Levi–Malcev decomposition

Remarks

The proof starts from the Levi decomposition of the Lie algebra, $\mathfrak{g} = \mathfrak{r} \rtimes \mathfrak{s}$ with $\mathfrak{s}$ a semisimple subalgebra. The quotient $G/R$ is simply connected with Lie algebra $\mathfrak{s}$, so the connected subgroup $S$ with Lie algebra $\mathfrak{s}$ maps isomorphically onto $G/R$. Thus $S$ is the image of a continuous section of $G \to G/R$, which gives 2. Part 3 comes from Malcev's theorem that two Levi subalgebras are conjugate by $\exp(\operatorname{ad} z)$ for some $z$ in $\mathfrak{r}$; then $x = \exp z$ works.

Example: the group $G$ of affine maps $v \mapsto Av + b$ of $\R^2$ with $A \in \mathrm{SL}_2(\R)$ has radical the translations $R \cong \R^2$ and Levi subgroup $\mathrm{SL}_2(\R)$. This $G$ is not simply connected; its universal cover is $\R^2 \rtimes \widetilde{\mathrm{SL}}_2(\R)$.

Simple connectedness is needed for 2. For a general connected Lie group, a Levi subgroup $S$ still exists and $G = RS$, but $R \cap S$ need not be trivial, and $S$ need not be closed.

Going deeper

For a student

The decomposition reduces many questions about Lie groups to two separate theories. Solvable groups are studied by induction on dimension, using Lie's theorem and the structure of nilpotent groups. Semisimple groups are studied with root systems, the Cartan and Iwasawa decompositions, and Weyl groups. The Lie algebra version is due to Levi, and the conjugacy statement to Malcev.