Schur Indices in Finite Quaternion-free Groups

Let G be a finite, quaternion-free group with exponent e, let F be a field of characteristic zero and let x be an absolutely irreducible character of G. Suppose that a Sylow 2-subgroup of the Galois group of F(J\) over F is cyclic. It is shown that if x is not real valued, then the Schur index of x over F is odd. Let G be a finite group with exponent e and let F be a field of characteristic zero. We denote the set of absolutely irreducible characters of G by Irr(G), and write "V(x) f°r tne Schur index over F of x G Irr(G). £„ denotes a primitive «th root of unity. We say that x G Irr(G) is real valued if the field Q(x) is embedded in R, the field of real numbers. Following [6] we call G a quaternion-free group if a generalized quaternion group is not involved in G. In [3], B. Fein proved that if -1 is a sum of two squares in F and a Sylow 2-subgroup of the Galois group Gal( F(£e)/F) is cyclic, then mF(x) is odd. Because of the well-known Brauer-Speiser theorem, we investigate mF(x) when x is not real valued. We prove the following. Theorem. Let G be a finite, quaternion-free group with exponent e, let F be afield of characteristic zero and let x G Irr(G). Suppose that a Sylow 2-subgroup of the Galois group Gal(F(^e)/F) is cyclic. If x is not real valued, then mF(x) & odd. The assumption that G is quaternion-free in the theorem cannot be dropped as the following example shows. Let G be a nilpotent group of order Sq where q is a prime such that q = 1 (mod 8), and assume that its Sylow 2-subgroup is generalized quaternion. Choose a faithful, complex irreducible character x of G and let F = Q(x)Clearly, F = Q(£q) and F(£e) = F(fÄ). To see that mF(x) = 2, we consider an F-triple (G, X, x) where X < G with \G: X\-2. Since q = 7 (mod 8), -1 is not a norm in F(vcT)/Pby [1, Note, p. 173], thus mF(x) = 2 by Theorem 2.2 of [2]. The proof of the theorem is based on reductions of G and x to an F-triple as in [2], and norm computations in reduced field extensions utilizing Theorem 2.2 of [2]. Lemma \.If(H, X,8) is an F-triple and K is a subfield ofF(O), then (H, X,6) is a K-triple. Proof. If K is a subfield of F, then (H, X, 0) is a ^-triple by Lemma 9.17 (b) of [1]. Since (H, X, 6) is an F(0)-triple, the result follows. Received by the editors December 27, 1981. 1980 Mathematics Subject Classification. Primary 20C15. ©1982 American Mathematical Society 0O02-9939/82/0OO0-O070/$01.75