Groups and Representations

Transcription

1 Groups and Representations Madeleine Whybrow Imperial College London These notes are based on the course Groups and Representations taught by Prof. A.A. Ivanov at Imperial College London during the Autumn term of 25.

2 Contents Linear Groups 3. Finite Fields Key Groups The General Linear Group The Projective General Linear Group The Special Linear Group The Projective Special Linear Group A Simple Group of Order A Simple Group of Order The Fano Plane The Automorphism Group of the Projective Plane Stabilisers of Subgroups in GL(V ) 8 2. Semidirect Products General Theory of Stabilisers

3 3 The Representation Theory of L 3 (2) Representation Theory The Dual of a Vector Space Counting Irreducible Representations Permutation Representations Submodules of Permuation Modules Permutation Modules of 2-Transitive Actions Permutation Modules of Actions on Cosets of Subgroups Representations of L 3 (2) over GF (2) The Permutation Module 2 Ω The Permutation Modules 2 Λ and Representations of L 3 (2) over C The Ordinary Character Table of L 3 (2) A Error Correcting Codes 42 2

4 Chapter Linear Groups In this chapter, we will introduce some specific examples of linear groups.. Finite Fields We first recall some key results concerning finite fields. Theorem.. Let p be a prime number and m be a positive integer. Then there exists a unique (up to isomorphism) field of order q = p m. We call it GF (q). If m = in the above theorem then GF (p) Z /pz. For m >, choose f(x) to be an irreducible polynomial of degree m in GF (p)[x] then we can construct GF (q) as GF (q) GF (p)[x] /f(x)gf (p)[x]. This construction is independant of the choice of f(x). It relies on the fact (which we will not prove) that such a polynomial exists for all p and m. Example.2. We will construct GF (4). We have GF (2) = Z /2Z = {, }. Take 3

5 f(x) = x 2 + x + then GF (4) = {,, x, x + }. Note that (GF (q), ) is isomorphic to the cyclic group of order q. We denote by V n (q) the n-dimensional vector space over GF (q)..2 Key Groups Recall that we define the centre of a group G as Z(G) = {z G zg = gz g G} and that we define the commutator subgroup as G = [g, h] g, h G where [g, h] = g h gh. Equivalently, the commutator subgroup can be defined as the minimal subgroup H such that G/H is abelian..2. The General Linear Group We define GL n (q) = the set of non-singular n n matrices over GF (q) = the set of linear bijections of from V n (q) to itself. It is possible to show that these two definitions are equivalent by choosing a basis of V n (q). The order of GL n (q) is GL n (q) = q n(n ) 2 n i= (q i ). If we let G = GL n (q) then Z(G) = {λi n λ GF (q) } 4

6 and G = {A det(a) = }. Note that the map det GL n (q) GF (q) is a homomorphism..2.2 The Projective General Linear Group We define Then P GL n (q) = GLn(q) /Z(GL n(q)). P GL n (q) = GL n(q). q.2.3 The Special Linear Group We note that the map det GL n (q) GF (q) is a homomorphism and define Then SL n (q) GL n (q) and SL n (q) = ker(det). so GL n (q) GF (q) SL n (q) SL n (q) = GL n(q). q.2.4 The Projective Special Linear Group We define P SL n (q) = SLn(q) /Z(GL n(q)) SL n(q). Then P SL n (q) = SL n(q) (n, q ) where (n, q ) denotes the highest common factor of n and q. We also denote P SL n (q) as L n (q). In the tables below we calculate the value of L 2 (q) for some small values of q. 5

7 q = L 2 (q) Note that L 2 (4) = L 2 (5) = A 5 = 6. In fact, these three groups are isomorphic to each other. We can also show that L 2 (7) = L 3 (2) = 68, again these groups are isomorphic to each other..3 A Simple Group of Order 6 In this section, we will prove that there exists a unique (up to isomorphism) simple group of order 6. We will require a few well known results in this proof: Theorem.3. The group A 5 is simple. Proof. See Problem Sheet. Definition.4. If G is of order p a m where p is a prime not dividing m, then a subgroup H of G of order p a is called a Sylow p-subgroup of G. The number of Sylow p-subgroups of G will de denoted n p. Theorem.5 (Sylow s Theorem). Let G be a group of order p a m, where p is a prime not dividing m. Then the following hold:. G has a Sylow p-subgroup. 2. If P and Q are distinct Sylow p-subgroups of G, then there exists some g G such that Q = gp g. That is, any two Sylow p-subgroups of G are conjugate in G. 3. The number of Sylow p-subgroups of G, n p satisfies n p mod p. Further, n p is the index of the normaliser N G (P ) of any Sylow p-subgroup P, and as such n p m. Corollary.6. If G is simple then n p. 6

8 Proof. Follows from the fact that any two Sylow p-subgroups are conjugate in G. Theorem.7. Suppose that H is a simple group of order 6. Then H A 5. Proof. Suppose that H is simple of order 6 = Then by Sylow s Theorem n 2 =, 3, 5 or 5; n 3 =, 4 or ; n 5 = or 6. As H is simple, Corollary.6 implies that n 5 = 6. Let K = {K () 5, K (2) 5, K (3) 5, K (4) 5, K (5) 5, K (6) 5 } be the set of Sylow 5-subgroups and let M = Sym(K) S 6. Then we have a map φ H M induced by the action of conjugation of H on K. We can show that φ is a homomorphism and that φ(h) is a bijection for all h H. As H is simple, the kernel of φ must be trivial (as it is a normal subgroup of H and φ is non-trivial). Thus H embeds into S 6. Moreover, φ(h) Alt(K) A 6, otherwise φ(h) Alt(K) would be an index 2 subgroup in φ(h) H which is a contradiction as index 2 subgroups are necessarily normal. We consider the action of Alt(K) on the cosets of φ(h). This gives a homomorphism µ Alt(K) Sym(L) S 6 where L is the set of cosets of φ(h) in Alt(K). However, by comparing orders, we see that µ is actually an isomorphism Alt(K) Alt(L). Moreover, φ(h) as a subgroup of Alt(K) is the stabiliser of the coset φ(h). so is isomorphic to a copy of A 5 in Alt(L). Thus we have that H A 5 is the only simple group of order 6. 7

9 .4 A Simple Group of Order 68 In this section, we will prove that there exists a unique (up to isomorphism) simple group of order The Fano Plane We consider the group L 3 (2) = P SL 3 (2) = SL 3 (2) = GL 2 (3) = P GL 3 (2). It is the automorphism group of V 3 (2). Note that for any vector space over a finite field In fact, for any subspace U of V v =. v V v =. v U For any v V, scalar multiplication is easy to determine as v = and v = v. In order to determine addition in V 3 (2), we consider its two dimensional subspaces. The number of two dimensional subspaces is [ 3 2 ] 2 = (23 )(2 3 2) (2 2 )(2 2 2) = 7 where [ n k ] q is defined to be the number of k-dimensional subspaces in V n(q) and is equal to [ n k ] q = (qn )(q n q)... (q n q k ) (q k )(q k q)... (q k q k ). Each such subspace U contains three non-zero vectors, say U = {, u, u 2, u 3 }. Thus, using the fact that v U v =, we can define u + u 2 to be u 3, the third non-zero vector in the 2-dimensional subspace containing u and u 2. Using this observation, we can express the two dimensional subspaces of V 3 (2) as the Fano plane (the projective plane of order 2). 8

10 Here, the points correspond to the vectors in V and the lines correspond to the 2-dimensional subspaces of V. We denote the set of points as P and the set of lines L. A point in P lies on a line in L if and only if the corresponding vector is contained in the corresponding 2-dimensional vector space. This gives a set of incidence relations, which we denote I. We can then define the Fano plane to be the triple Π = (P, L, I ). In fact any plane can be described in this form, with sets of points and lines and corresponding incidence relations. If Π = (P, L, I ) and Π = (P, L, I ) are two planes then a map ψ Π Π is a morphism of planes if it maps points to points, lines to lines and p l if and only if ψ(p) ψ(l) for and p P and l L. The incidence relations I can be equivalently described using an incidence matrix defined as M(Π) = (a lp ) l L,p P 9

11 where if p l a lp = if p l. For Π this gives the matrix below. The last row gives the vectors of V which each of the columns refer to. p p 2 p 3 p 4 p 5 p 6 p 7 l l 2 l 3 l 4 l 5 l 6 l 7 Note that the matrix above obeys the following rules: A There are 7 rows; A2 There are 7 columns; A3 There are 3 ones in each row; A4 There are 3 ones in each column; A5 The inner product of two distinct rows is ; A6 The inner product of two distinct columns is ; where the inner product of two rows or columns is their inner product as vectors i.e. coordinate-wise multiplication. We will now show that these rules are in fact sufficient to describe the Fano plane. Proposition.8. A plane whose incidence matrix obeys the axioms A - A6 is unique up to isomorphism.

12 Sketch of proof. Any plane is isomorphic to another plane whose incidence matrix has its ones as far upwards and leftwards as possible (as an isomorphism of a plane is effectively a relabelling of its points and lines). There is only one matrix in this form which obeys the axioms above, we call it M. Thus a plane whose incidence matrix obeys the rules above is isomorphic to the plane whose incidence matrix is equal to M. M = p p 2 p 3 p 4 p 5 p 6 p 7 l l 2 l 3 l 4 l 5 l 6 l 7 There is yet another equivalent way in which we can consider the Fano plane. We now take P = GF (7), Q = {α 2 α GF (7)} = {, 2, 4} and L = {Q + i i GF (7)}. Note that the set Q is a difference set. Proposition.9. The plane with points and lines described above is the Fano plane. Proof. Check that the incidence matrix obeys the axioms A-A6. Expressing elements of L as vectors in V 7 (2) where the ith coordinate of the vector corresponding to l L is if and only if p i l give vectors which all lie in the Hamming code Ham(3) (see Appendix A). If we include the zero vector we can extend our matrix to get

13 . This matrix is of the form N = M M where M is the incidence matrix of (P, L ) and M is its complement. The rows of this matrix, along with the vectors and give a Hamming Code Ham(3) which has been extended by a parity check bit. As usual, the columns of N correspond to vectors V = V 3 (2) and the rows correspond to U + v where U is a fixed 2-dimensional subspace of V and v U..4.2 The Automorphism Group of the Projective Plane If Π = (P, L ) is the Fano plane, recall that Aut(Π) = {φ P P a bijection such that φ(l) L l L }. 2

14 Labelling the points as p,..., p 7 means we can write such automorphisms as elements of S 7. Example.. φ = ()(2 4)(3 5)(7 6) Aut(Π) φ = ()(2 4)(3 5)(7)(6) Aut(Π) As well as all the methods mentioned in the previous section, we can also think of the columns (equivalently the points of the Fano plane) as the -dimensional subspaces in V 2 (7) with the following correspondences: Note that from here onwards, when we write a vector of L 2 (7), we are actually refering to the -dimensional subspace which it lies in. This correspondence allows us to consider L 2 (7) as acting on the Fano plane via its actions on the points. Proposition.. For each g L 2 (7), the map Π Π induced by the action of g is an automorphism. That is to say L 2 (7) Aut(Π). Proof. We consider the action of L 2 (7) on M, the incidence matrix of Π, and show that the image of M under an element of L 2 (7) obeys the axioms A-A6 so corresponds to a plane isomorphic to Π. The group L 2 (7) is generated by (representatives of) the matrices t =, s = 2 4, r = 6. 3

15 For α GF (7), the matrix t maps α α +. Rows of M can also be thought of as subsets of the form l = { + i, 2 + i, 4 + i} for i GF (7). Thus t simply permutes the rows of M, and so the image of M under t will obey the axioms, as required. Similarly, the matrix s maps α 2 4α = 2α. So will map l = { + i, 2 + i, 4 + i} to {2 + 2i, 4 + 2i, 8 + 2i} { + 2i, 2 + 2i, 4 + 2i}. So similarly s permutes the rows of M. The case of r is not so straightforward. It sends maps α α 6α. Check by hand that this maps, 6, = 3, = = 5, = 4, = and that the image of the matrix M under this map obeys axioms A-A6 so is still an incidence matrix of the Fano plane. Theorem.2. The Fano plane is uniquely determined by the choice of three non-collinear points. Proof. Suppose that we choose u, v, w three non-collinear points in Π. In the Fano plane, any two points lie in a unique line so the points u + v, u + w, v + w complete the three lines containing pairs of u, v and w. As u, v, w are noncollinear, these points are distinct. However, the points u + v, u + w, v + w are clearly collinear, give us a fourth line. The point u + v + w is again distinct and so the lines {u, v + w, u + v + w}, {v, u + w, u + v + w} amd {w, u + v, u + v + w} complete the Fano plane, as required. 4

16 Proposition.3. Aut(Π) 68 Proof. From Theorem.2 above, an automorphism of Π is determined by its action on any three non-collinear points of the Fano plane. Fix three such points. Then there are 7 choices for the image of the first point, 6 for the second and 4 for the third. Thus there are at most 68 automorphisms of Π. Theorem.4. L 2 (7) Aut(Π) Proof. From Proposition. and Proposition?? we know that L 2 (7) Aut(Π) and Aut(Π) 68. However, L 3 (2) = 68 so we must have L 3 (2) Aut(Π). Any element l of L clearly acts on U. Recall that we have a map β [ U ] V sending We can use this to define an action of l on V. We say that v l = β((β (v)) l ) giving the commutative diagram v β (v) v l (β (v)) l l β l β. 5

17 We define λ(l) = t l l where t v u u + v for u, v V. Then λ is our desired isomorphism. Example.5. Take ˆl = GL 2(7) and let l be its representative in L 2 (7). Then l maps l So we can write l as a permutation l = ( ). We have = β l = β =. Writing t l as a permutation of elements of V gives t l = ( ) ( 3) (2 4) (5 6). So λ(l) = t l l = ( ) ( ). Using the basis {(,, ), (,, ), (,, ) }, we can write λ(l) as the linear transformation λ(l) =. 6

18 Using this isomorphism and Theorem.4 we have L 3 (2) L 2 (7) Aut(Π). 7

19 Chapter 2 Stabilisers of Subgroups in GL(V ) In this section we will consider a vector space V and a subspace U of V. Our aim is to determine G(U) = Stab GL(V ) (U) = {g GL(V ) U g = U}. 2. Semidirect Products For a group G with a normal subgroup N G, we have the quotient G/N = {Ng g G}. Given a pair (N, G /N) up to isomorphism, is it possible to reconstruct G? In general, this is a very hard, but important, question to answer. Semidirect products are a special case of this problem where, given certain information about N and G /N, it is possible to reconstruct G. Example 2.. If N C 2 and G /N C 2 then G may be equal to C 4 or C 2 C 2. 8

20 Suppose that N G and K G. Then we have a map φ G G /N g Ng. When is φ K K K /K N an isomorphism? The kernel of φ K is K N so φ K is injective if and only if K N = ; We will have Im(φ K ) = Im(φ) if and only if g G, k K such that Ng = Nk. So φ K is surjective if and only if G = KN = kn k K, n N. Definition 2.2. Given G a group with N G and K G such that K N = and G = KN then G is said to be an internal semidirect product of N and K (denoted G = N K). As N is normal, we can take λ K Aut(N) k (n k nk). Thus, given the triple (N, K, λ), we can recover the group G = N K. general, given any two groups K and H and a homomorphism λ K Aut(N), we define the outer semidirect product of N and K with respect to λ to be the set with multiplication N λ K = {(k, n) k K, n N}. (k, n ) (k 2, n 2 ) = (k k 2, n λ(k) n 2 ). Theorem 2.3. The set N λ K as defined above is a group. In Note that if λ is the identity map then N λ K N K. Example 2.4 (The Stabiliser of the Extended Hamming Code). Let H be the extended Hamming code of degree 3 (which we first saw on page 2). It is a 9

21 subspace of V 8 (2) and has parity-check matrix H =. So the group GL 4 (2) acts on the Hamming code via its action on the columns of this matrix. However, this action may not necessarily preserve the Hamming code, but might send it to a different subspace of V 8 (2). This action will preserve the code if and only if it permutes the columns of H. Using the basis {e, e 2, e 3, e 4 } of V = V 4 (2), the columns of H form the following set X = {α e α 4 e 4 α 4 = }. This is not a subspace of V but the set Y = {α e α 4 e 4 α 4 = }. is a subspace. Moreover, V is the disjoint union of X and Y so the stabiliser of X in GL(V ) is equal to the stabiliser of Y in GL(V ). Thus the stabiliser consists of matrices which preserve the final basis vector of V and which acts as GL 3 (2) on the remaining basis vectors. Thus the stabiliser of the Hamming code in G = GL 4 (2) is G{H } = A A GL 3 (2), x, y, z GF (2) = N λ K x y z where N C2, 3 K GL 3 (2) and λ is the natural action of GL 3 (2) on V 3 (2). Note that the action of the subgroup L isomorphic to GL 3 (2) in this stabiliser preserves the first column of the parity check matrix, which corresponds to the first bit, or column, of the extended Hamming code itself. However we know that K L 2 (7) also acts on the columns of H via its action on the -dimensional subspaces of V 2 (7). It is clear that this action stabilises H but does not preserve any columns. Thus L and K are not conjugate in G{H } but are isomorphic. 2

22 To have two non-conjugate copies of L n (2) in 2 n L n (2) is unique to the case n = 3, it does not occur for any larger n. 2.2 General Theory of Stabilisers The above example is a specific case of the general theory of stabilisers which we will cover in this section. Given a vector space V = V n (q) and a subspace U of V, we wish to find the stabiliser of the U which is defined as G(U) = {g G U g = U} for G = GL(V ) GL n (q). Take B = {b,..., b n } be a basis of V and let U = Span(b,..., b m ) for some m n. Then we have X G(U) = P Y X GL(U), Y GL(W ), P M n m,m(q) where W = Span(b m+,..., b n ) and M n m,m (q) is the set of matrices with n m rows and m columns over GF (q). This gives G(U) = GL m (q) GL n m (q) q m(n m). It can also be deduced that GL(V ) = GL(U) [ n m ] q. Thus we have that where G(U) P λ K K = G m (q) G n m (q) corresponding to the matrices X and Y ; P = (C q ) (n m)m corresponding to (n m)m copies of the additive group of GF (q). 2

23 Chapter 3 The Representation Theory of L 3 (2) In this chapter we will construct a number of representations of the group G = L 3 (2). We will be working both with fields of characteristic (ordinary representation theory) and fields of positive characteristic (modular representation theory). We assume a basic knowledge of ordinary representation theory but spend the next two sections going over results which will be particularly key to our work. 3. Representation Theory Recall that given two modules V and V 2 of a group G along with representations ϕ G GL(V ) and ϕ 2 G GL(V 2 ), we define the direct sum of V and V 2 to be V V 2 = {(v, v 2 ) v V, v 2 V 2 } with representation ϕ = ϕ ϕ 2 such that ϕ (v, v 2 ) (ϕ (g)v, ϕ 2 (v 2 )). 22

24 If there exist submodules V and V 2 of V such that V = V V 2 and V V 2 = then V = V V 2. Our main strategy for constructing representations of L 3 (2) will be to construct a permuation module V (see below) of G and then to (attempt to) decompose it into the direct sum V = V... V k (3.) where V,..., V k are irreducible representations. In such a decomposition, the same submodule of V may occur more than once (or indeed not at all). Given a module V and a submodule U (which not necessarily irreducible) we say that the number of times U occurs as a decomposition factor of V is the multiplicity of U in V and is denoted m V (U). For ease of notation, if it is obvious to which module V we are referring, we write m(u). From this discussion, the equation 3. becomes V = l V mi i= i = V... V m times... V l... V l m l times where the V i are pair-wise non-isomorphic irreducible representations of G and m i = m V (V i ). Proposition 3.. If V is any module of a finite group and U is an irreducible module of the same group then m V (U) = χ V, χ U where, is the inner product on characters of G. Proof. We decompose V as V = U m(u) W Irr(G) W U W m(w ). We then take the inner product of both sides with U to get χ V, χ U = m(u) χ U, χ U χ W Irr(G) W U χ W, χ U. 23

25 However, recall that the irreducible representations are orthonormal with respect to, so Thus if W = U χ W, χ U = if W U. m V (U) = χ V, χ U. However, such a decomposition is not always possible. If a module V can be completely decomposed as a direct sum of irreducible modules then we say it is semisimple. Theorem 3.2 (Maschke s Theorem). Let V be a module of a finite group G over a field of characteristic p. Then V is semisimple if p does not divide the order of G. If a module V is not semisimple then it is possible that it contains two submodules V, V 2 of a module such that V /V V 2 but V / V V 2. In this case, we say that V is an indecomposable extension of V by V 2 which we denote V = V /V The Dual of a Vector Space We will frequently use the following definition in this section. Definition 3.3. If V is a vector space over the field k then the dual space of V is V = Hom k (V, k). Example 3.4. What is the dual space of V n (k) for a given field k? Given two vectors v, w V n (k), their dot product lies in k. So, for a given vector v V n (k), we can define a function ϕ v w v w. It is clear that ϕ v (V n (k)). In fact, the map v ϕ v induces an isomorphism V n (k) (V n (k)). 24

26 3..2 Counting Irreducible Representations We know that the number of ordinary representations of a group G is equal to the number of conjugacy classes of G. There is an analogue for modular representation theory: Theorem 3.5. The number of irreducible representations of a group G over a field of characteristic p is equal to the number of conjugacy classes of G whose elements have order coprime to p. Example 3.6. The group L 3 (2) has four conjugacy classes with elements of odd order; the class consisting of the identity element, the class with elements of order 3 and the two classes with elements of order 7. Thus L 3 (2) has four irreducible representations over fields of characteristic Permutation Representations Given a group G of permuations of a set Ω (i.e. a group which acts on a set), permutation representations give a very easy way of constructing representations of G. Although permutation representations are not in general irreducible, decomposing them may give a good way of constructing some irreducible representations. Given a finite field GF (q), we construct the permutation representation of G over Ω as the vector space whose basis is indexed by the elements of Ω. That is to say We can equivalently think of this as V = { α(ω)ω α(ω) GF (q)}. ω Ω GF (q) Ω = q Ω = {f Ω GF (q)} the set of functions from Ω to GF (q). Given an element ω Ω α(ω)ω of V, the map ω α(ω) uniquely determines an element of GF (q) Ω and vice versa. The group G acts on V and GF (q) Ω via its action on Ω. That is to say we have 25

27 a map G GL(V ) which sends g to λ(g) α(ω)ω α(ω)ω g ω Ω ω Ω f (f g ω ω g ). Example 3.7. The group S 5 has a 5-dimensional permutation over GF (2) with basis vectors {e, e 2, e 3, e 4, e 5 }. Choose g = ( 2 3)(4 5) and v = (,,,, ) then by permuting the basis vectors of V. g v (,,,, ) Lemma 3.8. The map g λ(g) is a representation (i.e. a homomorphism) of G into GL(V ). In practise, we consider λ(g) as acting on the basis vectors of V, then extend this action linearly. The vector space V q Ω is called the GF (q) permutation module of G acting on Ω. Proposition 3.9. If χ G k is the character of the permutation module of G acting on a set X then for any g G χ(g) = {x X gx = x}. That is to say, the value of a permuation character on an element g G is equal to the number of fixed points of g Submodules of Permuation Modules As usual, we say that a subspace U of V is a submodule if the map G GL(U) which send g λ(g) U is also a representation. We wish to study the submodule structure of permutation modules. To begin with, there are two obvious submodules which lie inside any permutation module: V () = { α(ω)ω α(ω) = α(δ) for some fixed δ Ω} = { the constant functions Ω GF (q)} V (n ) = { α(ω)ω α(ω) = }. 26

28 So V () is a -dimensional submodule and V (n ) is a (n )-dimensional submodule. Note that V () V (n ) if and only if ω Ω = = n i.e. if and only if p n where q = p m for some m N. We say that V (n ) /V () V (n ) is the heart of the representation. Theorem 3.. If G = Sym(Ω) then the heart of the representation q Ω irreducible. is Permutation Modules of 2-Transitive Actions Definition 3.. Suppose that G is a group acting on a set Ω. Then we say that the action of G on Ω is transitive if for all x, y Ω, there exists g G such that gx = y. Moreover, for an integer n Ω, we say that this action is n-transitive if for all subsets {x,..., x n } and {y,..., y n } of Ω such that the x i and y i are pairwise distinct, there exists g G such that gx i = y i for i n. Proposition 3.2. A group G acts 2-transitively on a set Ω if and only if it has exactly two orbits on Ω Ω. They are {(α, α) α Ω}; {(α, β) α, β Ω, α β}. Lemma 3.3 (Burnside s Lemma). Let G be a finite group acting on a set X then G {x X gx = x} = # orbits of G on X. g G This can of course be reformulated as G χ(g) = # orbits of G on X. g G where χ is the character of the permutation module of the action of G on X. 27

29 Proposition 3.4. If a group G acts on a set Ω then the permutation module over a field k of G acting on Ω decomposes as k Ω = V where is the trivial module and V Irr(G), V. Proof. Suppose that k Ω = n V mi i i= where V i Irr(G). Suppose that χ is the character of V and χ i is the character of V i for i n. Then However, we also have n χ, χ = m 2 i. i= χ, χ = χ χ, G where χ χ is the permutation character of the action of G on Ω Ω. Thus χ, χ = G (χ χ)(g) g G = # orbits of G on Ω Ω = 2. So the only possibilities for the values of the m i is that m = m 2 = and m i = for i, 2. One of the modules V, V 2 must be the trivial module and the other an irreducible module of dimension Ω Permutation Modules of Actions on Cosets of Subgroups Proposition 3.5. Suppose that G is a finite group with H G. Let Ω = G/H and let V = k Ω, the permutation module of G acting on Ω via right multiplication of cosets. Then for any irreducible submodule U of V, m V (U) = dim C U (H) where C U (H) = {v U hv = v h H}. 28

30 Proof. Using Propositions 3. and 3.9, we calculate χ V, χ U G = G χ V (g )χ U (g) = g G = G = G ω Ω g G gω=ω ω Ω h H = χ U, H H = m χ U (g) = G χ U (h) = H G g G ω Ω G(ω) ω Ω gω=ω χ U (h) h H where m is the multiplicity of the trivial module in U. χ U (g) χ U (g) Conversely, v V is contained in C U (H) if any only if the module v is isomorphic to the trivial module. Thus C U (H) = m i= V i where V i is isomorphic to the trivial module for i m. Thus dimc U (H) = m as required. Corollary 3.6. Suppose that V,..., V n are the irreducible representations of a group G. Suppose also that for i n, d i = dim(v i ). Then n G = d 2 i. i= Proof. We consider the action of G on itself by left multiplication (the regular action of G) and denotes its corresponding permuation module by k G. This can be equivalently thought of as the action of G on cosets of the trivial subgroup. So, for any V Irr(G), m k G(V ) = dim(c k G({e})) = dim(v ). So if V,..., V n are representatives of the isomorphism classes of the irreducible modules of G then k G = n V dim(vi) i. i= 29

31 Now let χ G be the character of k G. By the orthonormality of irreducible characters we have n χ G, χ G = (dim(v i )) 2. For g G G if g = e χ G (g) = if g e. i= So as required. χ G, χ G = G χ G (g )χ G (g) = G g G Corollary 3.7. A group G is abelian if and only if every irreducible representation is linear ( dimensional). Proof. Let n be the number of conjugacy classes of G. Then G is abelian if and only if n = G, with each class containing just one element. From Corollary 3.6, n G = d 2 i. and d i for i n. Thus n = G if and only if d i = for i n. i= 3.3 Representations of L 3 (2) over GF (2) First, note that any permutation representation over GF (2) of a group G via its action on Ω can be equivalently be thought of as the set of subsets of Ω i.e. {A A Ω}. This correspondence can be formalised by associating with each subset A Ω the vector α(ω)ω where α(ω) = if and only if ω A. We now return to the specific case of G = L 3 (2). We will consider the following three permuation modules: 3

32 . 2 Ω - the 7-dimensional permutation module based on the action of L 3 (2) on Ω = V 3 (2)/{}; the 8-dimensional permutation module based on the action of L 2 (7) on the -dimensional subspaces of V 2 (7); 3. 2 Λ - the 24-dimensional permutation module based on the action of L 3 (2) on Λ = G/S via right multiplication of cosets where S is a Sylow 7- subgroup of L 3 (2). We have already extensively studied the first two actions when looking at the Fano plane but this is the first time that we have seen the action of G on Λ. Our aim is to decompose these permutation modules into direct sums of the irreducible representations of G over GF (2). From Theorem 3.5, we know that there are four such representations. Note that 2 divides the order of G so these modules are not necessarily semisimple but there is still a fair amount that we can say about their internal structure The Permutation Module 2 Ω As per our initial discussion on general permutation modules, 2 Ω has the submodules V () and V (6). Moreover, since Ω = 7 is odd, these two submodules are disjoint and so we have 2 Ω = V (6) V (). In fact, the module V (6) has further structure V (6) = V 3 (2) /V 3 (2) where / denotes an indecomposable extension of modules and V 3 (2) is the dual of V 3 (2). We can think of V 3 (2) and V 3 (2) as V 3 (2) = a a 2 a 3 a, a 2, a 3 GF (2) V 3 (2) = {(b b 2 b 3 ) b, b 2, b 3 GF (2)}. 3

33 Moreover, it turns out that V 3 (2) and V 3 (2) are two of the irreducible modules of G over GF (2). We now have 2 Ω = (V 3 (2) /V 3 (2)). However, this is not a direct sum decomposition of 2 Ω as V 3 (2) /V 3 (2) / V 3 (2) V 3 (2) The Permutation Modules 2 Λ and 2 We now consider the permutation modules 2 Λ and 2. Our aim is to find the submodule decomposition of 2 Λ and to use this to show that the module contains a copy of the Golay code. We first note that 2 can equivalently be thought of as the permutation module of G acting by conjugation on its 8 Sylow 7-subgroups. Proposition 3.8. The permutation module 2 is a direct summand of 2 Λ. Proof. We will show that there exists a G-invariant surjective homomorphism φ 2 Λ 2. Then ker(φ) will be a submodule of 2 Λ such that 2Λ /ker(φ) 2. First take λ Λ. Then we define G(λ) = {g G g(λ) = λ} so that G(λ) is a conjugate of S and so is also a Sylow 7-subgroup of G. Then the map Λ sending λ G(λ) extends to a surjective G-invariant surjective homomorphism φ 2 Λ 2. Theorem 3.9. Let H be a Sylow 2-subgroup in G and let V be a module of G over GF (2). If U is a non-trivial submodule of V then H fixes at least one non-zero vector in U. Proof. If k is the dimension of U then U/{} = 2 k. 32

34 As U is stable under the action of G (and therefore the action of any of its subgroups) U is the union of H-orbits. Moreover every H-orbit has length 2 ei for some e i. Thus 2 k l = 2 ei i= and so it follows that e i = for some i. Then the corresponding orbit has length one so consists of a non-zero vector which is stabilised by H. We know that 2 is a submodule of 2 Λ but how do we go about finding other submodules? Note that u v u v 2 u v 3 V 3 (2) V 3 (2) = u 2 v u 2 v 2 u 2 v 3 u V 3 (2), v V 3 (2) M 3 (2) u 3 v u 3 v 2 u 3 v 3 and that G GL 3 (2) acts on V 3 (2) V 3 (2) via conjugation of matrices, turning it into a 9-dimensional module of G. It is easy to spot that V =, is a submodule of V 3 (2) V 3 (2). Its complement is V 8 = {M tr(m) = }. In fact this module V 8 is know as the Steinberg module of L 3 (2) and is an irreducible representation of G over GF (2). Proposition 3.2. where V () 8 and V (2) 8 are copies of V 8. 2 Λ = V () 8 V (2) 8 2 Proof. Consider S a Sylow 7-subgroup of G and suppose that S = s for some element s G. Then the characteristic polynomial p(s) is irreducible, so either p(s) = λ 3 + λ + or p(s) = λ 3 + λ

35 For any matrix, the coefficient of the second highest power in its characteristic equation is equal to its trace. Thus if χ(s) is equal to the first polynomial, s is traceless, else if it is equal to the second, it has trace. Moreover, if λ 3 + aλ 2 + bλ + c is the characteristic polynomial of s, then s has characteristic polynomial cλ 3 + bλ 2 + aλ +. On the other hand, s and s 2 have the same characteristic polynomial. It follows that there are exactly three elements of S whose corresponding matrices are traceless and are thus contained in a copy of V 8. Suppose without loss of generality that s is traceless. Then s 2 and s 4 are too. Thus C V8 (S) = {, s, s 2, s 4 } and m 2 Λ(V 8 ) = 2. Now, by comparing dimensions, along with Proposition 3.8, we have as required. 2 Λ = V () 8 V () 8 2. Note that there are actually three copies of V 8 in 2 Λ, the third of which we denote V (3) 8. If we index the basis vectors of V () 8 and V (2) 8 as v,..., v 8 and u,..., v 8 then V (3) 8 = {(v i, u i ) i 8}. Our next aim is to find an explicit basis of 2 Λ in terms of the right hand side of the equality proved in the previous result. We are trying to find a vector v s V () 8 V (2) 8 2 such that {v g s g G} forms a basis of 2 Λ. Chose a non-zero vector in 2. Then this vector is a Sylow 7-subgroup, call it S. Without loss of generality, we can assume that S = s where s is a traceless matrix of order 7. 34

36 Now let s and s 2 be the vectors corresponding to s in V () 8 and V (2) 8 respectively. We can take v s = (s, s i 2, S) for i =, 2 or 4. If i = then (s, s 2 ) V (3) 8 and v s V (3) 8 2. Thus i = 2 or 4. Without loss of generality, we can take i = 2 and v s = (s, s 2 2, s ). Then is a basis for 2. {v s s V 8, s = 7} All that remains is to show that 2 Λ contains a copy of the Golay code. Recall that the elements of index the columns of the extended Hamming code. Thus extended Hamming code is a submodule of 2 of dimension 4. Moreover, we know that L 2 (7) is the automorphism group of H. However, P GL 2 (7) also acts on H, and contains L 2 (7) as an index two subgroup. Take an element α P GL 2 (7)/L 2 (7) e.g. α =. We can also write this as a permutation of the columns of H to get α = ( )()( 6)(2 5)(3 4). Check that H α is an extended Hamming code which is L 2 (7)-invariant. Moreover H α H because α Aut(H) and H + H α is a submodule of 2 of codimension and H H α =. As the Golay code is 2-dimensional, it must contain one of the three Steinberg modules V (i) 8 for i =, 2, 3 and one of the two 4-dimensional submodules of 2 Λ, H and H α. In fact, all of these six different combinations of modules give us a submodule isomorphic to the Golay code. 3.4 Representations of L 3 (2) over C We start by deducing the dimensions of the irreducible representations. 35

37 The action of G L 3 (2) on the following sets is 2-transitive:, the -dimensional subspaces of V 2 (7); P, the points of the Fano plane; L, the lines of the Fano plane. So by Proposition 3.4, we have the following decompositions: C = V 7 ; C P = V () 6 ; C L = V (2) 6 ; for V 7, V () 6, V (2) 6 Irr(G). In fact, V () 6 V (2) 6. Now consider the action of L 3 (2) on the set of flags of the Fano plane: θ = {(p, l) p P, l L }. There are 2 incidence relations of the Fano plane so C θ is a 2-dimensional permutation module. It can be decomposed as C θ = V () 6 V (2) 6 V 8 where V 8 is an 8-dimensional irreducible module of G. When proving the simplicity and uniqueness of the simple group G of order 68 (which we now know to be L 3 (2)) we (implicitly) showed that G has 7 conjugacy classes, which we denote A, 2A, 3A, 4A, 7A, 7B. The notation NX is a standard way of representing a conjugacy class. The integer N refers to the order of the elements of the class. The smallest class with elements of order N is labelled NA, the next largest NB and so on. 36

38 Thus Irr(G) = #{ conjugacy classes of G} = 6. We already have four irreducible representations, one each of dimensions, 6, 7 and 8. Let x and y denote the dimensions of the remaining two irreducible representations. Then, from Corollary x 2 + y 2 = G = 68. Thus we have x 2 + y 2 = 8 and the only possibility is that x = y = 3 so that the final two irreducible representations of G have dimension The Ordinary Character Table of L 3 (2) We denote the irreducible representations of G as V, V () 3, V (2) 3, V 6, V 7, V 8. Theorem 3.2. The character table of G L 3 (2) L 7 (2) is A 2A 3A 4A 7A 7B χ χ () 3 3 b 7 b 7 χ (2) 3 3 b 7 b 7 χ χ 7 7 χ 8 8 where the second row gives the sizes of the conjugacy classes, and b 7 = and b 7 = 7. 2 Proof. Since we know the dimensions of the irreducible modules, we know the value which their characters take at e, giving the first column. The trivial character takes the value of one everywhere, giving the first row. 37

39 For the characters corresponding to the 6 and 8 dimensional representations, we recall that χ C P = χ C L = + χ 6 χ C θ = + 2χ 6 + χ 8. and that the value of a permutation character at an element g G is the number of fixed points of the permuation induced by g. Thus, for example, if g G then χ 6 (g) is equal to the number of points (or lines) of the Fano plane fixed by g minus. An element of order 2 in G acts on the Fano plane by stabilising a line l and by transposing the two points on each other line which do not also lie on l. Thus if g 2A, then g fixes three points (those on l) and 5 flags so χ 6 (g) = 3 = 2 and χ 8 (g) = 5 4 =. An element of order 3 in G acts on the Fano plane by fixing one point and acting as a cyclic permutation on the three lines which contain that point. Thus if g 3A then g fixes one point and no flags so χ 6 (g) = = and χ 8 (g) = =. An element of order 4 in G acts on the Fano plane by fixing one point, transposing two of the lines which is contained in and transposing the two other points on the third line in which it lies. Thus if g 4A, then g fixes one line and one flag (the fixed point and the line it is contained in which is not transposed with another) χ 6 (g) = = and χ 8 (g) = =. Finally, an element of order 7 in G acts cyclicly on the points and lines of the Fano planes. Thus if g 7A or g 7B, then g fixes no points and no flags and χ 6 (g) = = and χ 8 (g) = + 2 =. Let t =, s = 2 4, r = 6. 38

40 Then t is of order 2, s is of order 3 and r is of order 7. From the discussion following Propostition., we know that t fixes none of the -dimensional subspaces of V 2 (7); s fixes two such subspaces, those generated by and ; r fixes one such subspace, that generated by. As characters are constant on conjugacy classes, these observations give us the values of χ 7 for the conjugacy classes 2A, 3A and 7A. The remaining two values for χ 7 can be deduced from the orthogonality of the rows of the character table. We now turn our attention to the final two characters - those of dimension 3. Take g to be an element of the 7A conjugacy class. Then the restriction of the presentation to g will be the direct sum of three -dimensional representations (see Corollary 3.7). Thus the matrices of the image of g under each of the two 3-dimensional representations are of the form e m 7 e m2 7 e m3 7 for m i Z for i 3 and where e n = e 2πi n. Given an element h N L3(2)( g ) and an irreducible representation ρ of g the map ρ h x ρ(x h ) is also an irreducible representation. Thus it follows that we have an action of N L3(2)( g ) on the irreducible characters of g. If h g then h acts trivially (as characters are constant on conjugacy classes). If h N L3(2)( g )/ g then h induces a non-trivial permuation of the representations of g. However, as h L 3 (2), the action it induces on the characters of L 3 (2) must be trivial. Thus the action of N L3(2)( g ) on the representations in question must preserve the set {m, m 2, m 3 }. 39

41 In particular, N L3(2)( g ) is generated by g and a 3-cycle h such that h gh = g 2. Thus the possibilities for {m, m 2, m 3 } are {3, 5, 6} and {, 2, 4} giving character values of b 7 and b 7 for χ () 3 and χ (2) 3 respectively. The values of these characters on 7B can be deduced from the orthogonality of the columns of the character table. Moreover, the automorphism ρ () 3 (g) of V clearly has three distinct eigenvalues, e 7, e 2 7 and e 4 7 so we have a basis of V consisting of eigenvectors of ρ () 3 (g). As the element h permutes these eigenvectors, ρ () 3 (g) will be a matrix conjugate to. This argument also holds for ρ (2) 3 so we can conclude that the character values of the two 3-dimensional representations are on the conjugacy class of elements of order 3. We now claim that the image of L 3 (2) under either of the representations of dimension 3 is contained in SL 3 (C). Since L 3 (2) is simple and the representations in question are not trivial, both have trivial kernels. Thus they give embeddings of L 3 (2) into GL 3 (C). As SL 3 (C) is normal in GL 3 (C), the intersection of the image of L 3 (2) and SL 3 (C) is normal in the image of L 3 (2) and so must either be trivial, or must be the whole of the group. The intersection cannot be trivial as we have already shown that an element of order 7 in L 3 (2) is mapped to a matrix with determinant in GL 3 (2). Thus the image of L 3 (2) under the two representations of dimension 3 is contained in SL 3 (C), as required. Now consider an element x of order 2. By the same reasoning as in the case of the 7A conjugacy class, the image of x under one of the representations of dimension 3 the must be of the form e n 2 e n2 2 e n 2 for n i Z for i 3. We must have (n i, 2) = for some i 3, else the 4

42 representation would be trivial. Along with the fact that this matrix must have determinant, this implies that the only possibility for {n, n 2, n 3 } is {,, 2}. Thus both characters of dimension 3 take the value - on the conjugacy class of involutions. Finally, we consider an element y of order 4. As before, the image of y under either of the representations of dimension 3 must be of the form e p 4 e p4 4 e p 4 for p i Z for i 3. Moreover, we must have (p i, 4) = for at least one i 3, else the matrix would be of order or 2. In this case, N L3(2)( y ) is generated by y and z, an element of order 2 such that z yz = y 3. Using these observations, along with the fact that the image of y must be a matrix of determinant, we deduce that the only possibility for {p, p 2, p 3 } is {, 3, 4}. Thus both characters of dimension 3 take the value on the conjugacy class of elements of order 4. 4

43 Appendix A Error Correcting Codes Definition A.. A (binary) code of length n is a subset C of V n (2). The vectors in C are called codewords. The distance between two codewords x, y is d(x, y), the number of coordinates where x and y differ in value. Definition A.2. The minimum distance of a code C is d(c) = min{d(x, y) x, y C, x y}. Definition A.3. A code C is a linear code if it is a subspace of V n (2). The motivation for studying codes is that they offer a certain level of error correction. Suppose we transmit a codeword c C but e errors are made and we actually receive vector b C. Then, as long as e is sufficiently small, if we send b to its nearest vector in C, this vector will be our original codeword c. So we can think of each codeword as being a (binary) message which we want to transmit, along with a certain number of bits of extra information which will enable us to recover c if it is corrupted in transmission. Formally, we say that a code C V n (2) corrects e errors if for any c, c C and x V n (2), if d(c, x) e and d(c, x) e then c = c. Ideally, we are looking for codes which correct a high number of errors and have low dimension and minimum distance. 42

44 Definition A.4. If A M n (2) and C = {x V n (2) Ax = } then C is a linear code and A is said to be its check matrix. Proposition A.5. Let C be a code with check matrix A. Suppose. all columns of A are different; 2. no column of A is the zero vector. Then C corrects at least one error. Proposition A.6. Let d 2 and let C be a linear code with check matrix A. Assume that every set of d columns of A are linearly independent. Then d(c) d Definition A.7. Let k 2. A Hamming code Ham(k) is a code whose check matrix has for columns all non-zero vectors in V k (2). A Hamming code Ham(k) has minimum distance 3; has length 2 k ; has dimension 2 k k ; corrects one error. It is a perfect code which (informally) means it corrects the highest possible number of errors for a given dimension and minimum weight. There are only three possible types of perfect codes, the Hamming codes which corrects one error, the codes C = { n, n } which correct n 2 errors and a third, called the Golay code. The Golay code is actually constructed from the Hamming code. We start with a code H = Ham(3) with check matrix 43

45 and it s reverse K with check matrix. Then, to each codeword in H and K, we add a parity check bit. That is to say, given a vector (v,..., v 7 ) we add an eighth bit 7 v i i= v 8 = which is if there is an even number of s and if there are an odd number. This gives us two codes H and K of length 8. Definition A.8. The extended Golay code is defined as G 24 = {(a + x, b + x, a + b + x) a, b H, x K } V 24 (2) The Golay code G 23 consists of the codewords of the extended Golay code with the last bit removed. The Golay code is perfect and has minimum length 7; has length 23; has dimension 2; corrects 3 errors. 44