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Replacement of Factors by Subgroups in the Factorization of Abelian Groups
Author(s) -
Sands A. D.
Publication year - 2005
Publication title -
bulletin of the london mathematical society
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 2.396
H-Index - 48
eISSN - 1469-2120
pISSN - 0024-6093
DOI - 10.1112/s0024609304003686
Subject(s) - coset , mathematics , mathematical proof , abelian group , element (criminal law) , paragraph , order (exchange) , factorization , zero (linguistics) , discrete mathematics , combinatorics , pure mathematics , algebra over a field , law , algorithm , linguistics , philosophy , geometry , finance , political science , economics
The above‐titled paper of mine appeared in the Bulletin of the London Mathematical Society , 32 (2000) 297–304. Regrettably, there is a careless error in the proofs of Theorems 6 and 8. In line 6 of the proof of Theorem 6, it is claimed that a certain subset must be a subgroup. For this to hold, the subset must contain the zero element. This need not be the case; the true deduction is that the subset is a coset, say M + h , of a subgroup M . Now M and M + h contain the same number of elements, and so the deduction that M has p elements is still correct. Similarly, in the proof of Theorem 8, the subgroup M k must be replaced by a coset M k + h k . This is the only change needed in this proof, since the sum M k + h k +( nB ∩ H ) being direct implies that the sum M k +( nB ∩ H ) is also direct. Since the zero element does belong to the sets ( mA ∩ H ) and ( nB ∩ H ), the statements about these sets are correct. So the second paragraph of the Proof of Theorem 8 is correct, and is also a proof of Theorem 6. Now we present an example that, we hope, will clarify the situation, as well as showing that certain statements in the original ‘Proof’ of Theorem 6 not only could be wrong but actually are wrong. The smallest numerical example occurs with p = 2, m = 3, n = 5. Then G is a cyclic group of order 60, and may be represented as the integers modulo 60. Let A = {0, 1, 2, 3, 4} + {0, 15} and B = {0, 5, 10} + {0, 30}. It is easily verified that A + B = {0, 1,…, 59}. In the notation of Theorem 6, we see that H = {0, 15, 30, 45}, K = {0, 12, 24, 36, 48}, L = {0, 20, 40}, and M = {0, 30}. Now we see that mA = {0, 3, 6, 9, 12} + {0, 45} ≠ M + K , and that nB = {0, 25, 50} + {0, 30} ≠ M + L . We note, however, that A is a complete set of residues modulo 10; that is, that B can be replaced by M + L .