Free Groups and Nilpotency/Solvability (6.3.12)

Dummit and Foote Abstract Algebra, section 6.3, exercise 12:

Let $S$ be a set and let $c$ be a positive integer. Formulate the notion of a free nilpotent group on $S$ of nilpotence class $c$ and prove it has the appropriate universal property with respect to nilpotent groups of class $≤c$.

Proof: Let $N$ be any group of nilpotence class $≤c$ and let $ψ: S → N$ be a set map with $G = \langle~\text{img}~ψ~\rangle$. We have a unique homomorphism $φ : F(S) → N$ (fixing $S$) so that $F(S)/\text{ker}~φ ≅ G$. We prove that there is a unique homomorphism $\Phi : F(S)/F(S)^c → N$ such that $\Phi\mid_{\pi(S)} = ψ$ (where $\pi$ is the natural homomorphism from $F(S)$ to $F(S)/F(S)^c$). Assume $F(S)^c \not ≤ \text{ker}~φ$; we then have a contradiction: $$G^c≅(F(S)/\text{ker}~φ)^c=F(S)^c/\text{ker}~φ≠1$$ Therefore there is the desired homomorphism afforded by: $$(F(S)/F(S)^c)/(\text{ker}~φ/F(S)^c)≅F(S)/\text{ker}~φ≅G$$ Assume $\Phi_1 ≠ \Phi_2$ are two homomorphisms from $F(S)/F(S)^c$ to $N$ fixing $\pi(S)$; then $\Phi_1 \circ \pi ≠ \Phi_2 \circ \pi$ are two homomorphisms from $F(S)$ to $G$ fixing $S$, a contradiction. We thus have $\text{ker}~φ$ factors through $F(S)^c$, and the following diagram commutes:$\square$
Note that this theorem can be paralleled to produce a similar result regarding a free solvable group on a set $S$.