Abstract
It is well known that if the ground field K has characteristic zero and G is a connected algebraic group, defined over K, then the Lie algebra Lie(G′) of the commutant G′ of G coincides with the commutant Lie(G)′ of Lie(G). We show that this result is no longer true in the category of algebraic supergroups. We also construct a reductive supergroup H=X⋊G, where X and G are connected, reduced and abelian supergroups, such that Xu≠1 and (Hev)u is non-trivial connected (super)group. Quasi-reductive supergroups have been introduced in [10]. We prove that a supergroup H is quasi-reductive if and only if the largest even (super)subgroup of the solvable radical R(H) is a torus, H˜=H/R(H) contains a normal supersubgroup U, which is quasi-isomorphic to a direct product of normal supersubgroups Ui, and H˜/U is a triangulizable supergroup with odd unipotent radical. Moreover, for every i, Lie(Ui)=Ui⊗Sym(ni) are such that either ni=0 and Ui is a classical simple Lie superalgebra, or ni=1 and Ui is a simple Lie algebra.
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