Probabilistically nilpotent groups of class two

For $G$ a finite group, let $d_2(G)$ denote the proportion of triples $(x, y, z) \in G^3$ such that $[x, y, z] = 1$. We determine the structure of finite groups $G$ such that $d_2(G)$ is bounded away from zero: if $d_2(G) \geq \epsilon>0$, $G$ has a class-4 nilpotent normal subgroup $H$ such that $[G : H] $ and $|\gamma_4(H)|$ are both bounded in terms of $\epsilon$. We also show that if $G$ is an infinite group whose commutators have boundedly many conjugates, or indeed if $G$ satisfies a certain more general commutator covering condition, then $G$ is finite-by-class-3-nilpotent-by-finite.

Then G is nilpotent of class k if and only if d k (G) = 1, so class-k nilpotency degree is a statistical relaxation of class-k nilpotency.We may call G probabilistically nilpotent of class k if d k (G) is bounded away from zero (so G has statistically significant class-k nilpotency).
For example, d 1 (G) is the probability that two random elements of G commute, sometimes called the commuting probability of G.It is well-known that d 1 (G) ≤ 5/8 for any nonabelian group G. Another important result, less well known, is a theorem of Peter Neumann which states that if d 1 (G) is bounded away from zero then G The first author has received funding from the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme (grant agreement No. 803711).The second author was supported by FAPDF and CNPq-Brazil.has a subgroup H such that [G : H] and |H ′ | are both bounded.Thus a finite probabilistically abelian group is bounded-by-abelian-by-bounded. 1  It is natural to ask for an analogous qualitative description of probabilistically nilpotent groups of class k.Essentially this question has been asked by several authors, including Shalev [S], Martino,Tointon,Valiunas,and Ventura [MTVV,Question 1.10],and Green [personal communication].If G has boundedly many generators then it was shown by Shalev [S,Theorem 1.1] that G is class-k-bybounded.If not, the structure of G is much less transparent, and certainly any structure theorem must at least include the class of bounded-by-class-k-by-bounded groups.
In this paper we consider finite groups G with d 2 (G) bounded away from zero.We show that G need not be bounded-by-class-2-by-bounded (as might be hoped), nor even class-3-by-bounded, but we will show that G must be bounded-by-class-3by-bounded.
Theorem 1.1.Let G be a finite group such that d 2 (G) ≥ ǫ > 0. Then G has a subgroup H of nilpotency class at most 4 such that [G : H] and |γ 4 (H)| are both ǫ-bounded.
Further details about the structure of G can be extracted from the proof of the theorem.It is noteworthy that if T is the trilinear map (H/H ′ ) 3 → γ 3 (H)/γ 4 (H) induced by the triple commutator [x, y, z], then there is an expression for the map T of the form T (x, y, z) = A(a(x, y), z) + B(b(x, z), y) + C(c(y, z), x), (1.1) We will prove that a finite group with d 2 (G) bounded away from zero has a boundedindex subgroup satisfying the covering condition with bounded n and |S|, and conversely the covering condition implies d 2 (G) is bounded away from zero by a function of n and |S|, so for finite groups the condition that d 2 (G) is bounded away from zero and the covering condition are loosely equivalent.On the other hand the covering condition makes sense for infinite as well as finite groups, and is satisfied by groups with boundedly finite conjugacy classes of commutators (the case S = {1}), which were studied in [DS2].We will prove the following structure theorem for groups (finite or infinite) satisfying the commutator covering condition.
1 Here and throughout the paper we say "(a, b, c, . . .)-bounded" to mean "bounded by a function of a, b, c, . . .only".Where the parameters a, b, c, . . .are clear from the context we often just say "bounded".Also, if we say a group is A-by-B-by-   We do not know whether the techniques used in this paper can be adapted to handle groups G with d k (G) bounded away from zero for k ≥ 3 (that is, probabilistically nilpotent groups of class k), or even more specially groups G in which all weight-k commutators [x 1 , . . ., x k ] have boundedly many conjugates.It may be that any such group has a subgroup H for which [G : H] and |γ k+1 (H)| are bounded.However, where

Example
In this section we construct a family of finite groups G for which d 2 (G) is bounded away from zero but such that the largest class-3 subgroup of G does not have bounded index.A slight variant also gives an infinite group G in which commutators have boundedly many conjugates but such that G is not virtually class-3.In both cases G is nilpotent of class 4 and bounded-by-class-3.
Let K = F p where p is a small prime (say 2, 3, or 5) and let V be a vector space over K. Let f : V × V → K be a (generic) bilinear map.We define a graded unital and R k = 0 for k > 4. We must define the multiplication maps be an isomorphism.We define multiplication between R 1 and R 2 and between R 1 and R 3 by the rules for all x, y, z, w ∈ R 1 , it suffices to verify that (xyz)w = x(yzw) for x, y, z, w ∈ R 1 , and this is a straightforward check: Then G is a nilpotent group of class 4. The group commutator and Lie bracket are related by For any x, y ∈ L 1 the formula for [x, y, z] L shows that the set of z ∈ L 1 for which [x, y, z] L = 0 is a subspace of codimension at most 2. Hence the group of 1 + z ∈ G for which [1 + x, 1 + y, 1 + z] ∈ 1 + L 4 has index at most p 2 .Since |1 + L 4 | = |K| = p, it follows that [1 + x, 1 + y] has at most p 3 conjugates.Hence every commutator has boundedly many conjugates.On the other hand, the quadruple commutator is By our earlier remarks about [x, y, z, w] To give a concrete example let V be the F 2 -vector space with basis (e i , e ′ i : i ∈ I) (for some index set I) and let

Outline
In this section we give an outline of the proof and in the next we list some tools that we need.The proof of the main theorems then occupies the rest of the paper and consists of the following steps.
1.In Section 5 we abstract the method in the proof of Peter Theorem 4.1 (B.Neumann's BFC theorem [N1]; see also [R, 14.5.11]).Let G be a group in which |x G | ≤ n for every x ∈ G. Then |G ′ | is finite and n-bounded.
An explicit bound for |G ′ | was first established by Wiegold [W].Subsequently several people worked on the bound for |G ′ |.Guralnick and Maróti proved in [GM,Theorem 1.9 The second is the theorem of Peter Neumann about d 1 already mentioned in the introduction.
Theorem 4.2 (P.Neumann [N2]).Let G be a finite group such that This theorem bears roughly the same relation to Theorem 4.1 as Theorem 1.1 bears to Theorem 1.2.
Actually we will use an asymmetric version of Peter Neumann's theorem that considers not just d 1 (G) but the commuting probability between two normal subgroups A, B G. Theorem 4.3 (P.Neumann's theorem, asymmetric version).Let G be a finite group with normal subgroups A, B G such that P a∈A,b∈B ([a, b] To our knowledge this result does not appear in the literature, though it is similar to (and easier than) [DS1,Proposition 1.2].It can be established by adapting the proof in the symmetric case appropriately.This result is also a corollary of a more general result we will prove, so a proof will be given in Section 5. 4.2.Tools for reducing to the finite case.In Theorem 1.2 we allow G to be infinite, mostly because we can reduce to the finite case whenever G is locally residually finite.
(1) G is m-by-class-s-by-n if and only if every finitely generated subgroup of G is so.
. ., Γ k , and the hypothesis that every Γ ∈ I is m-by-class-s-by-n implies that C Γ = ∅.Hence the sets C Γ have the finite intersection property, so by so the forward implication is clear.Now suppose every finite quotient Q = G/N is m-by-class-sby-n.Let I be the set of finite quotients of G.A compactness argument as in the previous part establishes that there are subgroups We will also need to know that the covering condition (1.1) behaves well with respect to subgroups and quotients.
(1) If H ≤ G then H satisfies the covering condition with (n 2 , S ′ ) in place of (n, S) for some set S ′ of size at most |S|.
(2) If G is a homomorphic image of G then G satisfies the covering condition with (n, S) in place of (n, S). Proof.
(1) Let S ′ be a set containing one point of Bs ∩ H whenever Somewhat related to this lemma, the beautiful result [MTVV,Theorem 1.21] states that, for G finite, whenever N G.This shows that d k behaves very well with respect to normal subgroups and quotients.However, we do not need a result of this form because we work mainly with the covering condition instead of d 2 .
where for each I the functions g I and G I are multilinear maps and (or perhaps its logarithm) the rank of the expression (4.2), the conclusion of Theorem 4.6 is simply that there is an expression for F of (ǫ, k)-bounded rank.This terminology is reasonable given the connection to the theory of analytic rank, which is explained in [E3].
We need to probe the uniqueness of bounded-rank expression such as (4.2).Certainly the expression need not be unique, strictly speaking, but suppose F has two bounded-rank expressions for some sets I and J of nonempty subsets of [k].If I = J , then it is natural to ask whether there is a third expression, still of bounded rank, only involving terms which could have appeared in either sum.This may be true generally.We will establish it in the special case we need.
Lemma 4.7.Suppose F : both of complexity at most C, where Then there are C-bounded-index subgroups We will continue to use large Roman letters F, G, H, . . .for arbitrary multilinear maps and little Roman letters f, g, h, . . .for multilinear maps with bounded codomain.
Proof.Throughout the proof, "bounded" means "C-bounded".We will repeatedly use the observation that if we have boundedly many maps g 1 , . . ., g n of one variable and bounded codomain then we can pass to the kernel of all these maps to get rid of them.In particular we can immediately get rid of all terms of the forms G(g(x), y, z, w), G(x, g(y), z, w), G(x, y, g(z), w), G(x, y, z, g(w)), so forget those.Hence we may assume the I-expression takes the simpler form First consider the term F 2 (f 2 (x, z), y, w).Fix u = f 2 (x, z).By equating (4.3) with a J -expression we find an identity of the form where g u , g ′ u , g ′′ u have bounded codomain.Since | cod(f 2 )| is bounded, in fact we have (4.4) for every u in the group Pass to u∈U ker g ′ u ∩ ker g ′′ u to reduce to F 2 (u, y, w) = G u (g u (y, w)).
Now replace g u with f ′′ 2 = (g u : u ∈ U ), which has bounded codomain u∈U cod(g u ), and replace G u with G u • π u , where π u is the projection; thus 2 (y, w)).We may restrict the codomain of f ′′ 2 to f ′′ 2 (y, w) : y ∈ A 2 , w ∈ A 4 and we may restrict the domain of G u to cod(f ′′ 2 ).But now since F 2 (u, y, w) is u-linear, G u (v) must be u-linear for every v ∈ cod(g), so ), and by an analogous argument Finally, consider F 4 (f 4 (x, y, z), w).By comparing the Iand J -expressions, we get an identity of the form where H ′ (x, y, z, w) has an expression involving terms only of the forms G(g(x, z), g ′ (y, w)), G(g(y, z), g ′ (x, w)), G(g(x, w), y, z), G(g(y, w), x, z), G(g(z, w), x, y), G(g(x, y, w), z), G(g(x, z, w), y), G(g(y, z, w), x), G(g(x, y, z, w)).
Fix x, y and let u = h(x, y).Then by rearranging (4.5) we get an identity of the form . By arguing exactly as before we establish that H(h(x, y), z, w) = G(h(x, y), g(z, w)).Now fix x, y, z and let u = g(x, y, z).Then from (4.5) we have an expression of the form Thus the I-expression (4.3) is reduced to the desired form.

Neumann's theorem for metric entropy
Let G be a group.An invariant seminorm on G is a function ) is an invariant seminorm.This makes it reasonable to use metric language.
Example 5.1.The discrete norm is Example 5.2.The conjugacy class norm is x n ∈ X and some C-bounded n by Lemma 5.4 again.Then the subgroup Symmetrically, there is a subgroup K ≤ B of index at most 2C such that for every k ∈ K there is a subgroup Hence there is a C-bounded number D such that for every h ∈ H we have and for every k ∈ K we have It follows that if S is any set of (4D + 1)-separated points of Comm(H, K) we must have |S| ≤ D 2 .Hence there is a set S of size at most D 2 such that any point of Comm(H, K) is within distance 4D + 1 of a point of S.
For the converse, write B for the ball of radius C and S for a set of size at most C such that Comm(H, K) ⊆ BS.
Let h ∈ H.By the pigeonhole principle there is some s ∈ S such that This holds for every h ∈ H, so Letting D = 2C 3 , we get the claim.
By taking the discrete norm we recover the asymmetric version of Neumann's theorem stated earlier as Theorem 4.3.
Since both H 0 and K 0 have boundedly many generators and Comm(H, K) is bounded, the centralizers C H0 (K 0 ) and C K0 (H 0 ) have bounded indices in H 0 and K 0 , respectively.Now it follows from Baer's asymmetric version of Schur's theorem ( [R, 14.5 The main corollary we need in the rest of the paper is the conjugacy class norm case.
(1) If S = {1}, the covering condition states that the commutators in G have boundedly many conjugates.The main result of [DS2] states in this case that G is bounded-by-metabelian.We closely follow some parts of [DS2] to prove the more general but weaker result stated above.
(2) The proof actually shows that G is bounded-by-metabelian-by-class-2-bybounded.We do not need this more detailed information right now, and we will eventually prove the stronger result Theorem 1.2 anyway.(3) Results of Bors, Larsen, and Shalev (see [BS,Corollary 1.5] and references), relying on the classification of finite simple groups, imply that if G is finite and , where rad(G) is the soluble radical of G, and it follows from (4.1) that rad(G) has (k, ǫ)-bounded derived length.By comparison, in Proposition 6.1, we do not depend on the classification, the group G is allowed to be infinite, and there is a stronger conclusion (though it will be superseded by Theorem 1.2).Now assume the hypothesis of Proposition 6.1.We may assume 1 ∈ S and we will use inducton on |S|.Suppose there is some commutator x ∈ Comm(G, G) such that n < |x G | ≤ n 100 .By the covering condition x ∈ Bs for some nontrivial s ∈ S.This implies that |s G | ≤ n 101 .Then for any g ∈ Bs we have |g G | ≤ n 102 .Hence we can remove s from S at the cost of increasing n to n 102 and then apply induction on |S|.Hence we may assume B satisfies the following stability condition: (6.1) Let H = B .Given an element g ∈ H, we write l(g) for the minimal number l with the property that g can be written as a product of l elements of B.

Thus for any
, so the claim holds.
Let R = [H, x] G .By Lemma 6.4 the order of R is bounded.By Lemma 6.5, [H, u] ≤ R whenever u ∈ U and ux ∈ X.
Let U 0 be the normal core of U .Then [G : We may expand y has a product of four commutators: Since [x 1 , x 2 ] = x and [x, U 0 ] = 1, we may simplify this slightly to Each commutator involving a t i is in B 4 , and x ∈ B, so y ∈ B 13 .By (6.1), this implies y ∈ B. Applying Lemma 6.5, we find that Hence there are at most |S| commutators in U 0 / B 0 , so (U 0 / B 0 ) ′ has bounded order by Neumann's BFC theorem (Theorem 4.1).We can pass to a bounded-index subgroup of U 0 whose image in U 0 / B 0 is nilpotent of class at most 2. Hence G is bounded-by-metabelian-by-class-2-by-bounded.This proves Proposition 6.1.

From bounded derived length to bounded class
Let G be a soluble group satisfying the covering condition (1.1).In this section we show that G has a nilpotent subgroup of bounded index and class, with bounds depending on |S|, n, and the derived length l of G.
Proposition 7.1.Let G be a group satisfying the covering condition (1.1).Assume G is soluble of derived length l.Then G has a nilpotent subgroup H such that [G : H] and the class of H are both (|S|, n, l)-bounded.
The proof of this proposition fills the rest of this section.Throughout, we say "bounded" to mean "(|S|, n, l)-bounded".
7.1.Reduction to the finite metabelian case.Recall Hall's criterion for nilpotency (see [R, 5.2.10]): if N is a normal subgroup of G such that N and G/N ′ are both nilpotent then G is nilpotent with class bounded in terms of the classes of N and G/N ′ .
By induction on derived length, G ′ has a nilpotent subgroup A of bounded index and class.Replacing A with its normal core in G ′ , we may assume A is normal in G ′ .Then it follows from Fitting's theorem that the characteristic closure of A has bounded class, so we may assume A is characteristic in G ′ .If A ′ = 1 then by induction on the class of A there is a nilpotent subgroup of G/A ′ of bounded index and class.By Hall's criterion we are done in this case, so we may assume A is abelian.Since |G ′ /A| is bounded, we may replace G with a bounded-index subgroup whose image in G/A has class at most 2.
In particular, G is abelian-by-class-2.By another classical result of Hall, any abelian-by-nilpotent group is locally residually finite (see [R, 15.4.1]).Applying Lemmas 4.4 and 4.5, we may assume G is finite.
Let g ∈ G and assume g is central modulo A. Then the subgroup [A, g] is normal in G and consists of commutators, so [A, g] It follows that Applying Theorem 4.3, there are subgroups B g ≤ [A, g] and T g ≤ G, both normal in G, such that the indices [ [A, g] and If we can show that U/U ′′ has a subgroup H/U ′′ of bounded index and class then H has bounded index in G and it would follow from Hall's criterion that H has bounded class.Thus Proposition 7.1 is reduced to the finite metabelian case.
For the rest of this section we assume G is a finite metabelian group and A = G ′ .We will continue to refer to the subgroups B g ≤ [A, g] and T g ≤ G, now valid for any g ∈ G since G/A is abelian.7.2.The case of p-groups.We use the notation [x, l y] for the long commutator [x, y, . . ., y] where y is repeated l times.If l = 0 then [x, l y] = x.An element y of a group Γ is l-Engel if [x, l y] = 1 for all x ∈ Γ, and Γ is l-Engel if every y ∈ Γ is l-Engel.The following lemma will be useful.
Proof.Both subgroups Γ ′ a and Γ ′ b are normal in Γ and nilpotent of class at most l.By Fitting's theorem, Γ ′ a, b is nilpotent of class at most 2l and the lemma follows.Now let G be a finite metabelian p-group obeying the covering condition (1.1).Let A = G ′ and B g ≤ [A, g] and T g ≤ G be as in the previous subsection.Since |[B g , T g ]| is bounded, there is a bounded i such that Z i (G) contains [B g , T g ] for every g ∈ G. Passing to the quotient G/Z i (G), we may assume [B g , T g ] = 1 identically.We may also assume that Proof.Since G/T g has bounded order, G = x 1 , . . ., x k T g for some x 1 , . . ., x k ∈ G and bounded k.Since G is metabelian and l-Engel, A x i has class at most l for each i, so A x 1 , . . ., x k has bounded class by Fitting's theorem.In particular B g x 1 , . . ., x k has bounded class.Since [B g , T g ] = 1, this shows that B g is contained in Z j (G) for some bounded j.Since [A, g]/B g has bounded order, [A, g] is contained in Z j ′ (G) for bounded j ′ .Since [A, g] ∈ Z j ′ (G) for all g, γ 3 (G) ≤ Z j ′ (G), so G has class at most j ′ + 2.
Thus it suffices to find a bounded-index subgroup of G which is l-Engel for some bounded l.Since [A, g]/B g is bounded, there is a bounded number f such that [A, f g] ≤ B g for all g ∈ G.For g ∈ G and i ≥ 0 write B i,g for [B g , i g], and note Proof.We may assume K = 1 and |A| and |Γ| are coprime.Since A is the direct product of its Sylow subgroups, which are preserved by Γ, we may assume A is a p-group for some prime p, and so Γ is a p ′ -group.Now see [A, (24.5)].
We now consider the case of Proposition 7.1 in which G is finite and metabelian, G/G ′ is a p-group, and G ′ is a p ′ -group.As before let A = G ′ and let B g , T g be normal subgroups of G such that B g ≤ [A, g], the indices [[A, g] : B g ] and [G : T g ] and the order of [B g , T g ] are bounded.Then C G ([B g , T g ]) has bounded index in G, so we may additionally assume that T g centralizes [B g , T g ] by replacing T g with C Tg ([B g , T g ]) if necessary.Similarly we may assume T g centralizes [A, g]/B g .But then two applications of the lemma show Hence we may assume B g = [A, g] and T g = C G ([A, g]).
Choose g ∈ G so that [G : it follows that T g is nilpotent (and in fact abelian, since G has no nonabelian nilpotent subgroups).Since G/G ′ is finite and abelian it is the direct product of its Sylow p-subgroups.For each prime p ≤ n let G p be the preimage in G of the Sylow p-subgroup of G/G ′ .For p > n, the Sylow p-subgroups of G are contained in G e .Hence G = G e p≤n G p , so by Fitting's theorem it suffices to prove the result for G = G p .Thus we may assume G/G ′ is a p-group.
Let P be a Sylow p-subgroup of G. Then G = G ′ P .By the subsection on p-groups, we can replace P with a subgroup of bounded index and bounded class, so without loss of generality P has bounded class c say.Since the action of G on G ′ factors through P , it follows that [P ∩ G ′ , c G] = 1, so P ∩ G ′ ≤ Z c (G). Factoring out Z c (G), we may thus assume G ′ is a p ′ -group.By the result of the coprime section we are done.

Bounded-class groups
We need the following result from [FAM], which is a local version of a classical result of Baer.
where A, B, C, a, b, c are bilinear maps and a, b, c have ǫ-bounded codomains.Conversely, if G has this structure then d 2 (G) is bounded away from zero by a function of [G : H], |γ 4 (H)|, and the size of the codomains of a, b, c.Much of the proof of Theorem 1.1 consists of a study of groups satisfying a certain commutator covering condition.If X and Y are subsets of a group G, let Comm(X, Y ) denote the set of commutators {[x, y] : x ∈ X, y ∈ Y }.Let n ≥ 1, let B = {x ∈ G : |x G | ≤ n}, and let S ⊆ G ′ .Then we say G satisfies the commutator covering condition (with given n and S) if Comm(G, G) ⊆ BS.
(2) If G is residually finite, G is m-by-class-s-by-n if and only if every finite quotient of G is so.Proof.(1) Suppose [G : H] ≤ n and |γ s (H)| ≤ m.Let Γ ≤ G be a subgroup.Then clearly [Γ : H ∩ Γ] ≤ n and |γ s (H ∩ Γ)| ≤ m, so the forward implication is clear.Now suppose every finitely generated subgroup Γ ≤ G is m-by-class-s-by-n.Let I be the set of finitely generated subgroups of G. Since any d-generated group has at most n! d subgroups of index at most n, the product space

4. 3 .
Multilinear bias and a 4-linear rank-reduction lemma.We need the following structure theorem for biased multilinear maps.If A 1 , . . ., A k are groups and I ⊆ [k] = {1, . . ., k} we write A I = i∈I A i and we write x I for the projection of x ∈ A [k] to A I .We write I c for the complement [k] \ I.If g is a (multilinear) function we write cod(g) for its codomain.

Theorem 5. 3 .
Let G be a finite group with an invariant seminorm • .Let A, B G be normal subgroups and suppose P a∈A,b∈B ( [a, b] ≤ C) ≥ 1/C.Then there are subgroups H ≤ A and K ≤ B, both normal in G, such that [A : H] and [B : K] are C-bounded and such that Comm(H, K) is covered by C-boundedly many balls of C-bounded radius.Conversely, if there are subgroups H ≤ A and K ≤ B such that [A : H], [B : K] ≤ C and such that Comm(H, K) is covered by at most C balls of radius C, then there is some C-bounded number D such that P a∈A,b∈B ( [a, b] ≤ D) ≥ 1/D.We will use the following lemma from [E1] multiple times.Lemma 5.4 (See [E1, Lemma 2.1]).Let G be a finite group and X a symmetric subset of G containing the identity.Then X = X r for r = 3 ⌊|G|/|X|⌋.Proof of Theorem 5.3.Let X = {a ∈ A : P b∈B ( [a, b] ≤ C) ≥ 1/2C}.Then 1/C ≤ P a∈A,b∈B ( [a, b] ≤ C) ≤ P(a ∈ X) + 1/2C, so P(a ∈ X) ≥ 1/2C.Fix x ∈ X.Let Y x = {b ∈ B : [x, b] ≤ C}.Then P b∈B (b ∈ Y x ) ≥ 1/2C.It follows that K x = Y x has C-bounded index in B and every k ∈ K x is the product of C-boundedly many y ∈ Y x by Lemma 5.4.Suppose k Corollary 5.5(Neumann's theorem, asymmetric version).Let G be a finite group with normal subgroups A, B G such that P a∈A,b∈B([a, b]  = 1) ≥ 1/C.Then there are C-bounded-index subgroups H ≤ A and K ≤ B, both normal in G, such that |[H, K]| is C-bounded.Proof.By Theorem 5.3 applied to the discrete norm, there are subgroups H ≤ A and K ≤ B of C-bounded index such that Comm(H, K) has C-bounded size.Now we may apply [DS1, Lemma 2.1] to conclude that |[H, K]| is C-bounded, or we may argue directly as follows.Let [h 1 , k 1 ], . . ., [h n , k n ] be an enumeration of the elements of Comm(H, K Corollary 5.6.Let G be a finite group such that d 2 (G) ≥ 1/C.Then there is a subgroup H ≤ G of C-bounded index satisfying the covering condition (1.1) with n and |S| both C-bounded.Proof.By Theorem 5.3 applied with A = B = G and the conjugacy class norm x = log |x G |, there is a subgroup H of C-bounded index such that Comm(H, H) is covered by C-boundedly many balls of C-bounded radius.Let S be a set containing one point from each ball.Then |S| is C-bounded and for every x, y ∈ H there is some s ∈ S such that [x, y]s −1 is C-bounded, so [x, y]s −1 has C-boundedly many conjugates.6.From covering to bounded derived length Throughout this section we assume the covering condition (1.1), which states Comm(G, G) ⊆ BS, where Comm(G, G) is the set of commutators of G, B = {x ∈ G : |x G | ≤ n}, and S ⊆ G ′ is finite.We will show that G is bounded-by-derived-length-4-by-bounded, with a bound depending only on |S| and n.Proposition 6.1.Let G be a group satisfying the covering condition (1.1).Then G has a subgroup H such that [G : H] and |H Lemma 6.3 (See [DS2, Lemma 2.1]).Let K ≤ H be a subgroup of index m in H, and let b ∈ H. Then the coset Kb contains an element g such that l(g) ≤ m − 1. Lemma 6.4.For any x ∈ B the subgroup [H, x] has n-bounded order.Consequently, the order of [H, x] G is n-bounded.Proof.Let m = [H : C H (x)] = |x H |, so m ≤ n.By Lemma 6.3 we can choose b 1 , . . ., b m such that x H = {x b1 , . . ., x bm } and l( : B g ] and [G : T g ] and the order of [B g , T g ] are bounded.Write G ′ = b 1 , . . ., b s A, where s is bounded.Since G/A has class 2, b 1 , . . ., b s are central mod A. Write T i for T bi , and 3, H is nilpotent of bounded class, as required.7.3.The coprime case.Lemma 7.4.Let A be a finite nilpotent group and let Γ be a group acting on A with kernel K. Assume |A| and |Γ/K| are coprime.Then [A, Γ] = [A, Γ, Γ].

7. 4 .
Finite metabelian groups in general.Let G be a finite metabelian group satisfying (1.1).Let e = n!.Then G e centralizes B. The group G e /Z(G e ) has at most |S| commutators so it has a subgroup of bounded index of nilpotency class at most 2, so G e has a bounded-index subgroup of class at most 3.