Let $\mathfrak{g} \subset \mathfrak{gl}\_n$ be one of the classical real or complex semisimple Lie algebras. If $g \in \mathfrak{g}$, then $g$ has a Jordan decomposition $g = g\_s + g\_n$ with $g\_s$ semisimple and $g\_n$ nilpotent, and $[g\_s,g\_n]=0$. 


The elements $g\_s,g\_n$, which a priori are just in $\mathfrak{gl}\_n$, are both in $\mathfrak{g}$ again. There are various middle-brow general ways to see this (for one, use that $\mathfrak{g}$ is algebraic), but for concrete choices of $\mathfrak{g}$ it's basically elementary, as follows. One knows from the construction of the Jordan decomposition that $g\_s,g\_n$ are both polynomials in $g$ (different polynomials for different $g$, of course), and (EDIT) you can rig the construction so that these polynomials are odd. The Lie algebra $\mathfrak{g}$ is the subspace of $\mathfrak{gl}\_n$ cut out by conditions like $\mathrm{trace}(g)=0$, or $Jg = -g^{t} J$ for some matrix $J$, and so forth. The condition $\mathrm{trace}(g)=0$ is always true for $g\_n$, so it's true for $g\_s$ if true for $g$. The condition $Jg=-g^t J$ is visibly true for odd $p(g)$ if true for $g$, so if true for $g$ then it's true for both $g\_s$ and $g\_n$. Thus $g\_s$ and $g\_n$ visibly satisfy whatever conditions $g$ is required to satisfy, and so are contained in $\mathfrak{g}$.


(This might seem lowbrow but in fact I think this is basically the idea of the proof that Fulton-Harris give for general semisimple Lie algebras.)


Now suppose instead that $G$ is a real or complex linear Lie group with Lie algebra $\mathfrak{g}$. This time the Jordan decomposition is $g = g\_s g\_u$ with $g\_u$ unipotent, and indeed $g\_s$ and $g\_u$ are still in $G$. But if you try to make the same lowbrow argument as in the Lie algebra case, it appears to die horribly (a condition like $g^t = g^{-1}$ certainly need not be preserved by taking a polynomial in $g$). My question is, is there an elementary way to rescue it? (In particular, something other than just the general argument for algebraic groups.) Obviously you're fine for elements $g$ in the image of the exponential map, so the issue is passing to the whole group. A caveat is that I do $\textit{not}$ want to assume that $G$ is connected.



Jordan decomposition in a classical group
Give the proof in Humphreys' "Linear Algebraic Groups". It is essentially a context-free version of the argument you give, and hinges only on the fact that if $\rho\_g$ is right-translation by $g$ in $k[\operatorname{GL}\_n]$ and $I$ is the ideal defining $G$ in $\operatorname{GL}\_n$, then $g \in G$ if and only if $\rho\_g(I) \subset I$. It is a simple fact of linear algebra that the semisimple and unipotent parts of $\rho\_g$ stabilize any subspace which $\rho\_g$ itself stabilizes. The intuition, of course, is that $I$ consists of all the "equations" defining $G$ in $\operatorname{GL}\_n$.


This may be the general argument you said you didn't want, in which case I think you should reconsider it as being just the right amount of generality on top of what you have done for Lie algebras.


