z-logo
Premium
The Quotient of a Free Product of Groups by a Single High‐Powered Relator
Author(s) -
Howie James
Publication year - 1993
Publication title -
proceedings of the london mathematical society
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 1.899
H-Index - 65
eISSN - 1460-244X
pISSN - 0024-6115
DOI - 10.1112/plms/s3-66.3.538-s
Subject(s) - mathematics , combinatorics , quotient , word (group theory) , permutation (music) , product (mathematics) , cohomology , closure (psychology) , group (periodic table) , free product , wreath product , permutation group , normal subgroup , discrete mathematics , pure mathematics , geometry , market economy , chemistry , physics , organic chemistry , acoustics , economics
Let A, B be groups, and r ∈ A * B a cyclically reduced word of the form xUyU −l for some word U and letters x, y, where x 2 = y 3 = l. We call this form exceptional . If m = 4 or m = 5, and G is the one‐relator product group ( A * B )/ N ( r m ) (where N (·) denotes normal closure), then it was stated in [ 2 , 3 ] (Theorem E) that the cohomology of G is linked to that of A * B and of the finite triangle group G (2, 3, m ) = 〈 a, b | a 2 = b 3 = ( ab ) m = 1〉 by a certain Mayer–Vietoris sequence. The proof of this theorem involves so‐called spherical pictures over G that are induced from the finite subgroup G (2, 3, m ). This proof fails if r is multiply exceptional, in the sense that both r and some cyclic permutation of r have exceptional forms, because then more spherical pictures arise than are dealt with in [2, 3]. An example of such a word r is ababab −1 where A = 〈 a | a 2 = 1 〉 and B = 〈 b | b 3 = 1 〉, since both r and its cyclic permutation ab −l abab are exceptional. Theorem E of [2, 3] is correct only under the stronger hypothesis that r is uniquely exceptional (i.e. not multiply exceptional). This problem does not affect the other results of [2,3]. A more detailed discussion is given in [1].

This content is not available in your region!

Continue researching here.

Having issues? You can contact us here