Integral group rings for a series of p-groups.
by
Harald Weber
Universität Stuttgart

Let p be a prime and G the following semidirect product

G   = <a, b  |  apn+1=1,   bp=1,   b-1ab=apn+1>.

The aim is to describe the group ring Z G via the pullback, corresponding to the Idempotent e=1/p(1+a^(p^n)+ ... +a^((p-1)p^(n)). Now we have to concider the twisted group ring \Lambda : = \Z [ \zeta_pⁿ⁺¹ ] \sdp C_p, with \zeta_pⁿ⁺¹ a pⁿ⁺¹-th root of unity . It turns out that the 'Pascal'-matrix

Pa = æ ç ç ç ç ç ç ç è 1 0 0 1 -1 0 1 -2 1 0 1 -3 3 -1 0 : : : : ··· ö ÷ ÷ ÷ ÷ ÷ ÷ ÷ ø

appears twice. First, we conjugate some 'obvious' representations with the 'Pascal'-matrix to get the R:=Z[\zeta_pⁿ]-order \Lambda as suborder of \Gamma,

æ ç ç ç ç ç è R ... ... R \pi ··· : : ··· ··· : \pi ... \pi R ö ÷ ÷ ÷ ÷ ÷ ø =:\Gamma,

with \pi: = 1- \zeta_pⁿ. (Remark: In R is p totally ramified over \pi.)
Secondly, the 'Pascal'-matrix gives also a description of the congruences in \Lambda.

Date received: July 28, 2000