Proposition 2

Let $\mathfrak{g}$ be a Lie subalgebra of $\mathfrak{g}\mathfrak{l}(V)$ reductive in $\mathfrak{g}\mathfrak{l}(V)$, E a finite dimensional vector space and $\pi :\mathfrak{g}\rightarrow \mathfrak{g}\mathfrak{l}(E)$ a semi-simple linear representation of $\mathfrak{g}$ on E. Then:

(i) $\mathfrak{g}$ and $\pi (\mathfrak{g})$ are decomposable.

(ii) The semi-simple (resp. nilpotent) elements of $\pi (\mathfrak{g})$ are the images under $\pi$ of the semi-simple (resp. nilpotent) elements of $\mathfrak{g}$.

(iii) If $\mathfrak{h}$ is a decomposable subalgebra of $\mathfrak{g}\mathfrak{l}(V)$ contained in $\mathfrak{g},\pi (\mathfrak{h})$ is a decomposable subalgebra of $\mathfrak{g}\mathfrak{l}(E)$.

(iv) If $\mathfrak{h}'$ is a decomposable subalgebra of $\mathfrak{g}\mathfrak{l}(E),\pi^{-1}(\mathfrak{h}')$ is a decomposable subalgebra of $\mathfrak{g}\mathfrak{l}(V)$.

Let $\mathfrak{s}= [\mathfrak{g},\mathfrak{g}]$ and let $\mathfrak{c}$ be the centre of $\mathfrak{g}$. Then $\mathfrak{g}=\mathfrak{s}\times \mathfrak{c}$, and $\pi (\mathfrak{g}) =$ $\pi (\mathfrak{s})\times \pi (\mathfrak{c})$ by Chap. I, §6, no. 4, Cor. of Prop. 5. Let $y\in \mathfrak{s}, z\in \mathfrak{c}, y_s$ and $y_n$ the semi-simple and nilpotent components of $y$. Then $y_s, y_n\in \mathfrak{s}$ (Chap. I, §6, no. 3, Prop. $3$)$,y_s+z$ is semi-simple (Algebra, Chap. VII, §5, no. 7, Cor. of Prop. 16), and $y_n$ commutes with $y_s+z$. Hence, the semi-simple and nilpotent components of $y+z$ are $y_s+z$ and $y_n$. Thus, $\mathfrak{g}$ is decomposable. Since $\pi (\mathfrak{g})$ is reductive in $\mathfrak{g}\mathfrak{l}$(E), the same argument applies to $\pi (\mathfrak{g})$ and shows that $\pi (\mathfrak{g})$ is decomposable. Moreover, the nilpotent elements of $\mathfrak{g}$ (resp. $\pi (\mathfrak{g})$) are the nilpotent elements of $\mathfrak{s}$ (resp. $\pi (\mathfrak{s})$). Hence the nilpotent elements of $\pi (\mathfrak{g})$ are the images under $\pi$ of the nilpotent elements of $\mathfrak{g}$ (Chap. I, §6, no. 3, Prop. 4). The semi-simple elements of $\mathfrak{g}$ (resp. $\pi (\mathfrak{g})$) are the sums of the semi-simple elements of $\mathfrak{s}$ (resp. $\pi (\mathfrak{s})$) and the elements of $\mathfrak{c}$ (resp. $\pi (\mathfrak{c})$). Thus the semi-simple elements of $\pi (\mathfrak{g})$ are the images under $\pi$ of the semi-simple elements of $\mathfrak{g}$ (Chap. I, loc. cit.). Hence (ii).

Assertions (iii) and (iv) follow immediately from (i) and (ii).