... elements of A have orders in {n/2, n/3}. For A to be a basis, we must have an element of each of these orders. Therefore, 6 | n. Multiplying by a unit, wemay assume {0, 2, 3t} ⊆ A. Then A − 2 contains ... Swedenhegarty@chalmers.seSarada Herke‡Mathematics and StatisticsUniversity of Victoria, Victoria BC, Canada V8W3R4sarada@uvic.caSubmitted: Oct 5, 2009; Accepted: May 18, 2010; Published: May 25, 2010Mathematics ... n) and argue as before. Thiscompletes the proof of the lemma for bases {0, 1, t}.Now let us deal with the general case of a 3-element basis A = {0, a, b}. Again, fix k ∈ N, letn be very large...