SIMPLICITY OF PSL n (F )

Transcription

1 SIMPLICITY OF PSL n F KEITH CONRAD Introduction For any field F and integer n 2, the projective special linear group PSL n F is the quotient group of SL n F by its center: PSL n F SL n F /ZSL n F In 83, Galois claimed that PSL 2 F p is a simple group for any prime p > 3, although he didn t give a proof He had to exclude p 2 and p 3 since PSL 2 F 2 S 3 and PSL 2 F 3 A 4, and these groups are not simple It turns out that PSL n F is a simple group for any n 2 and any field F except when n 2 and F F 2 and F 3 The proof of this was developed over essentially 3 years, from 87 to 9: Jordan [4] for n 2 and F F p except n, p 2, 2 and 2,3 Moore [5] for n 2 and F any finite field of size greater than 3 Dickson for n > 2 and F finite [], and for n 2 and F infinite [2] We will prove simplicity of PSL n F using a criterion of Iwasawa [3] from 94 that relates simple quotient groups and doubly transitive group actions This criterion will be developed in Section 2, and applied to PSL 2 F in Section 3 and PSL n F for n > 2 in Section 4 2 Doubly transitive actions and Iwasawa s criterion An action of a group G on a set X is called transitive when, given any distinct x and y in X, there is a g G such that gx y We call the action doubly transitive if any pair of distinct points in X can be carried to any other pair of distinct points in X by some element of G That is, given any two pairs x, x 2 and y, y 2 in X X, where x x 2 and y y 2, there is a g G such that gx y and gx 2 y 2 Although the x i s are distinct and the y j s are distinct, we do allow an x i to be a y j For instance, if x, x, x are three distinct elements of X then there is a g G such that gx x and gx x Here x y x and x 2 x, y 2 x Necessarily a doubly transitive action requires #X 2 Example 2 The action of A 4 on {, 2, 3, 4} is doubly transitive Example 22 The action of D 4 on {, 2, 3, 4}, as vertices of a square, is not doubly transitive since a pair of adjacent vertices can t be sent to a pair of nonadjacent vertices Example 23 For any field F, the group AffF acts on F by a b x ax + b and this action is doubly transitive Example 24 For any field F, the group GL 2 F acts on F 2 { } by the usual way matrices act on vectors, but this action is not doubly transitive since linearly dependent vectors can t be sent to linearly independent vectors by a matrix Theorem 25 If G acts doubly transitively on X then the stabilizer subgroup of any point in X is a maximal subgroup of G

2 2 KEITH CONRAD A maximal subgroup is a proper subgroup contained in no other proper subgroup Proof Pick x X and let H x Stab x Step : For any g H x, G H x H x gh x For g G such that g H x, we will show g H x gh x Both gx and g x are not x, so by double transitivity with the pairs x, gx and x, g x there is some g G such that g x x and g gx g x The first equation implies g H x, so let s write g as h Then hgx g x, so g hgh x H x gh x Step 2: H x is a maximal subgroup of G The group H x is not all of G, since H x fixes x while G carries x to any element of X and #X 2 Let K be a subgroup of G strictly containing H x and pick g K H x By step, G H x H x gh x Both H x and H x gh x are in K, so G K Thus K G The converse of Theorem 25 is false If H is a maximal subgroup of G then left multiplication of G on G/H has H as a stabilizer subgroup, but this action is not doubly transitive if G has odd order because a finite group with a doubly transitive action has even order Theorem 26 Suppose G acts doubly transitively on a set X Any normal subgroup N G acts on X either trivially or transitively Proof Suppose N does not act trivially: nx x for some x X and some n in N Pick any y and y in X with y y By double transitivity, there is g G such that gx y and gnx y Then y gng gx gng y and gng N, so N acts transitively on X Example 27 The action of A 4 on {, 2, 3, 4} is doubly transitive and the normal subgroup {, 234, 324, 423} A 4 acts transitively on {, 2, 3, 4} Example 28 For any field F, let AffF act on F by a b x ax + b This is doubly transitive and the normal subgroup N { b : b F } acts transitively by translations on F Example 29 The action of D 4 on the 4 vertices of a square is not doubly transitive Consistent with Theorem 26, the normal subgroup {, r 2 } of D 4 acts on the vertices neither trivially nor transitively Here is the main group-theoretic result we will use to prove PSL n F is simple Theorem 2 Iwasawa Let G act doubly transitively on a set X Assume the following: For some x X the group Stab x has an abelian normal subgroup whose conjugate subgroups generate G 2 [G, G] G Then G/K is a simple group, where K is the kernel of the action of G on X The kernel of an action is the kernel of the homomorphism G SymX; it s those g that act like the identity on X Proof To show G/K is simple we will show the only normal subgroups of G lying between K and G are K and G Let K N G with N G Let H Stab x, so H is a maximal subgroup of G Theorem 25 Since N is normal, NH {nh : n N, h H} is a subgroup of G, and it contains H, so by maximality either NH H or NH G By Theorem 26, N acts trivially or transitively on X

3 SIMPLICITY OF PSL nf 3 If NH H then N H, so N fixes x Therefore N does not act transitively on X, so N must act trivially on X, which implies N K Since K N by hypothesis, we have N K Now suppose NH G Let U be the abelian normal subgroup of H in the hypothesis: its conjugate subgroups generate G Since U H, NU NH G Then for g G, gug gnug NU, which shows NU contains all the conjugate subgroups of U By hypothesis it follows that NU G Thus G/N NU/N U/N U Since U is abelian, the isomorphism tells us that G/N is abelian, so [G, G] N Since G [G, G] by hypothesis, we have N G Example 2 We can use Theorem 2 to show A 5 is a simple group Its natural action on {, 2, 3, 4, 5} is doubly transitive Let x 5, so Stab x A4, which has the abelian normal subgroup {, 234, 324, 423} The A 5 -conjugates of this subgroup generate A 5 since the 2,2-cycles in A 5 are all conjugate in A 5 and they generate A 5 The commutator subgroup [A 5, A 5 ] contains every 2,2-cycle: if a, b, c, d are distinct then Therefore [A 5, A 5 ] A 5, so A 5 is simple abcd abcabdabc abd 3 Simplicity of PSL 2 F Let F be a field The group SL 2 F acts on F 2 { }, but this action is not doubly transitive since linearly dependent vectors can t be sent to linearly independent vectors by a matrix We saw this for GL 2 F in Example 24, and the same argument applies for its subgroup SL 2 F Linearly dependent vectors in F 2 lie along the same line through the origin, so let s consider the action of SL 2 F on the linear subspaces of F 2 : let A SL 2 F send the line L F v to the line AL F Av Equivalently, we let SL 2 F act on P F, the projective line over F Theorem 3 The action of SL 2 F on the linear subspaces of F 2 is doubly transitive Proof An obvious pair of distinct linear subspaces in F 2 is F and F It suffices to show that, given any two distinct linear subspaces F v and F w, there is an A SL 2 F that sends F to F v and F to F w, because we can then use F and F as an intermediate step to send any pair of distinct linear subspaces to any other pair of distinct linear subspaces Let v a c and w b d Since F v F w, the vectors v and w are linearly independent, so D : ad bc is nonzero Let A /D, which has determinant ad/d b/dc c d/d D/D, so A SL 2 F Since A a c v and A b/d d/d /Dw, A sends F to F v and F to F /Dw F w We will apply Iwasawa s criterion Theorem 2 to SL 2 F acting on the set of linear subspaces of F 2 This action is doubly transitive by Theorem 3 It remains to check the following: the kernel K of this action is the center of SL 2 F, so SL 2 F /K PSL 2 F, the stabilizer subgroup of contains an abelian normal subgroup whose conjugate subgroups generate SL 2 F,

4 4 KEITH CONRAD [SL 2 F, SL 2 F ] SL 2 F It is only in the third part that we will require #F > 3 At some point we need to avoid F F 2 and F F 3, because PSL 2 F 2 and PSL 2 F 3 are not simple Theorem 32 The kernel of the action of SL 2 F on the linear subspaces of F 2 is the center of SL 2 F Proof A matrix a c d b SL 2F is in the kernel K of the action when it sends each linear subspace of F 2 back to itself If the matrix preserves the lines F and F then c and b, so a c d b a d a The determinant is, so d /a If /a preserves the line F then a /a, so a ± This means c d ± Conversely, the matrices ± both act trivially on the linear subspaces of F 2, so K {± } If a matrix a c d b is in the center of SL 2F then it commutes with and, which implies a d and b c check! Therefore a c d b a a Since this has determinant, a 2, so a ± Conversely, ± commutes with all matrices Let x F Its stabilizer subgroup in SL2 F is { Stab F A SL 2 F : A F { } SL d 2 F { : a F /a, b F This subgroup has a normal subgroup { } U which is abelian since λ µ λ+µ { λ } } : λ F, Theorem 33 The subgroup U and its conjugates generate SL 2 F More precisely, any matrix of the form is conjugate to a matrix of the form, and every element of SL 2 F is the product of at most 4 elements of the form or This is the analogue for SL 2 F of the 3-cycles generating A n Proof The matrix is in SL 2F and λ λ, so conjugates U { } to the group of lower triangular matrices { } Pick a c d b in SL 2F To show it is a product of matrices of type or, first suppose b Then c d If c then c d d /b a /c b c } a /b d /c

5 SIMPLICITY OF PSL nf 5 If b and c then the matrix is a /a, and a /a a/a a /a So far F has been any field Now we reach a result that requires #F 4 Theorem 34 If #F 4 then [SL 2 F, SL 2 F ] SL 2 F Proof We compute an explicit commutator: a b a b /a /a ba 2 Since #F 4, there is an a,, or in F, so a 2 Using this value of a and letting b run over F shows [SL 2 F, SL 2 F ] contains U Since the commutator subgroup is normal, it contains every subgroup conjugate to U, so [SL 2 F, SL 2 F ] SL 2 F by Theorem 33 Theorem 34 is false when #F 2 or 3: SL 2 F 2 GL 2 F 2 is isomorphic to S 3 and [S 3, S 3 ] A 3 In SL 2 F 3 there is a unique 2-Sylow subgroup, so it is normal, and its index is 3, so the quotient by it is abelian Therefore the commutator subgroup of SL 2 F 3 lies inside the 2-Sylow subgroup in fact, the commutator subgroup is the 2-Sylow subgroup Theorem 35 If #F 4 then the group PSL 2 F is simple Proof By the previous four theorems the action of SL 2 F on the linear subspaces of F 2 satisfies the hypotheses of Iwasawa s theorem, and its kernel is the center of SL 2 F 4 Simplicity of PSL n F for n > 2 To prove PSL n F is simple for any F when n > 2, we will study the action of SL n F on the linear subspaces of F n, which is the projective space P n F Theorem 4 The action of SL n F on P n F is doubly transitive with kernel equal to the center of the group and the stabilizer of some point has an abelian normal subgroup Proof For nonzero v in F n, write the linear subspace F v as [v] Pick [v ] [v 2 ] and [w ] [w 2 ] in P n F We seek an A SL n F such that A[v ] [w ] and A[v 2 ] [w 2 ] Extend v, v 2 and w, w 2 to bases v,, v n and w,, w n of F n Let L: F n F n be the linear map where Lv i w i for all i, so det L and on P n F we have L[v i ] [w i ] for all i In particular, L[v ] [w ] and L[v 2 ] [w 2 ] Alas, det L may not be For c F, let L c : F n F n be the linear map where L c v i w i for i n and L c v n cw n, so L L Then L c sends [v i ] to [w i ] for all i and det L c c det L, so L c SL n F for c / det L If A SL n F is in the kernel of this action then A[v] [v] for all nonzero v F n, so Av λ v v, where λ v F : every nonzero element of F n is an eigenvector of A The only matrices for which all vectors are eigenvectors are scalar diagonal matrices To prove this, use the equation Av λ v v when v e i, v e j, and v e i + e j for the standard basis e,, e n of F n The equation Ae i + e j Ae i + Ae j implies λ ei +e j e i + λ ei +e j e j λ ei e i +λ ej e j, so λ ei λ ei +e j λ ej Let λ be the common value of λ ei over all i, so Av λv when v runs through the basis By linearity, Av λv for all v F n, so A is a scalar diagonal matrix with determinant It is left to the reader to check that the center of SL n F is also the scalar diagonal matrices with determinant

6 6 KEITH CONRAD To show the stabilizer of some point in P n F has an abelian normal subgroup, we look at the stabilizer H of the point Pn F, which is the group of n n determinant matrices a M where a F, M GL n F, and is a row vector of length n For this to be in SL n F means a / det M The projection H GL n F sending a M onto M has abelian kernel { } 4 U : F I n n To conclude by Iwasawa s theorem that PSL n F is simple, it remains to show the subgroups of SL n F that are conjugate to U generate SL n F, [SL n F, SL n F ] SL n F This will follow from a study of the elementary matrices I n + λe ij where i j and λ F An example of such a matrix when n 3 is I 3 + λe 23 λ The matrix I n + λe ij has s on the main diagonal and a λ in the i, j position Therefore its determinant is, so such matrices are in SL n F The most basic example of such an elementary matrix in U is 42 I n + E 2 I n 2 Here are the two properties we will need about the elementary matrices I n + λe ij : For n > 2, any I n + λe ij is conjugate in SL n F to I n + E 2 2 For n > 2, the matrices I n + λe ij generate SL n F These properties imply the conjugates of I n + E 2 generate SL n F Since I n + E 2 U, the subgroups of SL n F that are conjugate to U generate SL n F, so the next theorem would complete the proof that PSL n F is simple for n > 2 Theorem 42 For n > 2, [SL n F, SL n F ] SL n F Proof We will show I n + E 2 is a commutator in SL n F Then, since the commutator subgroup is normal, the above two properties of elementary matrices imply that [SL n F, SL n F ] contains every I n + λe ij, and therefore [SL n F, SL n F ] SL n F

7 Set g SIMPLICITY OF PSL nf 7 and h Then ghg h, which is I 3 + E 2 For n 4, I n + E 2 is the block matrix I n 2 I n 3 g O O I n 3 h O O I n 3 g O h O O I n 3 O I n 3 All that remains is to prove the two properties we listed of the elementary matrices, and this is handled by the next two theorems Theorem 43 For n > 2, any I n + λe ij with λ F is conjugate in SL n F to I n + E 2 Proof Let T I n + λe ij For the standard basis e,, e n of F n, { e k, if k j, T e k λe i + e j, if k j We want asis e,, e n of F n in which the matrix representation of T is I n + E 2, ie, T e k e k for k 2 and T e 2 e + e 2 Define asis f,, f n of F n by f λe i, f 2 e j, and f 3,, f n is some ordering of the n 2 standard basis vectors of F n besides e i and e j Then T f λt e i λe i f, T f 2 T e j λe i + e j f + f 2, T f k f k for k 3, so relative to the basis f,, f n the matrix representation of T is I n + E 2 Therefore T AI n + E 2 A, where A is the matrix such that Ae k f k for all k Check T AI n + E 2 A by checking both sides take the same values at f,, f n There is no reason to expect det A, so the equation T AI n + E 2 A shows us T and I n + E 2 are conjugate in GL n F, rather than in SL n F With a small change we can get a conjugating matrix in SL n F, as follows For any c F, we have where T A c I n + E 2 A c, A c e k { f k, if k < n, cf n, if k n Check both sides of the equation T A c I n +E 2 A c are equal at f,, f n, cf n, where we need n > 2 for both sides to be the same at f 2 The columns of A c are the same as the columns of A except for the nth column, where A c is c times the nth column of A

8 8 KEITH CONRAD Therefore deta c c det A, so if we use c / det A then A c SL n F That proves T is conjugate to I n + E 2 in SL n F Example 44 Let T I 3 + λe 23 λ Starting from the standard basis e, e 2, e 3 of F 3, introduce a new basis f, f 2, f 3 by f λe 2, f 2 e 3, and f 3 e Since T f f, T f 2 f + f 2, and T f 3 f 3, we have λ λ λ, where the conjugating matrix λ has for its columns f, f 2, and f 3 in order The determinant of this conjugating matrix is λ, so it is usually not in SL 3 F If we insert a nonzero constant c into the third column then we get a more general conjugation relation between I 3 + λe 23 and I 3 + E 2 : λ c λ c λ The conjugating matrix has determinant λc, so using c /λ makes the conjugating matrix have determinant, which exhibits an SL 3 F -conjugation between I 3 + λe 23 and I 3 + E 2 Theorem 45 For n 2, the matrices I n + λe ij with i j and λ F generate SL n F Proof This will be a sequence of tedious computations By a matrix calculation, { E il, if j k, 43 E ij E kl δ jk E il O, if j k Therefore I n + λe ij I n + µe ij I n + λ + µe ij, so I n + λe ij λe ij, so the theorem amounts to saying that every element of SL n F is a product of matrices I n +λe ij We already proved the theorem for n 2 in Theorem 33, so we can take n > 2 and assume the theorem is proved for SL n F Pick A SL n F We will show that by multiplying A on the left or right by suitable elementary matrices I n + λe ij we can obtain lock matrix A Since this is in SL nf, taking its determinant shows det A, so A SL n F By induction A is a product of elementary matrices I n + λe ij, so A would be a product of block matrices of the form I n +λe ij, which is I n + λe i+ j+ Therefore product of some I n + λe ij Aproduct of some I n + λe ij product of some I n + λe ij, and we can solve for A to see that it is a product of matrices I n + λe ij

9 SIMPLICITY OF PSL nf 9 The effect of multiplying an n n matrix A by I n + λe ij on the left or right is an elementary row or column operation: a a n I n +λe ij A a i + λa j a in + λa jn ith row ith row of A + λjth row of A a n a nn and AI n + λe ij a a j + λa i a n a n a nj + λa ni a nn jth col jth col of A + λith col of A Looking along the first column of A, some entry is not since det A If some a k in A is not where k >, then 44 I n + a E k A a k If a 2,, a n are all, then a and I n + E 2 A a Then by 44 with k 2, I n + a E 2 a I n + E 2 A a, Once we have a matrix with upper left entry, multiplying it on the left by I n + λe i for i will add λ to the i, -entry, so with a suitable λ we can make the i, -entry of the matrix Thus multiplication on the left by suitable matrices of the form I n + λe ij produces lock matrix B whose first column is all s except for the upper left entry, which is Multiplying this matrix on the right by I n + λe j for j adds λ to the, j-entry without changing any column but the jth column With a suitable choice of λ we can make the, j-entry equal to, and carrying this out for j 2,, n leads to a block matrix A, which is what we need to conclude the proof by induction References [] L E Dickson, The analytic representation of substitutions on a power of a prime number of letters with a discussion of the linear group, Ann Math 897, 6 83 [2] L E Dickson, Theory of linear groups in an arbitrary field, Trans Amer Math Soc 2 9, [3] K Iwasawa, Über die Einfachkeit der speziellen projection Gruppen, Proc Imperial Acad Tokyo 7 94, [4] C Jordan, Traité des Substitutions, Gauthier-Villars, Paris, 87 [5] E H Moore, A doubly-infinite system of simple groups, Bull New York Math Soc 3 893, 73 78