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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!

2 Answers 2

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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.

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    thanks, that is actually a very nice way to think about it.2011-07-27
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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.

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    I just wanted to add that there is a very pretty proof of this case (when the subgroups are normal) in Segal's Polycyclic Groups, pg. 9.2011-07-25