Bulletin of the London Mathematical Society 2005 37(1):80; doi:10.1112/S0024609304003686
Replacement of Factors by Subgroups in the Factorization of Abelian Groups
Department of Mathematics, Dundee University Dundee, DD1 4HN, United Kingdom
Received 21 January 2004.
Abstract
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 Mk must be replaced by a coset Mk + hk. This is the only change needed in this proof, since the sum Mk+hk+(nB
H) being direct implies that the sum Mk+(nBH) is also direct. Since the zero element does belong to the sets (mAH) and (nBH), 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.