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All Lie groups here are assumed to be real. Let $M=G/K$ be a symmetric space of noncompact type and $H \subset G$. $H$ acts on $G/K$ by left-multiplication. If $H$ is a parabolic subgroup, then $H$ acts transitively on $M$.

Question: If $H$ acts transitively on $M$, is $H$ parabolic in $G$?

It is known that any proper subgroup $H \subset G$ is contained in a maximal reductive subgroup of $G$ or a maximal parabolic subgroup of $G$. All parabolic subgroups of $G$ act transitively on $M$. Reductive subgroups never act transitively on $M$, so their subgroups also do not act transitively.

That said, can there be a subgroup of a maximal parabolic which is not parabolic itself, but still acts transitively on $G/K$? I suspect the answer is yes but couldn't come up with an example.

user538044
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The answer to the question is: No. There exist groups acting transitively on $G/K$ that are not parabolic. As an example take a space $G/K$ with $$\mathfrak{k}_0 := \{ X \in \mathfrak{k} : [X,\mathfrak{a}]=0 \} \neq \{ 0\},$$ for example $G=SO^0_{r,r+n}$, $K=SO_r SO_{r+n}$ with $r \geq 2$, $n \geq 2$. (This has $\mathfrak{k}_0=\mathfrak{so}_n$ as listed in Berndt, Console, Olmos: Submanifolds and Holonomy, p. 339) Then the minimal parabolic of $\mathfrak{g}$ is $\mathfrak{q}=\mathfrak{k}_0 \oplus \mathfrak{a} \oplus \mathfrak{n}$ with corresponding Lie group $Q$. $AN$ acts transitively on $M$ but is not parabolic, as it does not contain a copy of the minimal parabolic.

Moreover: It seems that all groups acting transitively on $M$ are given as the connected subgroup with Lie algebra $\hat{\mathfrak{k}} \oplus \mathfrak{a} \oplus \mathfrak{n}$ where $\hat{\mathfrak{k}}$ is any subalgebra of $\mathfrak{k}$. I heard this fact but don't know of a proof.

user538044
  • 523