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Jul 3, 2011 at 15:33 comment added LSpice Sorry, after a bit more thought, it occurs to me that, since $p$-groups have centres, it's OK to have the Abelian group as sub-object. Then Theorem 1.8(c) of the paper by Kumar and Neeb shows that $\operatorname{Ext}^1_{\text{alg}}(G, \mathbb G_a) \cong H^2(\mathfrak g, \mathfrak{gl}_1)^{\mathfrak g}$; but I still can't see my way through to showing that the necessary map is surjective.
Jul 3, 2011 at 14:42 comment added S. Carnahan L. Spice, do you want to replace $GL_1$ with $\mathbb{G}_a$?
Jul 3, 2011 at 14:28 comment added LSpice By using a filtration of your $p$-group with all successive quotients $\mathbb F_p$, you can reduce the problem to showing that $\operatorname{Ext}^1_{\text{alg}}(\operatorname{GL}_1, G) \to \operatorname{Ext}^1_{\mathbb Z}(\mathbb F_p, G(\mathbb F_p))$ is surjective for all connected, unipotent $\mathbb F_p$-groups $G$. A quick Google search turned up “Extensions of algebraic groups” by Kumar and Neeb (#48 at math.unc.edu/Faculty/kumar), but the Abelian group by which you're extending there is the subobject, not the quotient.
Jul 3, 2011 at 13:46 history asked Georges CC BY-SA 3.0