April 12: Dickson's theorem: the list of finite subgroups of SL_2(K) Dickson proved: Theorem: Let K be a finite field of characteristic p, and G a subgroup of SL_2(K). Then: If |G| is relatively prime to p then it is either cyclic, the dihedral group , or the preimage in SL_2(K) of one of the groups A_4, S_4, A_5 described last week. If |G| is a multiple of p then either G is contained in a point stabilizer (in the action on P^1(K)), or p=2 and G is dihedral of order 2n with n odd, or p=3 and G is a preimage in SL_2(K) of A_5, or G is a subgroup SL_2(k) for some subfield k of K, or p>2 and G contains SL_2(k) with index 2 and is generated by SL_2(k) and diag(c,1/c) where c is a square root of a generator of k*. We follow the final section of Suzuki's _Group Theory I_ (Grundlehren der mathematischen Wissenschaften 247; Springer 1982; chapter 3, section 6, p.392 ff.). The theorem is stated as 6.17 (and describes all finite groups of SL_2(K) for K algebraically closed). It is due to Dickson, appearing in his 1901 _Linear Groups with an Exposition of the Galois Field Theory_ (reprinted by Dover 1958). Since a subgroup of PSL_2(K) is precisely a subgroup of SL_2(K) containing -I, this also gives the complete list of subgroups of PSL_2(K). For Galois' theorem we need only the first part: if |K|=p and |G| is a multiple of p then the image of G in PSL_2(K) has index prime to p, and thus equal at least p+1. We may also assume G contains -I (if p>2), because if not then G u -G is a subgroup that contains -I and has the same image in PSL_2. See Suzuki p.396 for the brief argument that shows that the list of subgroups containing -I also suffices to determine all subgroups of SL_2. Dickson's theorem is 6.17 in Suzuki. 6.25 answers our question directly: the subgroups of SL_2(K) are dihedral of order 2(q+1)/d and 2(q-1)/d and their subgroups, where d = gcd(2,q-1); an a^2 x + b group of order (q^2-q)/d and its subgroups, A_4, S_4, A_5 as described Friday, and PSL_2(k) or PGL_2(k) where k is a subfield (and [K:k] is even in the case of PGL_2). To save space we'll abbreviate the 2x2 matrix [a11 a12] [a21 a22] by [a11,a12; a21,a22]. (This is the format I'm used to from the computer algebra packagesgp; other such packages [Maxima, Mathematica, Magma, etc.] have similar conventions.) Also [a,0;0,b] is denoted diag(a,b), and is is SL_2(F) iff ab=1. 6.2, 6.3 We shall repeatedly use the following elements of SL_2(K). For c in F*, let d_c = diag(c,1/c). These form a subgroup D of SL_2(K) isomorphic with F*. In particular d_{-1} = -I, which for p>2 is the unique nontrivial element of the center Z of SL_2(K), and (as we saw Friday) the unique element of order 2 [while for p=2 the center is trivial and the order-2 elements are the conjugates of t_1]). Likewise let t_\lam = [1,0;\lam,1] ; these form a subgroup T isomorphic with (F,+), consisting of I and the q-1 transvections that fix 0 in P^1. [I don't know why Suzuki chose to fix 0 rather than infinity.] Then H = D T is the stabilizer of 0, and contains T as a normal subgroup. Finally let w = [0,1; -1,0]. Then w^2 = -1 and w^{-1} d_c w = d_{1/c} for all c. It is a basic fact of linear algebra that _over an algebraically closed field_ any 2*2 matrix is conjugate with some d_c or with \pm t_1 (Jordan canonical form). Thus the only elements of SL_2(K) with order divisible by p are the conjugates of \pm t_1, with order p for t_1 and 2p for -t_1 when p>2. 6.4-6.5 The centralizer of t_1 is T, and the centralizer of d_c is D for c not in {1,-1}. Hence if g in SL_2(K) is not in the center Z then its centralizer is abelian (because it's the intersection of SL_2(K) with the centralizer in SL_2(K') where K' is an algebraic closure, and even that centralizer is abelian). Moreover the normalizer of d_c with c not in {1,-1} is . 6.6-6.7 reviews what we know about 2- and 3-transitivity and fixed points of the action on P^1(K) [in particular any nontrivial element of PSL_2 has 0, 1, or 2 fixed points, with 1 iff it is a transvection.] 6.8 Let K be a field of characteristic p. Let G a finite subgroup of SL_2(K) containing Z, and M its set of maximal abelian subgroups (each of whom automatically contains Z). Then: (i) If x is in G but not in Z then its centralizer C_G(x) is in M. (ii) The intersection of any distinct A and B in M is Z. (iii) Any A in M is either cyclic of order prime to p or of the form Q x Z where Q is a p-Sylow subgroup of G. (iv) In the former case, the normalizer N_G(A) contains A with index 1 or 2. In the latter case N_G(A) contains y such that conjugation by y takes any x in A to x^{-1}. (v) In the latter case, if Q is not {1} then G has a cyclic subgroup K whose normalizer in Q is QK, and if |K| > |Z| then K is in M. Proof: (i) C_G(x) is certainly abelian because the centralizer in SL_2(K) is abelian. So it is contained in some A of M, and that A is abelian, and contains x because C_G does, so A = C_G(x). (ii) If x is in the intersection then its centralizer contains both A and B, so A and B don't commute else AB would be abelian and (as they're distinct) larger than both. So x has non-abelian centralizer, whence x is in Z. Thus the intersection is Z as claimed. (iii) If G has an element x not in Z then is abelian, so M cannot contain Z once G > Z. In this case A = C_G(x) for any x in A - Z. If x is not a transvection then its centralizer is isomorphic with a finite multiplicative subgroup of the algebraic closure, so is cyclic. If it is a transvection then the centralizer is Z x p-group. [See Suzuki for this case, which we don't need for Galois.] (iv) Over the algebraic closure A is a finite cyclic subgroup of D, so its normalizer even in SL_2(F) is which contains D with index 2, and conjugation by w inverts D. (v) [Again see Suzuki because in this case |G| is a multiple of p.] We now enumerate G in two different ways (double-counting is useful not just for combinatorial designs!). Let |Z| = e = 1 or 2, and |G| = e g so g is the order of the corresponding subgroup of PSL_2. [Suzuki lets q be the size of a p-Sylow Q of G, and |N_G(Q):Q| = ek. For us q=1. NB we're not using q=|F| here.] Consider the G-conjugacy classes in M [excluding those for which |A| is a multiple of p]. Let C_1, C_2, ..., C_s be the classes for which the normalizer index is 1, and C_{s+1}, ..., C_{s+t} be those for which the normalizer index is 2. Let e g_i = |A_i| with A_i in C_i. Now every element of G-Z is contained in some maximal abelian, which is unique by 6.8ii. The contribution of C_i to the count e(g-1) of non-central elements is e(g_i-1)g/g_i if i=1,2,...,s, and e(g_i-1)g/2g_i if i>s. So, 1 - 1/g is the sum of s terms (g_i-1)/g_i and t terms (g_i-1)/2g_i. But each (g_i-1)/g_i is at least 1/2, so 1 > s/2 + t/4 [and the same is true if q>1 because then the C_i account for even fewer elements of G]. Hence (s,t) is one of: I II III IV V VI (1,0) (1,1) (0,0) (0,1) (0,2) (0,3) In case I, (g_i-1)/g_i = 1-1/g g_i = g and G/Z is cyclic. [If q>1 then G is the normalizer of its p-Sylow.] In case II, 1/g_1 + 1/2g_2 = 1/2 + 1/g > 1/2. So g_1 = 2 or g_1=3. If g_1=2 then 2g_2 = g so G=N_G(A_2) and G contains a cyclic group with index 2. If g_1=3 then g_2=2 and g=12 and we soon get G/Z=A_4 (G itself is isomorphic with SL_2(Z/3Z)). [q>1 cannot arise.] Case III requires q>1 and G = QxZ. Case IV gives 1/2 + 1/2g_1 = 1/g + (q-1)/qk so again q>1 [and either p=2 and G is dihedral of singly even order or p=3 and G is again SL_2(Z/3)] Case V gives us 1/2g_1 + 1/2g_2 = 1/g which is impossible since the LHS exceeds 1/2g+1/2g = 1/g. [With q>1 this gives rise to a subgroup SL_2(F_q) or GL_2(F_q).] Case VI gives 1/2g_1 + 1/2g_2 + 1/2g_3 = 1/2 + 1/g [and q=1]. Hence g_1,g_2,g_3 are some permutation of 2,2,g_3, 2,3,3, 2,3,4, 2,3,5. These produce the corresponding triangle groups, except that 2,3,3 ends up in case II because the two generators of order 3 generate conjugate subgroups of G.