How big is the group ring $S$? (i.e how many elements does $S$ contain?)

Question

Given the abelian group $G={{1,x,y,xy,y^{2},xy^{2}}}$ with $x^{2}=1$ and $y^{3}=1$ i.e. ($G=C_{2} \times C_{3}$), I need to find how many elements are in the group ring $S=RG$ given $R=\mathbb{Z}_{4}$.

So far I have attemtped the following but am unsure as to whether or not it is correct:

Since $G$ is finite, we can write the elements of $S=\mathbb{Z}_{4}G$ as

$\lambda_{1}1+\lambda_{2}x+\lambda_{2}y+\lambda_{2}xy+\lambda_{3}y^{2}+\lambda_{2}x\lambda_{3}y$,

with $\lambda_{1},\lambda_{2},\lambda_{3}\in \mathbb{Z}_{4}.$

So does this mean there are $4^{2*3}$ elements???

Answer 1

So does this mean there are $4^{2*3}$ elements???

Yes, this simply follows from

$$|RG|=|R|^{|G|}$$

if $G$ is finite. The equality is easy to prove if you recall the definition

$$RG=\big\{f:G\to R\ |\ f(x)=0\text{ for almost all }x\in G\big\}$$

which is equal to $R^G$ if $G$ is finite.

Answer 2

It's important to know that $R[G]$ is a free $R$ module generated on the elements of $G$.

This can be done by viewing the elements of $R[G]$ as formal, finite $R$-linear combinations of elements of $G$, with the usual addition and multiplication. As such, it's just isomorphic to $\bigoplus_{g\in G} R$, which plainly has order $|R|^{|G|}$ in this case. This is basically for the reason you gave, that every element is determined by a unique list of coefficients of length $|G|$.

If you were taught the "functional" definition of $R[G]$, then it is well worth your while showing the equivalence of the definition using formal linear combinations since it makes this important property of $R[G]$ as an $R$ module plain.