- Thread starter
- #1

I am having some difficulties with computing an element in the Integral dihedral group with order 6.

Some background information for what is a group ring:

A group ring defined as the following from Dummit and Foote:

Fix a commutative ring $R$ with identity $1\ne0$ and let $G=\{g_{1},g_{2},g_{3},...,g_{n}\}$ be any finite group with group operation written multiplicatively. A group ring, $RG$, of $G$ with coefficients in $R$ to be the set of all formal sum

$a_1g_1+a_2g_2+\cdots+a_ng_n$, $a_i\in R$, $1\le i\le n$.

The addition is based on component addition. Multiplication for group ring is defined as $(ag_i)(bg_j)=(ab)g_k$, where the product $ab\in R$ and $g_ig_j=g_k$ is the product in the Group $G$, for the formal sum just add the distributive laws.

Here is question:

Let $G=D_6$ be the dihedral group of order 6 with the usual generators $r$,$s$ ($r^3=s^2=1$ and $rs=sr^{-1}$) and let $R=\Bbb{Z}$. The elements $\alpha=r+r^2-2s$ and $\beta=-3r^2+rs$ are typical members of $\Bbb{Z}D_6$. find the sum and product.

Work:

$\alpha + \beta=r-2r^2-2s+rs$

Here is where the problems are:

\begin{align*}\alpha\beta=&(r+r^2-2s)(-3r^2+rs)\\

&=r(-3r^2+rs)+r^2(-3r^2+rs)-2s(-3r^2+rs)\\

&=-3r^3+r^{2}s-3r^{4}+r^{3}s+6sr^2-2srs

\end{align*}

The end line is where I have trouble with the computations.

Thanks

Cbarker1