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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.
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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.
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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.
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