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Groups and Representations 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 2015. Contents 1 Linear Groups 3 1.1 Finite Fields . .3 1.2 Key Groups . .4 1.2.1 The General Linear Group . .4 1.2.2 The Projective General Linear Group . .5 1.2.3 The Special Linear Group . .5 1.2.4 The Projective Special Linear Group . .5 1.3 A Simple Group of Order 60 . .6 1.4 A Simple Group of Order 168 . .8 1.4.1 The Fano Plane . .8 1.4.2 The Automorphism Group of the Projective Plane . 12 2 Stabilisers of Subgroups in GL(V ) 18 2.1 Semidirect Products . 18 2.2 General Theory of Stabilisers . 21 1 3 The Representation Theory of L3(2) 22 3.1 Representation Theory . 22 3.1.1 The Dual of a Vector Space . 24 3.1.2 Counting Irreducible Representations . 25 3.2 Permutation Representations . 25 3.2.1 Submodules of Permuation Modules . 26 3.2.2 Permutation Modules of 2-Transitive Actions . 27 3.2.3 Permutation Modules of Actions on Cosets of Subgroups 28 3.3 Representations of L3(2) over GF (2) ................. 30 3.3.1 The Permutation Module 2Ω ................. 31 3.3.2 The Permutation Modules 2Λ and 2∆ ............ 32 3.4 Representations of L3(2) over C .................... 35 3.4.1 The Ordinary Character Table of L3(2) ........... 37 A Error Correcting Codes 42 2 Chapter 1 Linear Groups In this chapter, we will introduce some specific examples of linear groups. 1.1 Finite Fields We first recall some key results concerning finite fields. Theorem 1.1. Let p be a prime number and m be a positive integer. Then there exists a unique (up to isomorphism) field of order q ∶= pm. We call it GF (q). If m = 1 in the above theorem then GF (p) ≅ Z~pZ. For m > 1, 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 1.2. We will construct GF (4). We have GF (2) = Z~2Z = {0; 1}. Take 3 f(x) = x2 + x + 1 then GF (4) = {0; 1; x; x + 1}: Note that (GF (q)∗; ×) is isomorphic to the cyclic group of order q − 1. We denote by Vn(q) the n-dimensional vector space over GF (q). 1.2 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−1h−1gh. Equivalently, the commutator subgroup can be defined as the minimal subgroup H such that G~H is abelian. 1.2.1 The General Linear Group We define GLn(q) ∶= the set of non-singular n × n matrices over GF (q) ∶= the set of linear bijections of from Vn(q) to itself: It is possible to show that these two definitions are equivalent by choosing a basis of Vn(q). The order of GLn(q) is n n(n−1) i SGLn(q)S = q 2 M(q − 1): i=1 If we let G = GLn(q) then ∗ Z(G) = {λIn ∶ λ ∈ GF (q) } 4 and G′ = {A ∶ det(A) = 1}: ∗ Note that the map det ∶ GLn(q) → GF (q) is a homomorphism. 1.2.2 The Projective General Linear Group We define GL (q) P GLn(q) ∶= n ~Z(GLn(q)): Then SGL (q)S SP GL (q)S = n : n q − 1 1.2.3 The Special Linear Group ∗ We note that the map det ∶ GLn(q) → GF (q) is a homomorphism and define SLn(q) ∶= ker(det): Then SLn(q) ◁ GLn(q) and GL (q) n ≅ GF (q)× SLn(q) so SGL (q)S SSL (q)S = n : n q − 1 1.2.4 The Projective Special Linear Group We define SL (q) P SLn(q) ∶= n ~Z(GLn(q))∩SLn(q): Then SSL (q)S SP SL (q)S = n n (n; q − 1) where (n; q − 1) denotes the highest common factor of n and q − 1. We also denote P SLn(q) as Ln(q). In the tables below we calculate the value of SL2(q)S for some small values of q. 5 q = 2 3 4 5 6 7 8 9 SL2(q)S 6 12 60 60 - 168 8 ⋅ 63 360 Note that SL2(4)S = SL2(5)S = SA5S = 60. In fact, these three groups are isomorphic to each other. We can also show that SL2(7)S = SL3(2)S = 168, again these groups are isomorphic to each other. 1.3 A Simple Group of Order 60 In this section, we will prove that there exists a unique (up to isomorphism) simple group of order 60. We will require a few well known results in this proof: Theorem 1.3. The group A5 is simple. Proof. See Problem Sheet 1. Definition 1.4. If G is of order pam where p is a prime not dividing m, then a subgroup H of G of order pa is called a Sylow p-subgroup of G. The number of Sylow p-subgroups of G will de denoted np. Theorem 1.5 (Sylow's Theorem). Let G be a group of order pam, where p is a prime not dividing m. Then the following hold: 1. 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−1. That is, any two Sylow p-subgroups of G are conjugate in G. 3. The number of Sylow p-subgroups of G, np satisfies np ≡ 1 mod p. Further, np is the index of the normaliser NG(P ) of any Sylow p-subgroup P, and as such npSm. Corollary 1.6. If G is simple then np ≠ 1. 6 Proof. Follows from the fact that any two Sylow p-subgroups are conjugate in G. Theorem 1.7. Suppose that H is a simple group of order 60. Then H ≅ A5. Proof. Suppose that H is simple of order 60 = 22 ⋅3⋅5. Then by Sylow's Theorem n2 = 1; 3; 5 or 15; n3 = 1; 4 or 10; n5 = 1 or 6: As H is simple, Corollary 1.6 implies that n5 = 6. Let (1) (2) (3) (4) (5) (6) K ∶= {K5 ;K5 ;K5 ;K5 ;K5 ;K5 } be the set of Sylow 5-subgroups and let M = Sym(K) ≅ S6. 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 S6. Moreover, φ(H) ≤ Alt(K) ≅ A6, 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 homo- morphism µ ∶ Alt(K) → Sym(L) ≅ S6 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):1 so is isomorphic to a copy of A5 in Alt(L). Thus we have that H ≅ A5 is the only simple group of order 60. 7 1.4 A Simple Group of Order 168 In this section, we will prove that there exists a unique (up to isomorphism) simple group of order 168. 1.4.1 The Fano Plane We consider the group L3(2) = P SL3(2) = SL3(2) = GL2(3) = P GL3(2). It is the automorphism group of V3(2). Note that for any vector space over a finite field Q v = 0: v∈V In fact, for any subspace U of V Q v = 0: v∈U For any v ∈ V , scalar multiplication is easy to determine as 0v = 0 and 1v = v. In order to determine addition in V3(2), we consider its two dimensional sub- spaces. The number of two dimensional subspaces is 3 (23 − 1)(23 − 2) = = 7 2 2 2 2 (2 − 1)(2 − 2) where n is defined to be the number of k-dimensional subspaces in V (q) and k q n is equal to n (qn − 1)(qn − q) ::: (qn − qk−1) = : k k k k−1 k q (q − 1)(q − q) ::: (q − q ) Each such subspace U contains three non-zero vectors, say U = {0; u1; u2; u3}. Thus, using the fact that ∑v∈U v = 0, we can define u1 + u2 to be u3, the third non-zero vector in the 2-dimensional subspace containing u1 and u2. Using this observation, we can express the two dimensional subspaces of V3(2) as the Fano plane (the projective plane of order 2). 8 ⎛1⎞ ⎜ ⎟ ⎜0⎟ ⎝0⎠ ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜1⎟ ⎛1⎞ ⎜0⎟ ⎜ ⎟ ⎝0⎠ ⎜1⎟ ⎝1⎠ ⎝1⎠ ⎛0⎞ ⎛0⎞ ⎜ ⎟ ⎜ ⎟ ⎜1⎟ ⎛0⎞ ⎜0⎟ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ 0 ⎜1⎟ 1 ⎝1⎠ Here, the points correspond to the vectors in V and the lines correspond to the 2-dimensional subspaces of V .
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