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This is a continuation of my previous question. 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, $\mathrm{Pic}\ \mathrm{PGL}_n\neq 0$.

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I am convinced that this should be false in all non-special groups (so, for example, in $\mathrm{Spin}_n$ for all $n \ged 7$, essentially because if $G$ is a non-special group, the integral cohomology of the classifying space should have torsion, which should be detected by finite abelian subgroups. However, I am too lazy to check whether these results are actually known. –  Angelo Apr 7 '11 at 15:26
    
@Angelo: I'm not sure what "non-special" means in this context, but the formulation of Springer-Steinberg II, 5.8 seems precise. Their remark 5.13 points back to the papers by Borel-Serre (1953) and Borel (1961) on compact Lie groups for counterexamples related to elements of order equal to a torsion prime in the given commutative group. They view the translation of such examples to the algebraic group situation as immediate. One example is a product of copies of a cyclic group of order 2 for a Spin group. (But $SL_n, Sp_{2n}$ cause no trouble.) –  Jim Humphreys Apr 7 '11 at 17:07
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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.)

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Here is an explicit example. Let $G$ be a group of type $G_2$, over an algebraically closed field $k$ of characteristic not 2. Then $G$ is simply connected, and contains a maximal rank subgroup $M$ of type "$A_1 \times A_1$". As observed in [Springer-Steinberg, II.5.5], this semisimple subgroup is not simply connected. In fact, $M$ is isomorphic to $\operatorname{SL}_2 \times \operatorname{PGL}_2$.

Now, the subgroup $M$ is the centralizer of a finite order semisimple element $t$ of $G$. Moreover, by Angelo's Angelo's example for $\operatorname{PGL}_n$ there is a diagonalizable subgroup $D_1 \subset M$ of order 4 contained in no maximal torus of $M$.

Then the diagonalizable subgroup $D = \langle D_1,t\rangle$ of $G$ is not contained in any maximal torus of $G$.

For what it is worth, this example shows what went wrong with my (now deleted, and in hindsight, silly) argument in answer to the previous question. Namely, the centralizer of a semisimple element $t$ in a simply connected group is connected and reductive, but need not itself be simply connected. In fact the result found in [Springer-Steinberg, 5.3] gives more precise conditions -- related with torsion primes -- under which a group like $C_G(t)$ is simply connected.

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You write without proof or reference: "In fact, $M$ is isomorphic to $\mathrm{SL}_2 \times \mathrm{PGL}_2$". Could you please explain it? (My calculations show that M is isomorphic to $\mathrm{SO}_4$.) –  Mikhail Borovoi Apr 10 '11 at 15:52
    
@Mikhail: I am trusting the description found in exercise VIII.7 of "The book of involutions" (p. 508). [Well: already that exercise has at least one typo -- the comment "A2 x A2 in G2" should read "A1 x A1 in G2".] But is $\operatorname{SO}(4)$ really different from the group I describe? I may be thinking sloppily, but both seem to be quotients of the simply connected group $\operatorname{SL}_2 \times \operatorname{SL}_2$ by a central subgroup of order 2. –  George McNinch Apr 10 '11 at 17:39
    
@George: Your reference to "The book of involutions" seems to be correct, hence my calculations must contain a mistake. What about $\mathrm{SO}_4$, it is isomorphic to $\mathrm{SL}_2 \times \mathrm{SL}_2$ modulo $\mu_2$ embedded diagonally, and therefore $\mathrm{SO}_4$ has no normal subgroups isomorphic to $\mathrm{PGL}_2$, hence it is not isomorphic to $\mathrm{SL}_2 \times \mathrm{PGL}_2$. –  Mikhail Borovoi Apr 10 '11 at 19:09
    
@Mikhail: Let $G = \operatorname{SL}_2 \times \operatorname{SL}_2$ and $Z_0 \subset G$ be the diagonally embedded central $\mu_2$. Then $Z_0$ seems to be contained in the normal subgroup $H=\{(g,g) \mid g \in \operatorname{SL}_2\} \subset G$, and $\operatorname{PGL}_2 \simeq H/Z_0 \subset G/Z_0$. –  George McNinch Apr 10 '11 at 20:05
    
@George: This subgroup $H$ is not normal! –  Mikhail Borovoi Apr 11 '11 at 11:35
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