Self-dual modules for finite groups of odd order

Abstract

Let G be a finite group of odd order and let F be a finite field. Suppose that V is an F G-module which carries a G-invariant nondegenerate bilinear form which is symmetric or symplectic. We show that V contains a self-perpendicular submodule if and only if the characteristic polynomials of some specified elements of G (regarded as linear transformations of V ) are precisely squares. This result can be applied to the study of monomial characters if the form on V is symplectic, and self-dual group codes if the form is symmetric.

Metabolic modules of finite group algebras over finite fields of characteristic two

“…However, it seems more difficult and complicated to handle with the case where both the characteristic of the ground field and the order of the group are even. Dade's arguments used in [2] and ours in [11] are no longer valid in this "modular" case.…”

“…We remark that Loukaki's theorem mentioned above is an immediate consequence of Theorem C (along with an elementary fact that any symplectic module of finite groups of odd order is hyperbolic if and only if it is an evenmultiplicity module, see Corollary 3.8 in [11]). We may also use this theorem to study the real-valued complex characters or 2-Brauer characters of a finite group by taking F to be a splitting field for G of characteristic zero or two (see Theorem 6.2 below).…”

“…We now examine the duality of simple modules for F -special groups. The "cyclic case" is quite simple and already appeared in [11].…”

“…In the situation of the above Loukaki' theorem, we removed the assumption that the characteristic is odd, reduced the criterion to fewer elements, and showed one more criterion by characteristic polynomials. Moreover, our proof is no longer based on Dade's theorem, see Corollaries B and C in [11].…”

“…We will present examples in Section 5 to show that, in some sense, the set of the F -special subgroups of a finite group is "minimal" to guarantee that Theorem A holds. Another remark is that, of course, not every element of G has a square characteristic polynomial in general, even if V is hyperbolic and G is cyclic, as shown by [11,Example 2.5].…”