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Let $G$ be an algebraic (not necessarily linear) group and let $Z \subset G$ be a Zariski closed irreducible subset. Since tangent bundle of $G$ is trivial, we may identify tangent spaces at all points with $T_0 G$. Suppose that $T_z Z$ is the same subset of $T_0 G$ for any $z \in Z$, and is closed under Lie bracket. I want to prove that $Z$ is a coset of a closed subgroup of $G$.

The setting is the most classical possible: $Z$ is just a closed subset, i.e. has a reduced subscheme structure if you like, and the ground field is supposed algebraically closed.

It suffices to prove that such $Z$, if it contains $0$, is a closed subgroup. Analytically, one could take the exponent $exp(\mathfrak{z})$ of the Lie subalgebra $\mathfrak z$ corresponding to the tangent space of $Z$, and then use the fact that $Z$ is irreducible.

My question is how one could prove this algebraically.

Is this statement true in positive characteristic?

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After thinking on this question I have realised that I in fact cannot answer it even over complex numbers. Here is the problem. Suppose we are given a subalgebra $\mathfrak z$ of the Lie algebra of our group $G$, so we have an involutive subbundle of the tangent bundle; taking its non-vanishing sections gives us various vector fields, and solving the corresponding ODE we can find a submanifold passing through the identitiy and touching such a vector field. Now I am not sure which of these manifolds will be a subgroup; ... –  Dima Sustretov Jan 29 '14 at 17:00
... moreover, if it turns out that such manifold does not touch any constant vector field, then it obviously cannot be a closed subgroup. –  Dima Sustretov Jan 29 '14 at 17:01
these considerations in principle should lead to a counterexample, but I still cannot construct one. –  Dima Sustretov Jan 29 '14 at 17:03
Aw, too bad. If you post that as an answer, I could at least award the bounty to you without having it go to waste. –  Jesko Hüttenhain Jan 29 '14 at 17:05
Jesko, I also have some other considerations. For example, the statement is still true for Abelian varieties. The simplest case is when $Z$ is one-dimensional, then we can just notice that a curve with a trivial tangent bundle is an elliptic curve, and by a little argument involving functoriality of taking Albanese, an ellpitic curve embedded into an Abelian variety is a coset. –  Dima Sustretov Jan 29 '14 at 17:11

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