Maximal sum-free sets in abelian groups of order divisible by three: Corrigendum
Anne Street · Bulletin of the Australian Mathematical Society · 1972
The last step of the proof in [2] was omitted. To complete the argument, we proceed in the following way. We had shown that H = H(S) = H(S+S) = H(S-S), that |S-S| = 2|S| - |H| and hence that in the factor group G* = G/H of order 3m, the maximal sum-free set S* = S/H and its set of differences S* - S* are aperiodic, with so that By (1) and Theorem 2.1 of [1], S* - S* is either quasiperiodic or in arithmetic progression.