April 14: S_4 and A_5 twins in PSL_2(F); the twin A_5's in PSL_2(F_11) and the Paley biplane (= square 2-(11,5,2) design) We saw last week that PSL_2(F_q) contains S_4 if q is \pm 1 mod 8, and A_5 if q is not \pm 2 mod 5. Now we just finished (at least in sketch) showing where these fit in the list of all subgroups of PSL_2(F_q), and indeed of finite subgroups of PSL_2(F) for any F. The groups that appear for F=C, namely cyclic, dihedral, A_4, S_4, and A_5, are particularly important: versions of this list appear in contexts ranging from graph theory (the text's chapter on "graphs with least eigenvalue -2" is part of this story, and the -2 turns out to be there for a good reason) to reflection groups, quadratic forms, simple Lie groups (and thus finite simple groups like PSL_n and Sp_n coming from matrix groups), singularities of algebraic varieties, and at least a few other places whose names might not have much resonance with the undergraduate curriculum (tame quivers anyone?). For our main concern in 155r, we're particularly interested in the first cases where these groups A_4, S_4, A_5 first occur as proper subgroups of PSL_2(F_q). That's q=5 for A_4=PSL_2(F_3), q=7 and q=9 for S_4=PGL_2(F_3), and q=9 and q=11 for A_5 = PSL_2(F_5) = (P)SL_2(F_4). The action of PSL_2(F_7) on the 7 cosets of S_4 identifies it with Aut(Pi_2); you've seen a lot about this in the last and present homework. I already noted that the index-6 subgroup A_5 of PSL_2(F_9) identifies PSL_2(F_9) with A_6, and so gives another route to the outer automorphism of S_6, which we shall encounter again in connection with M_12. Today we'll show that the action of PSL_2(F_11) on the 11 cosets of A_5 identifies it with the Paley biplane, a square 2-(11,5,2) design we've constructed as a special case of a 2-(q,(q-1)/2,(q-3)/4) design for any prime power q == 3 mod 4 that usually has only the a^2 x + b group but here has 12 times as many automorphisms. (This happens also for the q=7 version of this design, which is just Pi_2 again; it is known that for q>11 there are no automorphisms beyond the a^2 x + b group and Aut(F), forming a semidirect product as usual.). You already showed that the Paley biplane is uniquely determined by its parameters up to isomorphism, and its automorphism group -- call it G for now -- has size 660 with point stabilizer A_5 acting transitively on the other 10 points. Since this design is square, the same is true of the dual design, which gives us another A_5 in G, namely the block stabilizer, which is not conjugate in G to the point stabilizer. This should look familiar: again the same is true for Aut(Pi_2), which has two conjugacy orbits of index-7 subgroups isomorphic with S_4. So there are two S_4's in PSL_2(F_7), and since we expect G to be SL_2(F_11) there should be two A_5's in PSL_2(F_11). This is true, and we'll use it to reconstruct the biplane. We can try to do this for any F_q with q odd, as follows: PSL_2(F_q) is an index-2 subgroup of PGL_2(F_q); given a copy of H=S_4 or A_5 in PSL_2(F_q) we can conjugate it by an element of PGL_2(F_q) to get a new copy H' of H in PSL_2(F_q). Can it be conjugate even in PSL_2, not just PGL_2? If so then by composing the two conjugations we find g in PGL_2(F_q) that is not in PSL_2(F_q) but for which conjugation by g takes H to itself. So this conjugation is an automorphism of H. Now for H=S_4 we know all automorphisms are inner. So, composing with conjugation by some element of H we get g', still in PGL_2(F_q) but not PSL_2(F_q), that centralizes H. But that's impossible because H is its own centralizer in PGL_2. (As we did last time we can check that the centralizer of any element of GL_2 is abelian, so so the centralizer in PGL_2 has an abelian subgroup of index at most 2, which H doesn't.) So H and H' are not conjugate in PSL_2. For H=A_5 the same argument almost works. It is not quite true that every automorphism of A_5 is inner. But an outer automorphism is conjugation by an odd element of S_5, and such an automorphism switches the two A_5-conjugacy classes of 5-cycles. But conjugation in GL_2 preserves the trace, so this can happen only if our two t-values coincide. Remember these are the roots of t^2 \pm t = 1. This can happen, but only in characteristic 5 (and 2, but there PGL_2=PSL_2 so there's no g to conjugate by). Indeed in char.5 there's only one conjugacy class of A_5 because we already know that A_5 is all of PSL_2(F_5) (and Aut(A_5)=S_5 is PGL_2(F_5)). But otherwise we again conclude that H and H' are not conjugate in PSL_2(F_q). In each case it follows that PSL_2(F_q) has two inequivalent actions on sets of size |PSL_2(F_q)| / |H|. For instance, Aut(Pi_2) acts on points and lines of Pi_2, and the point stabilizers are different (though interchanged by an automorphism of Aut(Pi_2), which comes from PGL_2(F_7) or the inverse-transpose automorphism for 3x3 matrices mod 2). [Aside: remarkably the cycle structures are the same: 1^7, 22111, 331, 421, 7 for elements of orders 1,2,3,4,7 respectively. This not only disproves a natural conjecture that the cycle structures determine the group up to conjugacy, but can also be exploited in related constructions of isospectral drums and non-isomorphic number fields of degree 7 with the same zeta function.] For PSL_2(F_11) we get two inequivalent actions on 11 points, the coset spaces for H and H'. Well H' is the point stabilizer on its 11-point set; how does it act on that of H? It cannot be transitive because 11 is not a factor of 60. Nor does it have a fixed point, else it would be contained in a conjugate of H, but then it would be a conjugate of H because |H'|=|H|, and we just showed that this doesn't happen. So it must have at least two orbits, each of which is the index of a proper subgroup of H. The only possibility is 5+6. So we get simultaneously the defining permutation representation of H=A_5 on 5 items, and the action on the 6 points of P^1(F_5) that we first saw in Aut(S_6). This also lets us prove that PSL_2(F_11) acts doubly transitively on each of our 11-point sets (as it should if it's going to be Aut(Paley)). That is, we show that the point stabilizer's orbit sizes are 1+10. The only other possibility is 1+5+5. But then a 3-cycle in H would have 5 fixed points, whereas one in H' has only 2 (since it has cycle structure 113 and 33 as a 5- and 6-point permutation respectively). But there's only one PSL_2(F_11)-conjugacy class of 3-cycles (they have trace \pm 1, which is not \pm 2, so we can apply the result proved at the end of the April 9 notes). So it can be placed in both H and H' and that's a contradiction. Doing this for each conjugate of H' gives us 11 five-point sets in the coset space PSL_2(F_11) / H, one for each of the 11 cosets in PSL_2(F_11) / H'. These are the blocks of a 2-design because H permutes them and acts doubly transitively. The design has the parameters of a Paley biplane, and we already showed that this determines the design uniquely, with a group G of 660 automorphisms. We now see that PSL_2(F_11) is contained in G, and by comparing sizes we conclude that G = PSL_2(F_11) as claimed. [BTW the two 11-point representations also have identical cycle structures, and can be put to the same uses as the two 7-point representations of the simple group of order 168.]