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) PSLn(q) ∶= n ~Z(GLn(q))∩SLn(q). Then SSL (q)S SPSL (q)S = n n (n, q − 1) where (n, q − 1) denotes the highest common factor of n and q − 1.
We also denote PSLn(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) = PSL3(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 . 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) ∶= (alp)l∈L ,p∈P
9 where ⎧ ⎪1 if p ∈ l alp = ⎨ ⎪ ⎩⎪0 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.
p1 p2 p3 p4 p5 p6 p7
l1 0 1 1 0 1 0 0
l2 0 0 1 1 0 1 0
l3 0 0 0 1 1 0 1
l4 1 0 0 0 1 1 0
l5 0 1 0 0 0 1 1
l6 1 0 1 0 0 0 1
l7 1 1 0 1 0 0 0 ⎛1⎞ ⎛0⎞ ⎛0⎞ ⎛1⎞ ⎛0⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜0⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜1⎟ ⎜1⎟ ⎜0⎟ ⎝0⎠ ⎝0⎠ ⎝1⎠ ⎝0⎠ ⎝1⎠ ⎝1⎠ ⎝1⎠
Note that the matrix above obeys the following rules:
A1 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 1;
A6 The inner product of two distinct columns is 1; 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 1.8. A plane whose incidence matrix obeys the axioms A1 - A6 is unique up to isomorphism.
10 Sketch of proof. Any plane is isomorphic to another plane whose incidence ma- trix 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.
p1 p2 p3 p4 p5 p6 p7
l1 1 1 1 0 0 0 0
l2 1 0 0 1 1 0 0 l 1 0 0 0 0 1 1 M = 3 l4 0 1 0 0 1 0 0
l5 0 1 0 0 1 0 1
l6 0 0 1 1 0 0 1
l7 0 0 1 0 1 1 0
There is yet another equivalent way in which we can consider the Fano plane. We now take P = GF (7), Q ∶= {α2 ∶ α ∈ GF (7)} = {1, 2, 4} and L = {Q + i ∶ i ∈ GF (7)}. Note that the set Q is a difference set.
Proposition 1.9. The plane with points and lines described above is the Fano plane.
Proof. Check that the incidence matrix obeys the axioms A1-A6.
Expressing elements of L as vectors in V7(2) where the ith coordinate of the vector corresponding to l ∈ L is 1 if and only if pi ∈ 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
11 ⎛0⎞ ⎛1⎞ ⎛0⎞ ⎛0⎞ ⎛1⎞ ⎛0⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜0⎟ ⎜0⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜1⎟ ⎜1⎟ ⎜0⎟ ⎝0⎠ ⎝0⎠ ⎝0⎠ ⎝1⎠ ⎝0⎠ ⎝1⎠ ⎝1⎠ ⎝1⎠ 1 0 1 1 0 1 0 0 1 0 0 1 1 0 1 0 1 0 0 0 1 1 0 1 1 1 0 0 0 1 1 0 1 0 1 0 0 0 1 1 1 1 0 1 0 0 0 1 . 1 1 1 0 1 0 0 0 0 1 0 0 1 0 1 1 0 1 1 0 0 1 0 1 0 1 1 1 0 0 1 0 0 0 1 1 1 0 0 1 0 1 0 1 1 1 0 0 0 0 1 0 1 1 1 0 0 0 0 1 0 1 1 1
This matrix is of the form ⎡ ⎤ ⎢ ⎥ ⎢ 1 M ⎥ N = ⎢ ⎥ ⎢ ′ ⎥ ⎣ 0 M ⎦ where M is the incidence matrix of (P, L ) and M ′ is its complement.
The rows of this matrix, along with the vectors 1 and 0 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 ∶= V3(2) and the rows correspond to U + v where U is a fixed 2-dimensional subspace of V and v ∉ U.
1.4.2 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 }.
12 Labelling the points as p1, . . . , p7 means we can write such automorphisms as elements of S7.
Example 1.10.
φ = (1)(2 4)(3 5)(7 6) ∉ Aut(Π) φ = (1)(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 1-dimensional subspaces in V2(7) with the following correspondences:
⎛0⎞ ⎛1⎞ ⎛0⎞ ⎛0⎞ ⎛1⎞ ⎛0⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜0⎟ ⎜0⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜1⎟ ⎜1⎟ ⎜0⎟ ⎝0⎠ ⎝0⎠ ⎝0⎠ ⎝1⎠ ⎝0⎠ ⎝1⎠ ⎝1⎠ ⎝1⎠
Õ Õ Õ Õ Õ Õ Õ Õ × × × × × × × × Ö Ö Ö Ö Ö Ö Ö Ö
⎛0⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎝1⎠ ⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ ⎝6⎠
Note that from here onwards, when we write a vector of L2(7), we are actually refering to the 1-dimensional subspace which it lies in.
This correspondence allows us to consider L2(7) as acting on the Fano plane via its actions on the points.
Proposition 1.11. For each g ∈ L2(7), the map Π → Π induced by the action of g is an automorphism. That is to say L2(7) ≤ Aut(Π).
Proof. We consider the action of L2(7) on M, the incidence matrix of Π, and show that the image of M under an element of L2(7) obeys the axioms A1-A6 so corresponds to a plane isomorphic to Π. The group L2(7) is generated by (representatives of) the matrices
⎛1 0⎞ ⎛2 0⎞ ⎛0 1⎞ t = , s = , r = . ⎝1 1⎠ ⎝0 4⎠ ⎝6 0⎠
13 For α ∈ GF (7), the matrix t maps
⎛1⎞ ⎛ 1 ⎞ ↦ . ⎝α⎠ ⎝α + 1⎠
Rows of M can also be thought of as subsets of the form l = {1 + 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
⎛1⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ ↦ = . ⎝α⎠ ⎝4α⎠ ⎝2α⎠
So will map l = {1 + i, 2 + i, 4 + i} to {2 + 2i, 4 + 2i, 8 + 2i} ≡ {1 + 2i, 2 + 2i, 4 + 2i}. So similarly s permutes the rows of M.
The case of r is not so straightforward. It sends maps
⎛1⎞ ⎛ α ⎞ ↦ . ⎝α⎠ ⎝6α⎠ Check by hand that this maps
⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛2⎞ ⎛1⎞ ⎛1⎞ ⎛3⎞ ⎛1⎞ ↦ , ↦ , ↦ = , ↦ = ⎝0⎠ ⎝0⎠ ⎝1⎠ ⎝6⎠ ⎝2⎠ ⎝6⎠ ⎝3⎠ ⎝3⎠ ⎝6⎠ ⎝2⎠ ⎛1⎞ ⎛4⎞ ⎛1⎞ ⎛1⎞ ⎛5⎞ ⎛1⎞ ⎛1⎞ ⎛6⎞ ⎛1⎞ ↦ = , ↦ = , ↦ = ⎝4⎠ ⎝6⎠ ⎝5⎠ ⎝5⎠ ⎝6⎠ ⎝4⎠ ⎝6⎠ ⎝6⎠ ⎝1⎠ and that the image of the matrix M under this map obeys axioms A1-A6 so is still an incidence matrix of the Fano plane.
Theorem 1.12. 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 non- collinear, 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.
14 Proposition 1.13. SAut(Π)S ≤ 168
Proof. From Theorem 1.12 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 168 automorphisms of Π.
Theorem 1.14. L2(7) ≅ Aut(Π)
Proof. From Proposition 1.11 and Proposition ?? we know that L2(7) ≤ Aut(Π) and SAut(Π)S ≤ 168. However, SL3(2)S = 168 so we must have L3(2) ≅ Aut(Π).
Any element l of L clearly acts on U. Recall that we have a map U β ∶ → V 1 sending
⎛0⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎝1⎠ ⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ ⎝6⎠ ↧ ↧ ↧ ↧ ↧ ↧ ↧ ↧ ⎛0⎞ ⎛1⎞ ⎛0⎞ ⎛0⎞ ⎛1⎞ ⎛0⎞ ⎛1⎞ ⎛1⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜0⎟ ⎜0⎟ ⎜1⎟ ⎜0⎟ ⎜1⎟ ⎜1⎟ ⎜1⎟ ⎜0⎟ . ⎝0⎠ ⎝0⎠ ⎝0⎠ ⎝1⎠ ⎝0⎠ ⎝1⎠ ⎝1⎠ ⎝1⎠
We can use this to define an action of l on V . We say that
vl = β((β−1(v))l) giving the commutative diagram
β−1 v β−1(v) l l vl (β−1(v))l β .
15 We define
λ(l) = t0l ⋅ l where tv ∶ u ↦ u + v for u, v ∈ V . Then λ is our desired isomorphism. Example 1.15. Take ⎛ ⎞ ˆ 1 1 l = ∈ GL2(7) ⎝0 1⎠ and let l be its representative in L2(7). Then l maps
∞ 0 1 2 3 4 5 6 ⎛0⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎝1⎠ ⎝0⎠ ⎝1⎠ ⎝2⎠ ⎝3⎠ ⎝4⎠ ⎝5⎠ ⎝6⎠ l ↧ ↧ ↧ ↧ ↧ ↧ ↧ ↧ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛1⎞ ⎛0⎞ ⎝1⎠ ⎝0⎠ ⎝4⎠ ⎝3⎠ ⎝6⎠ ⎝5⎠ ⎝2⎠ ⎝1⎠
So we can write l as a permutation
l = (∞ 1 4 5 2 3 6).
We have l ⎛0⎞ ⎛⎛0⎞ ⎞ ⎛⎛1⎞⎞ ⎜ ⎟ 0 = β ⎜ ⎟ = β = ⎜1⎟ . ⎝⎝1⎠ ⎠ ⎝⎝1⎠⎠ ⎜ ⎟ ⎝0⎠
Writing t0l as a permutation of elements of V gives
⎛⎛0⎞ ⎛0⎞⎞ ⎛⎛1⎞ ⎛1⎞⎞ ⎛⎛0⎞ ⎛0⎞⎞ ⎛⎛1⎞ ⎛1⎞⎞ ⎜⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟⎟ l = ⎜⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟⎟ ⎜⎜ ⎟ ⎜ ⎟⎟ t0 ⎜⎜0⎟ ⎜1⎟⎟ ⎜⎜0⎟ ⎜1⎟⎟ ⎜⎜0⎟ ⎜1⎟⎟ ⎜⎜1⎟ ⎜0⎟⎟ ⎝⎝0⎠ ⎝0⎠⎠ ⎝⎝0⎠ ⎝0⎠⎠ ⎝⎝1⎠ ⎝1⎠⎠ ⎝⎝1⎠ ⎝1⎠⎠ (∞ 1)(0 3)(2 4)(5 6).
So
λ(l) = t0l ⋅ l = (∞)(1 2 0 3 5 4 6). Using the basis {(1, 0, 0)⊺, (0, 1, 0)⊺, (0, 0, 1)⊺}, we can write λ(l) as the linear transformation ⎛1 0 1⎞ ⎜ ⎟ ( ) = ⎜ ⎟ λ l ⎜1 0 0⎟ . ⎝0 1 0⎠
16 Using this isomorphism and Theorem 1.14 we have
L3(2) ≅ L2(7) ≅ Aut(Π).
17 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 G(U) ∶= StabGL(V )(U) = {g ∈ GL(V ) ∶ U = U}.
2.1 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.
G Example 2.1. If N ≅ C2 and ~N ≅ C2 then G may be equal to C4 or C2 × C2.
18 Suppose that N ⊴ G and K ≤ G. Then we have a map
φ ∶ G → G~N g ↦ Ng.
K When is φSK ∶ K → ~K∩N an isomorphism?
• The kernel of φSK is K ∩ N so φSK is injective if and only if K ∩ N = 1;
• We will have Im(φSK ) = Im(φ) if and only if ∀g ∈ G, ∃k ∈ K such that
Ng = Nk. So φSK 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 = 1 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−1nk).
Thus, given the triple (N, K, λ), we can recover the group G = N ⋊ K. In 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
N ⋊λ K ∶= {(k, n) ∶ k ∈ K, n ∈ N}. with multiplication
λ(k1) (k1, n1) ⋅ (k2, n2) = (k1k2, n1 n2).
Theorem 2.3. The set N ⋊λ K as defined above is a group.
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 12). It is a
19 subspace of V8(2) and has parity-check matrix
⎛0 1 0 0 1 0 1 1⎞ ⎜ ⎟ ⎜ ⎟ ⎜0 0 1 0 1 1 0 1⎟ H = ⎜ ⎟ . ⎜ ⎟ ⎜0 0 0 1 0 1 1 1⎟ ⎝1 1 1 1 1 1 1 1⎠
So the group GL4(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 V8(2). This action will preserve the code if and only if it permutes the columns of H.
Using the basis {e1, e2, e3, e4} of V ∶= V4(2), the columns of H form the following set
X = {α1e1 + ... + α4e4 S α4 = 1}. This is not a subspace of V but the set
Y = {α1e1 + ... + α4e4 S α4 = 0}. 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 GL3(2) on the remaining basis vectors.
Thus the stabiliser of the Hamming code in G = GL4(2) is ⎧ ⎫ ⎪ ⎛ 0 ⎞ ⎪ ⎪ ⎜ ⎟ ⎪ ⎪ ⎜ ⎟ ⎪ ⎪ ⎜ A 0 ⎟ ⎪ G{H } = ⎨ ⎜ ⎟ ∶ A ∈ GL3(2), x, y, z ∈ GF (2) ⎬ = N ⋊λ K ⎪ ⎜ 0 ⎟ ⎪ ⎪ ⎜ ⎟ ⎪ ⎪ ⎪ ⎩⎪ ⎝ x y z 1 ⎠ ⎭⎪
3 where N ≅ C2 , K ≅ GL3(2) and λ is the natural action of GL3(2) on V3(2).
Note that the action of the subgroup L isomorphic to GL3(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 ≅ L2(7) also acts on the columns of H via its action on the 1-dimensional subspaces of V2(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.
20 n To have two non-conjugate copies of Ln(2) in 2 ⋊ Ln(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 = Vn(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 ) ≅ GLn(q).
Take B = {b1, . . . , bn} be a basis of V and let U = Span(b1, . . . , bm) for some m ≤ n.
Then we have ⎧ ⎫ ⎪⎛ X 0 ⎞ ⎪ G(U) = ⎨ ∶ X ∈ GL(U),Y ∈ GL(W ),P ∈ M (q)⎬ ⎪⎝ ⎠ n−m,m ⎪ ⎩⎪ PY ⎭⎪ where W = Span(bm+1, . . . , bn) and Mn−m,m(q) is the set of matrices with n−m rows and m columns over GF (q). This gives
m(n−m) SG(U)S = SGLm(q)S ⋅ SGLn−m(q)S ⋅ q .
It can also be deduced that n SGL(V )S = SGL(U)S ⋅ . m q
Thus we have that
G(U) ≅ P ⋊λ K where
• K = Gm(q) × Gn−m(q) corresponding to the matrices X and Y ;
(n−m)m • P = (Cq) corresponding to (n − m)m copies of the additive group of GF (q).
21 Chapter 3
The Representation Theory of L3(2)
In this chapter we will construct a number of representations of the group
G = L3(2). We will be working both with fields of characteristic 0 (ordinary representation theory) and fields of positive characteristic (modular representa- tion 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.
3.1 Representation Theory
Recall that given two modules V1 and V2 of a group G along with representations
ϕ1 ∶ G → GL(V1) and ϕ2 ∶ G → GL(V2), we define the direct sum of V1 and V2 to be
V1 ⊕ V2 ∶= {(v1, v2) ∶ v1 ∈ V1, v2 ∈ V2} with representation ϕ = ϕ1 ⊕ ϕ2 such that
ϕ ∶ (v1, v2) ↦ (ϕ1(g)v1, ϕ2(v2)).
22 If there exist submodules V1 and V2 of V such that V = V1 ∪ V2 and V1 ∩ V2 = 0 then V = V1 ⊕ V2.
Our main strategy for constructing representations of L3(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 = V1 ⊕ ... ⊕ Vk (3.1) where V1,...,Vk 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 mV (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 3.1 becomes
l ⊕mi V = ? Vi = V1 ...V1 ⊕ ... ⊕ Vl ...Vl i=1 ´¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¶ ´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶ m1 times ml times where the Vi are pair-wise non-isomorphic irreducible representations of G and mi ∶= mV (Vi).
Proposition 3.1. If V is any module of a finite group and U is an irreducible module of the same group then
mV (U) = ⟨χV , χU ⟩ where ⟨ , ⟩ is the inner product on characters of G.
Proof. We decompose V as
V = U ⊕m(U) ? W m(W ). W ∈Irr(G) W ≠U We then take the inner product of both sides with U to get
⟨χV , χU ⟩ = m(U)⟨χU , χU ⟩ Q ⟨χW , χU ⟩. χW ∈Irr(G) W ≠U
23 However, recall that the irreducible representations are orthonormal with re- spect to ⟨ , ⟩ so ⎧ ⎪1 if W = U ⟨χW , χU ⟩ = ⎨ ⎪ ⎩⎪0 if W ≠ U. Thus
mV (U) = ⟨χV , χU ⟩.
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 submod- ules V1,V2 of a module such that V ~V1 ≅ V2 but V ≅~ V1 ⊕ V2. In this case, we say that V is an indecomposable extension of V1 by V2 which we denote
V = V1V2.
3.1.1 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 ∶= Homk(V, k).
Example 3.4. What is the dual space of Vn(k) for a given field k? Given two vectors v, w ∈ Vn(k), their dot product lies in k. So, for a given vector v ∈ Vn(k), we can define a function
ϕv ∶ w ↦ v ⋅ w. ∗ It is clear that ϕv ∈ (Vn(k)) . In fact, the map v ↦ ϕv induces an isomorphism ∗ Vn(k) ≅ (Vn(k)) .
24 3.1.2 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 L3(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 L3(2) has four irreducible representations over fields of characteristic 2.
3.2 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 representa- tions of G. Although permutation representations are not in general irreducible, decomposing them may give a good way of constructing some irreducible rep- resentations.
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 V ∶= { Q α(ω)ω ∶ α(ω) ∈ GF (q)}. ω∈Ω We can equivalently think of this as
GF (q)Ω = qΩ ∶= {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 a map G → GL(V ) which sends g to
λ(g) ∶ Q α(ω)ω ↦ Q α(ω)ωg ω∈Ω ω∈Ω −1 f ↦ (f g ∶ ω ↦ ωg ).
Example 3.7. The group S5 has a 5-dimensional permutation over GF (2) with basis vectors {e1, e2, e3, e4, e5}. Choose g = (1 2 3)(4 5) and v = (1, 0, 1, 1, 0) then
g ∶ v ↦ (1, 1, 0, 0, 1) by permuting the basis vectors of 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 ≅ qΩ 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) = S{x ∈ X ∶ gx = x}S.
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.
3.2.1 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)SU 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 permuta- tion module:
V (1) ∶ = Q α(ω)ω ∶ α(ω) = α(δ) for some fixed δ ∈ Ω = { the constant functions Ω → GF (q)} V (n−1) ∶ = Q α(ω)ω ∶ Q α(ω) = 0 .
26 So V (1) is a 1-dimensional submodule and V (n−1) is a (n − 1)-dimensional sub- module.
(1) (n−1) Note that V ⊆ V if and only if ∑ω∈Ω 1 = 0 = n i.e. if and only if pSn where m q = p for some m ∈ N. We say that
(n−1) V ~V (1)∩V (n−1) is the heart of the representation.
Theorem 3.10. If G = Sym(Ω) then the heart of the representation qΩ is irreducible.
3.2.2 Permutation Modules of 2-Transitive Actions
Definition 3.11. 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 ≤ SΩS, we say that this action is n-transitive if for all subsets {x1, . . . , xn} and {y1, . . . , yn} of Ω such that the xi and yi are pairwise distinct, there exists g ∈ G such that gxi = yi for 1 ≤ i ≤ n.
Proposition 3.12. A group G acts 2-transitively on a set X if and only if it has a exactly two orbits on Ω × Ω. They are
• {(α, α) ∶ α ∈ Ω};
• {(α, β) ∶ α, β ∈ Ω, α ≠ β}.
Lemma 3.13 (Burnside’s Lemma). Let G be a finite group acting on a set X then 1 Q S{x ∈ X ∶ gx = x}S = # orbits of G on X. SGS g∈G
This can of course be reformulated as 1 Q χ(g) = # orbits of G on X. SGS g∈G where χ is the character of the permutation module of the action of G on X.
27 Proposition 3.14. If a group G acts on a set Ω then the permutation module over a field k of G acting on Ω decomposes as
kΩ = 1 ⊕ V where 1 is the trivial module and V ∈ Irr(G), V ≠ 1.
Proof. Suppose that n Ω ⊕mi k = ? Vi i=1 where Vi ∈ Irr(G). Suppose that χ is the character of V and χi is the character of Vi for 1 ≤ i ≤ n. Then n 2 ⟨χ, χ⟩ = Q mi . i=1 However, we also have
⟨χ, χ⟩ = ⟨χ × χ, 1G⟩ where χ × χ is the permutation character of the action of G on Ω × Ω. Thus 1 ⟨χ, χ⟩ = Q (χ × χ)(g) SGS g∈G = # orbits of G on Ω × Ω = 2.
So the only possibilities for the values of the mi is that m1 = m2 = 1 and mi = 0 for i ≠ 1, 2. One of the modules V1,V2 must be the trivial module and the other an irreducible module of dimension SΩS − 1.
3.2.3 Permutation Modules of Actions on Cosets of Sub- groups
Proposition 3.15. Suppose that G is a finite group with H ≤ G. Let Ω ∶= G~H and let V = kΩ, the permutation module of G acting on Ω via right multiplication of cosets. Then for any irreducible submodule U of V ,
mV (U) = dim CU (H) where
CU (H) ∶= {v ∈ U ∶ hv = v ∀h ∈ H}.
28 Proof. Using Propositions 3.1 and 3.9, we calculate
1 −1 1 ⟨χV , χU ⟩ = Q χV (g )χU (g) = Q Q χU (g) SGS g∈G SGS g∈G v∈V gv=v 1 1 = Q Q χU (g) = Q Q χU (g) SGS v∈V g∈G SGS v∈V G(v) gv=v 1 1 = Q Q χU (h) = Q χU (h) SGS v∈V h∈H SGS h∈H
= dim CU (H).
Corollary 3.16. Suppose that V1,...,Vn are the irreducible representations of a group G. Suppose also that for 1 ≤ i ≤ n, di ∶= dim(Vi). Then n 2 SGS = Q di . i=1
Proof. We consider the action of G on itself by left multiplication (the regular action of G) and denotes its corresponding permuation module by kG. This can be equivalently thought of as the action of G on cosets of the trivial subgroup. So, for any V ∈ Irr(G),
mkG (V ) = dim(CkG ({e})) = dim(V ).
So if V1,...,Vn are representatives of the isomorphism classes of the irreducible modules of G then n G ⊕dim(Vi) k = ? Vi . i=1 G Now let χG be the character of k . By the orthonormality of irreducible char- acters we have n 2 ⟨χG, χG⟩ = Q(dim(Vi)) . i=1 For g ∈ G ⎧ ⎪SGS if g = e χG(g) = ⎨ ⎪ ⎩⎪0 if g ≠ e. So 1 −1 ⟨χG, χG⟩ = Q χG(g )χG(g) = SGS SGS g∈G as required.
29 Corollary 3.17. A group G is abelian if and only if every irreducible represen- tation is linear (1 dimensional).
Proof. Let n be the number of conjugacy classes of G. Then G is abelian if and only if n = SGS, with each class containing just one element. From Corollary 3.16, n 2 SGS = Q di . i=1 and di ≥ 1 for 1 ≤ i ≤ n. Thus n = SGS if and only if di = 1 for 1 ≤ i ≤ n.
3.3 Representations of L3(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) = 1 if and only if ω ∈ A.
We now return to the specific case of G = L3(2). We will consider the following three permuation modules:
Ω 1.2 - the 7-dimensional permutation module based on the action of L3(2)
on Ω = V3(2){0};
∆ 2.2 - the 8-dimensional permutation module based on the action of L2(7)
on the 1-dimensional subspaces of V2(7);
Λ 3.2 - the 24-dimensional permutation module based on the action of L3(2) on Λ ∶= G~S via right multiplication of cosets where S is a Sylow 7-
subgroup of L3(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
30 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 semisim- ple but there is still a fair amount that we can say about their internal structure.
3.3.1 The Permutation Module 2Ω
As per our initial discussion on general permutation modules, 2Ω has the sub- modules V (1) and V (6). Moreover, since SΩS = 7 is odd, these two submodules are disjoint and so we have
2Ω = V (6) ⊕ V (1).
In fact, the module V (6) has further structure
(6) ∗ V = V3(2) V3(2)
∗ where denotes an indecomposable extension of modules and V3(2) is the dual ∗ of V3(2). We can think of V3(2) and V3(2) as
⎪⎧ ⎪⎫ ⎪⎛a1⎞ ⎪ ⎪⎜ ⎟ ⎪ V (2) = ⎨⎜a ⎟ ∶ a , a , a ∈ GF (2)⎬ 3 ⎪⎜ 2⎟ 1 2 3 ⎪ ⎪ ⎪ ⎩⎪⎝a3⎠ ⎭⎪ ∗ V3(2) = b1 b2 b3 ∶ b1, b2, b3 ∈ GF (2) .
∗ Moreover, it turns out that V3(2) and V3(2) are two of the irreducible modules of G over GF (2).
We now have Ω ∗ 2 = (V3(2) V3(2)) ⊕ 1.
However, this is not a direct sum decomposition of 2Ω as
∗ ∗ V3(2) V3(2)≅ ~ V3(2) ⊕ V3(2).
31 3.3.2 The Permutation Modules 2Λ and 2∆
We now consider the permutation modules 2Λ and 2∆. Our aim is to find the submodule decomposition of 2Λ 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 3.18. The permutation module 2∆ is a direct summand of 2Λ.
Proof. We will show that there exists a G-invariant surjective homomorphism Λ ∆ Λ Λ ∆ φ ∶ 2 → 2 . Then ker(φ) will be a submodule of 2 such that 2 ~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 ∶ 2Λ → 2∆.
Theorem 3.19. 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
SU{0}S = 2k − 1.
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 2ei for some ei. Thus l 2k − 1 = Q 2ei i=1 and so it follows that ei = 0 for some i. Then the corresponding orbit has length one so consists of a non-zero vector which is stabilised by H.
32 We know that 2∆ is a submodule of 2Λ but how do we go about finding other submodules?
Note that ⎪⎧ ⎪⎫ ⎪⎛u1v1 u1v2 u1v3⎞ ⎪ ⎪⎜ ⎟ ⎪ V (2) ⊗ V (2)∗ = ⎨⎜u v u v u v ⎟ ∶ u ∈ V (2), v ∈ V (2)∗⎬ ≅ M (2) 3 3 ⎪⎜ 2 1 2 2 2 3⎟ 3 3 ⎪ 3 ⎪ ⎪ ⎩⎪⎝u3v1 u3v2 u3v3⎠ ⎭⎪ ∗ and that G ≅ GL3(2) acts on V3(2)⊗V3(2) via conjugation of matrices, turning it into a 9-dimensional module of G.
It is easy to spot that ⎧ ⎫ ⎪⎛0 0 0⎞ ⎛1 0 0⎞⎪ ⎪⎜ ⎟ ⎜ ⎟⎪ V ∶= ⎨⎜0 0 0⎟ , ⎜0 1 0⎟⎬ 1 ⎪⎜ ⎟ ⎜ ⎟⎪ ⎪ ⎪ ⎩⎪⎝0 0 0⎠ ⎝0 0 1⎠⎭⎪ ∗ is a submodule of V3(2) ⊗ V3(2) . Its complement is
V8 ∶= {M ∶ tr(M) = 0}.
In fact this module V8 is know as the Steinberg module of L3(2) and is an irreducible representation of G over GF (2).
Proposition 3.20. Λ (1) (2) ∆ 2 = V8 ⊕ V8 ⊕ 2 (1) (2) where V8 and V8 are copies of V8.
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) = λ3 + λ + 1 or p(s) = λ3 + λ2 + 1.
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 1.
Moreover, if λ3 + aλ2 + bλ + c is the characteristic polynomial of s, then s−1 has characteristic polynomial cλ3 + bλ2 + aλ + 1. On the other hand, s and s2 have the same characteristic polynomial.
33 It follows that there are exactly three elements of S whose corresponding ma- trices are traceless and are thus contained in a copy V8(2). Suppose without loss of generality that s is traceless. Then s2 and s4 are too. Thus
2 4 CV8(2)(S) = {0, s, s , s } and m2Λ (V2(8)) = 2.
Now, by comparing dimensions, along with Proposition 3.18, we have
Λ ∆ 2 = V8 ⊕ V8 ⊕ 2 . as required.
Λ Note that there are actually three copies of V8 in 2 , the third of which we (3) (1) (2) denote V8 . If we index the basis vectors of V8 and V8 as v1, . . . , v8 and u1, . . . , v8 then (3) V8 = {(vi, ui) ∶ 1 ≤ i ≤ 8}.
Our next aim is to find an explicit basis of 2Λ in terms of the right hand side of the equality proved in the previous result. We are trying to find a vector (1) (2) ∆ vs ∈ V8 ⊕ V8 ⊕ 2 such that
g {vs ∶ g ∈ G} forms a basis of 2Λ.
From Theorem 3.19, G fixes at least one 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.
(1) (2) Now let s1 and s2 be the vectors corresponding to s in V8 and V8 respectively. We can take i vs = (s1, s2,S)
(3) (3) ∆ for i = 1, 2 or 4. If i = 1 then (s1, s2) ∈ V8 and vs ∈ V8 ⊕ 2 . Thus i = 2 or 4. Without loss of generality, we can take i = 2 and
2 vs = (s1, s2, ⟨s⟩).
34 Then
{vs ∶ s ∈ V8, S⟨s⟩S = 7} is a basis for 2∆.
All that remains is to show that 2Λ 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 L2(7) is the automorphism group of H. However, P GL2(7) also acts on H, and contains L2(7) as an index two subgroup.
Take an element α ∈ P GL2(7)L2(7) e.g.
⎛1 0 ⎞ α = . ⎝0 −1⎠
We can also write this as a permutation of the columns of H to get
α = (∞)(0)(1 6)(2 5)(3 4).
α Check that H is an extended Hamming code which is L2(7)-invariant. More- over Hα ≠ H because α ∉ Aut(H) and H + Hα is a submodule of 2∆ of codi- mension 1 and H ∩ Hα = 1.
As the Golay code is 12-dimensional, it must contain one of the three Steinberg (i) Λ modules V8 for i = 1, 2, 3 and one of the two 4-dimensional submodules of 2 , H and Hα. In fact, all of these six different combinations of modules give us a submodule isomorphic to the Golay code.
3.4 Representations of L3(2) over C
We start by deducing the dimensions of the irreducible representations.
The action of G ≅ L3(2) on the following sets is 2-transitive:
• ∆, the 1-dimensional subspaces of V2(7);
• P, the points of the Fano plane;
35 • L , the lines of the Fano plane.
So by Proposition 3.14, we have the following decompositions:
∆ • C = 1 ⊕ V7;
P (1) • C = 1 ⊕ V6 ;
L (2) • C = 1 ⊕ V6 ;
(1) (2) for V7,V6 ,V6 ∈ Irr(G). In fact,
(1) (2) V6 ≅ V6 .
Now consider the action of L3(2) on the set of flags of the Fano plane:
θ = {(p, l) ∶ p ∈ P, l ∈ L }.
θ There are 21 incidence relations of the Fano plane so C is a 21-dimensional permutation module. It can be decomposed as
θ (1) (2) C = 1 ⊕ V6 ⊕ V6 ⊕ V8 where V8 is an 8-dimensional irreducible module of G.
When proving the simplicity and uniqueness of the simple group G of order 168
(which we now know to be L3(2)) we (implicitly) showed that G has 7 conjugacy classes, which we denote1
1A, 2A, 3A, 4A, 7A, 7B.
Thus SIrr(G)S = #{ conjugacy classes of G} = 6.
We already have four irreducible representations, one each of dimensions 1, 6, 7 and 8. Let x and y denote the dimensions of the remaining two irreducible representations. Then, from Corollary 3.16
12 + 62 + 72 + 82 + x2 + y2 = SGS = 168.
1The 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 Thus we have x2 + y2 = 18 and the only possibility is that x = y = 3 so that the final two irreducible representations of G have dimension 3.
3.4.1 The Ordinary Character Table of L3(2)
We denote the irreducible representations of G as
(1) (2) V1,V3 ,V3 ,V6,V7,V8.
Theorem 3.21. The character table of G ≅ L3(2) ≅ L7(2) is
1A 2A 3A 4A 7A 7B 1 21 56 42 24 24
χ1 1 1 1 1 1 1 (1) ∗ χ3 3 −1 0 1 b7 −b7 (2) ∗ χ3 3 −1 0 1 −b7 b7
χ6 6 2 0 0 −1 −1
χ7 7 −1 1 −1 0 0
χ8 8 0 −1 0 1 1 where the second row gives the sizes of the conjugacy classes, and √ √ 1 + −7 1 − −7 b ∶= and b∗ ∶= . 7 2 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.
For the characters corresponding to the 6 and 8 dimensional representations, we recall that
χ P = χ L = 1 + χ C C 6
χ θ = 1 + 2χ + χ . C 6 8
37 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
χ6(g) is equal to the number of points (or lines) of the Fano plane fixed by g minus 1.
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
χ6(g) = 3 − 1 = 2 and χ8(g) = 5 − 1 − 4 = 0.
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
χ6(g) = 1 − 1 = 0 and χ8(g) = 0 − 1 = −1.
An element of order 4 in G acts on the Fano plane by fixing one point, transpos- ing 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)
χ6(g) = 1 − 1 = 0 and χ8(g) = 1 − 1 = 0.
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
χ6(g) = 0 − 1 = −1 and χ8(g) = 0 − 1 + 2 = 1.
Let ⎛1 0⎞ ⎛2 0⎞ ⎛0 1⎞ t = , s = , r = . ⎝1 1⎠ ⎝0 4⎠ ⎝6 0⎠ Then t is of order 2, s is of order 3 and r is of order 7. From the discussion following Propostition 1.11, we know that
• t fixes none of the 1-dimensional subspaces of V2(7);
38 ⎛0⎞ ⎛1⎞ • s fixes two such subspaces, those generated by and ; ⎝1⎠ ⎝0⎠
⎛0⎞ • r fixes one such subspace, that generated by . ⎝1⎠
As characters are constant on conjugacy classes, these observations give us the values of χ7 for the conjugacy classes 2A, 3A and 7A. The remaining two values for χ7 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 1-dimensional representations (see Corollary 3.17). Thus the matrices of the image of g under each of the two 3-dimensional representations are of the form
m1 ⎛e7 0 0 ⎞ ⎜ ⎟ ⎜ m2 ⎟ ⎜ 0 e7 0 ⎟ ⎝ m3 ⎠ 0 0 e7
2πi for mi ∈ Z for 1 ≤ i ≤ 3 and where en ∶= e n .
Given an element h ∈ NL3(2)(⟨g⟩) and an irreducible representation ρ of ⟨g⟩ the map
ρh ∶ x ↦ ρ(xh) is also an irreducible representation. Thus it follows that we have an action of
NL3(2)(⟨g⟩) on the irreducible characters of ⟨g⟩.
If h ∈ ⟨g⟩ then h acts trivially (as characters are constant on conjugacy classes).
If h ∈ NL3(2)(⟨g⟩)⟨g⟩ then h induces a non-trivial permuation of the represen- tations of ⟨g⟩. However, as h ∈ L3(2), the action it induces on the characters of
L3(2) must be trivial. Thus the action of NL3(2)(⟨g⟩) on the representations in question must preserve the set {m1, m2, m3}.
−1 2 In particular, NL3(2)(⟨g⟩) is generated by g and a 3-cycle h such that h gh = g .
Thus the possibilities for {m1, m2, m3} are {3, 5, 6} and {1, 2, 4} giving character ∗ (1) (2) values of b7 and −b7 for χ3 and χ3 respectively. The values of these characters
39 on 7B can be deduced from the orthogonality of the columns of the character table.
(1) Moreover, the automorphism ρ3 (g) of V clearly has three distinct eigenvalues, 2 4 (1) e7, e7 and e7 so we have a basis of V consisting of eigenvectors of ρ3 (g). As (1) the element h permutes these eigenvectors, ρ3 (g) will be a matrix conjugate to ⎛0 0 1⎞ ⎜ ⎟ ⎜ ⎟ ⎜1 0 0⎟ . ⎝0 1 0⎠
(2) This argument also holds for ρ3 so we can conclude that the character values of the two 3-dimensional representations are 0 on the conjugacy class of elements of order 3.
We now claim that the image of L3(2) under either of the representations of di- mension 3 is contained in SL3(C). Since L3(2) is simple and the representations in question are not trivial, both have trivial kernels. Thus they give embeddings of L3(2) into GL3(C). As SL3(C) is normal in GL3(C), the intersection of the image of L3(2) and SL3(C) is normal in the image of L3(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 L3(2) is mapped to a matrix with determinant 1 in GL3(2). Thus the image of L3(2) under the two representations of dimension 3 is contained in SL3(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
n1 ⎛e2 0 0 ⎞ ⎜ ⎟ ⎜ n2 ⎟ ⎜ 0 e2 0 ⎟ ⎝ n1 ⎠ 0 0 e2 for ni ∈ Z for 1 ≤ i ≤ 3. We must have (ni, 2) = 1 for some 1 ≤ i ≤ 3, else the representation would be trivial. Along with the fact that this matrix must have determinant 1, this implies that the only possibility for {n1, n2, n3} is {1, 1, 2}. Thus both characters of dimension 3 take the value -1 on the conjugacy class of
40 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
p1 ⎛e4 0 0 ⎞ ⎜ ⎟ ⎜ p4 ⎟ ⎜ 0 e4 0 ⎟ ⎝ p1 ⎠ 0 0 e4 for pi ∈ Z for 1 ≤ i ≤ 3. Moreover, we must have (pi, 4) = 1 for at least one
1 ≤ i ≤ 3, else the matrix would be of order 1 or 2. In this case, NL3(2)(⟨y⟩) is generated by y and z, an element of order 2 such that z−1yz = y3. Using these observations, along with the fact that the image of y must be a matrix of determinant 1, we deduce that the only possibility for {p1, p2, p3} is {1, 3, 4}. Thus both characters of dimension 3 take the value 0 on the conjugacy class of elements of order 4.
41 Appendix A
Error Correcting Codes
Definition A.1. A (binary) code of length n is a subset C of Vn(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 Vn(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 ⊂ Vn(2) corrects e errors if for any c, c ∈ C and ′ ′ x ∈ Vn(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 Definition A.4. If A ∈ Mn(2) and C = {x ∈ Vn(2) ∶ Ax = 0} 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
1. 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 − 1 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 Vk(2).
A Hamming code Ham(k)
• has minimum distance 3;
• has length 2k − 1;
• has dimension 2k − k − 1;
• 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 = { n n} n−1 corrects one error, the codes C 0 , 1 which correct 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
⎛1 1 1 0 1 0 0⎞ ⎜ ⎟ ⎜ ⎟ ⎜1 1 0 1 0 1 0⎟ ⎝1 0 1 1 0 0 1⎠
43 and it’s “reverse” K with check matrix
⎛0 0 1 0 1 1 1⎞ ⎜ ⎟ ⎜ ⎟ ⎜0 1 0 1 0 1 1⎟ . ⎝1 0 0 1 1 0 1⎠
Then, to each codeword in H and K, we add a “parity check bit”. That is to say, given a vector (v1, . . . , v7) we add an eighth bit
7 v8 = Q vi i=1 which is 1 if there is an even number of 1’s and 0 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
′ ′ G24 ∶= {(a + x, b + x, a + b + x) ∶ a, b ∈ H , x ∈ K } ⊆ V24(2)
The Golay code G23 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 12;
• corrects 3 errors.
44