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Is it true that every finite p-group can be realized as the group of rational points over $\mathbb{F_p}$ of some connected unipotent algebraic group defined over $\mathbb{F_p}$? For abelian p-groups, the answer is yes via Witt vectors, but is it true in general?

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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. – L Spice Jul 3 '11 at 14:28
    
L. Spice, do you want to replace $GL_1$ with $\mathbb{G}_a$? – S. Carnahan Jul 3 '11 at 14:42
    
@S. Carnahan, yes, thanks. – L Spice Jul 3 '11 at 14:51
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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. – L Spice Jul 3 '11 at 15:33

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