Nilpotent implies solvable

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.

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:

$$G_1 = G, G_i = [G,G_{i-1}]$$

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], $$G^{(n)} = [G^{(n-1)},G^{(n-1)}$$

Note: This is the most convenient definition here.

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 $$G$$ has class $$c$$ its solvable length is at most $$\log_2c + 1$$.
 * 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 $$l$$ could have arbitrarily large nilpotence class.

Proof using lower central series and derived series
Given: A nilpotent group $$G$$ with nilpotence class $$c$$.

To prove: $$G$$ is solvable with solvable length at most $$c$$.

Proof: The crucial fact used in the proof is the following lemma: For any group, the $$i^{th}$$ member of the derived series is contained in the $$(i+1)^{th}$$ member of the lower central series. We prove this lemma inductively.

Induction base case $$i=0$$: $$G^{(0)} = G = G_1$$. Thus, $$G^{(0)} \le G_1$$.

Induction step: Suppose $$G^{(m)} \le G_{m+1}$$. Then we have:

$$G_{m+2} = [G,G_{m+1}]; G^{(m+1)} = [G^{(m)},G^{(m)}]$$

Now, $$G^{(m)} \le G$$ and $$G^{(m)} \le G_{m+1}$$ (using the induction assumption). Thus, every commutator between $$G^{(m)}$$ and $$G^{(m)}$$ is also a commutator between $$G$$ and $$G_{m+1}$$. 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 $$[G,G_{m+1}]$$. Thus:

$$G^{(m+1)} \le G_{m+2}$$.

This completes the induction.

$$G^{(c)} \le G_{c+1}$$

Since $$G$$ has class $$c$$, $$G_{c+1}$$ is trivial, 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.