As the question says, is a connected unipotent subgroup $U$ of a linear algebraic group scheme $G$ always contained in a Borel subgroup of $G$? I have an argument for why the answer is yes, and I have a somewhat confusing example that the answer is no. Obviously one (or both!) of these is wrong but I can't spot the error.

Argument for yes: Unipotent subgroups are solvable and a solvable connected group must be contained in a maximal solvable connected group, i.e., a Borel subgroup.

That seems too simple to be false, so I suspect the problem is with the counterexample which is not so simple.

Counterexample: Let $G = \mathrm{PGL}_2$ over an algebraically closed field $k$ of characteristic $2$. The first Frobenius kernel $G_1 \leq G$ is a closed subgroup whose group ring $kG_1$, which is the dual of the coordinate ring $k[G_1]$, is isomorphic as a Hopf algebra to the universal enveloping algebra of $\mathrm{Lie}(G) = \mathfrak{pgl}_2$. This means any Lie subalgebra of $\mathfrak{pgl}_2$ which consists of nilpotent elements is, by Engel's theorem, a unipotent subgroup of $\mathrm{PGL}_2$. Well, take the subalgebra $$\mathfrak e = \begin{bmatrix}0 & k \\ k & 0\end{bmatrix}.$$ The enveloping algebra of $\mathfrak e$ is $k[x, y]/(x^2, y^2)$. This is self dual so as a group scheme $\mathfrak e$ is connected. This means it should lie in a Borel subgroup of $\mathrm{PGL}_2$, and hence lie in the Lie algebra of that Borel subgroup. But $\mathrm{PGL}_2$ is semisimple of type $A_1$, the unipotent radical of any Borel has a $1$-dimensional Lie algebra so $\mathfrak e$ can't be contained in it.

  • $\begingroup$ The answer in the affirmative is Corollary 11.3 (2) of Borel's book. In fact, the maximal such groups are each the unipotent part of a Borel subgroup and they are all conjugate. I don't know where the counterexample fails. $\endgroup$ – Stefan Dawydiak Aug 23 '17 at 16:41

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