For algebraic groups or Lie groups, the subject of Levi decompositions tends to be surrounded by some mystery in the literature (and in an older question raised here). While I postpone further my intention to post a careful note on my homepage someday, I'm curious about the status of disconnected algebraic groups in this discussion. Take the ground field $K$ to be algebraically closed of characteristic 0 throughout.
If $G$ is a disconnected linear algebraic group over $K$, of positive dimension, does it always have a Levi decomposition; if so, what can be said about conjugacy of Levi subgroups?
First I have to outline briefly the history of the concept of Levi decomposition. This was first carried out for a finite dimensional Lie algebra $\mathfrak{g}$ and makes sense over an arbitrary field of characteristic 0: Levi's theorem says that the solvable radical of $\mathfrak{g}$ has as a semidirect complement some semisimple subalgebra, while Mal'cev showed the conjugacy of any two such "Levi subalgebras" under special automorphisms of $\mathfrak{g}$ (which can be defined algebraically). This is treated efficiently in modern language by Bourbaki in Chapter I, 6.8 of their treatise Lie Groups and Lie Algebras.
To carry this over to connected Lie groups (real or complex), one then has to use the nice correspondence between Lie subgroups and Lie subalgebras. (But this is developed using exponentials.) In the 1950s Chevalley experimented with linear algebraic groups in characteristic 0, using a sort of formal exponentiation process to imitate the Lie group framework.
Borel works some of this out in somewhat more modern language in II.7.9 of his Benjamin notes, reproduced in his GTM 126 second edition. For the Lie algebra $\mathfrak{g}$ of a connected algebraic group $G$, the basic observation is that any Lie subalgebra equal to its derived algebra is algebraic, i.e., is the Lie algebra of some connected closed subgroup of $G$. (And this subgroup is unique, since we are in characteristic 0.) In particular, a semisimple Levi subalgebra of $\mathfrak{g}$ is the Lie algebra of a connected semisimple subgroup $H$ of $G$ whose intersection with the solvable radical is finite.
(In some sources the product of subgroups here is supposed to be semidirect, but this is artificial for cases like the general linear group.) Using Jordan decomposition, this becomes the more useful decomposition of $G$ into a semidirect product of a connected reductive subgroup and the unipotent radical.
The conjugacy theorem for $\mathfrak{g}$ then carries over to the group in a natural way. Unfortunately, there seems to be no complete account of this in a textbook, though Demazure-Gabriel translate some of it into scheme language in II, Section 6. In any event, my question above doesn't seem to be answered directly, since the characteristic 0 correspondence between closed subgroups and their Lie algebras requires connected subgroups.
Footnote: In prime charactertistic things tend to fall apart in general, as indicated in my 1967 note here. But a recent paper by George McNinch offers numerous useful refinements here.
Best Answer
See G. P. Hochschild: Basic Theory of Algebraic Groups and Lie Algebras (Graduate Texts in Mathematics)-Springer (1981) VIII, Theorem 4.3