Let $G$ be a Lie group and $\mathfrak{g}$ its Lie algebra. Fix $g \in G$. Do we have $\{gxg^{-1}: x \in \mathfrak{g}\} = \mathfrak{g}$? Thank you very much.
Edit: I think that the answer is yes. We have $\{gxg^{-1}: x \in \mathfrak{g}\} \subset \mathfrak{g}$ and for any $y \in \mathfrak{g}$, we have $y = g (g^{-1} y g) g^{-1} \in \{gxg^{-1}: x \in \mathfrak{g}\}$.