# Question on Wedderburn components of $\mathbb C[G]$

$G$ is a finite group. Wedderburn's theorem says that $\mathbb{C}G\cong R_1 \times\cdot \cdot \cdot R_r$ as rings where $R_i=M_{n_i}(D^i)$ is the ring of $n_i\times n_i$ matrices over a division ring $D^i$ for each $i$. Let $n=\sum_{i=1}^r n_i$.

Since $G$ is finite each $D^i$ is a vector space of finite $\mathbb C$-dimension.

How can I show that each ${\rm dim \;}_{\mathbb C} D^i$ is at most $n$? Please help.

• Any division ring which is also a finite-dimensional $\mathbb{C}$-algebra is actually one-dimensional over $\mathbb{C}$, as noted in this Wikipedia article. – Pierre-Guy Plamondon Oct 22 '14 at 12:37
• I imagine there are some typos here, or else it is just a very bizarrely stated problem. – rschwieb Oct 22 '14 at 13:14

Since then $D^i$ would be a finite dimensional $\Bbb R$ division algebra, the Frobenius theorem says $D^i$ has to be $\Bbb R$, $\Bbb C$ or $\Bbb H$.
But $\Bbb R$ and $\Bbb H$ are not $\Bbb C$ algebras because their centers are both $\Bbb R$, and so neither center can contain a copy of $\Bbb C$.
So $D^i=\Bbb C$ for every $i$. That means each matrix ring has $\Bbb C$ dimension $n_i^2$, and that the $\Bbb C$-dimension of the whole ring is exactly $\sum n_i^2$.