# Calculation in a Group Ring

I have some problems with the verification of the third equation in Lemma 1 in this paper.

First of all, one has to notice that there is at least one Error in the Definition of $a_{\kappa,\nu}$ above. I think it must be $a_{\kappa,\nu} = \sum\limits_{j=0}^{f-1} \zeta_e^{2^{j+\nu}} a_{2^j\kappa} \sigma^j$ (that means the exponent of $\sigma$ has to be $j$, not $\nu$. Otherwise even the first equation of Lemma 1 makes no sense).

Another mistake one can find in the defintion of the involutory antiautomorphism $\ast$ on the second page of the paper. It has to be $(\rho^i \gamma)^\ast = c^i \rho^i \gamma^{-1}$ (that means the $s$ has to be replaced by $\rho$).

Now i want to give you a little summary of the general settings in the Situation of Lemma 1. Let $G$ denote the direct product of a cyclic group of order 2 with generator $\rho$ and the Group $G_0$ generated by elements $\sigma$ and $\tau$, which satisfy the relations $\sigma^f = 1, \tau^e = 1, \sigma^{-1} \tau \sigma = \tau^{q_0}$, where $f > 1$ is odd, $e = 2^f - 1, q_0 = 2^{f_0}, f_0 \geq 1$.

We consider the group algebra $\bar{\mathbb{F}}_2[G]$ and the involutory antiautomorphism $\ast$ on $\bar{\mathbb{F}}_2[G]$, defined by linearly extending the mapping $(\rho^i \gamma)^\ast = c^i \rho^i \gamma^{-1}$, where $i = 0,1; \gamma \in G_0; c \in \bar{\mathbb{F}}_2[G], c^2 = 1$.

The Element $a_{\kappa,\nu}$ of $\bar{\mathbb{F}}_2[G]$ is given by $a_{\kappa,\nu} = \sum\limits_{j=0}^{f-1} \zeta_e^{2^{j+\nu}} a_{2^j\kappa} \sigma^j = \sum\limits_{j=0}^{f-1} \zeta_e^{2^j(2^\nu - \kappa i)} \tau^i \sigma^j$ where $\zeta_e$ is a primitive root in $\mathbb{\bar{F}_2}$ which generates a normal basis of the extension $\mathbb{F}_2(\zeta_e)/\mathbb{F_2}$.

In order to proof equation 3 of Lemma 1, the author gives two auxiliary equations $a^\ast_{\kappa,\nu} = a_{e-\kappa, \nu}$ and $a_{\kappa,\nu} \cdot a_{\kappa'} = \left\{\begin{array}{ll} a_{\kappa, \nu}, & \kappa = \kappa' \\ 0, & \kappa \neq \kappa'\end{array}\right.$, but unfortunately i am not able to show only one of these equations.

Here my attempt for the first equation: We have $a^\ast_{\kappa,\nu} = \big(\sum\limits_{j=0}^{f-1} \sum\limits_{i=0}^{e-1} \zeta_e^{2^j(2^\nu - \kappa i)} \tau^i \sigma^j \big)^\ast = \sum\limits_{j=0}^{f-1} \sum\limits_{i=0}^{e-1} \zeta_e^{2^j(2^\nu - \kappa i)} (\tau^i \sigma^j)^{-1}$

and because of $\sigma^{-j}\tau^i\sigma^j = \tau^{i q_0^j}$ we obtain $\sum\limits_{j=0}^{f-1} \sum\limits_{i=0}^{e-1} \zeta_e^{2^j(2^\nu - \kappa i)} \tau^{-iq_0^j} \sigma^{-j}$

In the next step i have made indextransformations $i = -iq_0^{-j}$ and $j = -j$.

This leads to $\sum\limits_{j=0}^{f-1} \sum\limits_{i=0}^{e-1} \zeta_e^{2^{-j}(2^\nu + \kappa iq_0^j)} \tau^i \sigma^j = \sum\limits_{j=0}^{f-1} \sum\limits_{i=0}^{e-1} \zeta_e^{2^{-j}(2^\nu - (e - \kappa)iq_0^j)} \tau^i \sigma^j$.

But i have absolutely no clue why this should be the same as $a_{e-\kappa, \nu} = \sum\limits_{j=0}^{f-1} \sum\limits_{i=0}^{e-1} \zeta_e^{2^j(2^\nu - (e - \kappa)i)} \tau^i \sigma^j$.

Did i make a mistake or can anyone see what must be done to get the solution?