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.

Property-theoretic 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

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:

$G_{1} = [G, G], G_{i} = [G, G_{i - 1}]$

here $[H, K]$ is the subgroup generated by all commutators between elements of $H$ and elements of $K$ .

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

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:

$G^{(1)} = [G, G], < m a t h > G^{(n)} = [G^{(n - 1)}, G^{(n - 1)}$

Note: This is the most convenient definition here.

Proof

Lemma for the proof

The crucial fact used in the proof is the following lemma: For any group, the $i^{t h}$ member of the derived series is contained in the $i^{t h}$ member of the lower central series. We prove this lemma inductively.

Induction base case $i = 1$ : Both $G^{(1)}$ and $G_{1}$ are the same, viz the commutator subgroup of $G$ . Thus $G^{(1)} \leq G_{1}$ .

Induction step: Suppose $G^{(m)} \leq G_{m}$ . Then we have:

$G_{m + 1} = [G, G_{m}]; G^{(m + 1)} = [G^{(m)}, G^{(m)}]$

Now, $G^{(m)} \leq G$ and $G^{(m)} \leq G_{m}$ (using the induction assumption). Thus, every commutator between $G^{(m)}$ and $G^{(m)}$ is also a commutator between $G$ and $G_{m}$ . Thus, we have a generating set for $[G^{(m)}, G^{(m)}]$ which is a subset of a generating set for $[G, G_{m}]$ .

From this, it follows that $[G^{(m)}, G^{(m)}]$ is a subgroup of $[G, G_{m}]$ . Thus:

$G^{(m + 1)} \leq G_{m + 1}$ .

Final proof

We now use the fact that for any group $G$ , $G^{(i)} \leq G_{i}$ . Suppose $G$ is a nilpotent group. Let $c$ be the nilpotence class of $G$ , viz the length of the lower central series of $G$ , or the smallest $c$ such that $G_{c}$ is the trivial group. Then, by the lemma:

$G^{(c)} \leq G_{c}$

and hence $G^{(c)}$ is trivial. This means that $G$ is solvable. Further, the smallest $l$ such that $G^{(l)}$ is trivial is at most $c$ . So the solvable length is at most equal to the nilpotence class.

Converse

Solvable does not imply nilpotent

Further information: Solvable not implies nilpotent

Solvable length need not equal nilpotence class

The solvable length may be much less than the nilpotence class. The only cases where we are guaranteed that the solvable length equals the nilpotence class are when the nilpotence class equals 1 or 2. We can construct groups of solvable length 2 (also called metabelian groups) with arbitrarily high nilpotence class.

Intermediate properties

Some properties between nilpotent and solvable, for finite groups:

Supersolvable group