let $k$ be any field of char 0. $G$ is split reductive algebriac group over k. Let p in G(k) be kdiagonalizable. Does there exist a split maximal torus of G(k) containing p? I know that is ture for lie algbera case.
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$\begingroup$ Please define "k"diagonalizable. If you mean $p$ lies in a split multiplicative type $k$subgroup (perhaps disconnected), a counterexample is $G = {\rm{PGL}}_2$ over $k = \mathbf{Q}$, with $p$ offdiagonal having offdiagonal entries equal 1 and $1$. Indeed, $p$ generates a $\mu_2$ inside $G$ and its "determinant" in $k^{\times}/(k^{\times})^2$ is 1 whereas the unique nontrivial 2torsion element in any 1dimensional split maximal $k$torus has "determinant" equal to $1$ (as may be checked by inspection of the diagonal one, due to $G(k)$conjugacy). Is your $G$ ss and simply connected? $\endgroup$ – user27056 Oct 24 '12 at 13:29

$\begingroup$ I forget to mention G is reductive but its derived subgroup (G,G) is simply connected. I mean diagonalizable for any representaions like acting on k[G]. But in your example p have eigenvalues $\pm i$? $\endgroup$ – user27501 Oct 24 '12 at 20:55

$\begingroup$ The eigenvalues in my example aren't $\pm i$ (mixing up GL$_2$ and PGL$_2$) since my $p$ has order 2 in $G(k)$. By your definition of "$k$diagonalizable" there's a closed immersion $G \hookrightarrow {\rm{GL}}(V)$ so that $p$ is diagonalizable on $V$ and hence lies in a split torus $T$ of GL($V$). Hence, $M := T \cap G$ is a split $k$subgroup of $G$ of mult. type with $p \in M(k)$, so your definition implies mine. The converse is easy (and my definition is intrinsic to $G$ and more robust). Thanks for clarifying that $(G,G)$ is simply connected. $\endgroup$ – user27056 Oct 25 '12 at 4:22

$\begingroup$ Thank you so much! I got it. How about simply connected case? $\endgroup$ – user27501 Oct 25 '12 at 21:44