How is the Lie algebra $\mathfrak{sl}_{2}(\mathbb{C})$ generated by the one element $(E_{11} - E_{22})$?

Question

In the article Classification of simple complex Lie algebras, 2nd paragraph of chapter 7 (p.15, bottom), the author considers the basis $$ x = E_{21} := \left[ \begin{array}{ll} 0 & 0\\ 1 & 0\\ \end{array} \right], \quad y = E_{12} := \left[ \begin{array}{ll} 0 & 1\\ 0 & 0\\ \end{array} \right], \quad h = \left[ \begin{array}{lr} 1 & 0\\ 0 & -1\\ \end{array} \right] $$ for the Lie algebra $\mathfrak{sl}_{2}(\mathbb{C})$. The following relations hold: $$ [h, x] = 2x, \quad [h, y] = -2y, \quad [x, y] = h. $$ The author asserts that $h$ alone generates the entire Lie algebra. I have tried to carry out the generation and got $$ hh = I, \quad {1 \over 2}(hh + h) = E_{11} := \left[ \begin{array}{ll} 1 & 0\\ 0 & 0\\ \end{array} \right], \quad {1 \over 2}(hh - h) = E_{22} := \left[ \begin{array}{ll} 0 & 0\\ 0 & 1\\ \end{array} \right], $$ the latter two expressions not even being Lie brackets. But I am not seeing how to generate $x$ or $y$ from $h$ exclusively. What am I missing? Sources are appreciated as well.

Answer 1

It's not true. Each element generates a $1$-dimensional abelian Lie algebra. However, if you read the proof, he clearly means the ideal generated by $h$, not the subalgebra. Indeed, he is trying to prove that $\mathfrak{sl}_2(\mathbb{C})$ is simple, so he wants ideals.

Also, $[h,h]=0$, not $[h,h]=I$.