Root reflections are induced by inner automorphisms

First, a vanishing criterion: if $X,x\in \mathfrak{g}$ satisfy $[X,x]=0$, then ${\mathrm{Ad}}_{{\exp }_{G}X}(x)=x$. Indeed the curve $t\mapsto {\mathrm{Ad}}_{{\exp }_G(tX)}$ is a homomorphism in $t$ with derivative satisfying $U^{\mathrm{\prime }}(t)={\mathrm{ad}}_X\circ U(t)$, by [L4] and the chain rule, and $U(0)=I$; hence $u(t)=U(t)x$ solves $u^{\mathrm{\prime }}={\mathrm{ad}}_{X}u$ with $u(0)=x$, and so does the constant curve $x$ because $[X,x]=0$. Uniqueness [L5] gives ${\mathrm{Ad}}_{{\exp }_{G}X}(x)=x$.

Similarly, if ${\mathrm{ad}}_X$ is nilpotent and $({\mathrm{ad}}_X{)}^n=0$, then ${\mathrm{Ad}}_{{\exp }_{G}X}=e^{{\mathrm{ad}}_X}={\sum }_{k<n}\frac{({\mathrm{ad}}_X{)}^k}{k!}\mathrm{.}$ Indeed the polynomial curve $E(t)={\sum }_{k<n}\frac{t^k({\mathrm{ad}}_X{)}^k}{k!}$ satisfies $E^{\mathrm{\prime }}(t)={\mathrm{ad}}_X\circ E(t)$ and $E(0)=I$ by termwise differentiation. Uniqueness [L5] therefore identifies it with $e^{t{\mathrm{ad}}_X}$, and setting $t=1$ gives the displayed formula.

${\tau }_{\alpha }$ is an inner automorphism: by [L4] each factor ${\mathrm{Ad}}_{{\exp }_G(\pm e_{\alpha })}$, ${\mathrm{Ad}}_{{\exp }_G(-f_{\alpha })}$ is an automorphism of $\mathfrak{g}$, and $\mathrm{Ad}$ is multiplicative, so ${\tau }_{\alpha }={\mathrm{Ad}}_g$ for $g={\exp }_G(e_{\alpha }){\exp }_G(-f_{\alpha }){\exp }_G(e_{\alpha })\in G$.

If $x\in \ker \alpha \subseteq \mathfrak{h}$, then $[e_{\alpha },x]=\alpha (x)e_{\alpha }=0$ and $[f_{\alpha },x]=-\alpha (x)f_{\alpha }=0$ by [L2], so step 1.1 applied to $X=e_{\alpha },-f_{\alpha }$ gives ${\tau }_{\alpha }(x)=x$.

The operator ${\mathrm{ad}}_{e_{\alpha }}$ is nilpotent on $\mathbb{C}{h}_{\alpha }\oplus \mathbb{C}{e}_{\alpha }$: indeed $[e_{\alpha },h_{\alpha }]=-2e_{\alpha }$ and $[e_{\alpha },e_{\alpha }]=0$ by [L1]; likewise ${\mathrm{ad}}_{-f_{\alpha }}$ is nilpotent on $\mathbb{C}{h}_{\alpha }\oplus \mathbb{C}{f}_{\alpha }$. Using step 1.2 we compute ${\mathrm{Ad}}_{{\exp }_G(e_{\alpha })}(h_{\alpha })=h_{\alpha }-2e_{\alpha }$, then ${\mathrm{Ad}}_{{\exp }_G(-f_{\alpha })}(h_{\alpha }-2e_{\alpha })=-h_{\alpha }-2e_{\alpha }$ from $[-f_{\alpha },h_{\alpha }]=-2f_{\alpha }$ and $[-f_{\alpha },e_{\alpha }]=h_{\alpha }$, and finally ${\mathrm{Ad}}_{{\exp }_G(e_{\alpha })}(-h_{\alpha }-2e_{\alpha })=-h_{\alpha }$; hence ${\tau }_{\alpha }(h_{\alpha })=-h_{\alpha }$.

Consequently ${\tau }_{\alpha }$ preserves $\mathfrak{h}=\ker \alpha \oplus \mathbb{C}{h}_{\alpha }$: it fixes $\ker \alpha$ pointwise by step 2.1 and negates $h_{\alpha }$ by step 2.2. For a root $\beta$ and $x\in {\mathfrak{g}}_{\beta }$, the element ${\tau }_{\alpha }(x)$ satisfies $[H,{\tau }_{\alpha }(x)]={\tau }_{\alpha }([{\tau }_{\alpha }^{-1}(H),x])=\beta ({\tau }_{\alpha }^{-1}(H)){\tau }_{\alpha }(x)$ for $H\in \mathfrak{h}$; since ${\tau }_{\alpha }^{-1}{\mathrm{\mid }}_{\mathfrak{h}}={\tau }_{\alpha }{\mathrm{\mid }}_{\mathfrak{h}}$ is the identity on $\ker \alpha$ and negation on $h_{\alpha }$, the functional $H\mapsto \beta ({\tau }_{\alpha }^{-1}(H))$ agrees with $\beta$ on $\ker \alpha$ and takes the value $-\beta (h_{\alpha })$ at $h_{\alpha }$, hence equals $\beta -\beta (h_{\alpha })\alpha =s_{\alpha }(\beta )$ because $\alpha (h_{\alpha })=2$ and $\alpha$ vanishes on $\ker \alpha$. Therefore ${\tau }_{\alpha }({\mathfrak{g}}_{\beta })\subseteq {\mathfrak{g}}_{s_{\alpha }(\beta )}$, and since ${\tau }_{\alpha }$ is an automorphism and $s_{\alpha }$ is an involution, dimensions agree and equality holds; the Axiom of Choice was used only through [L3] and [L4], that is, through [A1].