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[SOLVED] Group Ring Integral dihedral group with order 6

Cbarker1

Active member
Jan 8, 2013
236
Dear Every one,


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
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
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.
All that remains is to use the identities $r^3=s^2=1$ and $rs=sr^{-1} = sr^2$ to simplify that last line. For example, $-3r^3 = -3$ (which I would prefer to write as $-3e$ where $e$ is the identity element of $D_6$). Also, $6sr^2 = 6sr^{-1} = 6rs$, and $-2srs = -2s(sr^2) = -2r^2$. In that way, you can write $(r+r^2-2s)(-3r^2+rs)$ as a linear combination of the six elements of $D_6$, which are $e,r,r^2,s,rs,r^2s$.