Supersolvable implies nilpotent derived subgroup

Statement
The fact about::derived subgroup of a fact about::supersolvable group is a fact about::nilpotent group. Moreover, the fact about::nilpotence class of the commutator subgroup is bounded from above by the length of any normal series for the whole group with cyclic quotients.

Facts used

 * 1) uses::Commutator subgroup centralizes cyclic normal subgroup
 * 2) uses::Commutator subgroup is normal
 * 3) uses::Normality is strongly intersection-closed
 * 4) uses::Normality satisfies image condition
 * 5) uses::Second isomorphism theorem
 * 6) uses::Cyclicity is subgroup-closed
 * 7) uses::Commutator subgroup satisfies image condition: Under a surjective homomorphism, the image of the commutator subgorup equals the commutator subgroup of the image.

Proof
Given: A supersolvable group $$G$$ with a normal series $$1 = N_0 \le N_1 \le \dots N_c = G$$ with $$N_i/N_{i-1}$$ cyclic.

To prove: $$[G,G]$$ is nilpotent of class at most $$c$$.

Proof: We will prove that the series:

$$1 = N_0 \cap [G,G] \le N_1 \cap [G,G] \le \dots \le N_c \cap [G,G] = [G,G]$$

is a central series for $$[G,G]$$.


 * 1) $$[G,G]$$ is normal in $$G$$: This is fact (2).
 * 2) $$N_i \cap [G,G]$$ is normal in $$G$$ for all $$i$$: This follows from fact (3).
 * 3) $$(N_i \cap [G,G])/(N_{i-1} \cap [G,G])$$ is normal in $$G/(N_{i-1} \cap [G,G])$$ for all $$1 \le i \le c$$: This follows from fact (4).
 * 4) $$(N_i \cap [G,G])/(N_{i-1} \cap [G,G])$$ is cyclic: By the second isomorphism theorem (fact (5)), this quotient is isomorphic to $$N_{i-1}(N_i \cap [G,G])/N_{i-1}$$, which in turn is a subgroup of $$N_i/N_{i-1}$$. Fact (6) therefore yields that it is cyclic.
 * 5) $$[G,G]/(N_{i-1} \cap [G,G])$$ is the commutator subgroup of $$G/(N_{i-1} \cap [G,G])$$: This is direct from fact (7).
 * 6) $$(N_i \cap [G,G])/(N_{i-1} \cap [G,G])$$ is in the center of $$[G,G]/(N_{i-1} \cap [G,G])$$: By the previous three steps, $$[G,G]/(N_{i-1} \cap [G,G])$$ is cyclic normal in $$G/(N_{i-1} \cap [G,G])$$, and $$[G,G]/(N_{i-1} \cap [G,G])$$ is the commutator subgroup. So fact (1) yields that $$(N_i \cap [G,G])/(N_{i-1} \cap [G,G])$$ is in the center of $$[G,G]/(N_{i-1} \cap [G,G])$$.

Thus, the series is indeed a central series.