Semisimple algebras of simple Lie algebras and their quotients

Let $$\mathfrak s$$ be a complex semisimple Lie algebra, then $$\dim _\mathbb C\mathfrak s\geq 3$$. But, however, is it possible for $$\mathfrak s$$ to have a semisimple Lie subalgebra $$\mathfrak h$$ such that $$\dim_\mathbb C\mathfrak s/\mathfrak h=2$$?

On Lie groups level, do homogeneous spaces of the form $$\mathrm{SL}(n,\mathbb C)/H$$ where $$H$$ is a closed complex semisimple subgroup of $$\mathrm{SL}(n,\mathbb C)$$ have special properties, for example parallelizable, etc?

No, it's not. Indeed it's known that no simple Lie algebra of rang $$\ge 3$$ has a proper subalgebra of codimension 2; in rank 2, the only proper subalgebras of codimension $$\le 2$$ are parabolic of codimension 2, and in rank 1 the only proper subalgebras of codimension $$\le 2$$ are solvable (either parabolic or abelian).
So if $$\mathfrak{g}$$ is a semisimple Lie algebra and $$\mathfrak{h}$$ is a proper subalgebra of codimension $$\le 2$$ then there exists a direct decomposition $$\mathfrak{g}=\mathfrak{g}_1\times\mathfrak{g}_2$$ and $$\mathfrak{h}=\mathfrak{g}_1\times\mathfrak{h}_2$$ with either $$\mathfrak{h}_2$$ parabolic in $$\mathfrak{g}_2$$ (hence not semisimple), or $$\mathfrak{g}_2$$ has dimension 3 and $$\mathfrak{h}_2$$ is abelian (hence not semisimple).
Codimension 3 is possible, namely in $$\mathfrak{sl}_2^2$$.
• Do you know a reference for the fact that: A simple Lie algebra of rank $n$ doesn't have subalgebras of codimension less than $n$? Thanks! – user328669 Dec 20 '19 at 4:17