Let $G$ be a connected linear algebraic group
over an algebraically closed field $k$ of characteristic 0.
Let $D\subset G$ be a closed diagonalizable subgroup of $G$
(a subgroup of multiplicative type).
Is it true that $D$ is contained in some torus $T\subset G$?


This is so for $G=\mathrm{GL}\_n$.
Is this true for any connected linear $G$ (or any connected reductive $G$)?


I am stuck with this simple question...


**Edit.** The answer to the original question is NO, see Angelo's answer. 
However, is it true that any *cyclic* finite diagonalizable subgroup $C$ of $G$ 
is contained in some torus $T\subset G$?



Diagonalizable subgroups of a connected linear algebraic group
No. For example, $\mathrm{PGL}\_n$ contains a subgroup $G$ isomorphic to the product of two cyclic subgroups of order $n$, generated by the classes of the diagonal matrix whose entries are the powers of a fixed primitive $n^{\rm th}$ root of 1, and the permutation matrix corresponding to a cycle of length $n$. The inverse image of this subgroup in $\mathrm{GL}\_n$ is not commutative, while the inverse image of a maximal torus in $\mathrm{PGL}\_n$ is a maximal torus in $\mathrm{GL}\_n$, so $G$ is not contained in a torus.


