Nilpotency class of subgroup Let $G$ be a group of nilpotency class $n$. Every proper subgroup of $G$ would have nilpotency class at most $n$.
Do we know of an example where the commutator subgroup of a nilpotent group of class $n$ also has class $n$?
 A: No. Recall that the nilpotenct class of $G$ can be defined as the smallest natural number $n$ such that $\gamma_n (G)=0$, where $\gamma_1(G)=G$, $\gamma_{j+1}(G)=[G,\gamma_j(G)]$. Since $\gamma_1([G,G])=[G,G]=\gamma_2(G)$ it follows (by induction) that $\gamma_j([G,G]) \leq \gamma_{j+1}(G)$ for $j$ a natural number. This means that if $G$ has nilpotency class $n$, then $0\leq \gamma_{n-1}([G,G])\leq \gamma_n(G)=0$ and hence $[G,G]$ has nilpotency class at most $n-1$.
For the induction step above note that if $\gamma_{k}([G,G]) \leq \gamma_{k+1}(G)$, then since $[G,G]\leq G$ we have that $$\gamma_{k+1}([G,G])=[[G,G],\gamma_{k}([G,G])] \leq [G,\gamma_{k+1}(G)] = \gamma_{k+2}(G)$$
as required.
Edit 2015-12-05: We will improve the bound according to Derek Holt's comment (to the extent I was able).
We prove by induction on $k$ that $$[\gamma_k(G),\gamma_l(G)] \leq \gamma_{k+l}(G).\qquad\qquad(*)$$
Base case: By definition we know that for all $l$ 
$$[\gamma_1(G),\gamma_l(G)] = [G,\gamma_l(G)] = \gamma_{1+l}(G).$$
Now for the induction step. Suppose that for all $l$ and for $k\leq n$
$$[\gamma_k(G),\gamma_l(G)] \leq \gamma_{k+l}(G).$$ Using the three subgroups lemma for the first inequality and the inductive hypothesis for the second and third we obtain:
$$[\gamma_{n+1}(G),\gamma_l(G)] = [[G,\gamma_n(G)],\gamma_{l}(G)]\leq [\gamma_n(G),[G,\gamma_{l}(G)]][G,[\gamma_{n}(G),\gamma_l(G)]]\leq[\gamma_n(G),\gamma_{l+1}(G)][G,\gamma_{n+l}(G)]\leq \gamma_{n+1+l}(G)$$
as required.
Next we will prove that for all $j$ and $k$ that
$$\gamma_j(\gamma_k(G)) \leq \gamma_{jk}(G).\qquad\qquad \text{$($#$)$}$$
Since $\gamma_1(\gamma_k(G)) = \gamma_{k}(G)$ and if $\gamma_j(\gamma_k(G)) \leq \gamma_{jk}(G)$, then
$$\gamma_{j+1}(\gamma_k(G)) =[\gamma_k(G),\gamma_j(\gamma_k(G))]\leq [\gamma_k(G),\gamma_{jk}(G)] \leq \gamma_{k+jk}(G)=\gamma_{(j+1)k}(G),$$
using $(*)$ for the second inequality, $($#$)$ follows by induction.
Now suppose that $G$ is of nilpotency class $n$. Then for each $k$ if $m=\lceil \frac{n}{k}\rceil$ we have that $\gamma_m(\gamma_k(G)) \leq \gamma_{mk}(G)=0$ (since $mk\geq n$) so that the nilpotency class of $\gamma_k(G)$ is less than or equal to $m$. Setting $k=2$ gives a bound of $\lceil \frac{n}{2} \rceil$ for the nilpotency class of $[G,G]$.
