Proposition 9

Every automorphism of the Lie group G that induces the identity on T is of the form Int $t$, with $t\in T$.

Assume first of all that C(G) reduces to the identity element, in other words that X(T) = Q(R) (Prop. 8). Let $f$ be an automorphism of G inducing the identity on T, and $\varphi = L(f)_{(\mathbf{C})}$; then $\varphi$ is an automorphism of $\mathfrak{g}_{\mathbf{C}}$ inducing the identity on $\mathfrak{t}_{\mathbf{C}}$. By Chap. VIII, §5, no. 2, Prop. 2, there exists a unique homomorphism $\theta : Q(R)\rightarrow \mathbf{C}^*$ such that $\varphi$ induces on each $\mathfrak{g}^{\alpha}$ the homothety with ratio $\theta (\alpha )$. Since $\varphi$ leaves stable the real form $\mathfrak{g}$ of $\mathfrak{g}_{\mathbf{C}}$, it commutes with the conjugation $\sigma$ of $\mathfrak{g}_{\mathbf{C}}$ with respect to $\mathfrak{g}$; but $\sigma (\mathfrak{g}^{\alpha}) =\mathfrak{g}^{-\alpha}$, so $\theta (-\alpha ) =\overline{\theta(\alpha)}$ for all $\alpha \in R$. This implies that $\theta (\alpha )\theta (\alpha ) =\theta (\alpha )\theta (-\alpha ) = 1$. It follows that $\theta$ takes values in $\mathbf{U}$, and hence corresponds by duality to an element $t$ of T such that (Ad $t$)$_{(\mathbf{C})}=\varphi$, so Int $t=f$.

In the general case, the preceding applies to the group $G/C(G)$, whose centre reduces to the identity element, and to its maximal torus $T/C(G)$. It follows that, if $f$ is an automorphism of G inducing the identity on T, there exists an element $t$ of T such that $f$ and Int $t$ induce by passage to the quotient the same automorphism of $G/C(G)$. But, since the canonical morphism $D(G)\rightarrow G/C(G)$ is a finite covering (§1, no. 4, Cor. 1 of Prop. 4), $f$ and Int $t$ induce the same automorphism of D(G), hence of $D(G)\times C(G)$, and hence also of G (loc. cit.).