This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math. J. Okayama Univ. 44 (2002), 5156 ON THE NILPOTENCY INDEX OF THE RADICAL OF A GROUP ALGEBRA. XI Kaoru MOTOSE Let t ( G ) be the nilpotency index of the radical J ( KG ) of a group algebra KG of a finite psolvable group G over a field K of characteristic p &gt; 0. Then it is well known by D. A. R. Wallace [7] that p e t ( G ) e ( p 1) + 1 , where p e is the order of a Sylow psubgroup of G . H. Fukushima [1] characterized a group G of plength 2 satisfying t ( G ) = e ( p 1) + 1, see also [4]. Unfortunately, his characterization holds under a condition such that the ppart V = O p ,p ( G ) /O p ( G ) of G is abelian. In this paper, using Dickson near fields, we shall give an explicit example (see Example 1) such that a group G of plength 2 has the non abelian p part V and satisfies t ( G ) = e ( p 1)+1. This example will be new and have a contributions in our research. Example 2 is also very interesting because quite different objects (see [3] and [5]) are unified on the ground of Dickson near fields. Let H be a sharply 2fold transitive group on = { , 1 ,,,..., } (see [8, p. 22]). Let V = H be a stabilizer of 0, and let U be the set consisting of the identity and fixed pointfree permutations in H . Then U is an elementary abelian psubgroup of H with the order p s (see Lemma 1). Let be a permutation of order p on satisfying conditions H 1 H, p = 1 , (0) = 0 and (1) = 1 . Then it is easy to see U 1 U and V  1 V . We set W = h i and C V ( ) = { v V  v = v } . Assume that there exists a normal subgroup T of WV contained in V such that V is a semidirect product of T by C V ( ). We set G = h W,T,U i . Now, we present the well known results Lemmas 1 and 2 for completeness of this paper. Lemma 1. U is a normal and elementary abelian psubgroup of H . Proof. First we shall prove, for k * = \ { } , there exists only one u k U with u k (0) = k , equivalently, the following map from U to is bijective: : u u (0) . This paper was financially supported by the GrantinAid for Scientific Research from Japan Society for the Promotion of Science (Subject No. 1164003). 51 52 K. MOTOSE For U \{ } , there exists H with ( (0)) = k since (0) 6 = 0 and H is transitive on * . We set u k =  1 . Then u k U and u k (0) = k . Thus is surjective. It follows from definition of H and U that U = H \ a ( H a \ { } ) , ( H a \ { } ) ( H b \ { } ) = for a 6 = b. Using  H  =  H a  a H  =  H a   , where a H is an orbit of a , we can see  U  =   . Hence is injective. Assume has a fixed point for , U . Then we may assume = 0 since H is transitive on and U 1 = U for H . Thus =  1 follows from  1 U , (0) =  1 (0) and the above observation. This means U . Hence U is a normal subgroup of H because U 1...
View Full
Document
 Winter '11
 PhanThuongCang

Click to edit the document details