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Let $G$ be connected nilpotent complex Lie group and consider the adjoint representation $Ad:G\rightarrow GL(\mathfrak g)$. How to show that the eigenvalues of $Ad_g$ are one?

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Since all elements $g$ of a nilpotent Lie group $G$ are unipotent (Lie-Kolchin), we have for $g$ that $$ (Ad_g-Id)^k=0 $$ for some $k$. This is equivalent to saying that all eigenvalues of $Ad_g$ are equal to $1$.

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