Group acting nilpotently on another group Suppose $G$ is a group acting faithfully by automorphisms on a group $K$, and let $[k,g]=k^{-1}k^g$ for $k\in K$ and $g\in G$.  The subgroup generated by these elements we'll call $[K,G]_1$, and we define inductively $[K,G]_n = [[K,G]_{n-1},G]$.  If $[K,G]_n=1$ for some $n$, then $G$ is nilpotent, and in a book I am reading, it is claimed Hall showed the class of $G$ is bounded by $n(n-1)/2$.  But I remember reading somewhere else (and of course now I can't remember!) that, in fact, the class of $G$ is bounded by $n-1$.  Is this true? And if so, where can I find a proof (or could one be reproduced below)?
Thanks!
 A: This is proved with the bound $n-1$ (and attributed to Hall) in Satz 2.9, Kapitel III of "Endliche Gruppen" by B. Huppert. It is proved as corollary of the following result.
Let $N$ and $L$ be subgroups of a group $G$, and let $N=N_0 \ge N_1 \ge \cdots$ be a chain of normal subgroups of $N$ with $[N_i,L] \le N_{i+1}$ for all $i$. Define $L_j = \{g \mid g \in L, [N_i,g] \le N_{i+j} \forall i \}$ (so $L_1=L$). Then the $L_j$ are subgroups of $G$, such that $[L_j,L_m] \le L_{j+m}$ for all $i,j,m$. We then have $[N_i,K_j(L)]] \le N_{i+j}$ for all $i,j$, where $G=K_1(G) \ge K_2(G) \ge \cdots$ is the lower central series of $G$.
The proof consists mainly of commutator calculations, which I won't try and copy out right now.
A: I think you just use three subgroups lemma. For subgroups $A,B,C$: $[A,B]=[B,A]$ and $[A,B,C]$ is contained in the normal closure of $[B,C,A][C,A,B]$.  One views $K$ as a normal subgroup of the holomorph $KG$ (one should just be careful to take $K$-normal closures; I'll ignore this once or twice to avoid too much notation).
Lemma:  Let $K(n)$ be the $K$-normal closure of the subgroup generated by all commutators of weight $n$ with exactly one occurrence of an element of $K$.  Set $K[n]$ to be the $K$-normal closure of $[K,G]_{n-1} = [K,G,\dots,G]$.  Then $K(n) = K[n]$.
Proof:  This is clearly true for $n=1,2,3$; also $K[n] ≤ K(n)$ is clear.  Since $K(n)$ is (by definition) generated by the $K$-normal closures of $[K(n-i),G(i)]$ for all $i=1,\dots,n-1$, we need to prove that $$P(i): \qquad [K(n-i),G(i)] \leq K[n]$$ for $i=1,\dots,n-1$.  $P(1)$ is just the statement that $[K(n-1),G(1)] = [K[n-1],G] \leq K[n]$, so we may assume $i \geq 2$ and that (by induction) $P(i-1)$ is true.  Then $P(i)$ follows easily:
$$\begin{align*}
[ K(n-i), G(i) ]
&= [ G(i), K(n-i) ] \\
&= [ G(i-1), G, K(n-i) ] \\
&≤ [ G, K(n-i), G(i-1) ] [ K(n-i), G(i-1), G ] \\
&= [ K(n-i), G, G(i-1) ] [ K(n-i), G(i-1), G ] \\
&≤ [ K(n-i+1), G(i-1) ] [ K(n-1), G ] \\
&\leq K[n] K[n] = K[n]
\end{align*}$$
Corollary: In particular, $[G,G,\dots,G,K] ≤ ([K,G,G,\dots,G])^K$, and so if $[K,G]_n = 1$, then $[G,G]_{n-1}$ commutes with $K$, and so if $G$ acts faithfully, $[G,G]_{n-1} = 1$ and $G$ has nilpotency class at most $n-1$.
Probably it is a good idea to understand the case of $K$ an elementary abelian $p$-group, where this is basically just an exercise on upper triangular matrices with 1s on the diagonal.
