[proofplan] Fix a root $\alpha$ and use the standard compact rank-one subgroup theorem attached to the pair of root spaces $\mathfrak g_\alpha$ and $\mathfrak g_{-\alpha}$. This gives a compact rank-one subgroup $G_\alpha$ and a representative $n_\alpha$ which inverts the coroot circle. The kernel torus of $\alpha$ centralizes this rank-one subgroup, so $n_\alpha$ fixes the kernel directions and inverts only the coroot direction; this proves that $n_\alpha$ normalizes all of $T$. Finally we compute the induced action on $\mathfrak t_{\mathbb C}^*$ from this two-summand decomposition and obtain exactly $\beta\mapsto\beta-\beta(H_\alpha)\alpha$. [/proofplan] [step:Record the rank-one input attached to $\alpha$] For each root $\gamma\in\Phi$, define its root space by \begin{align*} \mathfrak g_\gamma:=\{Y\in\mathfrak g_{\mathbb C}:[X,Y]=\gamma(X)Y\text{ for every }X\in\mathfrak t_{\mathbb C}\}. \end{align*} We use the standard compact rank-one subgroup theorem for roots of compact connected Lie groups in the following form. If $G$ is compact and connected, $T\le G$ is a maximal torus, and $\alpha\in\Phi$, then there are elements $E_\alpha\in\mathfrak g_\alpha$ and $F_\alpha\in\mathfrak g_{-\alpha}$ such that \begin{align*} [E_\alpha,F_\alpha]=H_\alpha. \end{align*} The complex span \begin{align*} \mathfrak s_{\alpha,\mathbb C}:=\operatorname{span}_{\mathbb C}\{E_\alpha,F_\alpha,H_\alpha\} \end{align*} is an $\mathfrak{sl}_2(\mathbb C)$-subalgebra, with \begin{align*} [H_\alpha,E_\alpha]=2E_\alpha \end{align*} and \begin{align*} [H_\alpha,F_\alpha]=-2F_\alpha. \end{align*} Its compact real form inside $\mathfrak g$ integrates to a connected compact subgroup $G_\alpha\le G$. Moreover, if \begin{align*} S_\alpha:=G_\alpha\cap T, \end{align*} then $S_\alpha$ is the coroot circle, and there is an element $n_\alpha\in G_\alpha$ such that conjugation by $n_\alpha$ inverts $S_\alpha$ and its differential sends $H_\alpha$ to $-H_\alpha$: \begin{align*} \operatorname{Ad}_{n_\alpha}(H_\alpha)=-H_\alpha. \end{align*} The hypotheses of this theorem are exactly the standing hypotheses here: $G$ is compact and connected, $T$ is a maximal torus, and $\alpha$ is a root in $\Phi$. Fix such a choice of $G_\alpha$, $S_\alpha$, and $n_\alpha$ for the rest of the proof. [guided] We first isolate the only external structural input being used. For each root $\gamma\in\Phi$, the corresponding root space is \begin{align*} \mathfrak g_\gamma:=\{Y\in\mathfrak g_{\mathbb C}:[X,Y]=\gamma(X)Y\text{ for every }X\in\mathfrak t_{\mathbb C}\}. \end{align*} The standard compact rank-one subgroup theorem says the following in the present setting. Given a compact connected Lie group $G$, a maximal torus $T\le G$, and a root $\alpha\in\Phi$, the two root spaces $\mathfrak g_\alpha$ and $\mathfrak g_{-\alpha}$ can be completed to an $\mathfrak{sl}_2(\mathbb C)$-triple. Thus there are $E_\alpha\in\mathfrak g_\alpha$ and $F_\alpha\in\mathfrak g_{-\alpha}$ with \begin{align*} [E_\alpha,F_\alpha]=H_\alpha. \end{align*} They also satisfy \begin{align*} [H_\alpha,E_\alpha]=2E_\alpha \end{align*} and \begin{align*} [H_\alpha,F_\alpha]=-2F_\alpha. \end{align*} Consequently \begin{align*} \mathfrak s_{\alpha,\mathbb C}:=\operatorname{span}_{\mathbb C}\{E_\alpha,F_\alpha,H_\alpha\} \end{align*} is a copy of $\mathfrak{sl}_2(\mathbb C)$. The same theorem includes the compact-group integration statement needed here: the compact real form of this rank-one algebra inside $\mathfrak g$ integrates to a connected compact subgroup $G_\alpha\le G$. Its intersection \begin{align*} S_\alpha:=G_\alpha\cap T \end{align*} is the coroot circle. In this compact rank-one group, the nontrivial Weyl element has a representative $n_\alpha\in G_\alpha$ which inverts $S_\alpha$, and the infinitesimal form of that inversion is \begin{align*} \operatorname{Ad}_{n_\alpha}(H_\alpha)=-H_\alpha. \end{align*} The theorem applies because the standing assumptions give precisely its hypotheses: $G$ is compact and connected, $T$ is maximal, and $\alpha$ is a root. We now fix the resulting $G_\alpha$, $S_\alpha$, and $n_\alpha$. [/guided] [/step] [step:Show that the connected kernel of $\alpha$ centralizes the rank-one subgroup] Define the connected kernel torus by \begin{align*} K_\alpha:=\{t\in T:\alpha(t)=1\}^{\circ}. \end{align*} Its [Lie algebra](/page/Lie%20Algebra) is \begin{align*} \mathfrak k_\alpha:=\{X\in\mathfrak t:\alpha(X)=0\}. \end{align*} Let $X\in\mathfrak k_\alpha$. Since $E_\alpha\in\mathfrak g_\alpha$, the root-space relation gives \begin{align*} [X,E_\alpha]=\alpha(X)E_\alpha=0. \end{align*} Since $F_\alpha\in\mathfrak g_{-\alpha}$, it also gives \begin{align*} [X,F_\alpha]=-\alpha(X)F_\alpha=0. \end{align*} Finally, $[X,H_\alpha]=0$ because $X$ and $H_\alpha$ both lie in the abelian Lie algebra $\mathfrak t_{\mathbb C}$. Therefore $X$ centralizes $\mathfrak s_{\alpha,\mathbb C}$, and hence centralizes the compact real Lie algebra $\mathfrak s_\alpha:=\operatorname{Lie}(G_\alpha)$. For any $k\in K_\alpha$, choose a finite product of exponentials in $\mathfrak k_\alpha$ representing $k$, which is possible because $K_\alpha$ is a connected torus. Each such exponential commutes with every element of the connected subgroup $G_\alpha$, since its infinitesimal generator centralizes $\mathfrak s_\alpha$. Thus every $k\in K_\alpha$ centralizes $G_\alpha$. In particular, \begin{align*} n_\alpha k n_\alpha^{-1}=k \end{align*} for every $k\in K_\alpha$. [/step] [step:Deduce that $n_\alpha$ normalizes the full torus $T$] The real hyperplane $\mathfrak k_\alpha=\ker(\alpha:\mathfrak t\to i\mathbb R)$ together with the compact real coroot line $\operatorname{Lie}(S_\alpha)$ spans $\mathfrak t$. Hence the image of the multiplication map \begin{align*} m:K_\alpha\times S_\alpha\to T, \end{align*} with $m(k,s)=ks$, is a connected subtorus of $T$ whose Lie algebra is all of $\mathfrak t$. Therefore $m(K_\alpha\times S_\alpha)=T$. Let $t\in T$. Choose $k\in K_\alpha$ and $s\in S_\alpha$ such that $t=ks$. The preceding step gives $n_\alpha k n_\alpha^{-1}=k$, while the rank-one input gives $n_\alpha s n_\alpha^{-1}=s^{-1}$. Thus \begin{align*} n_\alpha t n_\alpha^{-1}=ks^{-1}. \end{align*} Since $ks^{-1}\in T$, we have $n_\alpha Tn_\alpha^{-1}\subset T$. Applying the same argument to $n_\alpha^{-1}$ gives $n_\alpha^{-1}Tn_\alpha\subset T$, equivalently $T\subset n_\alpha Tn_\alpha^{-1}$. Hence \begin{align*} n_\alpha Tn_\alpha^{-1}=T. \end{align*} Therefore $n_\alpha\in N_G(T)$. [guided] At this point $n_\alpha$ is known to normalize the rank-one torus $S_\alpha$, but the Weyl group $W(G,T)$ is defined using the normalizer of the whole torus $T$. We must therefore prove that conjugation by $n_\alpha$ preserves every direction in $T$. The torus directions split into two pieces. The first piece is the connected kernel torus \begin{align*} K_\alpha:=\{t\in T:\alpha(t)=1\}^{\circ}, \end{align*} whose Lie algebra is \begin{align*} \mathfrak k_\alpha:=\{X\in\mathfrak t:\alpha(X)=0\}. \end{align*} The second piece is the coroot circle $S_\alpha$. Because $\alpha(H_\alpha)=2$, the coroot direction is not contained in $\ker\alpha$, and the hyperplane $\mathfrak k_\alpha$ together with the coroot line spans $\mathfrak t$. Therefore the multiplication map \begin{align*} m:K_\alpha\times S_\alpha\to T \end{align*} with $m(k,s)=ks$ has image a connected subtorus of $T$ whose Lie algebra is all of $\mathfrak t$. A connected subtorus of $T$ with the same Lie algebra as $T$ is $T$ itself, so every $t\in T$ can be written as $t=ks$ with $k\in K_\alpha$ and $s\in S_\alpha$. Now we compute the conjugation action on such a product. The previous centralization argument proves that every $k\in K_\alpha$ commutes with every element of $G_\alpha$, hence with $n_\alpha$. The rank-one Weyl representative property says that $n_\alpha$ inverts the coroot circle, so $n_\alpha s n_\alpha^{-1}=s^{-1}$ for $s\in S_\alpha$. Therefore, if $t=ks$, then \begin{align*} n_\alpha t n_\alpha^{-1}=ks^{-1}. \end{align*} The element $ks^{-1}$ belongs to $T$, so $n_\alpha Tn_\alpha^{-1}\subset T$. Repeating the same argument with $n_\alpha^{-1}$ gives the reverse inclusion. Consequently \begin{align*} n_\alpha Tn_\alpha^{-1}=T, \end{align*} which is exactly the assertion that $n_\alpha\in N_G(T)$. [/guided] [/step] [step:Compute the induced action on every root] Since $n_\alpha\in N_G(T)$, conjugation by $n_\alpha$ induces a linear automorphism of $\mathfrak t_{\mathbb C}$. For a weight $\beta\in\mathfrak t_{\mathbb C}^*$, define the induced weight $n_\alpha\cdot\beta\in\mathfrak t_{\mathbb C}^*$ by \begin{align*} (n_\alpha\cdot\beta)(X):=\beta(\operatorname{Ad}_{n_\alpha^{-1}}X) \end{align*} for $X\in\mathfrak t_{\mathbb C}$. Let $X\in\mathfrak t_{\mathbb C}$. Since $\alpha(H_\alpha)=2$, there is a unique decomposition \begin{align*} X=X_0+cH_\alpha \end{align*} with $X_0\in\ker(\alpha:\mathfrak t_{\mathbb C}\to\mathbb C)$ and $c\in\mathbb C$, and applying $\alpha$ gives \begin{align*} c=\frac{\alpha(X)}{2}. \end{align*} The centralization of $K_\alpha$ proved above implies that $\operatorname{Ad}_{n_\alpha}$ fixes $\ker\alpha$ pointwise after complexification, and the rank-one computation gives $\operatorname{Ad}_{n_\alpha}(H_\alpha)=-H_\alpha$. The same two assertions hold for $\operatorname{Ad}_{n_\alpha^{-1}}$. Hence \begin{align*} \operatorname{Ad}_{n_\alpha^{-1}}X=X_0-cH_\alpha. \end{align*} Using $X=X_0+cH_\alpha$ and $c=\alpha(X)/2$, this is equivalently \begin{align*} \operatorname{Ad}_{n_\alpha^{-1}}X=X-\alpha(X)H_\alpha. \end{align*} Therefore, for every $\beta\in\mathfrak t_{\mathbb C}^*$, \begin{align*} (n_\alpha\cdot\beta)(X)=\beta(X)-\beta(H_\alpha)\alpha(X). \end{align*} Since this holds for every $X\in\mathfrak t_{\mathbb C}$, we have \begin{align*} n_\alpha\cdot\beta=\beta-\beta(H_\alpha)\alpha=s_\alpha(\beta). \end{align*} In particular, the normalizer action of $n_\alpha$ on $X^*(T)$ is the restriction of $s_\alpha$. [guided] The action on roots is now a direct linear calculation; no root-string theorem is needed. Since $n_\alpha$ normalizes $T$, its adjoint action preserves $\mathfrak t$ and therefore preserves $\mathfrak t_{\mathbb C}$. For a weight $\beta\in\mathfrak t_{\mathbb C}^*$, the induced action is defined by \begin{align*} (n_\alpha\cdot\beta)(X):=\beta(\operatorname{Ad}_{n_\alpha^{-1}}X) \end{align*} for $X\in\mathfrak t_{\mathbb C}$. We compute $\operatorname{Ad}_{n_\alpha^{-1}}X$. Because $\alpha(H_\alpha)=2$, the vector $H_\alpha$ is transverse to $\ker\alpha$. Thus every $X\in\mathfrak t_{\mathbb C}$ has a unique decomposition \begin{align*} X=X_0+cH_\alpha \end{align*} where $X_0\in\ker(\alpha:\mathfrak t_{\mathbb C}\to\mathbb C)$ and $c\in\mathbb C$. Applying $\alpha$ to this equality gives $\alpha(X)=2c$, so \begin{align*} c=\frac{\alpha(X)}{2}. \end{align*} The previous steps determine the adjoint action on both summands. The kernel part is fixed because $K_\alpha$ centralizes $G_\alpha$, and differentiating this pointwise centralization gives that $\operatorname{Ad}_{n_\alpha}$ fixes $\ker\alpha$; after complexification it fixes $\ker(\alpha:\mathfrak t_{\mathbb C}\to\mathbb C)$. The coroot part changes sign because the rank-one Weyl representative satisfies $\operatorname{Ad}_{n_\alpha}(H_\alpha)=-H_\alpha$. The inverse representative has the same action on these two summands. Therefore \begin{align*} \operatorname{Ad}_{n_\alpha^{-1}}X=X_0-cH_\alpha. \end{align*} Substituting $X=X_0+cH_\alpha$ and $c=\alpha(X)/2$ gives \begin{align*} \operatorname{Ad}_{n_\alpha^{-1}}X=X-\alpha(X)H_\alpha. \end{align*} Now apply the definition of the induced action on weights: \begin{align*} (n_\alpha\cdot\beta)(X)=\beta(X)-\beta(H_\alpha)\alpha(X). \end{align*} This identity holds for every $X\in\mathfrak t_{\mathbb C}$, so the two linear functionals are equal: \begin{align*} n_\alpha\cdot\beta=\beta-\beta(H_\alpha)\alpha=s_\alpha(\beta). \end{align*} Thus the normalizer element $n_\alpha$ induces exactly the reflection $s_\alpha$ on characters and on all weights. [/guided] [/step] [step:Conclude that the reflection belongs to the Weyl group and preserves the roots] The coset $n_\alpha T$ is an element of $N_G(T)/T=W(G,T)$ because $n_\alpha\in N_G(T)$. The preceding computation shows that this Weyl group element acts on $X^*(T)$ by the restriction of $s_\alpha$, so $s_\alpha\in W(G,T)$ in the usual faithful realization of the Weyl group on the character lattice. It remains only to record that $s_\alpha$ preserves $\Phi$. Let $\beta\in\Phi$ and let $Y\in\mathfrak g_\beta$ with $Y\ne 0$. For $X\in\mathfrak t_{\mathbb C}$, using the Lie algebra automorphism property of $\operatorname{Ad}_{n_\alpha}$ gives \begin{align*} [X,\operatorname{Ad}_{n_\alpha}Y]=\operatorname{Ad}_{n_\alpha}([\operatorname{Ad}_{n_\alpha^{-1}}X,Y]). \end{align*} Since $Y\in\mathfrak g_\beta$, the right-hand side equals \begin{align*} \beta(\operatorname{Ad}_{n_\alpha^{-1}}X)\operatorname{Ad}_{n_\alpha}Y. \end{align*} Therefore $\operatorname{Ad}_{n_\alpha}Y$ lies in the root space $\mathfrak g_{n_\alpha\cdot\beta}$. Because $\operatorname{Ad}_{n_\alpha}Y\ne 0$, this root space is nonzero, so $n_\alpha\cdot\beta\in\Phi$. By the previous step $n_\alpha\cdot\beta=s_\alpha(\beta)$, hence $s_\alpha(\Phi)\subset\Phi$. Applying the same argument to $n_\alpha^{-1}$ gives the reverse inclusion. Thus \begin{align*} s_\alpha(\Phi)=\Phi. \end{align*} This proves the theorem. [guided] We have constructed an element $n_\alpha\in N_G(T)$ and computed its action. Since $n_\alpha$ normalizes $T$, its coset \begin{align*} n_\alpha T\in N_G(T)/T=W(G,T) \end{align*} is a Weyl group element. The previous step proved that this element acts on every weight by \begin{align*} \beta\mapsto\beta-\beta(H_\alpha)\alpha. \end{align*} This is exactly the reflection $s_\alpha$. Hence $s_\alpha$ is represented by an element of $N_G(T)$, so $s_\alpha\in W(G,T)$ under the usual realization of the Weyl group on characters. We finally prove that this reflection carries roots to roots. Let $\beta\in\Phi$, and choose a nonzero vector $Y\in\mathfrak g_\beta$. Because $n_\alpha$ normalizes $T$, the operator $\operatorname{Ad}_{n_\alpha}$ preserves $\mathfrak t_{\mathbb C}$. For $X\in\mathfrak t_{\mathbb C}$, the Lie algebra automorphism identity gives \begin{align*} [X,\operatorname{Ad}_{n_\alpha}Y]=\operatorname{Ad}_{n_\alpha}([\operatorname{Ad}_{n_\alpha^{-1}}X,Y]). \end{align*} Since $Y$ is a $\beta$-root vector, the bracket inside the right-hand side is \begin{align*} [\operatorname{Ad}_{n_\alpha^{-1}}X,Y]=\beta(\operatorname{Ad}_{n_\alpha^{-1}}X)Y. \end{align*} Therefore \begin{align*} [X,\operatorname{Ad}_{n_\alpha}Y]=\beta(\operatorname{Ad}_{n_\alpha^{-1}}X)\operatorname{Ad}_{n_\alpha}Y. \end{align*} By definition of the induced action, the scalar multiplying $\operatorname{Ad}_{n_\alpha}Y$ is $(n_\alpha\cdot\beta)(X)$. Thus $\operatorname{Ad}_{n_\alpha}Y$ is a nonzero vector in the root space $\mathfrak g_{n_\alpha\cdot\beta}$, so $n_\alpha\cdot\beta\in\Phi$. Since $n_\alpha\cdot\beta=s_\alpha(\beta)$, this proves $s_\alpha(\Phi)\subset\Phi$. Applying the same argument to $n_\alpha^{-1}$ gives the reverse inclusion, and hence \begin{align*} s_\alpha(\Phi)=\Phi. \end{align*} This completes the proof. [/guided] [/step]