By Dan Segal

After a forty-year lull, the examine of word-values in teams has sprung again into existence with a few wonderful new leads to finite crew idea. those are principally inspired by means of purposes to profinite teams, together with the answer of an outdated challenge of Serre. This e-book offers a finished account of the identified effects, either previous and new. The extra undemanding tools are constructed from scratch, resulting in self-contained proofs and enhancements of a few vintage effects approximately endless soluble teams. this is often through an in depth creation to extra complicated subject matters in finite team concept, and a whole account of the functions to profinite teams. the writer provides proofs of a few very fresh effects and discusses open questions for additional study. This self-contained account is obtainable to investigate scholars, yet will curiosity all study employees in workforce thought.

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**Additional resources for Words: Notes on Verbal Width in Groups (London Mathematical Society Lecture Note Series)**

**Example text**

Then H = P1 × · · · × Pq where for each i the group Pi is a pi -group of ﬁnite rank ri , say, and p1 , . . , pq are distinct primes. Let πi : H → Pi denote the natural projection. Now let U be a ﬁnitely generated subgroup of H. Then U ≤ Y for some ﬁnite subset Y of S. Since Y πi is a pi -group, the generating set Y πi contains a minimal generating set of size at most ri , so Y contains a subset Y (i) of size at most ri such that Y πi = Y (i) πi . Putting Z = Y (1) ∪ . . ∪ Y (q) we then have Z πi = Y πi for each i.

Ek ) ≥ 2. Then every dth power is a w-value, so w(F ) contains the element n (xj y)(j −1)d (xy)d gn = (xy)d j =2 for each n ≥ 1. Now βr+ (gn−1 ) = 0 for each r, while r=1 n(d − 1) for 1 for r = d + 1, 2d + 1, (n − 1)d + 1 . βr+ (gn ) = 0 else Hence φ(gn ) ≥ n − 1 for each n. This shows that φ is unbounded on w(F ). Thus to show that w has inﬁnite width in F , it will suﬃce to establish that φ is bounded on each of the sets Fw∗m . Now w = xe1 1 . . xekk v1 . . vq where each vj is a commutator in Fk ; so if ui ∈ F (k ) (i = 1, .

We will need some elementary combinatorial results. 2 Let G be a nilpotent group of class at most c, and let n ∈ N. (i) If G = X and H = xn | x ∈ X then Gn c (c + 1)/ 2 ≤ H. (ii) If x, y ∈ G and xn = y n then (x−1 y)n = 1. c (iii) (G )n 2 ( c −1 ) ≤ (Gn ) . Proof. Write Gi = γi (G). (i) This is clear if c = 1. Let c > 1 and suppose c ( c −1 ) / 2 ≤ HGc . Now Gc is central in G, and is generated inductively that Gn c by elements [x1 , . . , xc ] with each xi ∈ X, and [x1 , . . , xc ]n = [xn1 , .

### Words: Notes on Verbal Width in Groups (London Mathematical Society Lecture Note Series) by Dan Segal

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