Bourbaki

Proposition 3

Tag 01E8, permanent. Cite it as [Bourbaki, Tag 01E8].

Lie Groups and Lie Algebras, chapter IX, § 5. Conjugacy classes

Let uu be an automorphism of G, and let xx be a point of G fixed by uu.

a) There exists an element aa of g\mathfrak{g}, fixed by L(u)L(u) and by Ad xx, such that xx exp aa is regular.

b) There exists a regular element gg of G fixed by uu and commuting with xx.

Let H be the group of fixed points of u,Su, S a maximal torus of Z(x)∩HZ(x)\cap H, and K the identity component of Z(S)∩Z(x)Z(S)\cap Z(x). This is a connected closed subgroup of G; further, by Cor. 5 of Th. 2 of §2, no. 2, there exist maximal tori of G containing S and xx, so K is of maximal rank and contains S and xx. On the other hand, K is stable under uu since S and xx are; denote by V the set of fixed points of uu in K. Then

S⊂V0⊂K∩H⊂Z(S)∩Z(x)∩H S\subset V_0\subset K\cap H\subset Z(S)\cap Z(x)\cap H

so V0V_0 is contained in the centralizer of S in (Z(x)∩H)0(Z(x)\cap H)_0; but the latter reduces to S (loc. cit., Cor. 6), hence finally V0=SV_0= S. Lemma 1 now implies that K is commutative, hence is a maximal torus of G (since it is connected and of maximal rank). It contains S and xx, and is equal to the identity component of Z(S)∩Z(x)Z(S)\cap Z(x); assertion a) now follows from Lemma 2. We deduce b) by taking g=xg=x exp aa.

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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