Jordan decomposition in a classical group 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.
 A: 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.
A: The proof of the Jordan decomposition for algebraic groups over perfect fields has two parts:
(a) Linear algebra: An automorphism of a vector space has a unique multiplicative Jordan decomposition, which is compatible with maps and tensor products...
(b) Some baby Tannakian stuff.
Most proofs in the literature mix the two parts, making the proof seem more difficult than it is. If you accept the baby Tannakian stuff, which everyone should know anyway, one is left with some easy linear algebra (see, for example, I Section 9 of my online notes).
