Let $\mathfrak g$ be a complex Lie algebra. If $\mathfrak g$ is semisimple then its real forms are completely understood. In particular there is a compact real form of $\mathfrak g$, which is unique up to inner automorphism.

Suppose $\mathfrak g$ is not semisimple and write $\mathfrak g = \mathfrak r \ltimes \mathfrak s$ using the Levi-Malcev theorem, where $\mathfrak r$ is the radical of $\mathfrak g$ and $\mathfrak s$ is a maximal semisimple subalgebra. Assume that $\mathfrak g$ has a real form $\mathfrak g_0$. My question is:

Can $\mathfrak g_0$ be taken so that $\mathfrak g_0 = \mathfrak r_0 \ltimes \mathfrak s_0$, $\mathfrak r_0$ the radical of $\mathfrak g_0$ and $\mathfrak s_0$ a maximal semisimple subalgebra of $\mathfrak g_0$ which is a compact real form of $\mathfrak s$?

Let $\mathfrak g$ be a complex Lie algebra. If $\mathfrak g$ is semisimple then its real forms are completely understood. In particular there is a compact real form of $\mathfrak g$, which is unique up to inner automorphism.

Suppose $\mathfrak g$ is not semisimple and write $\mathfrak g = \mathfrak r \ltimes \mathfrak s$ using the Levi-Malcev theorem, where $\mathfrak r$ is the radical of $\mathfrak g$ and $\mathfrak s$ is a maximal semisimple subalgebra. Assume that $\mathfrak g$ has a real form. My question is:

Can a real form $\mathfrak g_0$ of $\mathfrak g$ be taken so that $\mathfrak g_0 = \mathfrak r_0 \ltimes \mathfrak s_0$, $\mathfrak r_0$ the radical of $\mathfrak g_0$ and $\mathfrak s_0$ a maximal semisimple subalgebra of $\mathfrak g_0$ which is a compact real form of $\mathfrak s$?