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)
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Statement

A finite solvable group need not be a supersolvable group. In particular, a solvable group need not be supersolvable.

Note that for a finite group, being solvable is equivalent to being a Polycyclic group (?), so this also gives an example of a Polycyclic group (?) that is not supersolvable.

Definitions used

Term Definition used
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.

Proof

Further information: alternating group:A4, subgroup structure of alternating group:A4

Consider the alternating group of degree four on the set \{1,2,3,4 \}. This is a group of order 12: