Modular trace depends only on the p-regular part

Write $\mathrm{\mid }g\mathrm{\mid }=m{p}^a$ with $p\nmid m$. Choose $e,f\in \mathbb{Z}$ with $em+f{p}^a=1$, and set $u=g^{em}$, $s=g^{f{p}^a}$. Their product is $g$ and they commute; $u^{p^a}=s^m=1$. This also covers $g=1$ and $a=0$.

Put $N=\rho (u)-I$. The commuting binomial identity in characteristic $p$, iterated $a$ times, gives $N^{p^a}=\rho (u{)}^{p^a}-I=0$. Since $\rho (s)N=N\rho (s)$, the map $T=\rho (s)N$ satisfies $T^{p^a}=\rho (s{)}^{p^a}{N}^{p^a}=0$. Moreover $\rho (g)-\rho (s)=T$.

A nilpotent map $T$ has trace zero over $k$ itself. To see this, extend an independent list successively along $0=\ker T^0\subseteq \ker T\subseteq \cdots \subseteq \ker T^{p^a}=V$. At each stage, if the current list does not span that kernel, append a vector outside its span. The dimension bound forces this finite procedure to terminate. Since $T(\ker T^j)\subseteq \ker T^{j-1}$, its matrix in the resulting basis has zero diagonal. The trace is independent of basis: $\mathrm{tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}=\mathrm{tr}(BA)$, so $\mathrm{tr}(P^{-1}TP)=\mathrm{tr}(T)$.

By additivity of the diagonal sum, $\mathrm{tr}(\rho (g))-\mathrm{tr}(\rho (s))=\mathrm{tr}(T)=0$. For $V=0$ both sums are empty and zero. No assertion that $\rho (u)=I$ is needed.