14 the Special Linear Group SL(N, F)

Home , Center (group theory), Commutator, Linear group, Special linear group

14 The Special Linear Group SL(n, F )

First some notation:

Mn(R) is the ring of n × n matrices with coeﬃcients in a ring R.

GL(n, R) is the group of units in Mn(R), i.e., the group of invertible n × n matrices with coefficients in R. GL(n, q) denotes GL(n, GF (q)) where GF (q) denotes the Galois field of order q = pk. SL(n, F ) denotes the kernel of the homomorphism det : GL(n, F ) ³ F × = {x ∈ F | x 6= 0} where F is a field. Note that the determinant homomorphism has section s given by sending x ∈ F × to the diagonal matrix:   x 0 ··· 0  . 0 1 . s(x) = diag(x, 1, ··· , 1) =  . .   . ..  0 ··· 1 Thus: Theorem 14.1. If F is a field and n ≥ 1 then GL(n, F ) is a semidirect product GL(n, F ) = SL(n, F ) o F × Theorem 14.2. The center of GL(n, F ) is the group of all nonzero scalar multiples of the identity matrix, i.e., all diagonal matrices of the form

xIn = diag(x, x, ··· , x) for some x ∈ F ×.

−1 × Proof. If A = (aij) ∈ ZGL(n, F ) then s(x)As(x ) = A for all x ∈ F . But the entries in the top row of s(x)As(x−1) are equal to the corresponding entries in A multiplied by x (except for the term a11). So these must all be zero. Similarly, the entries in the ﬁrst column must all be zero except for a11. Conjugating by diag(1, x, 1, ··· , 1) we see that the second row and column must also be all zero except for the (2, 2) entry. Proceeding in this way we see that central elements must be diagonal. But now note that conjugation by the matrix   0 1 0 1 0 0  0 0 In−2

1 switches the (1, 1) and (2, 2) entries in a diagonal matrix. Thus the diagonal entries of a central element must be equal. Theorem 14.3. The center of SL(n, F ) consists of all multiples of the iden- n tity matrix xIn where x = 1.

To prove this we need to use the elementary matrices Xij(λ) whose entries are the same as that of the identity matrix In except for an λ in the (i, j) location. These lie in SL(n, F ) assuming that i, j are distinct positive integers ≤ n.

Lemma 14.4. Left multiplication by Xij(λ) changes the i-th row of a matrix by adding λ times the j-th row. Simimilarly, right multiplication by Xij(λ) changes the j-th column of a matrix by adding λ times the i-th column.

Proof of Theorem 14.3. If A ∈ SL(n, F ) commutes with Xij(1) then the i-th and j-th columns and rows must be all zeroes except for the (i, i) and (j, j) entries which must be equal.

Lemma 14.5. The elementary matrices Xij(λ) generate SL(n, F ). Proof. If n = 1 then SL(1,F ) = 1 is trivial. So suppose that n ≥ 2. Let A ∈ SL(n, F ) then it suffices to reduce A to the identity matrix by elementary row and column operations. Suppose first that a21 6= 0. Then the (1, 1) entry of X12(λ)A will be equal to 1 for some (unique) λ ∈ F . If a21 = 0 then we can make it nonzero by a row operation X2j(1) for some j. Thus we may assume that a11 = 1. Now multiply on the left by the elementary matrices Xi1(−ai1) and on the right µby X1j¶(−a1j). This will clear the first row and column of A making a matrix 1 0 where B ∈ SL(n − 1,F ) is a product of elementary matrices by 0 B induction on n.

Lemma 14.6. The matrices Xij(λ) are commutators in SL(n, F ) except in the case n = 2 and |F | = 2 or 3. Proof. If n ≥ 3 this is easy since there is a third index k and

[Xik(λ),Xkj(1)] = Xij(λ). If n = 2 we use the commutator relation: ·µ ¶ µ ¶¸ µ ¶ α 0 1 β 1 (α2 − 1)β , = 0 α−1 0 1 0 1 But given any λ ∈ F the equation

λ = (α2 − 1)β can be solved for β as long a there exists a unit α ∈ F × so that α2 6= 1 (i.e., α 6= ±1). This works as long as F × has at least three elements.

2 Deﬁnition 14.7. A group G is perfect if it is equal to its own commutator subgroup, i.e., G = G0. Note that nontrivial solvable groups cannot be perfect. Therefore we have the following immediate consequences of the above two lemmas. Theorem 14.8. The group SL(n, F ) is perfect, i.e., equal to its commutator subgroup, except in the cases SL(2, 2) and SL(2, 3). Corollary 14.9. If n ≥ 2 then SL(n, F ) is not solvable except in the cases SL(2, 2) and SL(2, 3).

Invertible Matrices as Automorphisms If V is an n-dimensional vector space over a field F then the group of F -linear automorphisms of V is isomorphic to GL(n, F ). However the isomorphism is not canonical. It depends on a choice of basis for V . More precisely, if ψ : V −→≈ F n is an F -linear isomorphism then an isomorphism ≈ ψ∗ : AutF (V ) −→ GL(n, F ) −1 n is given by ψ∗(f) = ψfψ ∈ AutF (F ) = GL(n, F ). For example we have: Theorem 14.10. If G is an elementary abelian p-group of order pn then Aut(G) ∼= GL(n, p) Theorem 14.11. ψ−1(SL(n, F )) is independent of the choice of ψ and con- sequently defines a well-defined subgroup of GL(V ) = AutF (V ) which we will call SL(V ). −1 n Proof. Suppose that ψ∗(f) = ψfψ ∈ SL(n, F ) and φ : V → F is another −1 −1 −1 −1 isomorphism. Then φ∗(f) = φfφ = (φψ )ψfψ (ψφ ) is a conjugate of ψ∗(f) and therefore has determinant 1. Theorem 14.12. The finite groups GL(n, q),SL(n, q) have orders:

n(n−1) n n−1 |GL(n, q)| = q 2 (q − 1)(q − 1) ··· (q − 1)

n(n−1) n n−1 2 |SL(n, q)| = q 2 (q − 1)(q − 1) ··· (q − 1) Proof. Let V be an n-dimensional vector space over F = GF (q). Then V has n q elements. Choose a basis (v1, v2, ··· , vn) for V . Then an automorphism n f of V is given by its value on the basis. There are q − 1 choices for f(v1). n 1 Given f(v1), there are q − q choices for f(v2) since it can be any vector not n 2 in the span of f(v1). There are q − q choices for f(v3) and so on. Thus |GL(V )| = (qn − 1)(qn − q) ··· (qn − qn−1) = q1+2+···+(n−1)(qn − 1) ··· (q − 1) And |SL(n, q)| = |GL(V )|/|F ×| = |GL(n, q)|/(q − 1).