I know that the Jordan-Chevalley decomposition for real Lie algebras only applies to semisimple Lie algebras, but in general the addititive J-C decomposition says that for ANY operator, we can decompose it as $X = X_{SS} + X_{N}$, where $X_{SS}$ is semi-simple and $X_{N}$ is nilpotent.
Similarly, the Levi decomposition of a Lie algebra $\mathfrak{g} = \mathfrak{r}\oplus\mathfrak{s}$ says I can write every element of that algebra as $X = X_{SS} + X_{R}$, where $X_{SS}$ is in the semisimple subalgebra and $X_{R}$ is in the radical subalgebra.
My questions deal with the following: given an element $X$ in a Lie algebra $\mathfrak{g}$ I can use the Levi decomposition of Lie algebras to write $\mathfrak{g} = \mathfrak{r} \oplus \mathfrak{s}$, and thus $X = X_{SS} + X_{R}$. I can also write $X = Y_{SS} + Y_{N}$ using the J-C decomposition just based on the fact that $X$ is a linear operator.
(1) Is it necessarily the case that $X_{SS} = Y_{SS}$ and $Y_N = X_R$? I don't see why it should be, but I know nilpotent subalgebras are solvable, which $\mathfrak{r}$ is. Is there any relationship between these two additive decompositions of the matrix?
(2) Is there a simple explanation for why the J-C decomposition doesn't result in a Lie algebra decomposition for all Lie algebras? That is, if I have a Lie algebra $\mathfrak{g}$ and I know I can write every element $X\in \mathfrak{g}$ as $X = X_{SS} + X_{N}$, then what prevents this from being a Lie algebra decomposition? Is it because the sets $\{X_{SS}\,:\, X\in \mathfrak{g}\}$ and $\{X_{N}\,:\,X \in \mathfrak{g}\}$ don't form Lie subalgebras?