subgroup of GL
2
. Thus, from the list of all conjugacy class representatives of such groups in Theorem 5.1, we obtain that H/Zi is isomorphic with Sym(4), Alt(4), or Alt(5), where Zi /Ni = Z
H/Ni  for i = 1 2. Moreover, since D is abelian, by Blichfeldt’s criterion of imprimitivity [4, Corollary 50.7] the subgroups
H ∩ DNi /Ni of H/Ni are contained in Zi /Ni for i = 1 2. Now, H has index 2 in G, and so we have G = HD or D ≤ H. Consider first the case when HD = G and thus G/D ∼ = H/H ∩ D. Comparing the possibilities for G/D with those for H/Zi , it follows that H/H ∩ D is isomorphic with Sym(4) or Alt(4), and that H ∩ D = Z1 = Z2 . Since HD = G, there exists d ∈ D such that Ud = W and thus N1d = N2 . As G acts faithfully on V , we have N1 ∩ N2 = 1. Moreover, N1 and N2 are contained in the abelian group D, whence it follows that N1 = N2 = 1. Now, Z
H = Z1 = Z2 , so that in fact Z
H = Z
G. Since !H D" is contained in H ∩ D = Z
H and also d 2 ∈ Z
H, it fol2 lows that h = hd =
h!h d"d = h!h d"2 , and so !h d"2 = 1 for every h ∈ H. Moreover, the map φ H → Z
H, defined by hφ = !h d" is a

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464

homomorphism, If !h d" = 1 for every h ∈ H, then U and W would be isomorphic as
H-modules. Moreover, U and W must be irreducible
Hmodules by Lemma 2.1, and so V would be a homogeneous
H-module. But then G would be reducible by Lemma 2.2. This contradiction shows that Im φ has exponent 2. Now, Z
H = Z
G is cyclic, and hence K = Ker φ is a subgroup of index 2 in H containing Z
H. Therefore, H/Z
H must be isomorphic to Sym(4). Moreover, by the definition of K U and W are isomorphic as
K-modules. Now, by Lemma 2.1, every one-dimensional
K-submodule of U or W would lead to an imprimitivity system of G which would be a proper refinement of U W . Therefore, U and W are irreducible as
K-modules, and it follows from Lemma 2.2 that G/K cannot be cyclic. Moreover, by Lemma 2.3, every subgroup M with K < M < G is reducible, and the
M-homogeneous components of V form a system of imprimitivity for G. By Corollary 5.2, U1  W1  is nonrefinable. Therefore, we may replace H by DK and we may replace U W  by U1  W1 . Hence it suffices to consider the case when D ≤ H. Comparing the possibilities for G/D with those for H/Z1 ∼ = H/Z2 , it follows that G/D ∼ = Sym
4 and H/Z1 ∼ = Alt
4; hence D = Z1 = Z2 . If N1 = 1 = N2 , then U and W are nonisomorphic as
D-modules; moreover by Clifford’s theorem, U and W cannot be inhomogeneous as
D-modules because H/D ∼ = Alt
4 does not have a subgroup of index 2. Thus, U and W are the
D-homogeneous components of V . In particular, each Vi
i = 1  4 is either contained in U or W , and so V1   V4  is a refinement of U W . This contradiction shows that N1 = 1 = N2 and that D ≤ Z
G. Since Z
G/D = 1, we have D = Z
G = Z
H, and H/Z
H ∼ = Alt
4. Proceeding as in the proof of Proposition 3.1, we obtain that G is linearly isomorphic with G∗ = G
L N α l, where L is a finite primitive subgroup of GL
2
, where N is a normal subgroup of L α ∈ Aut
L, and where 2 l ∈ L is a fixed pint of α such that xl = xα for every x ∈ L. Note that N = 1 because N1 = 1. Moreover, the image H ∗ of H in G∗ has the form,    x 0  ∗ x∈L  H = 0 xα  and it is easy to see that ∗

∗

Z
G  = Z
H  =



x 0

0 x

   x ∈ Z
L xα = x



In particular, L/Z
L is isomorphic with a factor group of H ∗ /Z
G∗  ∼ = H/Z
G ∼ = Alt
4, and since the latter has a trivial centre, we conclude that L/Z
L ∼ = H/Z
G. It follows that Z
G = Z
L and that α fixes Z
L elementwise. Moreover, G/Z
G has only one subgroup of index 2, whence by Theorem 2.4, V cannot have more than one imprimitivity system consisting of two subspaces of V .

finite nonmonomial linear groups

465

Now, by Theorem 5.1, L is conjugate in GL
2
 to L = L3 n or L3∗  n , and by Proposition 3.3, we may assume that L = L3 n or L = L3∗  n . 3 1 If L = L3 n , it follows from Theorem 6.1 that α = α3 n 1 1 and n ≡ 1 k 0 (mod 3). In the case when L = L3∗  n , we have α = α3∗  n 1 1 for some k ∈ K
3n k = 1 by Theorem 6.2. Moreover, k = n + 1 or k = 2n + 1 because α fixes the elements of Z
L. Now, if k = n + 1 then n
n + 2 = n2 + 2n =
n + 12 − 1 ≡ 0 (mod 3n), because k ∈ K
3n, and so n ≡ 1 (mod 3). Similarly, if k = 2n + 1, one obtains n ≡ 2 (mod 3). Conversely, let n be a positive even integer, and put G = G
L 1 α l, 3 1 3 1 where L = L3 n  α = α3 n 1 1  l ∈ 3 n 1 1 , if n ≡ 0 (mod 3). By t Theorem 6.1, we have l = ωn w1 for an integer t. If n ≡ 0 (mod 3), 1 k 1 k let L = L3∗  n  α = α3∗  n 1 1 , and l ∈ 3∗  n 1 1 , where k = n + 1 if n ≡ 1 (mod 3), and k = 2n + 1 if n ≡ 2 (mod 3). In particular, k ≡ 2 (mod 3), and by Theorem 6.2, there exists an integer t such that l = ωtn w1 . We have to show that, in each case, G is monomial. Let   r 0 ω3 s  g= u 0 ω2r 3 s where r = 1 if n ≡ 0 (mod 3) or n ≡ 1 (mod 3), and r = 2 if n ≡ 2 (mod 3), so that g ∈ G and the subgroup P = g has order 3. Now,
ωtn w1 α = ωtn w1 and
w1 w1s α = w1s , whence the matrix,      0 ωtn w1 w1 w1s 0 0 ωtn w1 w1s = h= 0 w1s w1 w1s 0 1L 0 s

belongs to G. Moreover, gh = g−1 because sw1 w1 = sw1s =
su −1 and w ws w1 1 1 = w1−1 . Let N = NG
P, then N contains Z
G, which is cyclic of order n because α fixes Z
L elementwise. Let H denote the subgroup of G stabilising the nonrefinable imprimitivity system of V for G consisting of two two-dimensional subspaces of V , where V is the underlying
-vector space of G. Then, H/Z
H ∼ = L/Z
L ∼ = Alt(4) and PZ
G/Z
G is a Sylow 3-subgroup of H/Z
G. Therefore,
N ∩ H/Z
G = PZ
G/Z
G and H  H ∩ N ≤ G  H = 2. Since h ∈ N, it follows that N = g h Z
G which therefore has index 4 in G. Let U ≤ V be an eigenspace of g for the eigenvalue λ. If u ∈ U, then uh−1 g = ugh h−1 = ug−1 h−1 = uh−1 λ−1 , so that Uh−1 is the eigenspace of g for the eigenvalue λ−1 . Since s has eigenvalues ω3 and ω23 , g has an eigenspace U of dimension 2 for the eigenvalue 1, and so Uh = U. Moreover, U is reducible as a
h-module. Let T be a one-dimensional
h-submodule of U, then T is in fact, a
N-module. Since GN = 4, it follows from Lemma 2.1 that G is monomial.

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9. THE FINAL LIST In view of Theorems 2.10 and 3.5, the information obtained in Sections 5–7, and the list of monomial groups in Section 8, we have thus arrived at our final list of finite imprimitive nonmonomial subgroups of GL
4
. Let d ∈ 3 3∗  4 4∗  5, j ∈ 1 2 3 3∗  4 4∗  5, and assume that n m t i, and k are positive integers with n = mt. Then, a complete and irredundant list of conjugacy class representatives of the finite irreducible subgroups of GL
4
 with exactly one system of imprimitivity consists of the groups G
L N α l defined in Section 3, where L = Ld n , i k N = Lj m , α = αd n j m , and i j k m, and t assume all possible values subject to the conditions set out in the rows of Tables V–IX correspondi k ing to the possible values of d, and l runs through d n j m . Note that the conditions below always imply that n is even. Recall from Section 6 that K
t = k  1 ≤ k ≤ t k2 ≡ 1
mod t. The symbol
a b denotes the greatest common divisor of the integers a and b. “− means that there is no additional restriction on the parameter. Moreover, a complete and irredundant list of conjugacy class representatives of the finite imprimitive nonmonomial subgroups of GL
4
 with more than one system of imprimitivity consists of the groups G
L x y defined in Section 2, where L = Ld n  d assumes the values 3 3∗  4 4∗  5, TABLE V Parameters for Groups with One Imprimitivity System for d = 3 j

m

t

i

k

1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3∗

1 1 1 2 2 >2, ≡ 1
2 >2, ≡ 1
2 >2, ≡ 0
2 >2, ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2

≡ 0 (4) ≡ 2
4 ≡ 0 (6) — ≡ 0
3 ≡ 0
2 ≡ 0
6 — ≡ 0
3 — — ≡ 0
3 ≡ 0
3 ≡ 0
9 ≡ 0
9 ≡ 3 6
9 — ≡ 0
3

1 2 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 4 3 4 5 6 5 6 7 1 1

∈ K
t\1 2t + 1 ∈ K
t\1 ∈K
t\1 ∈ K
t\1 ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ≡ 1
3 ∈ K
t ≡ 2
3 ∈ K
t ≡ 1
3 ∈ K
t ≡ 2
3 ∈ K
t ∈ K
t ∈ K
t

i k

d n j m 
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1 3
t k − 1
t k − 1
t k − 1 3
t k − 1
t k − 1 3
t k − 1 3
3t  k − 1
t k − 1
t k − 1

finite nonmonomial linear groups

467

TABLE VI Parameters for Groups with One Imprimitivity System for d = 3∗ i k

j

m

t

i

k

d n j m 

1 1 1 1 1 1 1 1 1 2 3∗ 3∗∗

1 1 1 1 1 1 2 >2, ≡ 1
2 >2, ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2

≡ 0 (12) ≡ 2
12 ≡ 4 (12) ≡ 6 (12) ≡ 8 (12) ≡ 10 (12) — ≡ 0
2 — — ≡ 1
3 ≡ 2
3

1 1 1 1 1 1 1 1 1 1 1 1

∈ K
3t\1 3t2 + 1 ∈ K
3t\1 2t + 1 ∈ K
3t\1 3t2 + 1 t + 1 ∈ K
3t\1 ∈ K
3t\1 3t2 + 1 2t + 1 ∈ K
3t\1 t + 1 ∈ K
3t\1 ∈ K
3t ∈ K
3t ∈ K
3t ∈ K
t ∈ K
t


t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
3t k − 1
t k − 1
t k − 1

TABLE VII Parameters for Groups with One Imprimitivity System for d = 4 j

m

t

i

k

1 1 1 1 1 1 1 2 2 3 3 3 3 4 4∗

1 1 2 2 >2 ≡ 1
2 >2 ≡ 0
2 >2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2

≡ 0 (4) ≡ 2
4 — ≡ 0 (2) ≡ 0
2 — ≡ 0
2 — ≡ 0
2 — ≡ 0
2 ≡ 2
4 ≡ 0
16 — ≡ 0
2

1 2

∈ K
t\1 2t + 1 ∈ K
t\1 ∈ K
t\1 ∈ K
t\1 ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t ∈ K
t

1 2 1 2 1 2 1 2 1 2 3 4 5 1 1

i k

d n j m 
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1 2
t k − 1 2
2t  k − 1
t k − 1 2
4t  k − 1
t k − 1
t k − 1

¨ fling burkhard ho

468

TABLE VIII Parameters for Groups with One Imprimitivity System for d = 4∗ i k

j

m

t

i

k

d n j m 

1 1 1 1 2 3 4∗

1 2 >2 ≡ 1
2 >2 ≡ 0
2 ≡ 0
2 ≡ 0
2 ≡ 0
2

≡ 0 (2) — ≡ 0
2 — — — ≡ 1
2

1 1 1 1 1 1 1

∈ K
2t\1 t + 1 ∈ K
2t\1 ∈ K
2t ∈ K
2t ∈ K
2t ∈ K
2t ∈ K
t


t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
2t k − 1
t k − 1

TABLE IX Parameters for Groups with One Imprimitivity System for d = 5 i k

j

m

t

i

k

d n j m 

1 1 1 1 1 5

1 1 2 >2 ≡ 1
2 >2 ≡ 0
2 ≡ 0
2

≡ 0 (4) ≡ 2
4 — ≡ 0
2 — —

1 2 1 2 1 2 1 2 1 2 1

∈ K
t\1 2t + 1 ∈ K
t\1 ∈ K
t\1 ∈ K
t ∈ K
t ∈ K
t


t k − 1
t k − 1
t k − 1
t k − 1
t k − 1
t k − 1

TABLE X Parameters for Groups with Three Imprimitivity Systems d 3 3 3 3 3 3∗ , 4, 4∗ , 5 3∗ , 4, 4∗ , 5

n ≡0 ≡0 ≡0 ≡0 ≡0 ≡0 ≡0


2
2
2
2
2
2
2

x

y

1L 1L ω2n 1L ω2n 1L ω2n u 1L ω2n 1L

1L u ω2n 1L u ω2n u 1L ω2n 1L

finite nonmonomial linear groups

469

n is a positive even integer, and
x y ∈ d n , where d n is the set defined in Lemma 5.3. Note that every such group has precisely three imprimitivity systems. The possible values for d n x, and y can be found in Table X. ACKNOWLEDGMENTS The author is indebted to Dr. L. G. Kov´acs for introducing him to the problem of classifying the finite nonmonomial imprimitive subgroups of GL
4
, for his advice, and for many helpful discussions concerning the topic. He would also like to express his gratitude toward the referee for his detailed comments on the article, which helped to eliminate a number of errors and to clarify the exposition. The computer algebra system GAP [11] was used to verify the lists of automorphisms in Sections 6 and 7 for small values of n.

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15. L. G. Kov´acs, On tensor induction of group representations, J. Austral. Math. Soc. Ser. A 49 (1990), 486–501. 16. G. A Miller, H. F. Blichfeldt, and L. E. Dickson, “Theory and Applications of Finite Groups (Reprint of the 1916 edition with corrections),” Stechert, New York, 1938. 17. D. A. Suprunenko, “Matrix groups,” in Translations of Mathematical Monographs, Vol. 45, Am. Math. Soc., Providence, RI, 1976.