arXiv2013
This work was inspired by two natural questions. The first question is when Lie(G')=Lie(G)', where G is a connected algebraic supergroup defined over a field of characteristic zero. The second question is whether the unipotent radical of any normal supersubgroup H of G coincides with the intersection of H and G_u, where G_u is the unipotent radical of G. Both questions have affirmative answers in the category of algebraic groups (in the second case one has to assume additionally that G and H are reduced whenever char K >0). Surprisingly, using the technique of Harish-Chandra superpairs and a complete description of an action of an algebraic supergroup on an abelian supergroups by supergroup automorphisms we found out rather simple counterexamples to both questions. Besides, the second counterexample shows that the reductivity of G does not imply that G_{ev} has even finite unipotent radical. On the other hand, if G_{ev} is reductive, then it is easy to see that G_u is finite (odd) supergroup. In other words, the reductivity of an algebraic supergroup does not correspond to the reductivity of its largest even subgroup in contrast to such properties as unipotency or solvability. In the last section of our article we describe reductive algebraic supergroups in terms of sandwich pairs and give necessary and sufficient conditions for an algebraic supergroup to be quasireductive. The last result complements the recent Serganova's description of quasireductive supergroups in terms of structural properties of their Lie superalgebras. Our approach is focused on the normal subgroup structure.