# a question about centralizers in semi-simple groups

I have a question concerning centralizers in real reductive groups. I'd like to know if the following property is available in any references.

Let $L\subset H\subset G$ be an inclusion chain of connected semi-simple algebraic groups defined over the real numbers $\mathbb{R}$. Assume that the centralizers $Z(L,H)$ and $Z(H,G)$ are both compact, can one show that $Z(L,G)$ is also compact? Here one may add the condition that $L$, $H$ are bth stable under a Cartan involution $\theta$ of $G$.

Or more algebraically, if the groups above are defined over a perfect field $k$ (say of characteristic zero), and that $Z(L,H)$ and $Z(H,G)$ are both $k$-anisotropic, in the sense that they do not contain split $k$-torus of dimension $>0$, then can one show that $Z(L,G)$ is also $k$-anisotropic?

Any examples and counter-examples are also welcome.

Many thanks

-
oh, I should have reduced to the case that $L$ is normal in $G$, and thus one may assume that $L$ is trivial, so that $Z(L,H)=H$ and $Z(L,G)=G$. The question is then as follows: for $H\subset G$, $H$ compact, and $Z(H,G)$ compact, can one show that $G$ is also compact? I do not see how to reduce for a second time to the case that $H$ is normal in $G$. for example, let $H$ be a maximal compact subgroup of $G$ (in the real case), then its normalizer is itself, and the centralizer is compact. but no compactness of $G$ is deduced. – genshin Sep 28 '10 at 16:39
Sorry, I misread the question and thought you were asking about normal subgroups. OK, so I have now deleted my previous (now useless) answer to the wrong question. So then what is the question, since as you note, the conclusion is frequently false? – BCnrd Sep 28 '10 at 17:31
Aside from the formulation, it's a good idea to add the tag lie-groups. – Jim Humphreys Sep 28 '10 at 17:55
In the case that $L$ is trivial, $G$ need not be compact. For example, if $G$ is a noncompact semisimple Lie group and $H$ is its maximal compact subgroup then $Z_G(H)=H,$ so it is compact, but $G$ is not (by assumption). – Victor Protsak Sep 28 '10 at 20:17
thanks for the comments. a lie-groups tag is added. and the counter example is mentioned in my comment above. – genshin Sep 29 '10 at 9:23