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15 Projective Special Linear Group P SL(n, F ) Definition 15.1. The projective linear group and projective special linear group are the quotients of GL(n, F ) and SL(n, F ) by their centers: GL(n, F ) GL(n, F ) = Z(GL(n, F )) F× SL(n, F ) SL(n, F ) P SL(n, F ) = = Z(SL(n, F )) {α ∈ F × | αn = 1} P GL(n, F ) = These groups act on the n − 1 dimensional projective space P n−1 (F ) (the set of one dimensional subspaces of F n ) by the following lemma. Lemma 15.2. The only elements of GL(n, F ) which stabilize every element of P n−1 (F ) are the scalar multiples of the identity matrix. Proof. If A ∈ GL(n, F ) stabilizes the spans F ei of the standard basis vectors ei of F n then Aei = λi ei for some λi ∈ F × so A = diag(λ1 , · · · , λn ). We claim that λi = λj for all i, j since, otherwise, A will not stabilize F (ei + ej ): A(ei + ej ) = λi ei + λj ej ∈ F (ei + ej ) ⇔ λi = λj Thus A = λIn . This implies that we get an induced action of P GL(n, F ) on P n−1 (F ) which is faithful in the sense that only the identity group element fixes every point in P n−1 (F ). We also get an induced action of P SL(n, F ) on P n−1 (F ). Identifying these groups with the corresponding symmetry groups on P n−1 (F ) we get: P SL(n, F ) ≤ P GL(n, F ). In this section we want to prove that P SL(n, q) is a simple group for n ≥ 2 with two exceptions. Following Alperin we use the fact that it acts doubly transitively on projective space. Definition 15.3. The action of a group G on a set X is called doubly transitive if G acts transitively on the set of all ordered pairs of distinct elements of X. In other words, given any x1 , x2 , y1 , y2 ∈ X so that x1 6= x2 and y1 6= y2 there is a g ∈ G so that gxi = yi for i = 1, 2. Lemma 15.4. If G acts doubly transitively on a set X then the stabilizer Hx of any x ∈ X is a maximal subgroup of G. Proof. Suppose that Hx is not maximal. Then there is a subgroup K so that G > K > Hx . Thus there is a g ∈ G, g ∈ / K and k ∈ K, k ∈ / Hx . Since g, k are not in Hx they do not stabilize x. Thus gx, kx are not equal to x. Since G acts doubly transitively on X, there is an h ∈ G so that hx = x and h(gx) = kx. But then h, k −1 hg ∈ Hx < K so g ∈ h−1 kHx ⊆ K which is a contradiction. 1 Lemma 15.5. If n ≥ 2, the action of SL(n, F ) and thus of P SL(n, F ) on projective space is doubly transitive. Proof. Suppose that L1 , L2 are distinct elements of P n−1 (F ) [Li ⊆ F n ]. Then any nonzero vectors vi ∈ Li will be linearly independent so can be extended to a basis for F n . Thus there is an A ∈ GL(n, F ) so that Aei = vi for i = 1, 2. Let λ = det(A). Then B = A diag(λ−1 , 1, · · · , 1) ∈ SL(n, F ) sends e1 to λ−1 v1 and e2 to v2 . So BF ei = Li for i = 1, 2. Lemma 15.6. Every Xij (α) ∈ SL(n, F ) is conjugate to an elementary matrix of the form X12 (β). Proof. Let σ ∈ Sn be a permutation of {1, · · · , n} so that σ(1) = i, σ(2) = j. Let ² be the sign of σ. Then there is an s ∈ SL(n, F ) so that s(e1 ) = ²ei and s(ek ) = eσ(k) for all 2 ≤ k ≤ n. Then sX12 (²α) = s(e1 , e2 + ²αe1 , e3 , · · · , en ) = (²ei , ej + αei , eσ(3) , · · · , eσ(n) ) = Xij (α)(²ei , ej , eσ(3) , · · · , eσ(n) ) = Xij (α)s. Consequently, s−1 Xij (α)s = X12 (²α). Theorem 15.7. For n ≥ 2, P SL(n, F ) is simple with the exception of P SL(2, 2) and P SL(2, 3). Proof. Let P be the stabilizer of F e1 in SL(n, F ). Then P has the following properties: 1. P is a maximal subgroup of SL(n, F ) by Lemmas 15.4 and 15.5. 2. For any s ∈ SL(n, F ), sP s−1 is the stabilizer of s(F e1 ). 3. The elements of P can be written uniquely in the form: µ ¶ a v 0 B where a ∈ F × , v ∈ F n−1 and B ∈ GL(n − 1, F ) with det(B) = a1 . To show that P SL(n, F ) is simple it suffices to show that any proper normal subgroup N of SL(n, F ) is contained in its center. Case 1. Suppose first that N ≤ P . Then N = sN s−1 ≤ sP s−1 for all s ∈ SL(n, F ). Consequently, N stabilizes every one dimensional subspace F v ∈ P n−1 (F ) and thus consists of scalar multiples of In by Lemma 15.2. Case 2. Now suppose that N is not contained in P . Then P N = SL(n, F ) by maximality of P . So for any K E P , we have KN E SL(n, F ) (since both P and N normalize KN ). Let K be the group of all matrices of the form: µ ¶ 1 w 0 In−1 2 This is a normal subgroup of P since: µ ¶µ ¶ µ ¶ µ ¶µ ¶ a v 1 w a v + aw 1 awB −1 a v = = 0 B 0 In−1 0 B 0 In−1 0 B K is also abelian, being isomorphic to the additive group F n−1 . Since KN E SL(n, F ) and every generator Xij (λ) of SL(n, F ) is conjugate to an element of K by Lemma 15.6, we must have KN = SL(n, F ). But then SL(n, F ) ∼ K = N K ∩N is abelian (since K ∼ = F n−1 is abelian). By Theorem 14.8 this is impossible except in the cases of SL(2, 2) and SL(2, 3). Other matrices Permutation matrices were used in the proof of Lemma 15.6. If σ ∈ Sn then the corresponding matrix φ(σ) ∈ GL(n, F ) is given by φ(σ)(ei ) = eσ(i) . In words this is the matrix obtained from the identity matrix In by permuting its rows by σ. Thus, e.g., 0 0 1 φ(123) = 1 0 0 0 1 0 This gives a homomorphism φ : Sn → GL(n, F ) whose image Alperin calls W because it is the Weyl group in this case. The (standard) Borel subgroup of GL(n, F ) is the group of invertible upper triangular matrices: ∗ ∗ ∗ B = 0 ∗ ∗ 0 0 ∗ Note that this includes all diagonal matrices and all Xij (λ) where i < j. Theorem 15.8. GL(n, F ) = BW B. Proof. Take any A ∈ GL(n, F ). Let anj be the first nonzero entry in the bottom row. Then by right multiplication by some b ∈ B we can change this entry to 1 and clean (make into 0) the rest of the bottom row. By left multiplication by some b0 ∈ B we can clean the rest of the j-th column. Now suppose that the first nonzero entry in the n − 1st row of b0 Ab is the (n − 1, k) entry. Then we can make it 1 and clean the rest of that row and column. Continuing in this way we find b1 , b2 ∈ B s.t. b1 Ab2 ∈ W so −1 A ∈ b−1 1 W b2 ⊆ BW B. 3 This theorem says that GL(n, F ) is a union of the double cosets BwB. The following lemma implies that these double cosets are disjoint. Lemma 15.9. Suppose that w1 , w2 ∈ W and b ∈ B so that w1 bw2 ∈ B. Then w2 = w1−1 . Proof. For each t ∈ F let b(t) be the matrix obtained from b by multiplying the off diagonal entries by t. Then w1 b(t)w2 ∈ B for all t ∈ F . In particular this is true for t = 0. But b(0) is a diagonal matrix so w1 b(0)w2 ∈ B implies w1 w2 = In . Theorem 15.10 (Bruhat decomposition). The double cosets BwB are disjoint. Consequently, GL(n, F ) can be expressed as a disjoint union: a GL(n, F ) = BwB w∈W Proof. Suppose that BwB intersects Bw0 B. Then there are elements bi of B so that b1 wb2 = b3 w0 b4 . But this implies 0 −1 w−1 b−1 1 w = b4 b2 ∈ B so w = w0 by Lemma 15.9. 4