## 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 X_{u}≠1 and (Hev)u is non-trivial connected (super)group. ⋊-reductive supergroups have been introduced in [10]. We prove that a supergroup H is ⋊-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 ⋊-isomorphic to a direct product of normal supersubgroups U_{i}, and H~/U is a triangulizable supergroup with odd unipotent radical. Moreover, for every i, Lie(Ui)=Ui⊗Sym(ni) are such that either n_{i}=0 and Ui is a classical simple Lie superalgebra, or n_{i}=1 and Ui is a simple Lie algebra.

Original language | English |
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Pages (from-to) | 448-473 |

Number of pages | 26 |

Journal | Journal of Algebra |

Volume | 452 |

DOIs | |

Publication status | Published - Apr 15 2016 |

Externally published | Yes |

## Keywords

- Algebraic supergroup
- Lie superalgebra
- Solvable radical
- Solvable supergroup
- Unipotent radical
- ⋊-reductive supergroup

## ASJC Scopus subject areas

- Algebra and Number Theory