Nilpotent implies solvable
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This article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property must also satisfy the second group property
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Statement
Verbal statement
Any nilpotent group is a solvable group. Further, the solvable length is bounded from above by the nilpotence class.
Propertytheoretic statement
The group property of being a nilpotent group is stronger than, or implies, the group property of beign a solvable group.
Definitions used
Nilpotent group
Further information: Nilpotent group
A group is said to be a nilpotent group if its lower central series terminates in finitely many steps at the trivial group. The lower central series is here defined as follows:
here is the subgroup generated by all commutators between elements of and elements of .
Note: This is the definition of nilpotent group that is convenient for proving this implication. The other definition, in terms of upper central series, is not convenient for our purpose.
Solvable group
Further information: Solvable group
A group is said to be a solvable group if its derived series terminate in finitely many steps at the trivial group. The derived series is defined as:
Note: This is the most convenient definition here.
Related facts
Stronger facts
The solvable length of a nilpotent group is not just bounded by the nilpotence class; it is also bounded by the logarithm of the nilpotence class:
 Solvable length is logarithmically bounded by nilpotence class. Specifically, this states that if has class its solvable length is at most .
 Second half of lower central series of nilpotent group comprises abelian groups
Converse
 Solvable not implies nilpotent: A solvable group need not be nilpotent.
 Solvable length gives no upper bound on nilpotence class: A nilpotent group of solvable length could have arbitrarily large nilpotence class.
Proof
Proof using lower central series and derived series
Given: A nilpotent group with nilpotence class .
To prove: is solvable with solvable length at most .
Proof: The crucial fact used in the proof is the following lemma: For any group, the member of the derived series is contained in the member of the lower central series. We prove this lemma inductively.
Induction base case : . Thus, .
Induction step: Suppose . Then we have:
Now, and (using the induction assumption). Thus, every commutator between and is also a commutator between and . Thus, we have a generating set for which is a subset of a generating set for .
From this, it follows that is a subgroup of . Thus:
.
This completes the induction.
Since has class , is trivial, and hence is trivial. This means that is solvable. Further, the smallest such that is trivial is at most . So the solvable length is at most equal to the nilpotence class.