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Partial Solutions to Homework 07

# Partial Solutions to Homework 07 - Partial Solution...

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Partial Solution Sketches, Math 74 HW7 (7.1a) Not associative: a - ( b - c ) = a - b + c = ( a - b ) - c . (7.1b) No inverse: 2 - 1 = 1 / 2 / Z . (7.1c) Group. (Check that the axioms hold. The most important one is closure under addition – how do you know a/b + c/d has an odd denominator if b, d are odd?) (7.3) Let a, b G . ( ab ) 2 = a 2 b 2 ( ab )( ab ) = abab abab = aabb groups are associative a - 1 ababb - 1 = a - 1 aabb - 1 group elements have inverses ba = ab Hence G is abelian. (7.4) Since a group must have an identity, we may write G = { e, x, y } . Trivially, x · e = e · x and y · e = e · y . Moreover, since the identity is unique, xy = x (as y = e ) and xy = y (as x = e ). Hence xy = e . By the same argument, yx = e , so xy = yx as desired. (7.5) Let G = { a 1 , . . . , a n } be a group of order n . For a i G , consider a i , a 2 i , a 3 i , . . . , a n +1 i , which are n + 1 elements among the n elements of G by closure of groups. By the pigeonhole principle, there exist t, s such that a t i = a s i and t > s . Then a t - s i = e ; define N i = t - s . Now consider N = n i N i . For each a i G , we then have a N i = a N 1 ··· N i ··· N n i = a N i · j = i N j i = ( a N
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