Lemma 1

Let $u$ be an automorphism of G, and H the set of its fixed points.

a) H is a closed subgroup of G.

b) If $H_0$ is central in G, then G is commutative (so G = T).

Assertion a) is clear. To prove b), we can replace G by D(G) (§1, Cor. 1 of Prop. 4), and hence can assume that G is semi-simple. Then, if $H_0$ is central in G, we have $L(H) =\{0\}$, so the endomorphism $L(u)-$Id of $\mathfrak{g}$ is bijective. Let $f$ be the endomorphism of the manifold G defined by $f(g) =u(g)^{-1}g$ for $g\in G$; it is étale, for if $g\in G$ and $x\in \mathfrak{g}$, we have $T(f)(xg) =u(g)^{-1}(x-L(u)(x))g$, so the tangent map of $f$ at $g$ is bijective. It follows that the image of $f$ is open and compact, hence coincides with G since G is connected. Now let E be a framing of G (§4, no. 10, Def. 3) and $u(E)$ its image under $u$. By Prop. 19 of loc. cit., there exists an element $h$ of G such that (Int $h$)$(E) =u(E)$. Let $g\in G$ be such that $h=f(g) =u(g)^{-1}g$; then

$u\circ$ Int $g=$ (Int $u(g)$)$\circ u=$ Int $g\circ$ (Int $h$)$^{-1}\circ u$,

so the framing (Int $g$)$(E)$ is stable under $u$. If (Int $g$)$(E) = (T_1,B,(U_{\alpha})_{\alpha\in B})$, then $\sum U_{\alpha}\in L(H)$; since $L(H) =\{0\}$, this implies that $B =\emptyset$, so $G = T_1$, and G is commutative.