Construction of finite matrix groups Robert A. Wilson School of Mathematics and Statistics, The University of Birmingham published in Birkh¨auserProgress in Math. 173 (1999), pp. 61{83 Abstract We describe various methods of construction of matrix representa- tions of finite groups. The applications are mainly, but not exclusively, to quasisimple or almost simple groups. Some of the techniques can also be generalized to permutation representations. 1 Introduction It is one thing to determine the characters of a group, but quite another to construct the associated representations. For example, it is an elementary exercise to obtain the character table of the alternating group A5 by first determining the conjugacy classes, then writing down the trivial character and the permutation characters on points and unordered pairs, and using row orthogonality to obtain the irreducibles of degree 4 and 5, and finally using column orthogonality to complete the table. The result (see Table 1) shows that there are two characters of degree 3, but how do we construct the corresponding 3-dimensional representations? In general, we need some more information than just the characters, such as a presentation in terms of generators and relations, or some knowledge of subgroup structure, such as a generating amalgam, or something similar. If we have a presentation for our group, then in a sense it is already de- termined, and there are various algorithms which in principle at least will construct more or less any desired representation. The most important and 1 Table 1: The character table of A5 Class name 1A 2A 3A 5A 5B Class size 1 15 20 12 12 χ1 1 1 1 1 1 χ2 3 −1 0 τ σ χ3 3 −1 0 σ τ χ4 4 0 1 −1 −1 χ5 5 1 −1 0 0 1 p 1 p τ = (1 + 5); σ = (1 − 5) 2 2 well-known is Todd{Coxeter coset enumeration, which converts a presenta- tion into a permutation representation, and is well described in many places, such as [11]. From a (faithful) permutation representation it is then (in prin- ciple) possible to obtain at least any irreducible representation over any finite field, by using `Meataxe' techniques described by Richard Parker [16]. A gen- eralization of these methods to characteristic zero has recently been described by Parker [17]. Basically these methods enable one to chop any given rep- resentation into its irreducible constituents, and then tensor representations together to produce new ones to chop up, and so on. What we are concerned with here, however, is something more basic, namely how to construct a matrix representation of a group from scratch, when no representation at all is known to begin with. We also assume that no presentation is known, or at least that no presentation can be used to produce a sufficiently small permutation representation. Note: the original title of my lectures was something like \Computer construction of matrix representations of sporadic simple groups over finite fields”, but it gradually became clear that almost every word of the title was redundant, and that \Construction of groups" was the most the various ideas had in common (and even the word \groups" was a trifle restrictive). The present title is a (somewhat unhappy) compromise between the two extremes. 2 A5 @ @ @ h(12)(34); (123)i =A4 S3 = h(123); (12)(45)i @ @ @ A3 = h(123)i Figure 1: Generating A5 from subgroups 2 A small example We illustrate the basic ideas by considering the 3-dimensional representations of A5 mentioned above. Now we know that A5 has a subgroup A4 obtained by fixing one of the five points, and that this subgroup is maximal since it has prime index. Fixing another point we obtain a subgroup A3, and by 2-transitivity there is an element of A5 interchanging these two points, and therefore normalizing A3 to S3. By maximality, these subgroups generate A5, and the situation is as shown in Fig. 1. In the figure, we also give generating permutations for these subgroups, although the figure can equally well be understood at the level of abstract groups. If we now wish to construct one of the 3-dimensional representations of A5, we can start by constructing its restriction to A4. For simplicity we assume that the characteristic of the underlying field is not 2 or 3. Then it is easy to see that the character restricts to A4 as the unique 3-dimensional irreducible, and the corresponding representation may be defined by 0 1 0 0 1 (12)(34) 7! @ 0 −1 0 A 0 0 −1 0 0 1 0 1 (123) 7! @ 0 0 1 A 1 0 0 (In general, an inductive approach allows us to assume anything we wish 3 about properties of the proper subgroups of our group.) 0 0 1 0 1 The next step is to adjoin an element of order 2 conjugating @ 0 0 1 A 1 0 0 0 1 0 0 1 to its inverse. One obvious possibility is @ 0 0 1 A, but this extends A4 0 1 0 to S4, not to A5. We actually need to look at all such involutions, in order to pick the right one(s). In this particular case, it is easy to show from 0 0 1 0 1 first principles that the elements inverting @ 0 0 1 A are exactly those of 1 0 0 0 α β γ 1 0 a b c 1 the form @ β γ α A. (To see this, note that if the matrix @ d e f A γ α β g h i conjugates the given element to its inverse, then 0 a b c 1 0 0 1 0 1 0 0 0 1 1 0 a b c 1 @ d e f A @ 0 0 1 A = @ 1 0 0 A @ d e f A g h i 1 0 0 0 1 0 g h i 0 c a b 1 0 g h i 1 ) @ f d e A = @ a b c A ; i g h d e f whence c = g = e, a = h = f, b = i = d, as required.) The extra condition that it is an involution gives α2 + β2 + γ2 = 1; and αβ + βγ + γα = 0; while the fact that all involutions in A5 have trace −1 gives α + β + γ = −1: In fact these three equations are not independent, and we obtain a one- parameter family of solutions. To find which ones actually give rise to A5, we need some more information. If the field is finite, we can of course try all cases until we have eliminated all but the correct answer(s). If the field is infinite, however, we need to use some more structure of the group. 4 In the present case, for example, we could use what we know about A5 to say that without loss of generality our extra element represents (12)(45), and 0 1 0 0 1 therefore has product of order 3 with our original element @ 0 −1 0 A, 0 0 −1 representing (12)(34). This product is 0 1 0 0 1 0 α β γ 1 0 α β γ 1 @ 0 −1 0 A @ β γ α A = @ −β −γ −α A ; 0 0 −1 γ α β −γ −α −β and if this is to have order 3 then it has trace 0. Thus we obtain α−γ−β = 0, 1 which together with α + β + γ = 1 gives α = β + γ = 2 , and substituting 2 2 2 1 into α + β + γ = 1 we obtain βγ = − 4 , whence β and γ are the roots of 2 1 1 the quadratic equation x − 2 x − 4 = 0, namely 1 p β; γ = (1 ± 5): 4 These two possibilities for the extra element, namely 0 1 σ τ 1 0 1 τ σ 1 1 1 σ τ 1 and τ σ 1 2 @ A 2 @ A τ 1 σ σ 1 τ p p 1 1 where τ = 2 (1 + 5) is the golden ratio, and σ = 2 (1 − 5), correspond to the two different 3-dimensional representations of A5. 3 Some variants We have already described all the important points in the method used to construct from scratch matrix representations of the sporadic groups, from Janko's construction of J1 published in 1965 [6], to my construction of the Monster in collaboration with Linton, Parker and Walsh [12], [27]. How- ever, to give more of the flavour of what was actually done in the finite-field cases, we describe slightly different methods for (a) obtaining the complete parametrization of all extending elements, and (b) determining which one(s) of these extending elements give rise to the desired group. We keep the same example as above. 5 Suppose we have two elements x and y which both invert the 3-cycle (123). Then xy−1 centralizes it. Thus the essential step is to find the centralizer of this 3-cycle (or of whatever group appears in its place in larger examples). Now, provided the field has characteristic not 3, and provided there is a primi- tive cube root ! of unity in the field, we find that the representation restricted to A3 is the direct sum of its three 1-dimensional representations, acting on the three 1-spaces spanned by the three eigenvectors (1; 1; 1), (1; !; !2) and (1;!2;!). By changing to a basis of eigenvectors, we immediately see that the centralizer consists of all diagonal matrices, constituting a 3-parameter family. Changing basis back again we obtain the centralizer as the set of matrices 0 α β γ 1 @ γ α β A β γ α and can continue as before.