March 24: The 5-(24,8,1) Steiner system D_24; its automorphism group M_24, and its subgroups M_23, M_22, and [PSL_3(F_4)=]M_21; the automorphism groups of D_23 and D_22 With the description of hyperovals and Baer subplanes we developed to construct and prove the uniqueness of D_22 and D_23, we now easily show D_24 is unique if it exists, and with a bit more effort verify existence and the number of automorphisms. We then describe Aut(D_24), which is M_24, the largest of Mathieu's 5 sporadic groups, and its k-point stabilizer M_{24-k} for k=1,2,3 (the last of which coincides with the normal subgroup PSL_3(F_4) of Aut(Pi_4)). Recall the intersection triangle of a 5-(24,8,1) Steiner system (Table 1.1 on p.21 of the text): 759 506 253 330 176 77 210 120 56 21 130 80 40 16 5 78 52 28 12 4 1 46 32 20 8 4 0 1 30 16 16 4 4 0 0 1 30 0 16 0 4 0 0 0 1 So in a 5-(24,8,1) design any two blocks meet in an even number of points (which also follows immediately from the odd-intersection property of the derived 4-(23,7,1) design. Fix three points \infty_1, \infty_2, \infty_3. [See the March 24 notes for the alternative I, II, III.] The doubly derived design is Pi_4, and we've already accounted for each of the singly derived designs last time. This tells us that the 759 blocks include 21: \infty_1 + \infty_2 + \infty_3 + line 3*56: \infty_j + \infty_k + hyperoval in equivalence class i ({i,j,k}={1,2,3}) 3*120: \infty_i + Bear subplane in equivalence class i That leaves 210, which must be 8-element subsets S of the points of Pi_4 that meet every line, hyperoval, and Baer subplane in an even number of points. Proposition: Suppose S is a set of 8 points in Pi_4. Then TFAE: (i) S intersects every line, hyperoval, and Baer subplane in an even number of points. (ii) S intersects every line in an even number of points. (iii) S is one of the Bin(21,2) = 210 symmetric differences of line pairs. Thus D_24 must use all 210 of them, so the structure of D_24 is completely determined by the choice of \infty_i and a bijection between those 3 points and the 3 equivalence classes of hyperovals and subplanes. _Proof of Proposition_: (i) ==> (ii) is obvious. (iii) ==> (i) is easy, because the intersection of every {line, hyperoval, subplane} with a line always has {odd, even, odd} size. This leaves (ii) ==> (iii). Applying (ii) to the five lines through a given p point of S, we see one of these lines has four points of S, and the others have two each. (If 7=8-1 is a sum of 5 odd natural numbers then one of them is 3 and the other four are 1.) Call that line l. Let p' be a point of S not in l. The same argument shows p' is contained in a line l' containing four points of S. The intersection of l and l', call it p", is not in S, else p" would lie on two four-point lines. Hence S is the symmetric difference of l and l', and we're done. Theorem: the resulting D_24 is indeed a (5,8,24) Steiner system. Proof: We show that any five points are contained in a unique block. If at least one of them is among the three \infty_i points then we have done this already in our analysis of the derived designs D_23, D_22, or [Pi_4=]D_21. So assume all five are in Pi_4. @ If all 5 on a line: must use that line plus all three \infty_i's. @ Four on line l, a fifth p is not on l: can't use a line, hyperoval, or subplane; so it must be a symmetric difference, and we readily see that the symm.diff. of l and the line through p and the missing point of l works and is unique. @ No four of a line, but three on each of two lines l and l': must be the Baer subplane they span (and whichever \infty_i goes with it). @ No four on a line, and exactly one l containing three of them: must be the symmetric difference of l and the line through the other 2. Finally, @ No three on a line: it's a Type I oval, so can't be a line or symm.diff. or subplane, but it's contained in a unique hyperoval, so it's that hyperoval (and whichever two of the \infty_i go with it). QED! Since D_24 exists and is uniquely determined by its design property there is an unambiguous M_24 = Aut(D_24). This is the largest of the Mathieu groups. We next give some of its properties (but do not prove the simplicity -- that will be next week, together with the other M_n). First we finally make explicit the relation among @ P\Gamma L_3(F_4), @ its index-2 normal subgroup PGL_3(F_4), and @ the index-3 normal subgroup PSL_3(F_4) of PGL_3(F_4). Namely: Lemma: (i) PSL_3(F_4) is a normal subgroup of P\Gamma L_3(F_4), and PSL_3(F_4) is a normal subgroup of P\Gamma L_3(F_4). (ii) In both cases the quotient subgroup is isomorphic to S_3. The three objects permuted can be identified with the three equivalence classes of B's (or O's). Proof: We start with the linear and conjugate-linear transformations of V = F_4^3. Choosing coordinates, we have the linear maps, which form the group GL_3(F_4) of invertible 3*3 matrices, and the conjugation c = c^{-1}, taking x = (x0,x1,x2) to (x0^2, x1^2, x2^2). These generate \Gamma L_3(F_4) as follows. The group consists of invertible matrices A and transformations cA where A is invertible. It's clear how to compose A B and cA B. As for A cB and cA cB, the key fact is that for any vector x we have Ac(x) = c(cAc)x = cA'x where A' is the matrix obtained by applying c to (i.e. squaring) each entry of A. So Ac = cA', whence A cB = Ac B = c A'B and similarly cA cB = A'B. This gives us the group law in \Gamma_3(F_4). Since 1'=1 the group still has 1 for an identity element (as it must -- it's still the identity transformation of V), and the inverse of cA is (A^{-1})c = cA'^{-1} = c(A^{-1})'. This is basically the special case H = {id,'} of the construction of the semidirect product of a group G with a group H of automorphisms of G. Now det(A') = det(A)' = 1/det(A). So the transformations A and cA with det(A)=1 form a subgroup, called (what else?) \Sigma L_3(F_4). It contains SL_3(F_4) with index 2, so SL_3 is normal in \Sigma L_3(F_4). But \Sigma L_3(F_4) is *not* normal in \Gamma L_3(F_4). Indeed let A, B in GL_3(F_4) have det(A)=1 but det(B) *not* 1. Then cA is in \Sigma L_3(F_4). I claim that its conjugate by B is not: B^{-1} cA B = c (B^{-1})' A B and so it's of the form cM with det(M) = det(B)/det(B)' which is not 1. On the other hand SL_3(F_4) *is* normal. Suppose det(A)=1 and B in GL_3(F_4). Then certainly det(B^{-1} A B) = 1 -- that's just the fact that SL_3 is normal in GL_3 -- and the new issue is whether (cB)^{-1} A cB also has determinant 1. But the inverse of cB is c(B')^{-1}, so the conjugate is c(B')^{-1} A cB = c(B')^{-1} c A' B = B^{-1} A' B (note that we undid the ' when conjugating (B')^{-1} by c) which has determinant det(A') = det(A)' = 1. Now that SL_3(F_4) is normal in \Gamma L_3(F_4), what's the quotient? I claim it's S_3, arising as \Gamma L_1(F_4), a.k.a. the semidirect product of the 3-element group F_4^* with '. Indeed I have a map from \Gamma L_3(F_4) to this group, which is the determinant on GL_3(F_4), and takes cA to c*det(A). This is a surjective homomorphism and its kernel is SL_3(F_4), so it identifies \Gamma L_3(F_4) / SL_3(F_4) with the target group S_3 as claimed. [Note that this used very little about F_4 and nothing special to 3*3 matrix, and can be adapted directly GL_n(k) for any field k with an involution, and with a bit of change to GL_n(k) for any field k with some automorphism group, which is inherited by GL_n in the same way. Even the next paragraph works more generally though in one place we exploit the fact that all the invertible scalar 3*3 matrices over F_4 happen to have determinant 1.] So this does (i) and (ii) for SL_3. What about PSL_3? That was the quotient of SL_3 by the scalar matrices. These matrices still form a normal subgroup of \Gamma_3(F_4): the conjugate-linear transformations don't commute with scalars but do conjugate each scalar to its inverse. (This also means that P\Gamma L_3 respects the construction of Pi_4 as the quotient of V-{0} by scalars.) So the normality and quotient are inherited by the projective groups PSL_3 and P\Gamma L_3. This proves (i) and (ii) for PSL_3(F_4). For (iii), it is now enough to produce two inequivalent ovals switched by c, and that's easy: use the two ovals that contain (1:0:0), (0:1:0), and (0:0:1) but do not contain (1:1:1). QED We'll also need the action of PGL_2(F_4) and P\Gamma L_2(F_4) on P^1(F_4). Since |P^1(F_4)| = 4+1 = 5 these map to S_5. Lemma: The map is an isomorphism from PGL_2(F_4) to A_5 and from P \Gamma L_2(F_4) to A_5. (So we have another of our exceptional isomorphisms -- one of the easier ones.) Proof: We have seen that the action of PGL_2(k) on P^1(k) is faithful and simply 3-transitive. For |k|=4 this forces its image to be A_5. Since P \Gamma L_2(F_4) contains a simple transposition, its image must be S_5. Remark: Likewise PGL_2(k) is S_3 and S_4 for |k|=2 and |k|=3 respectively, and you saw that for |k|=5 it's the transitive S_5 in S_6. Finally, we know already that PGL_3(F_4) acts 2-transitively on Pi_4 -- indeed PGL_n(k) acts 2-transitively on projective (n-1)-space over k for any field k and any n>1. We'll need the fact that PSL_n(k) acts 2-transitively as well. This is easy: 2-transitivity of PGL_n amounts to the fact that for any distinct 1-dimensional subspaces l1,l2 of k^n we can find a basis v1,v2,...,vn such that v1 is in l1 and v2 in l2; for PSL_n we need only show that this basis can be chosen so that det(v1,v2,...,vn) = 1 -- but this is easy because we can replace v1 by c*v1 for any nonzero scalar c, and that lets us adjust the determinant. We can now prove: Theorem: M_24 acts 5-transitively on the points of the design, and transitively on the 759 blocks. The point stabilizer has order 48. The block stabilizer acts on the block as A_8, and on the complement as affine GL_4(F_2). The stabilizer of a block and a point outside it acts as A_8 on the block and as GL_4 on the remaining 24 - 8 - 1 = 15 = |P^3(F_2)| points. Proof: We already have 3-transitivity, and the 3-point stabilizer is PSL_3(F_4). Since that's 2-transitive, we have 5-transitivity, and the 5-point stabilizer in M_24 is the 2-point stabilizer in PSL_3(24), which has order #PSL_3(24) / (21*20) = 16*9/3 = 48. Transitivity on blocks follows from 5-transitivity, because any 5 points determine a unique block. Since all blocks are equivalent, we can study the block stabilizer by choosing any convenient block B, and we choose B = \infty_1 + \infty_2 + \infty_3 + the line at infinity l0 of Pi_4. Let G be the stabilizer of B, and H the image of G in the group S_8 of permutations of B (i.e. H is the group of permutations of B that come from G). Then H is 5-transitive, and the stabilizer of {\infty_1, \infty_2, \infty_3} is the stabilizer of l0 in PSL_3(4), which is PGL_2(k), which is A_5. So H is A_8 as claimed. Now the action of G on the remaining 24-8=16 points must preserve the set of blocks disjoint from B. The intersection triangle promises us 30 such blocks. Our description of blocks disjoint from {\infty_1, \infty_2, \infty_3} tells us they are the symmetric differences of lines that meet in l0. But the complete of l0 is an affine plane over F_4, and lines of Pi_4 that meet in l0 are parallel in the affine plane. Well affine F_4^2 is also a 4-dimensional affine space over F_2, and a little thought and/or counting shows that our 30 blocks are just the 3-design of affine hyperplanes in F_2^4. So the action of G must preserve that structure, whence its image is contained in AGL_4(F_2) (the semidirect product of GL_4(F_2) with F_2^4). But G can barely fit in AGL_4(F_2): |G| = |M_24| / 759 = 16 |GL_4(F_2)| = |AGL_4(F_2)| and G must act faithfully on those 16 points (proof: if g is a non-identity element of G that fixes all of them then g(x)=y for some distinct x,y in B; choose some other z in B, and a block B' meeting B only in {x,z} to conclude that g takes B' to B' which is a contradiction). So G = AGL_4(F_2), and the block+point stabilizer is GL_4(F_2), QED. Corollary: GL_4(F_2) =~= A_8 ! Cf. the outer isomorphism S_6 --> S_6 we found in M_12. (We'll also find an outer automorphism of M_12 in M_24.) Note that PSL_3(F_4) also happens to have 20160 = 8!/2 elements, but its turns out that PSL_3(F_4) is *not* isomorphic with A_8. For example A_8 has elements of order 15 but PSL_3(F_4) does not.