Binary Operations ProofDate: 02/26/2009 at 22:31:35 From: David Subject: Binary operations Let s be a finite set with an associative binary operation * (it is not given that the operation has an identity). Prove that there is an element such that a * a = a. I know how to show this for a case but how do you prove it in general? I have tried setting up a table to show examples of where s is associative and each element * each element is itself. Date: 02/27/2009 at 06:23:07 From: Doctor Jacques Subject: Re: Binary operations Hi David, As the operation is associative, we may write "powers" of an element, i.e., expressions like a^n (where n is a strictly positive integer), without ambiguity. Furthermore, the usual "law of exponents" holds: a^(m+n) = (a^m)(a^n) although simplification is not allowed; if a^(m+k) = a^(n+k) where k is a positive integer, we are not allowed to conclude that a^m = a^n. Pick any element "a" in S and consider the infinite sequence of elements of S: a, a^2, a^3, ... As S is a finite set, there is at least one element b that appears infinitely often in the sequence. The means that there are at least two positive integers m and n, with m < n, such that: b = a^m = a^n [1] Note that [1] implies that: a^(m+k) = a^(n+k) [2] for all k >= 0. Can you continue from here? Don't forget that we must have k >= 0 in [2], since simplification is not allowed. Please feel free to write back if you require further assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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