Lemma 63.15.3. Let $f : P\to P$ be an endomorphism of the finite projective $\Lambda [G]$-module $P$. Then

$\text{Tr}_{\Lambda }(f; P) = \# G \cdot \text{Tr}_\Lambda ^ G(f; P).$

Proof. By additivity, reduce to the case $P = \Lambda [G]$. In that case, $f$ is given by right multiplication by some element $\sum \lambda _ g\cdot g$ of $\Lambda [G]$. In the basis $(g)_{g \in G}$, the matrix of $f$ has coefficient $\lambda _{g_2^{-1}g_1}$ in the $(g_1, g_2)$ position. In particular, all diagonal coefficients are $\lambda _ e$, and there are $\# G$ such coefficients. $\square$

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