No more absolute simples than $p$-regular conjucacy classes: elementary proof?

Question

I am wondering whether anyone knows an elementary (no number fields, no Brauer characters; ideally, not even passing to algebraic extensions) proof of the following fact:

Let $G$ be a finite group. Let $F$ be a field, and let $p = \operatorname{char} F$. We say that an element $g \in G$ is $p$-regular if the order of $g$ in $G$ is coprime to $\begin{cases} p, & \text{ if } p > 0 ; \\ 1, & \text{ if } p = 0 \end{cases}$. (Thus, for $p = 0$, each element of $G$ is $p$-regular.) We say that a conjugacy class $C$ of $G$ is $p$-regular if and only if each element of $C$ is $p$-regular (or, equivalently, at least one element of $C$ is $p$-regular).

A simple $FG$-module $U$ is said to be absolutely simple if $\operatorname{End}_{FG} U \cong F$ as $F$-algebras.

Theorem 1. The number of pairwise non-isomorphic absolutely simple $F$-modules is at most the number of $p$-regular conjugacy classes of $G$.

Theorem 1 is attributed to Brauer in Mark Wildon's Representation theory of the symmetric group, where it is used to verify that the simple $F S_n$-modules constructed as quotients of Specht modules (for $p$-regular partitions) comprise the whole list of absolutely simple $F S_n$-modules. Of course, when $p = 0$, Theorem 1 reduces to the well-known fact that the number of pairwise non-isomorphic absolutely simple $F$-modules is at most the number of conjugacy classes of $G$; this is not hard to prove elementarily (by arguing that their characters are orthonormal and therefore linearly independent). But how would one prove Theorem 1 in the general case?

(This is, of course, inspired by math.stackexchange question #2212663; but it is meant to be independent from it.)

Question 2. What if we replace "absolutely simple" by "simple"?

Answer 1

Berman proved the following general result. If $F$ is a field of characteristic $p$ and $G$ is a finite group let $n$ be the lcm of the orders of the $p$-regular elements of $G$. Let $\zeta$ be a primitive $n^{th}$ root of unity in an algebraic closure of $F$. The Galois group of $F(\zeta)$ over $F$ can be identified with a subgroup $T$ of $(\mathbb Z/n)^*$. Then call two $p$-regular elements $g,h\in G$ $F$-conjugate if $xgx^{-1}=h^j$ with $j\in T$ abusing notation in the obvious way. This is an equivalence relation on $G$. Then the number of simple $FG$-modules is the number if $F$-conjugacy classes of $p$-regular elements.

Proofs can be found in Curtis and Reiner for characteristic 0 and in http://www.ams.org/journals/proc/1964-015-05/S0002-9939-1964-0168665-8/S0002-9939-1964-0168665-8.pdf for positive characteristic.

Answer 2

In the Book "A first course in ring noncommutative ring theory" by Lam, theorem 7.17 says that if the field is a splitting field the number of simples (then absolutely simply because of the splitting) is equal to the dimension of A/(radA+[A,A]) for a general finite dimensional algebra A over a splitting field. In chapter 8 this is used to give a proof of theorem 1 (see theorem 8.9 there). In the book there is also some discussion of the non-absolutely simple case, which also covers the "at most" case as you stated it, see corollary 7.18. A similar elementary approach can be found in chapter 21 in www.minet.uni-jena.de/algebra/skripten/dt/dt-2010/dt.pdf