Do $F$-traces of simple modules at $p'$-classes uniquely determine the module?

Question

Let $G$ be a finite group and let $p$ be a prime number such that $p\mid |G|$.

Let $\text{IBr}(G)$ denote the set of irreducible Brauer characters of $G$ for the prime $p$.

Assume $\mathbb{F}_{q}$ is a splitting field for $G$ where $q=p^f$ for some positive integer $f$.

Set $r:=|\text{IBr}(G)|$.

Let $\{\rho_1, ..., \rho_r\}$ be the a set of representatives of all simple $\mathbb{F}_{q}G$-modules up to isomorphism.

It is well-known that $\text{IBr}(G)$ is linearly independent.

Now, take the $\mathbb{F}_{q}G$-traces of $\rho_j$ at the $p'$-classes and write the results in a vector, for each $j$.

Does the list of vectors obtained in that way always have the property that there are no repeated rows?

Example:

Doing the computations for $G=A_5$, the alternating group acting on $5$ symbols, for the prime number $p=2$ yields the following:

$[ Z(2)^0, Z(2)^0, Z(2)^0, Z(2)^0 ]$

$[ 0*Z(2), Z(2)^0, Z(2^2)^2, Z(2^2) ]$

$[ 0*Z(2), Z(2)^0, Z(2^2), Z(2^2)^2 ]$

$[ 0*Z(2), Z(2)^0, Z(2)^0, Z(2)^0 ]$

The list of vectors obtained in that way has the property that no two rows are identical.

Is this always the case (for any finite group G and for any prime number p)?

A reference would be cool.

Answer 1

It is still the case that the $\mathbb{F}_{q}$-valued trace functions of the (say) $\ell$ non-isomorphic simple $\mathbb{F}_{q}$-modules $V_{1},V_{2}, \ldots V_{\ell}$, are linearly independent, where $G$ has $\ell$ $p$-regular classes.

This was known to Brauer and Nesbitt, and a proof may be found (for example) in the book of Curtis and Reiner (1962).

The usual proof is to note that for each $i$ we may take an $\mathbb{F}_{q}$-linear combination of group elements which is represented by an idempotent of trace $1$ in ${\rm End}_{\mathbb{F}_{q}}(V_{i})$ and represented by the zero matrix in ${\rm End}_{\mathbb{F}_{q}}(V_{j})$ for all $j \neq i$.

Since (over $\mathbb{F}_{q}$) the trace of $g$ in its action on any $V_{k}$ is the trace of its $p^{\prime}$-part $g_{p^{\prime}}$, the claimed result follows.