Commutators in the symplectic group

Transcription

1 - 1 Arch. Math., Vol. 50, (1988) X/88/ $2.70/ Birkh/iuser Verlag, Basel Commutators in the symplectic group By R. Gow To Professor Bertram Huppert on his sixtieth birthday Let K be a conjugacy class in a group G and let K 2 denote the subset of G consisting of all elements expressible as a product of two members of K. According to [I], J. G. Thompson has suggested that if G is a finite simple group, there should be a conjugacy class K in G with K z = G. If such a class K exists, it is straightforward to prove that each element of G is a commutator. Thus, Thompson's conjecture is a more refined version of the well-known conjecture of Ore, that every element in a finite simple group is a commutator. In [1], Thompson's conjecture has been verified for several sporadic simple groups and some families of finite simple groups. Now let k be a field and let G denote the symplectic group Sp(2n, k) of degree 2n over k. The primary purpose of this paper is to show that if k has characteristic different from 2 and - 1 is a square in k there is a conjugacy class K in G whose elements have square equal to -I such that K2= G. Consequently, if G denotes the projective symplectic group obtained by factoring out the central subgroup generated by -I, there is a conjugacy class L, say, of involutions in G with L 2 = G. In the particular case that k = GF(q), with q - 1 (mod4), we have thus verified Thompson's conjecture for the simple group PSp(2n, q). Although not relevant to Thompson's conjecture, we show by way of contrast that if is not a square in k, there is no conjugacy class K with K z = Sp(2, k). Finally, we prove that if k is a field in which - 1 is not expressible as a sum of two squares, - I is not a commutator in Sp(4m + 2, k). This fact has already been observed by R. C. Thompson, [3], in the case that m = 0. We begin by sketching some details on the symplectic group. Let V denote a vector space of dimension 2 n over k and let f: V x V--* k be a non-degenerate alternating bilinear form. The symplectic group Sp (2 n, k) consists of those automorphisms cr of V that satisfy f (~u, ~v) = f (u, v) for all u, v in V. An automorphism r of V is said to be skew-symplectic with respect to f if we have f('cu, "cv) = - f(u, v) for all u, v in V.

2 Vol. 50, 1988 Commutators in the symplectic group 205 Lemma 1. Let k be a field of characteristic different from 2 in which - 1 is a square. Then there is a single conjugacy class K of elements a in Sp(2n, k) satisfying a 2 = - I. P r o o f. As - 1 is a square in k, there is a primitive element co of order 4 in k. The eigenvalues of an element a satisfying 0"2 = _ I are co and co3, and as a is semisimple, we can write v=gew, where U is the co-eigenspace, W the co3-eigenspace of ~r. Now if u, v ~ U, we have f (u, v) = f (0"u, av) = f (cou, coy) = - f (u, v). Thus f(u, v)= 0 and we see that U is totally isotropic, as is W. Since the maximum dimension of a totally isotropic subspace of V is n, we have dim U = dim W = n. Choose a basis ul,..., u, of U and a basis wl,..., w, of Wdual to this basis. Then we have ~u i = coui, awj = co3wi, f(u~, wj) = c~ij, where 61~ is Kronecker's delta. Thus we have obtained a canonically defined action of a on 1/; which proves the lemma. Our main theorem that, under the conditions of Lemma 1, K 2 = Sp(2n, k) follows easily from a result of Wonenburger, [4]. Theorem. Let k be a field of characteristic different from 2 in which - 1 is a square. Let K be the eonjugacy class in Sp(2n, k) consisting of those elements cr that satisfy 0"2 = I. Then we have K 2 = Sp(2n, k). P r o o f. Let ~ be any element in Sp(2n, k). By Wonenburger's theorem, [4, Theorem 2], we can write ~=st where s and t are skew-symplectic involutions. Thus, we have ~ = O"C where a = cos, r = co3t, and co is a primitive 4-th root of unity in k. We can check that a and r are both elements in Sp(2n, k) satisfying 0.2 =.g2 = I. The theorem now follows from Lemma 1. Corollary 1. Assume that the conditions on k described in Theorem 1 hold. Then there exists a conjugacy class L of involutions in the projective symplectic group PSp(2n, k) with L 2 = PSp(2n, k). Our next result examines Theorem 1 more closely in the special case of Sp(2, k).

3 206 R. Gow ;~xch. ~TH. Theorem 2. Let k be a field of characteristic different from 2. If - 1 is no~ a square in k, there is no conjugacy class K with K 2 = Sp(2, k). If - 1 is a square in k, there is a unique conjugacy class K with K 2 = Sp(2, k) and this is the class described in Theorem 1. P r o o f. We begin by recalling that Sp(2, k) = SL(2, k). Write G = Sp(2, k) and suppose that K 2 = G for some conjugacy class K. Then as - I is a product of two elements of K, we have -I = xy-lxy for some x ~ K, y ~ G. Thus y-txy= _x -1. Let the eigenvalues of x (in some extension field of k) be 2,/2. As x has determinant 1, /z = 2-1. Moreover, as x is conjugate to - x- 1, and - x- 1 has eigenvalues - 2-1, _ 2, we must have 2 = Thus 2 2 = - 1, and since x therefore has distinct eigenvalues, x is semisimple and satisfies x 2 = - I. It follows from Lemma 1 that if - 1 is a square in k, K can only be the class described in Theorem 1. If - 1 is not a square in k, we argue that K 2 cannot equal G. For let u be an element in G and suppose that we can write U~WZ for w, zek. Thus, As w z = z 2 = - I, we have W I:--W, Z-I ~--Z. W-Iuw z ZW ~ Z-1W -1 ~ U-I~ and we see that u is inverted by w. However, if we choose in G, any matrix t that inverts u must have the form with a, b e k, a + 0. Clearly, if t is in G, -- a 2 = J and thus -- i is a square in k. Consequently, we cannot have K 2 = G if - I is not a square in k. Before beginning the proof of our final result, we compile some facts that we will need to use. Given an element x of Sp(2n, k), it is shown in [2, Satz 1.5] that x and x 1 have the same characteristic polynomial. Thus we have the following result. Lemma 2. tf 2 is an eigenvalue of x ~ Sp(2n, k), 2- i is also an eigenvalue with the same multiplicity. (We allow )L to tie in some extension field of k.)

4 Vol. 50, 1988 Commutators in the symplectic group 207 For any polynomial f of degree d in k[z] with f(0) + 0, define f* by f*(z) = zaf(z - 1). We say that f is symmetric if f* = if polynomial of x, we have Now if x ~ Sp(2n, k) and m is the minimal m* = m(0) m by [2, Satz 1.5]. Thus we can write m = + h (PiPff) t~ FI q~j i=1 j=l with pairwise relatively prime irreducible polynomials Pi, P*, q~ in k[z], the qj being symmetric. There corresponds an orthogonal decomposition of V into x-invariant subspaces v= ul 177 u~ wl 177 w~, where Ui is the kernel of (pip*)ti(x) and Wj is the kernel of q'jj(x). This follows from the proof of Satz t.7 of [2]. As a consequence of this discussion, we have the following result. Lemma 3. Let x ~ Sp(2n, k), where k is a field of characteristic different from 2. Then i and - 1 both occur with even multiplicity as eigenvalues of x. P r o o f. Let the multiplicity of 1 as an eigenvalue of x be c. Then the generalized eigenspace W= {v e V:(x - I) c v = 0} is an orthogonal direct summand of V by our discussion above. As W carries therefore a non-degenerate alternating form and has dimension c, c must be even. The same argument can be applied to the eigenvalue - 1. It is also convenient to introduce a further idea to simplify the proof of our last theorem. Let x be an element of the general linear group GL(2n, k) over k. Considering x as a matrix over the algebraic closure ~: of k we have a unique Jordan decomposition in X ~ XsX u ~ XuXs~ where xs, xu are the semisimple and unipotent parts of x, respectively. Now if k is perfect (in particular, if k has characteristic zero), we can apply Galois theory, in conjunction with the uniqueness of the Jordan decomposition, to deduce that xs and x, are in GL(2n, k). Theorem 3. Let k be a field in which - 1 is not a sum of two squares. Then - I is not a commutator in Sp(4m + 2, k). P r o o f. It is well known that - I is a sum of two squares in GF(p), for any prime p, and hence in any field of prime characteristic. Thus our hypothesis implies that k has characteristic zero and it follows that our remark above about the Jordan decomposition

5 208 R. Gow ARCH. MATH. applies to invertible matrices over k. We will prove the theorem by induction on m, the case m = 0 being a result of R. C. Thompson, [3, Theorem 1]. Let G = Sp(4m + 2, k) and suppose that we have for x, y in G. Then - I = x-ty-lxy - x = y-lxy and we see that x has the same eigenvalues as - x. We also know from Lemma 2 that x has the same eigenvalues as x- 1. Thus, given an eigenvalue 2 of x, with 2 4: _+ 1, we see that 2, 2-1, _ 2, are all eigenvalues of x with the same multiplicity. Similarly, we see that both 1 and - 1 must occur as eigenvalues of x with the same multiplicity, and this multiplicity is even by Lemma 3. Consequently, as 4,f dim V, there must be an eigenvalue co 4= _+ 1, occurring with odd multiplicity, r say, such that co, co- 1, _ co, - co- 1 are not all distinct. This forces CO =-- -1 CO and so co2 = _ 1. Thus co is a primitive 4-th root of 1. As - 1 is certainly not a square in k, the polynomial z is irreducible in k[z]. Our argument of the previous paragraph shows that z is an irreducible symmetric factor of the minimal polynomial of x. Thus if we define W= {v~ V:(x 2 + I)'v = 0} our discussion after Lemma 2 shows that W is an x-invariant orthogonal direct summand of V of dimension 2 r. In particular, 4 ~/dim W. As we have - x = y-lxy, y commutes with x 2 and it then follows that W is y-invariant. On W,, we have - I w = xwlywlxwyw, where Xw, Yw denote the restrictions of x, y to W. Since 4 X dim W and Xw, Yw preserve the non-degenerate alternating form defined on W, our induction hypothesis yields a contradiction, unless W= V. Thus we may assume that W= V. In a similar manner, since we also have x y x -1 ~ -- y, we can assume that the minimal polynomial of y is a power of z 2 + t. tn this case, if x s, Ys are the semisimple parts of x, y, respectively, we have implying that 2 2 x~ +/=O=y~ +I, 2! y2.

6 Vol. 50, 1988 Commutators in the symplectic group 209 Now from we obtain - x = y-lxy - xsx u = y-lxsxuy, and the uniqueness of the Jordan decomposition implies that - xs = y-lxsy. Similarly, we obtain Ys = Xs Ys Xs We now have the relations 2 2 Xs = Ys =--I, Xsys = -- ysxs, where x~, y~ are matrices defined over k. We see thus that x~, y~ generate a quaternion group H of order 8 and V is a faithful kh-module. By Maschke's theorem, V is a direct sum of faithful irreducible kh-modules. By elementary representation theory, it is known that, up to isomorphism, H has a unique faithful irreducible module M over any field F of characteristic zero, and M has dimension 2 or 4. Moreover, direct calculation shows that it is only possible to define a faithful irreducible FH-module of dimension 2 if - 1 is a sum of two squares in F. Consequently, in our field k, V is a sum of isomorphic 4-dimensional kh-modules, which is impossible as 4,~dim V. Our theorem follows by induction. Note. In [3], it is proved that -I is a product of two commutators in SL(2, k) = Sp(2, k). Since we can embed the direct product of 2m + 1 copies of Sp(2, k) into G = Sp(4m + 2, k), - I is certainly a product of two commutators in G. As a converse to Theorem 3, if - 1 is a sum of two squares in k, we can use properties of a quaternion group to show that -I is a commutator in G. Finally, for any field k of characteristic not 2, - I is a commutator in Sp(4m, k). References [I] Z. ARAD and M. HERZOG (Eds.), Products of conjugacy classes in groups. LNM 1112, Berlin- Heidelberg-New York [2] B. HUVVERT, Isometrien von Vektorr/iumen I. Arch. Math. 35, (1980). [3] R.C. THOMPSON, Commutators in the special and general linear groups. Trans. Amer. Math. Soc. 101, (1961). [4] M.J. WO~ENBb2GER, Transformations which are products of two involutions. J. Math. and Mechanics 16, (1966). Anschrift des Autors: Roderick Gow Department of Mathematics University College Belfield, Dublin 4 Ireland *) Eine Neufassung ging am ein. Eingegangen am *) Archiv der Mathematik 50 14