Supersolvable not implies nilpotent

From Groupprops
Jump to: navigation, search
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., supersolvable group) need not satisfy the second group property (i.e., nilpotent group)
View a complete list of group property non-implications | View a complete list of group property implications
Get more facts about supersolvable group|Get more facts about nilpotent group

Statement

It is possible for a group to be a supersolvable group but not a nilpotent group.

Definitions used

Term Definition used
nilpotent group The upper central series terminates at the whole group. For a finite group, this is equivalent to being the direct product of its Sylow subgroups (see finite nilpotent group, equivalence of definitions of finite nilpotent group
supersolvable group There is a normal series where all the successive quotient groups are cyclic groups.

Related facts

Converse

We have finite nilpotent implies supersolvable. Note that infinite nilpotent groups, and even infinite abelian groups, need not be abelian, because of finiteness problems. For instance, the group of rational numbers (additive) is not supersolvable.

Similar facts

Opposite facts

Proof

Further information: symmetric group:S3

The symmetric group of degree three is an example of a finite supersolvable group that is not nilpotent:

  • The group has a normal subgroup A3 in S3 such that both the subgroup and the quotient are cyclic groups, so it is supersolvable.
  • The group is centerless, so it is not nilpotent.

Observations related to search for examples

class="sortable" border="1" Fact Nature of significance Details
prime power order implies nilpotent where not to look if you want to avoid nilpotent to make sure the group is non-nilpotent, do not look at prime powers. They will not work.
equivalence of definitions of finite nilpotent group where to look and where not to look if you want to avoid nilpotent any nilpotent group must be the direct product of its Sylow subgroups, or equivalently, all its Sylow subgroups are normal. Hence, we must look for groups that are not direct products of their Sylow subgroups, or equivalently, that have non-normal Sylow subgroups.
nilpotent of cube-free order implies abelian where to look if you want to avoid nilpotent any example of a supersolvable non-abelian group where the order is a cube-free number automatically gives an example of a supersolvable non-nilpotent group. This is because if the group were nilpotent, it would be abelian on account of its cube-free order.
nilpotency is subgroup-closed, supersolvability is finite direct product-closed how to construct bigger examples from smaller if G is a non-nilpotent supersolvable group, and H is any supersolvable group (possibly nilpotent, possibly not nilpotent), G \times H is also a non-nilpotent supersolvable group.

Related specific information