Bourbaki

Proposition 14

Tag 0AFN, permanent. Cite it as [Bourbaki, Tag 0AFN].

Lie Groups and Lie Algebras, chapter III, § 6. Real and complex Lie groups

Let GG be a Lie group and HH an integral subgroup whose topology admits a countable base. The following conditions are equivalent: (i) HH is normal in GG; (ii) L(H)L(H) is invariant under Ad(G)\mathrm{Ad}(G). If further GG is connected, these conditions are equivalent to the following: (iii) L(H)L(H) is an ideal of L(G)L(G). If further GG is simply connected and L(H)L(H) is of finite codimension in L(G)L(G), these conditions imply that HH is a Lie subgroup of GG and that G/HG/H is simply connected.

The equivalence (i) ⇔\Leftrightarrow (ii) follows from Lemma 3. If further GG is connected, condition (ii) is equivalent to saying that L(H)L(H) is stable under ad L(G)\mathrm{ad}\, L(G) (no. 5, Corollary 1 to Proposition 13 and § 3, no. 12, Proposition 44).

Suppose that GG is simply connected and that L(H)L(H) is an ideal of finite codimension in L(G)L(G). By Theorem 3 of no. 3, there exists a simply connected Lie group G′G' such that L(G′)L(G') is isomorphic to L(G)/L(H)L(G)/L(H). There exists a continuous morphism ff of L(G)L(G) onto L(G′)L(G') with kernel L(H)L(H). By Theorem 1 of no. 1, there exists a morphism ϕ\phi of GG into G′G' such that L(ϕ)=fL(\phi) = f. This morphism is a submersion and hence its kernel NN is a Lie subgroup of GG such that L(N)=Ker f=L(H)L(N) = \mathrm{Ker}\, f = L(H). Hence HH is the identity component of NN and is therefore a Lie subgroup of G. Let ψ\psi be the morphism of G/HG/H into G′G' derived from ϕ\phi when passing to the quotient. This morphism is étale; since G is simply connected, ψ\psi is an isomorphism of G/HG/H onto G′G'.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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