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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?

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    $\begingroup$ 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? $\endgroup$ – William McGonagall Nov 28 '18 at 16:16
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    $\begingroup$ @WilliamMcGonagall I interpreted $\mathbb Z/2$ to be $\mathbb Z/2\mathbb Z$ so that $G$ is the Klein 4-group. $\endgroup$ – rschwieb Nov 28 '18 at 16:28
  • $\begingroup$ @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. $\endgroup$ – William McGonagall Nov 28 '18 at 16:33
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    $\begingroup$ @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. $\endgroup$ – rschwieb Nov 28 '18 at 16:49
  • $\begingroup$ @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. $\endgroup$ – William McGonagall Nov 28 '18 at 16:55
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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.


Since a group ring of a nontrivial group is never simple (the augmentation ideal is a nontrivial ideal) a group ring over such a group can never be a matrix ring over a field (which is always simple.)

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  • $\begingroup$ Is there a more elementary solution? For example, what if I don't know those facts about simplicity and I don't want to prove them? $\endgroup$ – Wesley Nov 28 '18 at 15:57
  • $\begingroup$ @Wesley Almost certainly not. Simplicity of matrix rings is among the most elementary facts in noncommutative algebra. That you are learning about group rings without knowing that fact is a little alarming. You should probably just go learn about it here or in any basic book on ring theory. $\endgroup$ – rschwieb Nov 28 '18 at 16:06
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    $\begingroup$ Oh wait, yes facepalm, in this case there is another way. If the group had not been abelian, I would have stood by my assessment that simplicity was the obvious way to go. $\endgroup$ – rschwieb Nov 28 '18 at 16:09
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    $\begingroup$ Right! I don't know how I missed that either. Seems very obvious now. Thanks $\endgroup$ – Wesley Nov 28 '18 at 16:17

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