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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)?


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up vote 5 down vote accepted

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! I was worried that the $[K,G]_n$ only being subnormal would screw things up, but I think taking normal closures the argument is pretty much the same. – user641 Jul 22 '11 at 20:12
by the way, the book I mentioned in the question is Isaacs's FGT; I find it weird that the book is relatively recent, and discusses the three subgroup lemma at length, yet doesn't include a variant of this short proof (it even gives Hall's bound, but not this sharper one). Do you know a book/article reference for this, or did you come up with it on the spot? I'd like to have a reference for when I inevitably forget how to do this :) – user641 Jul 25 '11 at 23:14
@Steve: I just came up with it on the spot by working the examples for n=3,4,5 and thinking about the elementary abelian case; since being the identity is closed under normal-closure, the elementary abelian case and the general case are about the same due to Hall's identity aka three subgroups. I've read Huppert and Hall many times and spent many hours doing commutator calculus, so I don't claim any originality. – Jack Schmidt Jul 26 '11 at 16:42
thanks, that is actually a very nice way to think about it. – user641 Jul 27 '11 at 4:31

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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My only worry is that the $[K,G]_n$ don't have to be normal in $K$, do they? – user641 Jul 22 '11 at 20:10
[K,G]2 does not have to be normal in K (of course [K,G](n+1) is normal in [K,G](n), with [K,G](0)=K, so we always have subnormal). You can construct counterexamples with |K|=32 with: – Jack Schmidt Jul 22 '11 at 21:34
gap> for K in AllSmallGroups(32,IsAbelian,false) do G:=AutomorphismGroup(K);; KG:=SemidirectProduct(G,K);; G:=Image(Embedding(KG,1));; K:=Image(Embedding(KG,2));; if not ForAll(Subgroups(G), G-> IsNormal(K,CommutatorSubgroup(CommutatorSubgroup(K,G),G))) then Print(IdGroup(K)," ",IdGroup(G),"\n"); fi; od; – Jack Schmidt Jul 22 '11 at 21:35
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. – user641 Jul 25 '11 at 23:15

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