# Subgroup containing a given torus as a maximal torus

Let T be a non-trivial torus in a semisimple group G of adjoint type defined over Q. I was wondering whether there exists a reductive subgroup H of G (defined over Q !) such that T is a maximal torus of H and such that T is not in the centre of H.

• At least in the split case, I think the answer to your question is sometimes no (but would have to analyze it using the standard structure theory). It's useful to distinguish the two possible cases, where your given torus is regular or singular. In the singular case I don't see how a subgrooup $H$ can exist, even in the first nontrivial case of type $A_2$. Oct 28 '13 at 23:21

Consider a supplement $T'$ to $T$, so that $T \times T'$ is a maximal torus of $G$. The centraliser $H_0$ of $T'$ in $G$ is then a reductive group, defined over $Q$ if $T'$ is, whose maximal subgroup is $T \times T'$. Taking $H$ the derived group of $H_0$ will kill the center of $H_0$. $T' \times T$ meets $H$ against a maximal torus of $H$, which is the intersection of the roots of $G$ whose kernel contains $T$.
Using this process, you do not really get $T$ as maximal torus, but rather a kind of “arithmetic closure” of it. Is this close enough?
• This construction does not seem to work. Let $S$ be a maximal torus in $G$ and let $\lambda:{\rm{GL}}_1 \rightarrow S$ be a "generic" cocharacter. Then $Z_G(\lambda) = S$ and its derived group is 1. So if one begins with $T = \lambda({\rm{GL}}_1)$ and chooses $T'$ an isogeny complement to $T$ in $S$ then $T' \cdot T = S$, etc. Am I misunderstanding the proposed answer? Oct 28 '13 at 15:18
• @Marguax You are right, the construction I proposed only works in the case of where $T$ is degenerated. Oct 28 '13 at 15:27