# Group ring isomorphic to matrix ring

I want to figure out whether $$R$$ is isomorphic to $$S$$, where $$R = \mathbb{R}[G]$$, where $$G = \mathbb{Z}/2 \times \mathbb{Z}/2$$, and $$S = M_4(\mathbb{R})$$.

It seems that they might not be isomorphic, since the obvious isomorphism $$\phi \left( \begin{matrix} a & b \\ c & d \end{matrix}\right) = a(0,0)+b(0,1)+c(1,0)+d(1,1)$$ doesn't work because multiplication is not preserved, I think. Am I doing something wrong or are these actually not isomorphic?

• Please excuse me for my ignorance. What is $\mathbb Z/2$? Do you mean $\mathbb Z/2\mathbb Z$? Why is $G$ a multiplicative group? – William McGonagall Nov 28 '18 at 16:16
• @WilliamMcGonagall I interpreted $\mathbb Z/2$ to be $\mathbb Z/2\mathbb Z$ so that $G$ is the Klein 4-group. – rschwieb Nov 28 '18 at 16:28
• @rschwieb Oh, I see. So, the additive group $G$ is viewed as a multiplicative group in the definition of $R$. Thanks so much for the explanation. – William McGonagall Nov 28 '18 at 16:33
• @WilliamMcGonagall Quite frankly, there is no such thing as an "additive group" or "multiplicative group". "additive/multiplicative" are not adjectives modifying what the group is, but rather they describe the notation used for the operation in the group. So... there is simply no need to mention that when considering a group ring. – rschwieb Nov 28 '18 at 16:49
• @rschwieb Yes, I know, but I was confused because I thought the OP was using the multiplication in the ring $\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z$ to define $R$, and $(G,\cdot)$ is not a group. – William McGonagall Nov 28 '18 at 16:55

A group ring with a commutative ring and an abelian group is obviously a commutative ring. $$M_2(\mathbb R)$$ is obviously not a commutative ring.