Lemma 63.15.1. Let $e\in G$ denote the neutral element. The map

$\begin{matrix} \Lambda [G] & \longrightarrow & \Lambda ^{\natural } \\ \sum \lambda _ g\cdot g & \longmapsto & \lambda _ e \end{matrix}$

factors through $\Lambda [G]^\natural$. We denote $\varepsilon : \Lambda [G]^\natural \to \Lambda ^\natural$ the induced map.

Proof. We have to show the map annihilates commutators. One has

$\left(\sum \lambda _ g g\right)\left(\sum \mu _ g g\right)-\left(\sum \mu _ g g\right)\left(\sum \lambda _ g g\right) = \sum _ g\left(\sum _{g_1g_2=g} \lambda _{g_1}\mu _{g_2}-\mu _{g_1}\lambda _{g_2}\right)g$

The coefficient of $e$ is

$\sum _ g\left(\lambda _ g\mu _{g^{-1}}-\mu _ g\lambda _{g^{-1}}\right) = \sum _ g\left(\lambda _ g\mu _{g^{-1}}-\mu _{g^{-1}}\lambda _ g\right)$

which is a sum of commutators, hence it zero in $\Lambda ^\natural$. $\square$

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