Finite solvable not implies supersolvable
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This article gives the statement and possibly, proof, of a non-implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., finite solvable group) need not satisfy the second group property (i.e., supersolvable group)
View a complete list of group property non-implications | View a complete list of group property implications
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|supersolvable group||there is a normal series for the group (from the trivial subgroup to the whole group) where all successive quotients are cyclic groups. Here, normal series means that each member of the series is a normal subgroups of the whole group.|
|polycyclic group||there is a subnormal series for the group (from the trivial subgroup to the whole group) where all successive quotients are cyclic groups. Here, subnormal series means that each member of the series is a normal subgroup of its immediate successor.|
|solvable group||there is a subnormal series for the group where all successive quotient groups are abelian groups (in fact, this is equivalent to there existing a normal series with all successive quotient groups abelian).|
|finite solvable group||a finite group that is solvable, or equivalently, a finite group that is polycyclic.|
Consider the alternating group of degree four on the set . This is a group of order 12:
- The group is solvable and polycyclic: The derived subgroup, V4 in A4, is isomorphic to Klein four-group, which is a finite abelian group, and hence the group is solvable. We can also construct a polycyclic series as follows: . This is a subnormal series where all the successive quotients are cyclic: cyclic group:Z2, cyclic group:Z2, and cyclic group:Z3.
- This group is not supersolvable: The group has no cyclic normal subgroup, so there is no way it can have a normal series where all successive quotients are cyclic.