Linear Groups INTRODUCTION Robert A

Lecture 2: Linear groups INTRODUCTION Robert A. Wilson

Queen Mary, University of London

LTCC, 13th October 2008

Classical groups Finite ﬁelds

The six families of classical finite simple groups are all essentially matrix groups over finite fields: A field is a set F with all the usual arithmetical operations

I the projective special linear groups PSLn(q); and rules.

I the projective special unitary group PSUn(q); I {F, +, −, 0} is an abelian group; ∗ ∗ I the projective symplectic groups PSp2n(q); I {F , ., /, 1} is an abelian group, where F = F \{0};

I three families of orthogonal groups I x(y + z) = xy + xz

I PΩ2n+1(q); Example: the integers modulo p, where p is a prime. + I PΩ2n(q); − I PΩ2n(q). More ﬁnite ﬁelds

I In any finite field F, the subfield F0 generated by 1 has prime order, p.

I F is a vector space, of dimension d, over F0, so has pd elements. LINEAR GROUPS I In fact, there is exactly one ﬁeld of each such order pd .

I To make such a ﬁeld, pick an irreducible polynomial f of degree d, and construct the quotient F0[x]/(f ) of the polynomial ring. 2 I Example: p = 2, f (x) = x + x + 1, gives a ﬁeld of 2 order 4 as F4 = {0, 1, ω, ω} with ω = ω and ω + ω = 1.

The general linear group The orders of the linear groups

I GLn(q) is the group of all invertible n × n matrices I How many invertible matrices are there? with entries in the ﬁeld F = Fq of order q. I Choose each row to be linearly independent of the I The scalar matrices form a normal subgroup Z of previous rows. order q − 1. I The ﬁrst i rows span an i-dimensional space, which I the projective general linear group has qi vectors. PGL (q) = GL (q)/Z . n n Therefore there are qn − qi choices for the (i + 1)th ∗ I I The determinant map det : GLn(q) → F is a group row. homomorphism. n n n n−1 I Hence |GLn(q)| = (q − 1)(q − q) ··· (q − q ). I Its kernel is the special linear group SLn(q). I |SLn(q)| = |PGLn(q)| = |GLn(q)|/(q − 1). I The projective special linear group I |PSLn(q)| = |SLn(q)|/gcd(n, q − 1). PSLn(q) = SLn(q)/(Z ∩ SLn(q)). An example: GL2(2) More examples

I F = F2 = Z/2Z = {0, 1} with 1 + 1 = 0. 2 PGL (3) ∼= S , permuting the four 1-dimensional I The 2-dimensional vector space F has four vectors, I 2 4 (0, 0), (0, 1), (1, 0), (1, 1). subspaces; PSL (3) ∼= A ; I The ﬁrst row can be any of the 3 non-zero vectors. I 2 4 PSL (4) ∼= A , permuting the ﬁve 1-dimensional I The second row can be any of the remaining 2 I 2 5 non-zero vectors. subspaces; ∼ ∼ I PSL (5) = A , and PGL (5) = S . I Hence |GL2(2)| = 6. 2 5 2 5 GL (2) acts on the vector space by permuting the I In fact, PSLn(q) is a simple group except for the I 2 ∼ ∼ three non-zero vectors in all possible ways. cases PSL2(2) = S3 and PSL2(3) = A4. ∼ I Hence GL2(2) = S3.

Iwasawa’s Lemma Proof of Iwasawa’s Lemma

I Otherwise, choose a normal subgroup K with 1 < K < G.

I Choose a point stabilizer H with K 6≤ H. The easiest way to prove simplicity of PSLn(q) is to use: Hence G = HK since H is maximal. Theorem (Iwasawa’s lemma) I Any g ∈ G can be written g = hk. If G is a ﬁnite perfect group acting faithfully and primitively I on a set Ω, and the point stabilizer H has a normal I Any conjugate of A is −1 −1 −1 −1 abelian subgroup A whose conjugates generate G, then g Ag = k h Ahk = k Ak ≤ AK . G is simple. I Therefore G = AK . ∼ I Now G/K = AK /K = A/(A ∩ K ) is abelian.

I But G is perfect, so has no nontrivial abelian quotients.

I Contradiction. Simplicity of PSLn(q) Simplicity of PSLn(q), cont.

I To prove that PSLn(q) is perfect it sufﬁces to prove I Let PSLn(q) act on the 1-dimensional subspaces of that these matrices (transvections) are commutators. F n. I If n ≥ 3 observe that I This action is 2-transitive, so primitive.       I The stabilizer of the point h(1, 0, 0,..., 0)i consists 1 0 0 1 0 0 1 0 0 λ 0 1 1 0 , 0 1 0 =  0 1 0 (modulo scalars) of matrices . v M 0 0 1 0 x 1 −x 0 1 It has a normal abelian subgroup consisting of I If q ≥ 4 then has an element x with x 3 6= x, so 1 0 I Fq matrices . v In−1 1 0 x 0 1 0 , = I These matrices encode elementary row operations, y 1 0 x −1 y(x 2 − 1) 1 and you learnt in linear algebra that every matrix of

determinant 1 is a product of such matrices. I Hence by Iwasawa’s Lemma, PSLn(q) is simple whenever n ≥ 3 or q ≥ 4.

SUBGROUPS OF COFFEE BREAK GENERAL LINEAR GROUPS Subspace stabilizers Imprimitive subgroups

I The stabilizer of a subspace of dimension k looks like Suppose V = V1 ⊕ · · · ⊕ Vk is a direct sum of k this: subspaces each of dimension m. k → GL (q) 0  k The stabilizer of this decomposition of the vector   I n − k → qk(n−k) GL (q) space has a normal subgroup GLm(q) × · · · × GLm(q) n−k acting on V ,..., V separately. 1 k Ik 0 I The subgroup of matrices of shape is a I There is also a subgroup Sk permuting these k AIn−k subspaces. normal abelian subgroup. I Together these generate a wreath product I The quotient by this subgroup is isomorphic to GLm(q) o Sk . GLk (q) × GLn−k (q).

Tensor products Wreathed tensor products

If A = (aij ) ∈ GLk (q) and B = (bij ) ∈ GLm(q) then the following matrix is in GLkm(q): Repeating this construction with m copies of GL (q) we   k a11B a12B ··· a1k B get a B a B ··· a B PGL (q) × · · · × PGL (q) < PGL m (q)  21 22 2k  k k k A ⊗ B =  .   .  We can also permute the m copies of PGLk (q) with a ak1B ak2B ··· akk B copy of Sm. Together these give a wreath product If we multiply A by a scalar λ, and B by the inverse λ−1, then this matrix does not change. PGLk (q) o Sm < PGLk m (q) Factoring out by the scalars we have

PGLk (q) × PGLm(q) < PGLkm(q) Extraspecial groups Extraspecial groups, cont.

Suppose r is an odd prime, and α is an element of order r in the ﬁeld F (so r is a divisor of |F| − 1). A slightly different construction is required for the prime 2 I Let R be the group generated by the r × r matrices I Both D8 and Q8 have 2-dimensional representations     α 0 0 ··· 0 0 1 0 ··· 0 I Tensoring them together gives extraspecial groups of 0 α2 0 ··· 0  0 0 1 0 1+2k k     order 2 , with representations of degree 2  .  and  .   .   .  I In fact D8 ⊗ D8 = Q8 ⊗ Q8, so there are just two     0 0 ··· 0  0 0 0 ··· 1 extraspecial groups of each order. 0 0 ··· αr−1 1 0 0 ··· 0 I If the ﬁeld contains elements of order 4, you can adjoin these scalars to make a bigger group which 3 I Then R is non-abelian of order r contains both extraspecial groups. I Taking the tensor product of k copies of R gives an extraspecial group of order r 1+2k acting on a space of dimension r k .

Almost quasi-simple groups The Aschbacher–Dynkin theorem

Every subgroup of GLn(q) which does not contain SLn(q) 0 I A group G is quasi-simple if G = G and G/Z (G) is is contained in the stabilizer of one of the following simple I a subspace of dimension k Example: SL (q) is quasi-simple except for SL (2) I n 2 I a direct sum of k subspaces of dimension m, where and SL2(3) n = km A group G is almost quasi-simple (for us - this is not k m I I a tensor product F ⊗ F , where n = km entirely standard terminology) if G/Z is almost a tensor product of m copies of F k , where n = k m simple, where Z is a suitable group of scalar I 1+2k k matrices. I an extraspecial group r , where n = r I an almost quasi-simple subgroup THE END