15 Projective Special Linear PSL(n, F )

Definition 15.1. The projective and projective are the quotients of GL(n, F ) and SL(n, F ) by their centers: GL(n, F ) GL(n, F ) P GL(n, F ) = = Z(GL(n, F )) F × SL(n, F ) SL(n, F ) PSL(n, F ) = = Z(SL(n, F )) {α ∈ F × | αn = 1} These groups act on the n − 1 dimensional 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 .

Proof. If A ∈ GL(n, F ) stabilizes the spans F ei of the standard basis vectors n × ei of F then Aei = λiei 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) = λiei + λjej ∈ 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 PSL(n, F ) on P n−1(F ). Identifying these groups with the corresponding groups on P n−1(F ) we get: PSL(n, F ) ≤ P GL(n, F ). In this section we want to prove that PSL(n, q) is a 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 tran- sitive 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 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 −1 h(gx) = kx. But then h, k hg ∈ Hx < K so −1 g ∈ h kHx ⊆ K which is a contradiction.

1 Lemma 15.5. If n ≥ 2, the action of SL(n, F ) and thus of PSL(n, F ) on projective space is doubly transitive.

n−1 n Proof. Suppose that L1,L2 are distinct elements of P (F )[Li ⊆ F ]. Then any nonzero vectors vi ∈ Li will be linearly independent so can be extended n to a basis for F . 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 )

−1 sends e1 to λ 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 ma- trix 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.

−1 Consequently, s Xij(α)s = X12(²α). Theorem 15.7. For n ≥ 2, PSL(n, F ) is simple with the exception of PSL(2, 2) and PSL(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.

−1 2. For any s ∈ SL(n, F ), sP s is the stabilizer of s(F e1). 3. The elements of P can be written uniquely in the form: µ ¶ a v 0 B

× n−1 1 where a ∈ F , v ∈ F and B ∈ GL(n − 1,F ) with det(B) = a . To show that PSL(n, F ) is simple it suffices to show that any proper N of SL(n, F ) is contained in its . Case 1. Suppose first that N ≤ P . Then N = sNs−1 ≤ sP s−1 for all s ∈ SL(n, F ). Consequently, N stabilizes every one dimensional subspace n−1 F v ∈ P (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 PN = 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 Alperin calls W because it is the in this case. The (standard) 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 ) = BWB.

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 b0Ab 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. b1Ab2 ∈ W so −1 −1 A ∈ b1 W b2 ⊆ BWB.

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 w1bw2 ∈ B. −1 Then w2 = w1 . Proof. For each t ∈ F let b(t) be the matrix obtained from b by multiplying the off diagonal entries by t. Then w1b(t)w2 ∈ B for all t ∈ F . In particular this is true for t = 0. But b(0) is a so w1b(0)w2 ∈ B implies w1w2 = In. Theorem 15.10 (). The double cosets BwB are disjoint. Consequently, GL(n, F ) can be expressed as a disjoint union: a GL(n, F ) = BwB w∈W

0 Proof. Suppose that BwB intersects Bw B. Then there are elements bi of B 0 so that b1wb2 = b3w b4. But this implies

−1 −1 0 −1 w b1 w = b4b2 ∈ B so w = w0 by Lemma 15.9.

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