Exercise 12 of Chapter 7 in Brian Hall's Lie groups, Lie algebras and their representation. I am stuck on the following exercise in Hall's book:

Let $\mathfrak{g}$ be a complex simple Lie algebra with complex structure denoted by J. Let $\mathfrak{g}_\mathbb{R}$ denote the Lie algebra $\mathfrak{g}$ viewed as a real Lie algebra with twice the dimension. Now let $\mathfrak{g}'$ be the complexification of $\mathfrak{g}_\mathbb{R}$ with the complex structure on $\mathfrak{g}'$ denoted by $i$.
(a) Show that $\mathfrak{g}'$ decomposes as a Lie algebra direct sum $\mathfrak{g}'=\mathfrak{g}_1\oplus\mathfrak{g}_2$ with $\mathfrak{g}_1\cong\mathfrak{g}_2\cong\mathfrak{g}$.
(b) Show that $\mathfrak{g}_\mathbb{R}$ is simple as a real Lie algebra.

For part (a), following the hint I have managed to show that $\mathfrak{g}'=\mathfrak{g}_1\oplus\mathfrak{g}_2$ where $\mathfrak{g}_1=\{X+iJX|X\in\mathfrak{g}_\mathbb{R}\}$ and $\mathfrak{g}_2=\{X-iJX|X\in\mathfrak{g}_\mathbb{R}\}$. Also one can show that the map $\pi:\mathfrak{g}_2\to\mathfrak{g}$ defined by $\pi(X-iJX)=2X$ is an isomorphism. However, I haven't managed to find an isomorphism between $\mathfrak{g}_1$ and $\mathfrak{g}$. The map defined by $\pi(X+iJX)=2X$ is not complex linear since $\pi(iX-JX)=-2JX\neq J\pi(X+iJX)$.
For part (b), I tried to prove it by contradiction: Assume $\mathfrak{h}$ is a non-trivial ideal of $\mathfrak{g}_\mathbb{R}$. I attempted to construct some non-trivial ideal of $\mathfrak{g}_1$ using the complexification of $\mathfrak{h}$. But I was not able to proceed further to arrive at a contradiction.
Any help or hints are appreciated.
 A: For part (a), it is easy to see that $\mathfrak{g}_1\cong\overline{\mathfrak{g}}$, where $\overline{\mathfrak{g}}$ is the complex conjugate Lie algebra of $\mathfrak{g}$. On the other hand, $\mathfrak{g}_2\cong\mathfrak{g}$. In general, $\overline{\mathfrak{g}}\cong\mathfrak{g}$ iff $\mathfrak{g}$ admits a real form. As Torsten pointed out in the comment below, $\mathfrak{g}$ has a real form, which implies that $\mathfrak{g}_1\cong\overline{\mathfrak{g}}\cong\mathfrak{g}\cong\mathfrak{g}_2$. Still, it would be better if we can show this directly.
For part (b), we can prove it in the following way: Let $\mathfrak{h}$ be a non-trivial ideal of $\mathfrak{g}_\mathbb{R}$. From my other question, we know that $\mathfrak{g}_\mathbb{R}$ is semisimple, then $\mathfrak{h}$ is also semisimple. Now consider the derived algebra $[\mathfrak{h},\mathfrak{h}]$. Since $\mathfrak{h}$ is semisimple, then $[\mathfrak{h},\mathfrak{h}]=\mathfrak{h}$. It follows that $J\mathfrak{h}=J[\mathfrak{h},\mathfrak{h}]=[J\mathfrak{h},\mathfrak{h}]\subset\mathfrak{h}$. So $\mathfrak{h}$ is also a complex subspace of $\mathfrak{g}$, and in particular an ideal of $\mathfrak{g}$. Since $\mathfrak{g}$ is simple, it follows that $\mathfrak{h}=\mathfrak{g}$. So $\mathfrak{g}_\mathbb{R}$ has no nontrivial ideal, thus simple.
