A finiteness property for semi-simple algebraic groups Let $G$ be a semi-simple algebraic group over a field $K$, I am considering a question about whether there exists a finite set of semi-simple $K$-subgroups, say $H_1,...,H_r$, such that for any  semi-simple $K$-subgroup $H\subset G$, $H$ is conjugate to one of the $H_i$ by an element in $G(K)$?
I know the answer is "no" for $K=\mathbb{Q}$. In fact, for any quaternion algebra $\mathcal{D}$ over $\mathbb{Q}$, the group $\mathrm{SL}(\mathcal{D})$ can be embedded in $\mathrm{SL}_4(\mathbb{Q})$. As $\mathcal{D}$ varies, we got infinitely many semi-simple subgroups $\mathrm{SL}(\mathcal{D})$ of $\mathrm{SL}_4(\mathbb{Q})$ that are not isomorphic to each other, therefore not conjugate to each other.
Now I would like to ask what happens if $K=\mathbb{R}$, I think the answer is yes, but how to prove it? And what about $K=\mathbb{Q}_p$ or some other fields?
Thanks very much!
 A: In case it is useful to the OP, here is a reference: 
Richardson, R. A rigidity theorem for subalgebras of Lie and associative algebras. Illinois J. Math. 11 1967 92–110. 
Proposition 12.1 seems to be the relevant result.
A: For fields of characteristic zero one can argue as follows: Assume first that $K$ is algebraically closed. Since semisimple subgroups of $G$ correspond bijectively to semisimple subalgebras of $\mathfrak g={\rm Lie}\,G$ it suffices to consider the same problem in $\mathfrak g$. The advantage is that the subalgebras of $\mathfrak g$ form the points of a variety $SA(\mathfrak g)$ (a closed subvariety of the Grassmannian) on which $G$ acts. Now we use that semisimple subalgebras are rigid, i.e., if two are "nearby" they are conjugated. The proof uses the following idea: the tangent space of $SA(\mathfrak g)$ in some $\mathfrak h$ can be interpreted as the space of cocycles while the tangent space to the orbit as coboundaries. Vanishing of some cohomology group (that's where semisimplicity is used) yields equality of both tangent spaces. I remember a survey of Kraft on this topic. This means now that the orbits of semisimple algebras are open in $SA(\mathfrak g)$. Clearly, there are only finitely many of them. That $G$ is semisimple is immaterial. So we get: any linear algebraic group contains only finitely many conjugacy classes of semisimple subgroups.
If $K$ is not algebraically closed then the same result would follow if $(G/H)(K)$ decomposes into finitely many $G(K)$-orbits for any semisimple $H$. This is known for local fields by Borel-Serre.
In positive characteristic, the finiteness result above is wrong: there are continuous families of pairwise non conjugate semisimple subgroups. But it may be still true that there are only finitely many classes of maximal subgroups. This probably follows from works of Seitz et at. 
