1
$\begingroup$

$G$ 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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Pierre-Guy Plamondon Oct 22 '14 at 12:37
  • $\begingroup$ I imagine there are some typos here, or else it is just a very bizarrely stated problem. $\endgroup$ – rschwieb Oct 22 '14 at 13:14
2
$\begingroup$

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 $\sum n_i^2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.