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# practice1 - MATH 4107 PRACTICE PROBLEMS WITH SOLUTIONS Here...

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PRACTICE PROBLEMS WITH SOLUTIONS January 17, 2007 Here are a few practice problems on groups. Try to work these WITHOUT looking at the solutions! After you write your own solution, you can compare to my solution. Your solution does not need to be identical—there are often many ways to solve a problem—but it does need to be CORRECT. 1. Suppose that G is a set and that * is an operation on G that satisFes the following conditions: a. G is closed under * . b. * is associative. c. There exists an element e G such that a * e = a for every a G . d. ±or each a G there exists an element b G such that a * b = e . Show that G is a group. NOTE: You do NOT know that G is a group, that is what you are trying to prove. You are given that SOME of the requirements of a group are satisFed, and you have to prove that the remaining requirements are also satisFed. You do NOT know that each element has an inverse, that is part of what you are trying to prove. Solution Here, and in most instances involving generic group notation, I’ll omit writing the symbol * ” explicitly. The hypotheses of this problem tell us that most of the properties required for G to be a group are satisFed. We just have to show that the remaining properties are also satisFed. SpeciFcally, we have to show the following two things. (1) ea = a for every a G . (2) ±or each a G , the element b G which satisFes ab = e ALSO satisFes ba = e . Let us work on statement (2) Frst. Let a be any element of G . Then we know by assumption c that ae = a , and we know by assumption d that there is an element b G such that ab = e . ±urther, by assumption d there is an element c such that bc = e . However, we do NOT yet know whether c = a ! We cannot assume this! Instead, using the associativity of the operation, we compute that ba = ( ba ) e = b ( ae ) = b ( a ( bc )) = b (( ab ) c ) = b ( ec ) = ( be ) c = bc = e. This establishes statement (2). Now let us prove that statement (1) is true. Let a G be given. We have to show that ea = a . This follows from the calculation ea = ( ab ) a = a ( ba ) = ae = a. ± 1

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practice1 - MATH 4107 PRACTICE PROBLEMS WITH SOLUTIONS Here...

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