I want to show:
If $\mathfrak{g}$ is a Lie algebra that has an abelian subalgebra $\mathfrak{h}$ such that $\mathfrak{g}$ has a Cartan decomposition $\mathfrak{g}=\mathfrak{h}\oplus(\bigoplus_{\alpha \in \mathfrak{h}^*} \mathfrak{g}_\alpha)$, then $\mathfrak{g}$ is semisimple and $\mathfrak{h}$ is a Cartan subalgebra.
$\mathfrak{g}_\alpha$ denote the root space; for any $H\in \mathfrak{h}$ and $X\in \mathfrak{g}_\alpha$, $\text{ad}(H)(X)=\alpha(H)\cdot X$. If $R$ denotes the set of roots, you can also verify that:
- $\dim \mathfrak{g}_\alpha=1$
- $R$ generates a lattice $\subset \mathfrak{h}^*$ with rank $=\dim \mathfrak{h}$.
- If $\alpha \in R$, then $-\alpha \in R$.
I also know $[\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}]\neq 0$ and $[[\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}],\mathfrak{g}_\alpha]\neq 0$ for $\alpha$ being a root.
For the first part I am not sure how to show $\mathfrak{g}$ is semisimple. Could anybody walk me through this? For the second part I need $\mathfrak{h}$ is a maximal toral abelian subalgebra. Clearly it is maximal abelian because $[\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}]\neq 0$ for all the other root spaces. How do I show it is a maximal toral subalgebra?