This is a continuation of [my previous question](http://mathoverflow.net/questions/60781).
Let $G$ be a connected reductive group over an algebraically closed field $k$ of characteristic 0.
We assume that $\mathrm{Pic}\ G=0$.
This is the same as to say that the derived group $G^{\mathrm{der}}$ of $G$ (which is semisimple) is simply connected.
In particular, if $G$ is simply connected semisimple, then $\mathrm{Pic}\ G=0$.



> 
> Let $D$ be a diagonalizable subgroup in $G$.
>  Is it true that $D$ is contained in some torus $T\subset G$ ?
> 
> 
> 


Angelo's answer to my previous question shows that this is *not* true for $G=\mathrm{PGL}\_n$.
Of course, $\mathrm{Pic}\ \mathrm{PGL}\_n\neq 0$.



Diagonalizable subgroups in a simply connected group
The answer seems to be no, according to II, 5.8 (and following material) in the Springer-Steinberg lecture notes (Lect. Notes in Math 131, 1970), though I might be overlooking something in your question. Of course, sometimes this type of embedding is possible as pointed out in those lecture notes, but there are problems with torsion primes. This kind of question goes back to work of Borel-Serre on compact Lie groups, and there are further details in Steinberg's 1975 paper in *Advances in Mathematics*. At this point I'm not aware of any significant improvement to the results written down by Springer-Steinberg.


In any case, the central torus of a reductive group plays little role in the question, so it's essentially about simply connected semisimple groups.


P.S. To clarify terminology, a *torsion prime* is a prime dividing some coefficient of the highest coroot for a simple algebraic group. So the possibilities are limited to $2,3,5$ (and there are no torsion primes for types $A, C$). There is an indirect connection with fundamental group orders in the topological setting. (Also, the results of Springer-Steinberg are formulated over an algebraically closed field of any characteristic and adapted somewhat to groups which are not simply connected.)


