MHB Group Ring Integral dihedral group with order 6

cbarker1
Gold Member
MHB
Messages
345
Reaction score
23
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
 
Physics news on Phys.org
Cbarker1 said:
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 $e,r,r^2,s,rs,r^2s$.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...

Similar threads

Back
Top