Section 1.3. Groups 21 obtain an element of G which we call an ultimate product. For example, consider the expression abcd. We may first multiply ab, obtaining (ab)cd, an expression with three factors, namely, ab, c, d. We may now choose either the pair c, d or the pair ab, c in either case, multiply these, obtaining expressions with two factors: (ab)(cd) having factors ab and cd or ((ab)c)d having factors (ab)c and d. The two factors in either of these last expressions can now be multiplied to give an ultimate product from abcd. Other ultimate products derived from the expression abcd arise from multiplying bc or cd as the first step. It is not obvious whether the ultimate products from a given expression are all equal. Definition. Let G be a set with a binary operation. An expression a1a2 · · · an in G needs no parentheses if all its ultimate products are equal elements of G. Theorem 1.20 (Generalized Associativity). If G is a group having elements a1,a2,...,an, then the expression a1a2 · · · an needs no parentheses. Proof. The proof is by induction on n 3. The base step holds because the operation is associative. For the inductive step, consider two ultimate products U and V obtained from a given expression a1a2 · · · an after two series of choices: U = (a1 · · · ai)(ai+1 · · · an) and V = (a1 · · · aj)(aj+1 · · · an) the parentheses indicate the penultimate products displaying the last two factors that multiply to give U and V , respectively there are many parentheses inside each of these shorter expressions. We may assume that i j. Since each of the four expressions in parentheses has fewer than n factors, the inductive hypothesis says that each of them needs no parentheses. It follows that U = V if i = j. If i j, then the inductive hypothesis allows the first expression to be rewritten as U = (a1 · · · ai) ([ai+1 · · · aj][aj+1 · · · an]) and the second to be rewritten as V = ([a1 · · · ai][ai+1 · · · aj]) (aj+1 · · · an), where each of the expressions a1 · · · ai, ai+1 · · · aj, and aj+1 · · · an needs no parenthe- ses. Thus, these expressions yield unique elements A, B, and C in G, respectively. The first expression gives U = A(BC) in G, the second gives V = (AB)C in G, and so U = V in G, by associativity. Corollary 1.21. (i) If a1,a2,...,ak−1,ak are elements in a group G, then (a1a2 · · · ak−1ak)−1 = ak −1ak−1 −1 · · · a2 −1a1 −1. (ii) If a G and k 1, then (ak)−1 = a−k = (a−1)k. Proof. (i) The proof is by induction on k 2. Using generalized associativity, (ab)(b−1a−1) = [a(bb−1)]a−1 = (a1)a−1 = aa−1 = 1
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