Let $G$ be a finite group and consider $k[G]$ where $k$ is a field. In the scenario where $\mathrm{char}(k)$ divides $|G|$, how can one show that the dimension of $Z(k[G]/\operatorname{rad}k[G])$ is strictly less than dimension of $Z(k[G])$?
I'm trying to use this to show that in the case where char(k) divides $|G|$, the number of simple modules (up to isomorphism) is strictly less than no. of conjugacy classes of $G$. I'm aware of the fact that this can be answered using some knowledge of characters but I'm wondering if the above line of attack is also possible?
Remark: In the case where $\operatorname{char}(k)$ does not divide $|G|$, we have that $\operatorname{rad}(k[G])$ is trivial since $k[G]$ is semisimple and therefore $Z(k[G]/\operatorname{rad}k[G]) = Z(k[G])$ and it can be shown that the number of simple modules is at most the number of conjugacy classes by considering dimension of $Z(k[G])$.
Thank you in advance for any light that can be shed on the matter! :)
