# Wedderburn decomposition of $D_{5}$

This is crossposted from MSE. The question:

Find the Wedderburn decomposition of $D_{5},$ the dihedral group of order 10, over the ﬁeld $\mathbb{F}_{3}.$

I have shown that the irreducible representations of $D_{5}$ over $\mathbb{F}_{3}$ are the two trivial 1-dimensional representations, and two 4-dimensional representations which derive from considering the action of $D_{5}$ on the group of $5^{\text{th}}$ roots of unity in $\mathbb{F}_{3}.$

Now, we can decompose the group ring $\mathbb{F}_{3}(D_{5})$ into a direct sum of two copies of $\mathbb{F}_{3}$ and two copies of $\mathbb{F}_{3}^{4}$. However, it seems that the two copies of $\mathbb{F}_{3}^{4}$ combine to form the matrix ring $M_{2\times 2}(\mathbb{F}_{3^{2}})$, giving the final Wedderburn decomposition into matrix rings.

Is there a nice explanation for this last step? I would appreciate any help in understanding this.

• – Rasmus Oct 13 '13 at 19:21
• Really $D_5$ is acting on the set of 5th roots of unity inside $\mathbf{F}_{81}$. Let $z$ be a primitive 5th root of unity in there, so $\mathbf{F}_{81} = \mathbf{F}_9[z]$ is 2-dimensional over $\mathbf{F}_9$ with basis $\{1,z\}$. Now you can imitate the action of $D_5 = \langle \sigma,\tau\rangle$ on $\mathbf{C}$ as a 2-dimensional vector space over $\mathbf{R}$ via $\sigma:1 \rightarrow z, z \mapsto z^2$ and $\tau:1 \mapsto 1, z \mapsto 1/z$. That defines $\mathbf{F}_3[D_5]\rightarrow {\rm{Mat}}_2(\mathbf{F}_9)$. – Marguax Oct 13 '13 at 19:30
• As an exercise, consider what happens over $\mathbf{F}_p$ when $p \equiv -1 \bmod 5$ (so $5$ is a square mod $p$ but $\mathbf{F}_p$ has no primitive 5th root of unity), and $p \equiv 1 \bmod 5$ (so $\mathbf{F}_p$ contains a primitive 5th root of unity). – Marguax Oct 13 '13 at 19:32
• Another simple reason to explain this is $D_5$ is not commutative, thus $F_3(D_5)$ is not a commutative algebra. – Xiaolei Wu Nov 30 '15 at 16:28