# Why is the Wedderburn formula in this case wrong?

in this question counterexample: degree of representation $\leq$ index of normal subgroup there was the answer (in the second comment under the answer), that the dihedral group $D_5$ hat exactly 3 irreducible representations over $\mathbb{F}_3$ (or $\mathbb{F}_{13}$): Two with dimension 1 and one faithful irreducible with dimension 4. But according to Wedderburn I should have $$10 = dim_{\mathbb{F}_3} \mathbb{F}_3 [D_5] = 2 \cdot 1 + 1 \cdot 4$$ (two 1-dimensional and one 4-dimensional). Where is my mistake?

Khanna

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as Jack's answer below showed, your faithful irreducible has dimension 2, rather than 4; and thus there are 2 (isomorphic) copies of such irreducibles in the decomposition of the algebra. – Aaron Aug 2 '12 at 17:07

Wedderburn showed that a finite dimensional semi-simple $K$-algebra has a decomposition as a direct product of matrix rings over division $K$-algebras. In this case, the decomposition is:

$$\mathbb{F}_3[D_5] = M_1(\mathbb{F}_3) \oplus M_1(\mathbb{F}_3) \oplus M_2(\mathbb{F}_9)$$

This has dimension

$$10 = 1\cdot 1^2 + 1 \cdot 1^2 + 2\cdot 2^2$$

where in general the matrix ring $M_n(\Delta)$ has dimension $[\Delta:K] \cdot n^2$.

Over finite fields and often in general one can predict $[\Delta:K]$ from the so-called “$K$-conjugacy classes” as described in this question. In full generality, one also needs the Schur index.

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