446 APPENDIX A. APPENDIX Reflexivity : Let (a1, a2, a3) ∈ Z3. Then since a1 + a2 + a3 = a1 + a2 + a3, we use the lexicographic order to compare the entries but all entries are equal and this satisfies the lexicographic order. Hence (a1, a2, a3) ≼ (a1, a2, a3). Antisymmetry : Suppose that (a1, a2, a3) ≼ (b1, b2, b3) and (b1, b2, b3) ≼ (a1, a2, a3). Assume that a1 + a2 + a3 ̸= b1 + b2 + b3, then a1 + a2 + a3 < b1 + b2 + b3 and b1 + b2 + b3 < a1 + a2 + a3, which is a contradiction. Hence we must have a1 + a2 + a3 = b1 + b2 + b3. Thus, we compare the triples by lexicographic order. However, the lexicographic order is a partial order, which is antisymmetric so we can conclude that (a1, a2, a3) = (b1, b2, b3). Transitivity : Suppose that (a1, a2, a3) ≼ (b1, b2, b3) and (b1, b2, b3) ≼ (c1, c2, c3). We can break the situation into three cases. If a1 + a2 + a3 = b1 + b2 + b3 = c1 + c2 + c3, then we compare the triples by lexicographic order, which is transitive so (a1, a2, a3) ≼ (c1, c2, c3). If a1 + a2 + a3 = b1 + b2 + b3 ̸= c1 + c2 + c3 (resp. a1 + a2 + a3 ̸= b1 + b2 + b3 = c1 + c2 + c3) then a1 + a2 + a3 < c1 + c2 + c3 so (a1, a2, a3 ≼ (c1, c2, c3). Finally, if all thee of the sums are different, then we deduce that a1 + a2 + a3 < b1 + b2 + b3 < c1 + c2 + c3 so we can conclude that (a1, a2, a3 ≼ (c1, c2, c3). We conclude that ≼ is transitive. Finally, to show that ≼ is a total order, let (a1, a2, a3), (b1, b2, b3) ∈ Z3 be arbitrary, distinct elements. There are two cases. If a1 + a2 + a3 ̸= b1 + b2 + b3, then either a1 + a2 + a3 < b1 + b2 + b3, which implies that (a1, a2, a3) ≼ (b1, b2, b3), or the reverse is true. If a1 + a2 + a3 = b1 + b2 + b3, then we compare the triples by the lexicographic order. In this case, let i be the least index for which ai = ̸ bi. Then (a1, a2, a3) ≼ (b1, b2, b3) if ai < bi and (b1, b2, b3) ≼ (a1, a2, a3) if bi < ai. Hence, every pair of elements is comparable and the relation is a total order. Exercise: 9 Section A.4 Question: Let (Ai, ≼i) be posets for i = 1, 2, . . . , n and define ≼lex as the lexicographic order on A1×A2×· · · An. Prove that ≼lex is a total order if and only if ≼i is a total order on Ai for all i. Solution: First suppose that ≼i is a total order on Ai for all i. Let (a1, a2, . . . , an) and (b1, b2, . . . , bn) be elements in A1 × A2 × · · · An. If ai = bi for i = 1, . . . , n, then (a1, a2, . . . , an) ≼lex (b1, b2, . . . , bn). Otherwise, let j be the least index for which ai = bi. Since all ≼i are total orders, then ≼j is a total order. Thus, either aj ≼j bj or bj ≼j aj. Hence, (a1, a2, . . . , an) ≼lex (b1, b2, . . . , bn) or the reverse is true. Thus ≼lex is a total order. Now suppose that ≼lex is a total order. Let j be any index with 1 ≤ j ≤ n and let aj, bj ∈ Aj. Consider a pair of n-tuples (a1, a2, . . . , an) and (b1, b2, . . . , bn) such that ai = bi for all i < j. Since (a1, a2, . . . , an) and (b1, b2, . . . , bn) are comparable in ≼lex, then aj ≼j bj or bj ≼j aj. Since aj, bj were arbitrary, then ≼j is a total order. Thus, we conclude that all the partial orders ≼i are partial orders. Exercise: 10 Section A.4 Question: Let ≼ be the lexicographic order on R3, where each R is equipped with the usual≤ . Prove or disprove the following statement: For all vectors a⃗, b⃗, c⃗, d⃗, if a⃗ ≼ b⃗ and c⃗ ≼ d⃗, then a⃗ + c⃗ ≼ b⃗ + d⃗. Solution: We first prove that if a⃗ ≼ b⃗, then a⃗ + c⃗ ≼ b⃗ + c⃗. With the lexicographic order, a⃗ ≼ b⃗ if and only if the first index j ∈ {1, 2, 3} for which aj ̸= bj has aj < bj. For any c⃗ ∈ R3, for all indices ai = bi if and only ai + ci = bi + ci. Hence the first index for with ai + ci ̸= bi + ci is j and then we have aj + cj < bj + cj. Thus a⃗ + c⃗ ≼ b⃗ + c⃗. Since the above result was arbitrary, a⃗ ≼ b⃗ implies a⃗+c⃗ ≼ b⃗+c⃗ and c⃗ ≼ d⃗ implies c⃗+b⃗ ≼ d⃗+b⃗. By transitivity of ≼, we deduce that a⃗ + c⃗ ≼ b⃗ + c⃗ ≼ b⃗ + d⃗. Exercise: 11 Section A.4 Question: Answer the following questions pertaining to the poset described by the Hasse diagram below. 1. List all the minimal elements. 2. List all the maximal elements. 3. List all the maximal elements in the subposet with {a, b, c, d, e, f, g}. 4. Determine the length of the longest chain and find all chains of that length. Solution Manual for Abstract Algebra A First Course (Textbooks in Mathematics) 2nd Edition by Stephen Lovett