[proofplan]
Fix a root $\alpha$ and choose a non-zero vector in the root space $\mathfrak{g}_{\alpha}$. The Killing form pairs $\mathfrak{g}_{\alpha}$ nondegenerately with $\mathfrak{g}_{-\alpha}$, so we choose an opposite-root vector with non-zero pairing. Invariance of the Killing form identifies their bracket as a non-zero scalar multiple of $t_{\alpha}$, and rescaling makes the bracket equal to $h_{\alpha}$. The root-space definitions and the normalization of $h_{\alpha}$ then give the two remaining $\mathfrak{sl}_2$ bracket relations.
[/proofplan]
[step:Choose opposite-root vectors with non-zero Killing pairing]
Fix $\alpha\in\Phi$. Since $\alpha$ is a root, the root space $\mathfrak{g}_{\alpha}$ is non-zero. Choose
\begin{align*}
x\in\mathfrak{g}_{\alpha}\setminus\{0\}.
\end{align*}
By the [Orthogonality Of Root Spaces](/theorems/TEMP-12) and the [Pairing Of Opposite Root Spaces](/theorems/TEMP-13) for the root-space decomposition of a semisimple Lie algebra, the restriction of the Killing form to
\begin{align*}
\mathfrak{g}_{\alpha}\times \mathfrak{g}_{-\alpha}
\end{align*}
is nondegenerate. The hypotheses apply because $\mathfrak{g}$ is semisimple, $\mathfrak{h}$ is the Cartan subalgebra used to define $\Phi$, and $\alpha\in\Phi$. Hence there exists
\begin{align*}
y\in\mathfrak{g}_{-\alpha}
\end{align*}
such that $\kappa(x,y)\ne 0$.
[/step]
[step:Compute the bracket $[x,y]$ by Killing form invariance]
We first show that $[x,y]=\kappa(x,y)t_{\alpha}$. Define the zero-weight space for the adjoint action of $\mathfrak{h}$ by
\begin{align*}
\mathfrak{g}_{0}:=\{z\in\mathfrak{g}: [h,z]=0\text{ for every }h\in\mathfrak{h}\}.
\end{align*}
By the [Root Space Bracket Rule](/theorems/TEMP-11),
\begin{align*}
[\mathfrak{g}_{\alpha},\mathfrak{g}_{-\alpha}]\subseteq \mathfrak{g}_{\alpha+(-\alpha)}=\mathfrak{g}_{0},
\end{align*}
and by the [Root Space Decomposition](/theorems/TEMP-10) attached to $\mathfrak{h}$ one has $\mathfrak{g}_{0}=\mathfrak{h}$. These hypotheses apply because $\alpha$ and $-\alpha$ are roots and $\mathfrak{h}$ is the Cartan subalgebra defining the root spaces. Thus $[x,y]\in\mathfrak{h}$. For every $h\in\mathfrak{h}$, the [Invariance of the Killing Form](/theorems/3808) and symmetry of the Killing form give
\begin{align*}
\kappa([x,y],h)
&=\kappa(x,[y,h])\\
&=\kappa(x,\alpha(h)y)\\
&=\alpha(h)\kappa(x,y)\\
&=\kappa(\kappa(x,y)t_{\alpha},h).
\end{align*}
The restriction $\kappa|_{\mathfrak{h}\times\mathfrak{h}}$ is nondegenerate by the Cartan-subalgebra nondegeneracy consequence of the [Root Space Decomposition](/theorems/TEMP-10) and the [Cartan Semisimplicity Criterion](/theorems/3813); its hypotheses apply because $\mathfrak{g}$ is semisimple and $\mathfrak{h}$ is the Cartan subalgebra in the root-space decomposition. Therefore equality of pairings against every $h\in\mathfrak{h}$ implies
\begin{align*}
[x,y]=\kappa(x,y)t_{\alpha}.
\end{align*}
[guided]
Recall the data chosen in the previous step: $x\in\mathfrak{g}_{\alpha}\setminus\{0\}$ and $y\in\mathfrak{g}_{-\alpha}$ with $\kappa(x,y)\ne 0$. We want to identify the bracket $[x,y]$ as an element of $\mathfrak{h}$. Define the zero-weight space by
\begin{align*}
\mathfrak{g}_{0}:=\{z\in\mathfrak{g}: [h,z]=0\text{ for every }h\in\mathfrak{h}\}.
\end{align*}
Because $x\in\mathfrak{g}_{\alpha}$ and $y\in\mathfrak{g}_{-\alpha}$, the [Root Space Bracket Rule](/theorems/TEMP-11) applies to the pair of roots $\alpha$ and $-\alpha$ and gives
\begin{align*}
[x,y]\in \mathfrak{g}_{\alpha+(-\alpha)}=\mathfrak{g}_{0}.
\end{align*}
For the [Root Space Decomposition](/theorems/TEMP-10) attached to the Cartan subalgebra $\mathfrak{h}$, the zero-weight space is exactly $\mathfrak{h}$. Thus $[x,y]\in\mathfrak{h}$.
To determine which element of $\mathfrak{h}$ it is, pair it with an arbitrary $h\in\mathfrak{h}$. The [Invariance of the Killing Form](/theorems/3808) says that
\begin{align*}
\kappa([a,b],c)=\kappa(a,[b,c])
\end{align*}
for all $a,b,c\in\mathfrak{g}$. Applying this with $a=x$, $b=y$, and $c=h$, we obtain
\begin{align*}
\kappa([x,y],h)=\kappa(x,[y,h]).
\end{align*}
Since $y\in\mathfrak{g}_{-\alpha}$, we have $[h,y]=-\alpha(h)y$, hence $[y,h]=\alpha(h)y$. Therefore
\begin{align*}
\kappa([x,y],h)
&=\kappa(x,\alpha(h)y)\\
&=\alpha(h)\kappa(x,y).
\end{align*}
By definition of $t_{\alpha}$, $\kappa(t_{\alpha},h)=\alpha(h)$ for every $h\in\mathfrak{h}$, so
\begin{align*}
\alpha(h)\kappa(x,y)=\kappa(\kappa(x,y)t_{\alpha},h).
\end{align*}
Thus $[x,y]$ and $\kappa(x,y)t_{\alpha}$ have the same Killing pairing with every element of $\mathfrak{h}$. The Cartan-subalgebra nondegeneracy consequence of the [Root Space Decomposition](/theorems/TEMP-10) and the [Cartan Semisimplicity Criterion](/theorems/3813) says that $\kappa|_{\mathfrak{h}\times\mathfrak{h}}$ is nondegenerate; its hypotheses apply here because $\mathfrak{g}$ is semisimple and $\mathfrak{h}$ is precisely the Cartan subalgebra used in the root-space decomposition. Hence the difference $[x,y]-\kappa(x,y)t_{\alpha}\in\mathfrak{h}$ pairs to zero with all of $\mathfrak{h}$, so the difference is zero:
\begin{align*}
[x,y]=\kappa(x,y)t_{\alpha}.
\end{align*}
[/guided]
[/step]
[step:Rescale the vectors so their bracket is $h_{\alpha}$]
We have $\kappa(x,y)\ne 0$ by construction. Also $\kappa(t_{\alpha},t_{\alpha})\ne 0$ by the root non-isotropy consequence of the [Pairing Of Opposite Root Spaces](/theorems/TEMP-13); equivalently, the coroot element $h_{\alpha}=2t_{\alpha}/\kappa(t_{\alpha},t_{\alpha})$ is well-defined. Therefore the following rescaling is legitimate. Define
\begin{align*}
e_{\alpha}:=\frac{2}{\kappa(x,y)\kappa(t_{\alpha},t_{\alpha})}x,
\qquad
f_{\alpha}:=y.
\end{align*}
Since $\mathfrak{g}_{\alpha}$ and $\mathfrak{g}_{-\alpha}$ are vector spaces, $e_{\alpha}\in\mathfrak{g}_{\alpha}$ and $f_{\alpha}\in\mathfrak{g}_{-\alpha}$. Using bilinearity of the Lie bracket and the bracket computation above,
\begin{align*}
[e_{\alpha},f_{\alpha}]
&=\frac{2}{\kappa(x,y)\kappa(t_{\alpha},t_{\alpha})}[x,y]\\
&=\frac{2}{\kappa(x,y)\kappa(t_{\alpha},t_{\alpha})}\kappa(x,y)t_{\alpha}\\
&=\frac{2t_{\alpha}}{\kappa(t_{\alpha},t_{\alpha})}\\
&=h_{\alpha}.
\end{align*}
[/step]
[step:Use the root-space action of $h_{\alpha}$ to obtain the remaining brackets]
Because $e_{\alpha}\in\mathfrak{g}_{\alpha}$, the definition of $\mathfrak{g}_{\alpha}$ gives
\begin{align*}
[h_{\alpha},e_{\alpha}]=\alpha(h_{\alpha})e_{\alpha}.
\end{align*}
By the definition of $h_{\alpha}$ and $t_{\alpha}$,
\begin{align*}
\alpha(h_{\alpha})
&=\kappa(t_{\alpha},h_{\alpha})\\
&=\kappa\left(t_{\alpha},\frac{2t_{\alpha}}{\kappa(t_{\alpha},t_{\alpha})}\right)\\
&=2.
\end{align*}
Hence
\begin{align*}
[h_{\alpha},e_{\alpha}]=2e_{\alpha}.
\end{align*}
Similarly, since $f_{\alpha}\in\mathfrak{g}_{-\alpha}$,
\begin{align*}
[h_{\alpha},f_{\alpha}]
=(-\alpha)(h_{\alpha})f_{\alpha}
=-2f_{\alpha}.
\end{align*}
[guided]
The remaining two brackets come directly from the defining property of root spaces. Since $e_{\alpha}\in\mathfrak{g}_{\alpha}$, every $h\in\mathfrak{h}$ acts on $e_{\alpha}$ by the scalar $\alpha(h)$:
\begin{align*}
[h,e_{\alpha}]=\alpha(h)e_{\alpha}.
\end{align*}
We apply this with $h=h_{\alpha}$. This gives
\begin{align*}
[h_{\alpha},e_{\alpha}]=\alpha(h_{\alpha})e_{\alpha}.
\end{align*}
Now compute the scalar. By definition, $t_{\alpha}$ represents the functional $\alpha$ under the Killing form on $\mathfrak{h}$, so $\alpha(h)=\kappa(t_{\alpha},h)$ for every $h\in\mathfrak{h}$. Therefore
\begin{align*}
\alpha(h_{\alpha})
&=\kappa(t_{\alpha},h_{\alpha})\\
&=\kappa\left(t_{\alpha},\frac{2t_{\alpha}}{\kappa(t_{\alpha},t_{\alpha})}\right)\\
&=2.
\end{align*}
Thus
\begin{align*}
[h_{\alpha},e_{\alpha}]=2e_{\alpha}.
\end{align*}
The same argument applies to the opposite root. Since $f_{\alpha}\in\mathfrak{g}_{-\alpha}$, every $h\in\mathfrak{h}$ acts on $f_{\alpha}$ by the scalar $-\alpha(h)$. Taking again $h=h_{\alpha}$ gives
\begin{align*}
[h_{\alpha},f_{\alpha}]
=(-\alpha)(h_{\alpha})f_{\alpha}
=-2f_{\alpha}.
\end{align*}
[/guided]
[/step]
[step:Identify the generated subalgebra with $\mathfrak{sl}_2(k)$]
Let
\begin{align*}
\mathfrak{s}_{\alpha}:=\operatorname{span}_k\{e_{\alpha},h_{\alpha},f_{\alpha}\}\subseteq\mathfrak{g}.
\end{align*}
The three bracket identities
\begin{align*}
[e_{\alpha},f_{\alpha}]=h_{\alpha},
\qquad
[h_{\alpha},e_{\alpha}]=2e_{\alpha},
\qquad
[h_{\alpha},f_{\alpha}]=-2f_{\alpha}
\end{align*}
show that $\mathfrak{s}_{\alpha}$ is closed under the Lie bracket, hence is a Lie subalgebra of $\mathfrak{g}$.
Define a $k$-[linear map](/page/Linear%20Map)
\begin{align*}
\varphi:\mathfrak{sl}_2(k)&\to\mathfrak{s}_{\alpha}\\
\begin{pmatrix}0&1\\0&0\end{pmatrix}&\mapsto e_{\alpha}\\
\begin{pmatrix}1&0\\0&-1\end{pmatrix}&\mapsto h_{\alpha}\\
\begin{pmatrix}0&0\\1&0\end{pmatrix}&\mapsto f_{\alpha}.
\end{align*}
The standard basis of $\mathfrak{sl}_2(k)$ satisfies
\begin{align*}
[h,e]=2e,
\qquad
[h,f]=-2f,
\qquad
[e,f]=h.
\end{align*}
The defining bracket relations for $e_{\alpha},h_{\alpha},f_{\alpha}$ therefore imply that $\varphi$ preserves brackets.
It remains to check bijectivity. The map $\varphi$ is surjective by the definition of $\mathfrak{s}_{\alpha}$ as the span of $e_{\alpha},h_{\alpha},f_{\alpha}$. For injectivity, suppose $a,b,c\in k$ satisfy
\begin{align*}
ae_{\alpha}+bh_{\alpha}+cf_{\alpha}=0.
\end{align*}
Here $e_{\alpha}\in\mathfrak{g}_{\alpha}$ is non-zero because it is a non-zero scalar multiple of $x$, $f_{\alpha}\in\mathfrak{g}_{-\alpha}$ is non-zero because $\kappa(x,y)\ne0$, and $h_{\alpha}\in\mathfrak{h}=\mathfrak{g}_{0}$ is non-zero because $\kappa(t_{\alpha},t_{\alpha})\ne0$. The root-space decomposition is direct, so the three summands lie in the direct sum
\begin{align*}
\mathfrak{g}_{\alpha}\oplus\mathfrak{g}_{0}\oplus\mathfrak{g}_{-\alpha}.
\end{align*}
Therefore $ae_{\alpha}=0$, $bh_{\alpha}=0$, and $cf_{\alpha}=0$, hence $a=b=c=0$. Thus $e_{\alpha},h_{\alpha},f_{\alpha}$ are linearly independent, so $\varphi$ is injective. Hence $\varphi$ is a Lie algebra isomorphism, and
\begin{align*}
\mathfrak{s}_{\alpha}\cong \mathfrak{sl}_2(k).
\end{align*}
This proves the theorem.
[/step]
