Premium
Complements of Normal Subgroups in Infinite Groups
Author(s) -
Dixon John D.
Publication year - 1968
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/s3-18.4.768-s
Subject(s) - counterexample , mathematics , lemma (botany) , mathematical proof , group (periodic table) , combinatorics , normal subgroup , discrete mathematics , ecology , chemistry , geometry , poaceae , organic chemistry , biology
A study group at the University of Utah led by Professors William Scott and Fletcher Gross with Mitchell Billis, Kendell Hyde, and Nelson Dinerstein taking part has discovered a number of non‐trivial errors in this paper. I am indebted to the members of this study group for the following corrections. 1. In the statements of Lemma 1 and Theorem 1 the hypothesis ‘ A possesses unique n th roots’ should be replaced by ‘ A is a π‐subgroup and n is a π′‐number’. All proofs remain essentially unchanged, and the new hypothesis is satisfied in all applications made of these results in the paper. The error in the proof of Lemma 1 as it stood occurs in the assumption that a factor group of a group possessing unique n th roots also possesses unique nth roots. Billis and Hyde have given a counterexample to the original formulation of Lemma 1, but they point out that Theorem 1 in its original formulation can be proved by different means. In particular, part (i) of Theorem 1 may be proved by a minor modification of the proof of Theorem 9.3.5 in W. R. Scott, Group theory (Prentice‐Hall, 1964). 2. There is an error in the proof of Theorem 6, where the expression for k should read | N M (G * ): A * N M (K * )|. With this correction the proof of part (i) is valid. However, part (ii) of Theorem 6 is false; as a counterexample we have G = A × B with A ≃ Z (p ∞ ) and | B | = p . 3. The claim made in Note 1 on p. 441 that part (i) of Theorem 6 remains true when H is locally nilpotent is probably incorrect. At any rate the suggested proof is incomplete since Theorem 1 is not available in this case. Similarly, the validity of Example 5 on p. 443 is doubtful because L 0 U is not a subgroup as claimed. 4. It is worth noting in Theorem 5 that if either π or π′ is finite then we can conclude that the π′‐subgroups of Aut A are also finite. Moreover, the proof of Theorem 7 is unduly complicated; we can simply define α on A as the identity and on K by Ax α = Ax ( x ɛ K, x α ɛ L ), and then verify that α extends to an automorphism of G .