In Weyl's book The classical groups, it is said the regular representations of a division algebra are faithful and irreducible.
The key step is to show the ideal of the division algebra is $\{0\}$ and itself, i.e. if we have $u^{-1}$, we can prove that any element of the whole algebra belongs to the ideal. Thus the subspace of left regular representation is the whole space.
My question is: we know that the regular representation of a finite group is reducible, which also have the inverse of an element $u^{-1}$. Seems that it's very close to a division algebra. Why do we get a different result?
The key difference is this: a division ring over a field $k$ is already a $k$-vector space, while a group needs to be turned into a vector space to become a representation.
If we take a group $G$, then $G$ is simple as a $G$-set, in particular it is irreducible. But there is no reason to conclude that $k[G]$ should be irreducible as a $G$-representation—all we can conclude is that any decomposition must involve some of the field structure, i.e. we cannot get a decomposition by just bunching together coordinates.
For example, take $G=C_2 = \langle g \mid g^2 = 1\rangle$. The natural invariant basis of $\mathbb{C}[G]$ is $\{e+g,e-g\}$, which involves structure not visible within $G$ itself.