[proofplan] We prove that the reflection $s_\alpha$ carries each root $\beta$ to another root by reducing the computation to the $\alpha$-string through $\beta$. The two roots parallel to $\alpha$ are handled directly. For every other root $\beta$, the [Root String Theorem](/theorems/4693) gives a finite string $\beta - p\alpha, \ldots, \beta + q\alpha$ and identifies $\langle \beta,\alpha^\vee\rangle$ with $p-q$; substituting this identity shows that $s_\alpha(\beta)$ is exactly the member of the same string at offset $q-p$. [/proofplan]

[step:Handle the roots equal to $\alpha$ and $-\alpha$] Fix $\alpha \in \Phi$. Since $\Phi$ is the [root system of a semisimple Lie algebra](/theorems/4697), $\Phi$ is closed under negation, so $-\alpha \in \Phi$. For $\beta = \alpha$, the defining property of the coroot gives $\langle \alpha,\alpha^\vee\rangle = 2$, hence \begin{align*} s_\alpha(\alpha) &= \alpha - \langle \alpha,\alpha^\vee\rangle \alpha \\ &= \alpha - 2\alpha \\ &= -\alpha \in \Phi . \end{align*} For $\beta = -\alpha$, bilinearity of the pairing gives $\langle -\alpha,\alpha^\vee\rangle = -2$, hence \begin{align*} s_\alpha(-\alpha) &= -\alpha - \langle -\alpha,\alpha^\vee\rangle \alpha \\ &= -\alpha + 2\alpha \\ &= \alpha \in \Phi . \end{align*} Thus $s_\alpha(\beta) \in \Phi$ whenever $\beta \in \{\alpha,-\alpha\}$. [/step]

[step:Use the $\alpha$-string through a nonparallel root]Let $\beta \in \Phi$ satisfy $\beta \notin \{\alpha,-\alpha\}$. Define the $\alpha$-string through $\beta$ to be the set \begin{align*} S_{\alpha,\beta} := \Phi \cap \{\beta + k\alpha : k \in \mathbb{Z}\}. \end{align*} By the Root String Theorem (citing a result not yet in the wiki: Root String Theorem), there exist integers $p,q \in \mathbb{Z}_{\ge 0}$ such that \begin{align*} S_{\alpha,\beta} = \{\beta + k\alpha : k \in \mathbb{Z},\ -p \le k \le q\}, \end{align*} and \begin{align*} \langle \beta,\alpha^\vee\rangle = p-q . \end{align*} Substituting this identity into the definition of $s_\alpha$ gives \begin{align*} s_\alpha(\beta) &= \beta - \langle \beta,\alpha^\vee\rangle \alpha \\ &= \beta - (p-q)\alpha \\ &= \beta + (q-p)\alpha . \end{align*} The integer $q-p$ lies in the interval $[-p,q]$, because $q-p \ge -p$ is equivalent to $q \ge 0$, and $q-p \le q$ is equivalent to $-p \le 0$. Therefore $\beta + (q-p)\alpha \in S_{\alpha,\beta} \subset \Phi$. Hence $s_\alpha(\beta) \in \Phi$.[/step]

[guided]Let $\beta \in \Phi$ be a root which is not equal to $\alpha$ or $-\alpha$. The useful object is the full line of possible roots obtained from $\beta$ by adding integer multiples of $\alpha$. Define \begin{align*} S_{\alpha,\beta} := \Phi \cap \{\beta + k\alpha : k \in \mathbb{Z}\}. \end{align*} This is the $\alpha$-string through $\beta$. We now invoke the Root String Theorem (citing a result not yet in the wiki: Root String Theorem). In the present setting, its hypotheses are satisfied because $\Phi$ is the root system of a finite-dimensional complex semisimple Lie algebra, $\alpha,\beta \in \Phi$, and $\beta \notin \{\alpha,-\alpha\}$. The theorem gives integers $p,q \in \mathbb{Z}_{\ge 0}$ such that the entire $\alpha$-string through $\beta$ is exactly \begin{align*} S_{\alpha,\beta} = \{\beta + k\alpha : k \in \mathbb{Z},\ -p \le k \le q\}, \end{align*} and it also gives the numerical identity \begin{align*} \langle \beta,\alpha^\vee\rangle = p-q . \end{align*} This identity is precisely what converts the reflection formula into an element of the known string: \begin{align*} s_\alpha(\beta) &= \beta - \langle \beta,\alpha^\vee\rangle \alpha \\ &= \beta - (p-q)\alpha \\ &= \beta + (q-p)\alpha . \end{align*} It remains only to check that the offset $q-p$ is one of the allowed offsets in the string. Since $p,q \ge 0$, we have \begin{align*} -p \le q-p \le q . \end{align*} Therefore the root $\beta + (q-p)\alpha$ belongs to $S_{\alpha,\beta}$. Since $S_{\alpha,\beta} \subset \Phi$ by definition, we conclude that $s_\alpha(\beta) \in \Phi$.[/guided]

[step:Conclude that the reflection preserves the root system] For the fixed root $\alpha \in \Phi$, the first step proves $s_\alpha(\beta) \in \Phi$ when $\beta = \alpha$ or $\beta = -\alpha$, and the second step proves $s_\alpha(\beta) \in \Phi$ for every remaining root $\beta \in \Phi$. Hence \begin{align*} s_\alpha(\Phi) \subset \Phi . \end{align*} Equivalently, for every $\beta \in \Phi$, the reflected functional \begin{align*} s_\alpha(\beta)=\beta-\langle \beta,\alpha^\vee\rangle\alpha \end{align*} is again a root. This proves the theorem. [/step]