Is there a Levi decomposition for Lie group and algebraic group?

Let $G$ be a Lie group and $R$ be the largest connected solvable normal subgroup of $G$.

Question 1

Is there a Lie subgroup $S$ such that: (1) $G=SR$; (2) every real representation of $S$ is semisimple?

Question 2

Is there a Lie subgroup $S$ such that: (1) $G=SR$; (2) every complex representation of $S$ is semisimple?

Let $G$ be an algebraic group and $R$ be the largest connected solvable normal subgroup of $G$. Is there a algebraic subgroup $S$ such that: (1) $G=SR$; (2) every representation of $S$ is semisimple?

I want to know the formal statement and references.

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I believe the answer to (1) is yes; see the Springer Encyclopedia of Mathematics, eom.springer.de/L/l058590.htm, Lie groups, section on their global structure. There it states that $S \cap R$ is trivial if $G$ is simply connected. Also, maybe you should edit the third question and make a Part (3). –  Qayum Khan Sep 11 '11 at 0:10
Question $1$: The theorem of Mostow says that every connected algebraic group $G$ over a field $K$ of characteristic zero has a Levi decomposition. This means, $G$ has a reductive algebraic subgroup $S$ such that $G$ is the semidirect prodcut of $S$ and $R_u(G)$, the unipotent radical of $G$. Moreover, any reductive algebraic subgroup $S'$ of $G$ is conjugate to $S$. The result is not true in general in prime characteristic. There is a nice article by Jim Humphreys, EXISTENCE OF LEVI FACTORS IN CERTAIN ALGEBRAIC GROUPS, in the pacific journal of mathematics $1967$.
Question $2$: The Levi decomposition for Lie groups was shown by Levi and Malcev.