# Wedderburn components of group algebra over $\mathbb R$

Let $G = \langle\sigma\rangle$ be the cyclic group of order $4$ and consider the group algebra $\mathbb RG$. By Maschke's theorem, $\mathbb RG$ is semisimple and so by Wedderburn's theorem, it is a product of matrix rings over some division rings.

For a tutorial problem, I'm required to determine the Wedderburn components. Since $\mathbb R$ is not algebraically closed, I don't know how to proceed. I think, based on the proof of Wedderburn's theorem, I need to start by finding the decomposition of $\mathbb RG$ into a sum of simple rings, but I don't even know how to do that. I know $\mathbb RG \cong \mathbb R \oplus \mathbb R\sigma \oplus \mathbb R\sigma^2 \oplus \mathbb R\sigma^3$ as $\mathbb R$-spaces but those summands aren't ideals of $\mathbb RG$ so I don't know where to go from here.

A couple of points

• The only division algebras that can occur in the Wedderburn decomposition of $\Bbb RG$ are $\Bbb R$, $\Bbb C$ and $\Bbb H$ (the quaternions).

• If $G$ is Abelian, $\Bbb RG$ is commuative, so we cannot have $\Bbb H$ appearing, nor can we have a matrix algebra of dimension greater than $1$.

So $\Bbb RG$ is a product of copies of $\Bbb R$ and $\Bbb C$ in this example. For each copy there will be a corresponding group homomorphism from $G$ to $\Bbb R^\times$ or $\Bbb C^\times$. What can they be?

• If the group homomorphism is denoted $\phi$ then we need the order of $\phi(\sigma)$ to be 4, so $\sigma$ can only map to i or -i. So do we have $\mathbb RG \cong \mathbb C \times \mathbb C$ by the map $\sigma \mapsto (i,-i)$? – IAlreadyHaveAKey Apr 22 '17 at 12:50
• @IAlreadyHaveAKey You could investigate the kernel and/or image of that map (which is certainly an algebra homomorphism). – Lord Shark the Unknown Apr 22 '17 at 12:53
• It's not surjective. Right we can also map $\sigma$ to $1$ or $-1$ since the corresponding group homomorphisms you mentioned don't need to be isomorphisms. So we have $\mathbb RG \cong \mathbb R \times \mathbb R \times \mathbb C$ via the map $\sigma \mapsto (1,-1,i)$ which is an isomorphism since the matrix representing it (over the reals) is invertible. I guess my question now is, we know that $\mathbb RG$ is isomorphic to this but how do we know this is the isomorphism that arises from Wedderburn's theorem? – IAlreadyHaveAKey Apr 22 '17 at 13:10
• @IAlreadyHaveAKey That's better isn't it? The Wedderburn isomorphism is essentially unique so that must be it! – Lord Shark the Unknown Apr 22 '17 at 13:11
• Oh yeah true. Thanks for your help :) – IAlreadyHaveAKey Apr 22 '17 at 13:13

Here is a somewhat different way to answer this sort of question: we have the isomorphisms (via CRT) \begin{align} \mathbb{R}[C_4]=\mathbb{R}[X]/(X^4-1)\simeq \mathbb{R}[X]/(X-1)\times\mathbb{R}[X]/(X+1)\times\mathbb{R}[X]/(X^2+1)\simeq \mathbb{R} \times \mathbb{R} \times \mathbb{C} \end{align}

This is a decomposition of $\mathbb{R}[C_4]$ into a product of fields, in particular of matrix rings, hence it must be the unique decomposition.