Let G be a group a1 a2 ak G Prove by induction that the

Let G be a group, a₁, a₂, . . . , a_k G. Prove by induction that the value of a₁a₂ . . . a_k is independent of how the expression is bracket.

Solution

Let P_k denotes an expression that we get by bracketing the k elements, keeping the order of elements same. We need to show all expressions P_k are same for all k. We will use induction over k. For k = 1 and k = 2, we have only one expression possible, so trivially all expressions are same. For k = 3, we have two ways of bracketing a_1a_2a_3, i.e a_1 ·(a_2 · a_3) and (a_1 · a_2)· a_3. Since the binary operation · is satisfying associativity( since, G is group), therefore both expression are same. Next suppose the result is true for k i1. We need to show that the result is equally true for k = i. Let P_i be some expression. Then it must be product of some two shorter expressions, i.e P_i = Q_ · R_(i), where < i. But all expressions with number of elements less than i are same. So Q_ = a1 · Q\'_( 1) , where Y\'_( 1) is some expression consists of 1 elements. Similarly, R_(i) = a_(+1) · Z\'_(i1) . Thus we have P_i = Q_ · R_(i) = (a_1 · Y\'_(1) ) · (a_(+1) · R\'_(i1) ) = a_1 · P\'_(i1) , where P\'_(i1) is some expression of i 1 terms. But all expressions containing i 1 terms are same (order of elements is assumed to remain same). But then all expression having i elements turns out to be equal to a_1 · X\'_(i1) . Thus all expressions having i elements are same. The induction hypothesis implies result is valid for all k belogs to N. Hence the result.