Algebraic subgroups of GL 2 (C)

Indag. Mathem., N.S., 19 (2, 287 297 June, 2008 Algebraic subgroups of GL 2 (C by K.A. Nguyen, M. van der Put and J. Top Department of Mathematics, University of Groningen, P.O. Box 07, 9700 AK Groningen, The Netherlands Communicated by Prof. M.S. Keane ABSTRACT In this note we classify, up to conjugation, all algebraic subgroups of GL 2 (C. 1. INTRODUCTION Although the classification, up to conjugation, of the algebraic subgroups of SL 2 (C ([3, Theorem.12], [6, Theorem.29], and the classification of subgroups of GL 2 over a finite field ([1], [8, Theorem 6.17] are well known, it seems that the determination of all algebraic subgroups of GL 2 (C is not presented well in the literature. In this paper we give this classification, including full proofs. The final result is Theorem. We note that C can be replaced everywhere by any algebraically closed field of characteristic zero. Notation. μ n C denotes the nth roots of unity and ζ n denotes a primitive nth root of unity. Let β : GL 2 (C PGL 2 (C = PSL 2 (C, γ : SL 2 (C PSL 2 (C denote the canonical projections. For any algebraic subgroup H PSL 2 (C we write H SL 2 = γ 1 (H SL 2 (C.Further {( a b B := a C 0 a 1,b C} E-mails: k.a.nguyen@math.rug.nl (K. Nguyen, mvdput@math.rug.nl (M. van der Put, j.top@math.rug.nl (J. Top. 287

and {( {( c 0 D := c C } 0 d 0 c 1 d C } d are the Borel subgroup and the infinite dihedral subgroup of SL 2 (C. We first recall the classification of all algebraic subgroups of PGL 2 (C. Theorem 1. Let H be an algebraic subgroup of PGL 2 (C. Then, up to conjugation, one of the following cases occurs: (1 H = PGL 2 (C; (2 H is a subgroup of the group γ(b; (3 H = γ(d ; ( H = D n (the dihedral group of order, A (the tetrahedral group, S (the octahedral group, or A 5 (the icosahedral group. The above theorem reduces the problem to describing the algebraic groups in GL 2 (C mapping to a given subgroup G PGL 2 (C. Each example is therefore a central extension of G and corresponds to an element in H 2 (G, μ,whereμ is either C or a finite cyclic subgroup of C. The first case defines the Schur multiplier of G. In the interesting cases, μ is a finite group and the Schur multiplier does not provide information because the canonical map H 2 (G, μ H 2 (G, C is not injective (see also Remark 3. We note that Theorem 1 is a corollary of the following two well-known theorems. Theorem 2 (Klein []. A finite subgroup of PGL 2 (C is isomorphic to one of the following polyhedral groups: a cyclic group C n ; a dihedral group D n of order, n 2; the tetrahedral group A of order 12; the octahedral group S of order 2; the icosahedral group A 5 of order 60. Up to conjugation, all of these groups occur as subgroups of PGL 2 (C exactly once. In Theorem 1, the cyclic groups C n happen to be subgroups of γ(b. Theorem 3 ([3, Theorem.12]; [6, Theorem.29]. Suppose that G is an algebraic subgroup of SL 2 (C. Then, up to conjugation, one of the following cases occurs: (1 G = SL 2 (C; (2 G is a subgroup of the Borel group B; 288

(3 G is not contained in the Borel group B and is a subgroup of the infinite dihedral group D ; ( G is one of the groups A SL 2,S SL 2,A SL 2 5. 2. ALGEBRAIC SUBGROUPS OF GL 2 (C Given a group H PGL 2 (C as in Theorem 1, we will determine all algebraic subgroups G GL 2 (C such that β(g = H. We first observe that there is only one maximal group with this property, namely H max := β 1 (H.AnyG with β(g = H satisfies C G = C H SL 2 = H max. By the Noetherian property, G contains a minimal algebraic subgroup with image H. We will denote any such minimal subgroup by H min.anyg with β(g = H has the form μ k H min or C H min = H max. Our problem now remains to determine all minimal groups H min (up to conjugation. We will proceed case by case based on Theorem 1. 2.1. H = PGL 2 (C Proposition 1. For H = PGL 2 (C the only minimal group is SL 2 (C. Proof. Clearly H max = GL 2 (C.LetG be a minimal group with β(g = PGL 2 (C. The latter group is equal to its commutator subgroup and therefore β([g, G] = H. Since G is minimal, one has G =[G, G] and G SL 2 (C. By Theorem 3, G cannot be a proper subgroup of SL 2 (C. 2.2. H is a subgroup of the group γ(b Then H = γ(f for some algebraic subgroup F of B SL 2 (C. The algebraic subgroups of the Borel group B SL 2 (C are listed below: {( a 0 B; G m = 0 a 1 {( F1 k ξ c = 0 ξ 1 F l 2 = {( ξ 0 0 ξ 1 a C } ; G a = ξ k = 1,c C}, with k Z 1 ; ξ l = 1}, with l Z 1. We note that μ l = F l 2 G m B and F 1 1 = G a F k 1 B. 2.2.1. H = γ(b Proposition 2. For H = γ(bthe minimal groups are H k,l = {( a b a k c l = 1} 0 c with k,l Z satisfying k + l 0 and gcd(k, l = 1. {( } 1 b b C ; 0 1 289

Proof. Let G H max ={ ( ab 0 c a,b,c C, ac 0} be minimal with β(g = H. Then G contains an element of the form A = α (1 1 with α C. The unipotent component A u = ( 11 of the multiplicative Jordan decomposition of A belongs to G. ThenG contains the normal subgroup N := { ( 1 b b C} and G/N is a proper subgroup of H max /N = G m G m. It follows that G ={ ( ab 0 c a k c l = 1} for a certain pair (k, l (0, 0. This group has projective image γ(b precisely when k + l 0. By minimality gcd(k, l = 1. 2.2.2. H = γ(g m Proposition 3. In this case, the minimal groups are {( a 0 a k b l = 1} 0 b with k,l Z satisfying k + l 0 and gcd(k, l = 1. Proof. A minimal subgroup G is a proper subgroup of H max ={ ( a 0 a,b C } with image G m in PGL 2 (C. Therefore it is of dimension one, hence it has the form { ( a 0 a k b l = 1} for some pair of integers (k, l (0, 0. This group has image G m in PGL 2 (C, if and only if k + l 0. SinceG is minimal one moreover has gcd(k, l = 1. Remark 1. Two pairs (k, l and (m, n define conjugated minimal subgroups of GL 2 (C for Proposition 2 if and only if (k, l =±(m, n. For Proposition 3 the two pairs define conjugated groups if and only if (k, l {±(m, n, ±(n, m}. 2.2.3. H = γ(g a In this case, we have H SL 2 ={±1} G a and H max = C G a. Proposition. In this case, the only minimal group is G a. Proof. Let G be minimal. Then G contains an element of the form A = α (1 1 01 with α C. The unipotent component A u = ( 11 of the multiplicative Jordan decomposition of A also belongs to G and thus G { ( 1 a a C} =Ga.By minimality G = G a. 2.2.. H = γ(f1 k The group H is topologically (for the Zariski topology generated by the images of the elements ( ζk 2 0 ( 0 1 and 11 in PGL2 (C (where ζ k is a primitive kth root of the identity. Let G denote a minimal subgroup with β(g = H. As before one concludes that G { ( 1 a a C}=Ga.Moreover,G is (topologically generated by G a and an element of the form A := α (ζ k 2 0 0 1 with α C.Ifα is not a root of unity, then the group, topologically generated by A and G a, contains C and is equal to H max. By the minimality of G we have that α is some primitive nth root of 290

unity. We define s by s = k/2 if k is divisible by 2 and s = k otherwise. For every prime number p, not dividing s, we may consider the subgroup of G generated by A p and G a. This group maps surjectively to H. Thus, by minimality, this group is equal to G and p does not divide the order n of α. We find that every prime divisor of n is also a prime divisor of s. Define, for any positive integer n with this property, and every primitive nth root of unity ζ n, the group H(ζ n as generated by ζ n (ζ k 2 0 0 1 and Ga. This group H(ζ n depends on the choice of the primitive nth root of unity ζ n.furtherβ(h(ζ n = H. The group H(ζ n is minimal since any proper subgroup of H(ζ n, containing G a, is contained in the group generated by (ζ n (ζ k 2 0 0 1 p and G a, where the prime p divides s. The latter group does not map surjectively to H. Moreover we found G H(ζ n for some n. Thus we found all minimal groups, namely the groups H(ζ n. Proposition 5. For H = γ(f1 k the minimal groups are the H(ζ n, generated by ζ n (ζ k 2 0 ( 0 1 and { 1 a a C}=Ga, where every prime divisor of the positive integer n divides k if k is odd and divides k/2 if k is even. Remark 2. One has H(ζ n o = G a and the order of the cyclic group H(ζ n /H (ζ n o is the smallest common multiple of n and k (for k odd and that of n and k/2 (if k is even. Moreover, if H(ζ n is conjugated to H m,thenn = m. Howevertheconverse is not true in general. 2.2.5. H = γ(f l 2 Similarly to Section 2.2. one finds the following proposition: Proposition 6. For H = γ(f2 l the minimal groups are the cyclic groups generated by ζ n (ζ l 2 0 0 1 where n is a positive integer such that every prime divisor of n is a prime divisor of l if l is odd or of l/2 if l is even. 2.3. H = γ(d Let G be minimal with β(g = H.ThenGis a proper subgroup of H max = C D. The component of the identity G o G has the form { ( a 0 a k b l = 1} for some (k, l with gcd(k, l = 1. Consider an element B G with image (the class of ( 01 H. Thus B = β (0 1 for some β C.FromBG o B 1 = G o it follows that k = l and thus G o ={ ( a 0 ab = 1}. By the minimality of G one has that B 2 = β 2 is a root of unity. The subgroup of G, generated by G o and B k,wherek is any odd integer, is also mapped surjectively to H. The minimality of G implies that β 2 is a primitive 2 n th root of unity for some n 0.LetH n be the group generated by { ( a 0 ab = 1} and B n := ζ 2 n+1 (0 1. This group does not depend on the choice of ζ2 n+1 since one may replace B n by any odd power of B n.furtherg H n for some n. The group G must contain { ( a 0 ( 0 b ab = 1} and some element λ 01. The latter element has the form ( a 0 0 b (ζ2 n+1 (0 1 p with ab = 1 and p Z. One concludes that a = b =±1 and p is odd. It follows that G = H n and we conclude: {H n n 0} is the collection of the minimal groups. 291

2.. H = D n,a,s or A 5 We first note that if H PGL 2 (C is a finite subgroup, then every H min GL 2 (C is also finite. Indeed, it is clear that H SL 2 is finite. Because H min C H SL 2,we see that H min is finite. 2..1. H = D n We write D n = a,b a n = b 2 = 1,ba= a 1 b PGL 2 (C. (i n odd and n 3. In this case, we may choose for a and b the images in PGL 2 (C of the matrices ( ζ n 0 ( 0 ζ, 01 n, with 1 ζn a primitive nth root of unity. Let G be a minimal group. As G is finite and generated by preimages of a,b D n one has that ( ( 0 1 G = A = λ 0 ζn 1,B= μ for certain roots of unity λ, μ. WehaveA n = λ n, B 2 = μ 2, BA = λ 2 A 1 B.Every element of G has the form ta k,orta k B, k = 0, 1,...,n 1, with t λ 2,λ n,μ 2 = λ,μ 2. Hence G C = λ,μ 2. Since both λ ( ζ n 0 0 ζ G and λ G, we can write n 1 ( G = A = 0 ζn 1,B= μ ( 0 1 The subgroup of G generated by A and B m,wherem 1 is odd, also maps surjectively to D n. By the minimality of G, this implies that the order of μ is 2 k for some k 0.Nowdefine H k := ( 0 ζ 1 n,ζ 2 k ( 0 1, for k 0. This group H k does not depend on the choice of the primitive 2 k th root of unity because one can replace the second generator by any odd power of itself. The groups H k are the only candidates for minimal groups. We now show that H k is indeed minimal. For k = 0, 1, the groups H 0 = ( 0 ζ 1 n, ( 0 1., H 1 = ( 0 ζ 1 n, ( 0 1 are ( minimal since they have order. The two groups are conjugated by the matrix 0 i. We note that H2 = D SL 2 n.fork 2, we see that H k C = ζ 2 2 k. Suppose that D is a subgroup of H k which maps surjectively to D n,then ( ( 0 1 D = 0 ζn 1,tζ 2 k for some t ζ 2 2 k. Since the order of tζ 2 k is also 2 k, one has D = H k and thus H k is minimal. For k 1, the order of H k is 2 k n. Thus two minimal groups H k and H l with k,l 1 are conjugated only if k = l. 292

(ii n even and n>2. A minimal G can be written as ( ( ζ 0 0 i G = A = λ 0 ζ 1,B= μ, for certain roots of unity λ,μ. We have A n = λ n, B 2 = μ 2, BA= λ 2 A 1 B. As before, this implies that G C = λ 2, λ n, μ 2 = 1,λ 2,μ 2. One can replace A and B by c 1 A and c 2 B with c 1,c 2 1,λ 2,μ 2. For a good choice of c 1,c 2, the group c 1 A,c 2 B will be a proper subgroup unless there exists an integer N with λ,μ μ 2 N. Thus the latter holds by the minimality of G. Then 1,λ,μ =μ 2 m+1 for some m 0. For m = 0,wehaveG C = μ 2 and this leads to only one group, namely ( ( ζ 0 0 i 0 ζ 1, = D SL 2 n. This group is clearly minimal. For m 1, one has G C = μ 2 m and this leads to the three groups given by the table: λ μ H 1,m ζ 2 m+1 1 H 2,m ζ 2 m+1 ζ 2 m+1 H 3,m 1 ζ 2 m+1 They all are minimal and have order 2 m. However H 1,m and H 2,m are conjugated. Indeed, ( ζ 0 ( ζ 1 01 H1,m = H2,m because ( ( ζ 0 0 i 0 1 = ζ 2 2 m+1 [ ζ 2 m+1 0 01 ( ζ 1 0 0 1 ( ζ 0 0 ζ 1 ] [ ( ] 0 i ζ 2 m+1. (iii n = 2. As in (ii. In this case also H 1,m and H 3,m are also conjugated, namely by a matrix of the form ( 0 a 1 1. 2..2. H = A Let G H max = C A SL 2 be a minimal group. Consider G + C A SL 2,the preimage of G under the obvious map α : C A SL 2 C A SL 2. We note that the kernel of α is {(1, ( ( 01, ( 1, 0 1 }. Sinceβ(G = A, there exists for every a A SL 2 an element (λ, a G +.Letμ k := {λ C (λ, 1 G + }. Then we obtain a homomorphism h : A SL 2 C /μ k given by h(a = λ mod μ k if (λ, a G +. 293

This homomorphism factors as A SL 2 h 1 C 3 C /μ k,wherec 3 ={1,σ,σ 2 } is the quotient of A SL 2 by its commutator subgroup. If h 1 is trivial, then G + contains {(1,a a A SL 2 } and by minimality G = A SL 2. By Theorem 3, the latter group of order 2 is minimal. Now we suppose that h 1 is not trivial. Write k = 3 r l with gcd(l, 3 = 1. Forany a A SL 2 there exists an element (λ, a G + with λ 3 μ 3 r l and λ can be multiplied by any element in μ 3 r l. Thus there exist a pair (λ, a G + with λ μ 3 r+1. Now G + (μ 3 r+1 A SL 2 is a subgroup of G + mapping surjectively to A. The minimality of G implies that l = 1 and G + μ 3 r+1 A SL 2.Moreover, μ 3 r G + and the map G + G is bijective. Then G has the form μ 3 r {δ(aa a A SL 2 }, whereδ =: A SL 2 δ 1 C 3 {1,ζ 3 r+1,ζ 2 3 r+1} for some map δ 1 which lifts the homomorphism h 1 : C 3 μ 3 r+1/μ 3 r C /μ 3 r. There are two possibilities for nontrivial homomorphism h 1 (and thus for δ 1 and δ and we find therefore two subgroups of GL 2 (C, lying in μ 3 r+1 A SL 2. The last group is contained in μ 3 r+1 S SL 2. Conjugation by an element τ S \ A induced on C 3 = A /[A,A ] the only non trivial automorphism and permutes the two possibilities for h 1.One lifts τ to an element τ S SL 2. Conjugation by τ permutes the two possibilities for h 1 and therefore the above two groups are conjugated. It suffices to consider the group H r := μ 3 r {δ(aa a A SL 2 } with δ 1 given by δ 1 (1 = 1, δ 1 (σ = ζ 3 r+1, δ 1 (σ 2 = ζ 2 3 r+1. The order of H r is 3 r 2. The group H 0 is isomorphic to A SL 2, but not conjugated to A SL 2. The minimality of H 0 follows from the fact that A does not have a faithful two-dimensional representation. Finally, we will show that H r is minimal for r 1. Suppose that D is a subgroup of H r with β(d = A.Letτ A SL 2 be an element of order 3. Then D contains an element d = λδ(ττ for some λ {±1} μ 3 r.nowδ(τ {ζ 3 r+1,ζ 2 3r+1} and d 3 D C has order 3 r or 2 3 r. Thus D contains μ 3 r and it follows that D = H r. Thus we found: There are two minimal groups for A with order 2 and for every r 1 there is one minimal group of order 3 r 2. Remark 3. A minimal subgroup G for H = A yields a central extension 1 μ k G A 1 for some k. The corresponding element ξ of H 2 (A,μ k has, by the minimality of G, the property that ξ does not lie in the image of H 2 (A,μ d for a proper divisor d of k. Since the order of A in 12, we only have to consider the groups H 2 (A,μ 2 a 3b. The minimal groups that we found above correspond to all the cases (a, b = (1,r. The central extensions with a 1 produce, apparently, groups which do not have a faithful representation of degree two. 2..3. H = S Let G H max = C S SL 2 be a minimal group. Consider G + C S SL 2,the preimage of G under the obvious map α : C S SL 2 C S SL 2. The kernel of α is {(1, ( ( 01, ( 1, 0 1 }. Sinceβ(G = S, there exists for every a S SL 2 29

an element (λ, a G +.Letμ k := {λ C (λ, 1 G + }. Then we obtain a homomorphism h : S SL 2 C /μ k given by h(a = λ mod μ k if (λ, a G +.This homomorphism factors as S SL 2 C 2 h 1 C /μ k,wherec 2 ={1,σ} is the quotient of S SL 2 by its commutator subgroup. If h 1 is trivial, then G + contains {(1,a a S SL 2 } and by minimality G = S SL 2. According to Theorem 3, the latter group of order 8 is minimal. Now we suppose that h 1 is not trivial. Write k = 2 r l with l odd. For any a S SL 2 there exists an element (λ, a G + with λ μ 2 r+1. NowG + (μ 2 r+1 S SL 2 is a subgroup of G + mapping surjectively to S. The minimality of G implies that l = 1 and G + μ 2 r+1 S SL 2.Defineδ: S SL 2 δ 1 C 2 {1,ζ 2 r+1} by δ 1 (1 = 1 and δ 1 (σ = ζ 2 r+1. All the elements of μ 2 r+1 have the form ζ ε 2r+1 λ with ε {0, 1} and λ μ 2 r. From this it follows that G + ={(δ(aλ, a a S SL 2,λ μ 2 r } and one concludes that G = H r := μ 2 r {δ(aa a S SL 2 }. The group H r has order 2 r 8. We note that H 0 is equal to S SL 2 and is minimal. Let r>0 and let D H r be a subgroup satisfying β(d = S.Letτ S SL 2 be an element with image the permutation (1, 2 S.ThenD contains an element of the form d =±λδ(ττ with λ μ 2 r.thend 2 = λ 2 ζ 2 r D C has order 2 r. Thus D contains μ 2 r and it follows easily that D = H r. Hence every H r is minimal and we conclude that: There is for every r 0 a unique minimal group of order 2 r 8. 2... H = A 5 Let G GL 2 (C be a minimal for H.SinceA 5 =[A 5,A 5 ], the group [G, G] also satisfies β([g, G] = H. By minimality G =[G, G] and thus G SL 2 (C. This implies that G A SL 2 5. Since, by Theorem 3, the latter group is minimal, we find that A SL 2 5 is the only minimal group. In summary, we obtain the following result. Theorem. The list of all minimal groups, up to conjugation, for each algebraic subgroup H PGL 2 (C (see Theorem 1 and Section 2.2is: (1 H = PGL 2 (C: the only minimal group is SL 2 (C. (2 H is a subgroup of the group B: (a H = γ(b: for each pair of integers (k, l with k + l 0 and gcd(k, l = 1 there is a minimal one, namely { ( ab 0 c a k c l = 1}. (b H = γ(g m : for each pair of integers (k, l with k +l 0 and gcd(k, l = 1 there is a minimal group, namely { ( a 0 a k b l = 1}.Further(k, l and (l, k define conjugated groups. (c H = γ(g a : there is only one minimal group, namely G a. and (d H = γ(f1 k: the minimal ones are the H(ζ n, generated by ζ n (ζ k 2 0 0 1 { ( 1 a a C}, where every prime divisor of the positive integer n divides k if k is odd and divides k/2 if k is even. 295

(e H = γ(f2 l: the minimal ones are the groups generated by ζ n (ζ l 2 0 0 1,where every prime divisor of the positive integer n divides l if l is odd and divides l/2 if l is even. (3 H = γ(d : the minimal groups are H n with n 0, whereh n is generated by { ( a 0 0 b ab = 1} and ζ2 n+1 (0 1 with ζ2 n+1 a primitive 2 n+1 th root of unity. ( H finite: (a H = D n : (i n 3 odd: For every k 1, there is one minimal group ( 0 ζn 1,ζ 2 k ( 0 1, with ζ 2 k a primitive 2 k th root of unity; (ii n>2 even: For k 1, the minimal ones H 1,k,H 2,k,H 3,k have the form ( ( ζ 0 0 i A = λ 0 ζ 1,B= μ, for certain roots of unity λ,μ which are given in the table: λ μ H 1,k ζ 2 k+1 1 H 2,k ζ 2 k+1 ζ 2 k+1 H 3,k 1 ζ 2 k+1 They all have order 2 k.furtherh 1,k and H 2,k are conjugated. For k = 0, there is only minimal group, namely the group ( ( ζ 0 0 i 0 ζ 1, = D SL 2 n of order n; (iii n = 2:Asin(ii, but now H 1,k,H 2,k,H 3,k are all conjugated. (b H = A : there are two minimal groups of order 2. For every n>0 there is one minimal group of order 3 n 2. (c H = S : For every n 0 there is a minimal group of order 2 n 8. (d H = A 5 : There is only minimal group, namely A SL 2 5. REFERENCES [1] Dickson L.E. Linear Groups, with an Exposition of the Galois Field Theory, Teubner, Leipzig, 1901. [2] Humphreys J.E. Linear Algebric Groups, Springer-Verlag, New York, 1981. [3] Kaplansky I. An Introduction to Differential Algebra, Hermann, Paris, 1957. [] Klein F. Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree, Dover, New York, 1956, revised ed. (Translated into English by George Gavin Morrice. 296

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