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 or- der q = pk. SL(n; F ) denotes the kernel of the homomorphism det : GL(n; F ) ³ F £ = fx 2 F j x 6= 0g where F is a field. Note that the determinant homomorphism has section s given by sending x 2 F £ to the diagonal matrix: 0 1 x 0 ¢ ¢ ¢ 0 B .C B0 1 .C s(x) = diag(x; 1; ¢ ¢ ¢ ; 1) = B . C @ . .. A 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 2 F £. ¡1 £ Proof. If A = (aij) 2 ZGL(n; F ) then s(x)As(x ) = A for all x 2 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 @1 0 0 A 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(¸) whose en- tries are the same as that of the identity matrix In except for an ¸ 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(¸) changes the i-th row of a matrix by adding ¸ times the j-th row. Simimilarly, right multiplication by Xij(¸) changes the j-th column of a matrix by adding ¸ times the i-th column. Proof of Theorem 14.3. If A 2 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(¸) generate SL(n; F ). Proof. If n = 1 then SL(1;F ) = 1 is trivial. So suppose that n ¸ 2. Let A 2 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(¸)A will be equal to 1 for some (unique) ¸ 2 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 2 SL(n ¡ 1;F ) is a product of elementary matrices by 0 B induction on n. Lemma 14.6. The matrices Xij(¸) are commutators in SL(n; F ) except in the case n = 2 and jF j = 2 or 3. Proof. If n ¸ 3 this is easy since there is a third index k and [Xik(¸);Xkj(1)] = Xij(¸): If n = 2 we use the commutator relation: ·µ ¶ µ ¶¸ µ ¶ ® 0 1 ¯ 1 (®2 ¡ 1)¯ ; = 0 ®¡1 0 1 0 1 But given any ¸ 2 F the equation ¸ = (®2 ¡ 1)¯ can be solved for ¯ as long a there exists a unit ® 2 F £ so that ®2 6= 1 (i.e., ® 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 à : V ¡!¼ F n is an F -linear isomorphism then an isomorphism ¼ ä : AutF (V ) ¡! GL(n; F ) ¡1 n is given by ä(f) = Ãfà 2 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. á1(SL(n; F )) is independent of the choice of à and con- sequently defines a well-defined subgroup of GL(V ) = AutF (V ) which we will call SL(V ). ¡1 n Proof. Suppose that ä(f) = Ãfà 2 SL(n; F ) and Á : V ! F is another ¡1 ¡1 ¡1 ¡1 isomorphism. Then Á¤(f) = ÁfÁ = (Áà )Ãfà (ÃÁ ) is a conjugate of ä(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 jGL(n; q)j = q 2 (q ¡ 1)(q ¡ 1) ¢ ¢ ¢ (q ¡ 1) n(n¡1) n n¡1 2 jSL(n; q)j = 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 jGL(V )j = (qn ¡ 1)(qn ¡ q) ¢ ¢ ¢ (qn ¡ qn¡1) = q1+2+¢¢¢+(n¡1)(qn ¡ 1) ¢ ¢ ¢ (q ¡ 1) And jSL(n; q)j = jGL(V )j=jF £j = jGL(n; q)j=(q ¡ 1). 3.