Corollary

Tag 00V7, permanent. Cite it as [Bourbaki, Tag 00V7].

Lie Groups and Lie Algebras, chapter VII, § 3. Conjugacy theorems

$E(\mathfrak{h}) = E(\mathfrak{h}')$ .

Let $u, u'$ be as in Prop. 2. Then

E(\mathfrak{h}) =uE(\mathfrak{h})u^{-1}= E(u(\mathfrak{h})) = E(u'(\mathfrak{h}')) =u'E(\mathfrak{h}')u^{'-1}= E(\mathfrak{h}')

hence the corollary.

Because of this result, if $k$ is algebraically closed we shall denote simply by E the group $E(\mathfrak{h})$ , where $\mathfrak{h}$ is a Cartan subalgebra of $\mathfrak{g}$ .

In general, Aut $_e(\mathfrak{g})\not= E$ (for example, if $\mathfrak{g}$ is nilpotent, E reduces to the identity element, even though non-trivial elementary automorphisms exist provided $\mathfrak{g}$ is non-commutative). However, it can be shown (Chap. VIII, §10, Exerc. 5) that Aut $_e(\mathfrak{g}) = E$ for $\mathfrak{g}$ semi-simple.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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