$\mathbb{F}_pG/J(\mathbb{F}_pG)$ is a separable $\mathbb{F}_p$-algebra 
Let $k$ be a field and $G$ a fnite group. Suppose that $char(k) = p > 0$. Denote by $\mathbb{F}_p$ the prime field of $k$. Every finite extension field $E$ of $\mathbb{F}_p$ is separable over $\mathbb{F}_p$. It follows that $J(EG) = E\otimes _{\mathbb{F}_p}J(\mathbb{F}_pG)$. Thus $\mathbb{F}_pG/J(\mathbb{F}_pG)$ is a separable $\mathbb{F}_p$-algebra.

I don't quite follow the last "thus" statement. Why do we need to discuss $E$? Any hints would be appreciated!
 A: I believe the author is applying the following alternate definition: an algebra $A$ over a field $k$ is separable iff the extension of scalars $A \otimes_k L$ is semisimple for all finite extensions $L$ of $k$. What you'd like to do is to be able to conclude that $\mathbb{F}_p[G]/J(\mathbb{F}_p[G])$ is semisimple by writing its extension of scalars to some finite extension $E$ of $\mathbb{F}_p$ as $E[G]/J(E[G])$, and to do that you'd need to know that $J(E[G]) \cong E \otimes J(\mathbb{F}_p[G])$, which apparently is automatic if $E$ is separable (I believe it but I'm not familiar with this fact).
But I don't understand why the author is doing this, because it's already clear that $\mathbb{F}_p[G]/J(\mathbb{F}_p[G])$ is a finite-dimensional semisimple $\mathbb{F}_p$-algebra, hence (by Artin-Wedderburn) a finite direct product of matrix algebras over finite fields, hence separable. The same argument shows that over any perfect field a finite-dimensional algebra is separable iff it's semisimple (iff it has trivial Jacobson radical).
