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# HW 01 - MAT 322 Algebra with Galois Theory Claus Sorensen...

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MAT 322: Algebra with Galois Theory Claus Sorensen Problem Set 1 Rik Sengupta September 25, 2009 1. Let G be a group. Prove the generalized associative law: Given n elements a 1 , . . . , a n G , all ways of bracketing this ordered sequence to give it a value by calculating a succession of binary products yield the same value, denoted a 1 . . . a n . How many ways can this be done? We will prove the result by induction. The case n = 3 is obvious, because the two ways of bracketing the sequence are ( a 1 a 2 ) a 3 and a 1 ( a 2 a 3 ), which are equal, by assumption. We denote this by a 1 a 2 a 3 , and the product is unambiguously defined. Suppose the result holds for all n up to, say, m - 1. We will show that the result also holds for n = m . To this end, suppose we multiply in two ways and end up with the two results: ( a 1 a 2 . . . a i )( a i +1 . . . a m ), and ( a 1 a 2 . . . a j )( a j +1 . . . a m ), where i 6 = j . Note that this can always be done, because each of the terms within the parentheses contains fewer than m of the a k , so by our inductive hypothesis, their product is unambiguously defined. Note that we have taken i 6 = j because i = j simply implies the two products are equal, and there remains nothing to prove. So suppose without loss of generality that i < j . Then, note that, using our inductive hypothesis, ( a 1 a 2 . . . a i )( a i +1 . . . a m ) = ( a 1 a 2 . . . a i )(( a i +1 . . . a j )( a j +1 . . . a m )) and ( a 1 a 2 . . . a j )( a j +1 . . . a m ) = (( a 1 a 2 . . . a i )( a i +1 . . . a j ))( a j +1 . . . a m ) where the expressions on the right are equal from the associative law from the definition of a group (i.e. A * ( B * C ) = ( A * B ) * C ). 1

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So, since the expressions on the right are equal, it follows that the two results are in fact the same, i.e. multiplying the elements in different ways by changing the sequence of parentheses yields the same result in each case. We denote this by a 1 a 2 . . . a m . Hence, inductively, it is obvious that all ways of bracketing this ordered sequence to give it a value by calculating a succession of binary products yield the same value, denoted a 1 a 2 . . . a m . Lemma. The n th Catalan number C n is given by C n = (2 n )! ( n +1)! n ! Proof. Let the n th Catalan number be defined as the number of distinct ways in which an ordered sequence of n + 1 elements can be bracketed completely. It is rather obvious that C 0 = 1, and C n +1 = n i =0 C i C n - i for n 0. The recurrence can simply be obtained by noticing that the last pair of elements to be bracketed consists of two terms, one of length i and the other of length n - i , where i runs from 1 to n. The generating function for the Catalan numbers is defined by c ( x ) = n =0 C n x n . From the conditions and the recurrence relation given, we expand both sides to power series to readily obtain c ( x ) = 1 + xc ( x ), whence we get c ( x ) = 1 ± 1 - 4 x 2 x where the plus sign can be easily eliminated because it has a pole at 0. Thus, c ( x ) = 1 - 1 - 4 x 2 x .
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HW 01 - MAT 322 Algebra with Galois Theory Claus Sorensen...

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