- 1 Arch. Math., Vol. 50, 204-209 (1988) 0003-889X/88/5003-0204 $2.70/0 9 1988 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.
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).
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 = - 2-1. 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.)
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
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, - 2-1 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 2 + 1 is irreducible in k[z]. Our argument of the previous paragraph shows that z 2 + 1 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.
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 -1 -- 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 1985. [2] B. HUVVERT, Isometrien von Vektorr/iumen I. Arch. Math. 35, 164-176 (1980). [3] R.C. THOMPSON, Commutators in the special and general linear groups. Trans. Amer. Math. Soc. 101, 16-33 (1961). [4] M.J. WO~ENBb2GER, Transformations which are products of two involutions. J. Math. and Mechanics 16, 327-338 (1966). Anschrift des Autors: Roderick Gow Department of Mathematics University College Belfield, Dublin 4 Ireland *) Eine Neufassung ging am 9. 2. 1987 ein. Eingegangen am 1.10. 1986 *) Archiv der Mathematik 50 14