I would like to show that $M_\alpha = \langle x, y, [x, y] \rangle_\mathbb{C}$ is a subalgebra of a semisimple complex Lie algebra $L$ such that $\dim_\mathbb{C} M_\alpha = 3$. $H$ is the Cartan subalgebra of $L$, $\Phi$ is the set of roots of $L$ relative to $H$, $\alpha \in \Phi$, $x \in L_\alpha$, $y \in L_{-\alpha}$ such that $x \neq 0$, $y \neq 0$ and $[x, y] \neq 0$.
It is obvious that $M_\alpha \subset L$. I have taken two elements of $M_\alpha$, $z_1 = a_1x + b_1y + c_1[x, y]$ and $z_2 = a_2x + b_2y + c_2[x, y]$ (for $a_i, b_i, c_i \in \mathbb{C}$) and tried to calculate $[z_1, z_2]$ to show it lies in $M_\alpha$, but I am stuck at this step:
$[z_1, z_2] = ... = (a_1b_2 - b_1a_2)[x, y] + (a_1c_2 - c_1a_2)[x, [x,y]] + (b_1c_2 - c_1b_2)[y, [x,y]]$.
Clearly the first term is okay, but I'm not sure how to show $[x, [x,y]]$ and $[y, [x,y]]$ both end up as some multiple of $x$, $y$ and/or $[x, y]$ to finish the proof that $[z_1, z_2] \in M_\alpha$.
As for showing the dimension is three, I assume this will be showing that $x \neq y \neq [x, y]$ and thus there are three linearly independent elements in the basis of $M_\alpha$ over $\mathbb{C}$, but I'm not entirely sure if that's correct.
Any help is appreciated, thanks.
Okay so I have an answer to this for anyone who was wondering.
As $x \in L_\alpha$, $y \in L_{-\alpha}$ we know $[x, y] \in H$ (as $[L_\alpha, L_{-\alpha}] \subset L_{\alpha + -\alpha} \subset H$).
So from the definition of $L_\alpha$, $[x, [x, y]] = \alpha([x, y]) x$ and similarly for $y$.
$\alpha([x, y]) \in \mathbb{C}$ and so the whole of $[z_1, z_2] \in M_\alpha$.
For the last part I'm not sure if this justification is entirely valid but as $x \in L_\alpha$, $y \in L_{-\alpha}$ and $[x, y] \in H$ we get that they are pairwise not equal. As none of these three elements are zero, they are all in the basis of $M_\alpha$ over $\mathbb{C}$ so the dimension is 3.