April 9: When does PSL_2(F) contain the exceptional triangle groups A_4, S_4, A_5? As announced last time: Proposition. Let F be a finite field of q elements. Then PSL_2(F) contains... ...A_4 (n=3), iff q is not an odd power of 2; ...S_4 (n=4), iff q is odd and congruent to 1 or -1 mod 8; ...A_5 (n=5), iff q is congruent to 0, 1, or -1 mod 5. We saw last time that each of these groups is contained in an arbitrary group H iff H contains elements b, c, d of order 2, 3, and n with bcd=1, with n in {3,4,5} as above. So we begin by describing the elements of order 2, 3, 4, 5 in PSL_2(F). Lemma: Let F be any field, and let g be a non-identity element of PSL_2(F) of trace \pm t. Then g is: i) A transvection iff t = \pm 2; ii) of order 2 iff t = 0; iii) of order 3 iff t = \pm 1; iv) of order 4 iff t is nonzero and t^2 = 2; v) of order 5 iff t^2+t-1=0 or t^2-t-1=0. [Note: since g is a 2x2 matrix A defined up to A <--> -A, its trace t=tr(A) is also defined only up to t <--> -t. Comparison of (i) with (ii), (iii), (v) shows that in characteristic p=2, 3, 5 respectively g is a transvection iff it is of order p; this is true in general. Comparing (ii) with (iv) yields the corollary that in characteristic 2 there are no elements of order 4; likewise in general PSL_2(F) has no elements of order p^2 if F has characteristic p.] Proof: Recall (or check directly) that any 2*2 matrix A satisfies its characteristic equation A^2 - tr(A)*A + det(A)*I = 0. If A is in SL_2(F) has trace t then this simplifies to (*) A^2 - t A + I = 0 (i) We show that a 2*2 matrix A is a transvection iff tr(A)=2 and A \neq I, from which the claim will follow. Recall that A is a transvection iff A-I has rank 1 and (A-I)^2 = 0. For a 2x2 matrix, (A-I)^2 = 0 already implies A-I has rank 0 or 1, and 0 is impossible unless A=I. So A is a transvection iff it satisfies (*) with t=2, and of course (*) can hold for only one value of t (else A=0 but then (*) does not hold at all), so A is a transvection iff tr(A)=2 as claimed. (ii) If t=0 then (*) becomes A^2 = -I, so the image of A in PSL_2(F) is a square root of the identity. Conversely, if A is not a multiple of the identity but A^2 = tA - I is then t=0. (Note that this also means that, except in characteristic 2, the matrix -I is the only order-2 element of SL_2(F): any other matrix with A^2=cI has c=-1.) (iii) We compute A^3 using (*), which tells us A^2 = tA - I so A^3 = A (tA-I) = tA^2 - A = t(tA-I) - A = (t^2-1) A - tI. Thus if A is not a multiple of identity then A^3 is a multiple of I iff t^2 = 1. [The "if" part can also be seen directly from the familiar factorizations A^3+I = (A+I)(A^2-A+I), A^3-I = (A-I)(A^2+A+I).] (iv) Likewise computing A^4 yields (t^3-2t) A + (1-t^2) I so A^4 is a multiple of I iff t^3-2t=0, and we already know that t=0 would entail A^2=I. (v) And A^5 = (t^4-3t^2+1)A + (2t-t^3), with the leading coefficient factoring as (t^2-1)^2-t = (t^2+t+1)(t^2-t-1). QED Remark: this can be interpreted in terms of the eigenvalues of A, which I suppose in this course are called r and s. So rs=1 and r^n = s^n = 1 or -1. For n=2 this means {r,s} = {i,-i} so r+s=0. For n=3, r and s are conjugate 3rd or 6th roots of unity, which have real part cos(2*Pi/3)=-1/2 or cos(Pi/3)=1/2 so r+s=-1 or 1. For n=4 they're 8th (but not 4th) roots of unity so we get 2*cos(Pi/4) or 2*cos(3*Pi/4), i.e. sqrt(2) or -sqrt(2). For n=5 we recover the identities relating cos(Pi/5) and cos(2*Pi/5) to the Golden Ratio. Similarly for n=6 we'll get cos(Pi/6) and cos(5*Pi/6), i.e. t^2=3, from a leading coefficient t(t^2-1)(t^2-3). Using this Lemma we already see that some of the conditions claimed in our Proposition are at any rate necessary: for S_4, the field must have odd characteristic and contain a square root of 2, so q is 1 or -1 mod 8; and for A_5 the field must contain a solution of t^2-t-1, so either q=2^f with f even or q is odd and F contains a square root of 5, which in either case is equivalent to q = 1 or -1 mod 5 (or 0 mod 5). (I used Quadratic Reciprocity here; in this case we can use the root-of-unity interpretation to prove the necessary cases of QR, and indeed one can use roots of unity to prove QR in general starting with these examples.) As noted last time, these conditions are also almost equivalent to the Lagrange condition that #(PSL_2(F)) be a multiple of the size of A_4, S_4, or A_5 respectively. Indeed #PSL_2(F) is q^3-q if q=2^f and (q^3-q)/2 otherwise. This is always a multiple of 3 [Fermat], and is a multiple of 5 iff q is 0, 1, or -1 mod 5. This leaves only divisibility by 4 or 8. For q odd, divisibility by 4 is automatic because q^2 is always 1 mod 8, and (q^3-q)/2 is a multiple of 8 iff (q-1)(q+1) is a multiple of 16 -- and since both factors are even and only one is a multiple of 4, this happens iff q-1 or q+1 is a multiple of 8. We can now dispose of the case q=2^f of the Proposition. We've already taken care of S_4. For A_5 the congruence condition on q means f must be even. But then F contains F_4, so PSL_2(F) contains PSL_2(F_4), and we already saw that this group is isomorphic with A_5. This also gives us A_4 when 2|f: it's the point stabilizer. Putting the stable point at infinity we see that the normal subgroup V of A_4 consists of 1 and the three transvections x --> x+1, x+a, x+b that commute with each other and multiply to 1. Conversely: two transvections commute iff they fix the same point. ("If" is easy. "Only if": else put the fixed point at 0 and infinity and calculate that the two products of [1,a;0,1] and [1,0;b,1] differ by diag(ab,-ab) so cannot agree.) Put that point at infinity. The other elements of A_4 must fix infinity as well because they conjugate each of the three transvections to another one. So it's in the ax+b group with a^3=1, whence 3|2^f-1 and 2|f as claimed. To go further we must also address the condition bcd=1. If bcd=1 then also b'c'd'=1 where b',c',d' are the conjugates of b,c,d by the same element of PSL_2(F). It turns out all candidate c elements are conjugate in PSL_2(F) [see below] so we may as well take c=[1,-1;1,0]. We show that over a finite field any of our values of t can be attained as trace(bc) for some b of order 2. By part (ii) of our opening lemma, b ranges over 2*2 matrices of the form [x,y;z,-x] for some x,y,z, and then det(b)=1 implies z=-(x^2+1)/y. The trace of bc is then (x^2+xy+y^2+1)/y, so we must solve x^2 + xy + y^2 + 1 = ty with y nonzero. Since we're in odd characteristic we may complete the square: let x = x1 - y/2 to get x1^2 = -(3*y^2/4 + t*y - 1). In characteristic 3 this simplifies to x1^2 = 1-t*y so there's always a solution (e.g. take x1=0 and y=1/t, or if t were zero take x1=1 and y arbitrary). Otherwise we complete the square on the right-hand side too, getting x1^2 = -(3/4)*y1^2 - 1 + t^2/3. So there's always a solution because t^2 isn't 3 (if it were we'd have n=6), by a trick we've already seen: as x1 and y1 vary over a finite field of odd order q, the LHS and RHS take (q+1)/2 different values, so there must be overlap. The only issue is that y might vanish, but then y1 doesn't, so changing y1 to -y1 finally yields the desired solution. This completes the proof of the Proposition. P.S. As promised: Lemma: Let F be a finite field. If A,A' in SL_2(F) are conjugate then tr(A)=tr(A'). This necessary condition is also sufficient unless tr(A) = 2 or -2. Remarks: 1) Neither condition (F finite, trace not 2 or -2) can be dropped. Over R, clockwise and counterclockwise rotations by the same angle are conjugate in GL_2(R) but not in SL_2(R). Even over a finite field, we certainly must exclude tr(A) = 2 or -2 because a transvection T and the identity I both have trace 2 and are not conjugate, and likewise -T and -I have trace -2; and as we've seen it is also possible for two transvections not to be conjugate. 2) It follows immediately from this criterion that two elements g,g' of PSL_2(F) that are not of trace \pm 2 (i.e. neither the identity nor a transvection) are conjugate iff tr(g) = \pm tr(g'). Proof: The condition is certainly necessary by general linear algebra. The converse is also standard linear algebra if we require only conjugacy in GL_2(F) because the eigenvalues are distinct. So we can conjugate A' to A with a matrix of some nonzero determinant D, and then we need only show that there's a matrix of the same determinant that conjugates A to itself, i.e. that commutes with A. We'll use matrices of the form Ax+Iy, whose determinant is x^2+txy+y^2. So we must solve x^2+txy+y^2=D in F. (Again we see how this can fail if F=R; it can also fail in a finite field of odd characteristic if t is 2 or -2.) To do this in characteristic 2, just let x=0 and make y the square root of D. Otherwise complete the square to get x^2+rz^2=D where r=(t/2)^2-1 is nonzero because t is not 2 or -2. Then use the fact that each of x^2 and rz^2 takes (q+1)/2 distinct values so the sets of possible x^2 and D-rz^2 must overlap and we're done.