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 $j$.



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



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> Is this always the case (for any finite group G and for any prime number p)?
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A reference would be cool.



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


