Corollary 2

*Let G be a finite-dimensional simply connected Lie group, H a normal connected Lie subgroup of G and $\pi$ the canonical morphism of G onto G/H.

(i) There exists an analytic mapping $\varphi$ of G/H into G such that $\pi \circ \varphi = \mathrm{Id}_{G/H}$.

(ii) For every mapping $\varphi$ with the properties of (i), the mapping $(h, m) \mapsto h \varphi(m)$ of $H \times (G/H)$ into G is an isomorphism of analytic manifolds.

(iii) H and G/H are simply connected.

Let $n = \dim G - \dim H$. The corollary is obvious for $n = 0$. We argue by induction on n.

Suppose that there exists an ideal of L(G) containing L(H) distinct from L(G) and L(H). Let H' be the corresponding connected Lie subgroup of G. Let $\pi_1 : G \to G/H'$ and $\pi_2 : H' \to H'/H$ be the canonical morphisms. By the induction hypothesis, there exist analytic mappings $\rho_1 : G/H' \to G$, $\rho_2 : H'/H \to H'$ such that $\pi_1 \circ \rho_1 = \mathrm{Id}_{G/H'}$, $\pi_2 \circ \rho_2 = \mathrm{Id}_{H'/H}$. Let

be the canonical morphism. If $x \in G/H$ and y is a representative of x in G, y and $\rho_1(\pi_3(x))$ have the same canonical image modulo H' and hence $x^{-1} \pi(\rho_1(\pi_3(x))) \in H'/H$. Let

Clearly $\rho$ is an analytic mapping of G/H into G and

If now the only ideals of L(G) containing L(H) are L(G) and L(H), the Lie algebra $L(G)/L(H)$ is either 1-dimensional or simple. In both cases, $L(G)$ is the semi-direct product of a subalgebra by $L(H)$; this is obvious in the first case and in the second this follows from Chapter I, § 6, Corollary 3 to Theorem 5. Assertion (i) then follows from Corollary 1.

Assertion (ii) is obvious. Assertion (iii) follows from (i) and (ii).

The conclusions of Corollary 2 are no longer necessarily true when G is infinite-dimensional or when H is not normal (Exercises 8 and 15).