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# aab - contact email donsen2 at hotmail.com Contemporary...

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contact email: donsen2 at hotmail.com Contemporary abstract algebra Contents 1 Finite groups 1 2 Isomorphisms 3 3 Cosets and Lagrange’s Theorem 4 4 External direct products 4 5 Normal subgroups and Factor groups 5 6 Group homomorphisms 6 7 Introduction to rings 7 8 Integral domains 9 1 Finite groups Page67:2 Let Q be the group of rational numbers under addition and let Q * be the group of nonzero rational numbers under multiplication. In Q, list the ele- ments in < 1 2 > . In Q * , list the elements in < 1 2 > . < 1 2 > = { . . . , - 3 1 2 , - 2 1 2 , - 1 1 2 , 0 , 1 1 2 , 2 1 2 , 3 1 2 . . . } in Q < 1 2 > = { . . . , ( 1 2 ) - 3 , ( 1 2 ) - 2 , ( 1 2 ) - 1 , ( 1 2 ) 0 , ( 1 2 ) 1 , ( 1 2 ) 2 , ( 1 2 ) 3 , . . . } in Q * Page67:4 Prove that in any group, an element and its inverse have the same order. Assume that G is a group and a G . Then we separate the discussion by two parts case, 1: finite group and case 2: infinite group. Let’s see case 1 first. a has finite order (say) n . It means that a n = e a n = e = ( a n * a - n ) = a n ( a - 1 ) n It gives us that a - 1 have at most order n . If we let a - 1 have k order such that k < n , then 1

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e = e k = ( a k * a - k ) = a k ( a - 1 ) k = a k But we know that k cannot be the order of a . Hence | a - 1 | = n . Next, see infinite order case. Let a has infinite order and a - 1 dose not, then we can say that | a - 1 | = n . Moveover finite inverse of a - 1 that is ( a - 1 ) - 1 has same number of order. But this cannot happen. Thus a - 1 has infinite order. Page67:6 Let x belong to a group. If x 2 6 = e and x 6 = e , prove that x 4 6 = e and x 5 6 = e . What can we say about the order of x ? Obviously, x 6 = e because x n = e for all x Z . Then we can determine that x 6 = e = x 4 · x 2 = x 2 if x 4 = 2. Also x 6 = x 5 · x = x = e if x 5 = e . Those cases are not true so that x 4 6 = e and x 5 6 = e . Further we can say x 3 = e and x 6 = e . That’s, x has order of 3 either 6. Page67:10 Prove that an Abelian group with two elements of order 2 must have a sub- group of order 4. Let G be an Abelian group with distinct elements a, b such that a 2 = b 2 = e . Then the set of H = { e, a, b, ab } has order 4 and it is the subgroup of G by Finite subgroup test. Page67:12 Suppose that H is a proper subgroup of Z under addition and H contains 18 , 30, and 40. Determine H . As it stated, H is closed under addition, H must be linear combination of 18 , 30, and 40. We know that gcd (18 , 30 , 40) = 2 so that 2 = 18 r + 30 s + 40 t for some integers r, s, and t . This means that 2 H but H 6 = Z . That’s H = 2 Z . Page67:15 Let G be a group. Show that Z ( G ) = T a G C ( a ) . Suppose x Z ( G ). Then x commutes with every a G and x C ( a ) for all a G . This means that x T a G C ( a ) and Z ( G ) T a G C ( a ). Conversely, suppose x T a G C ( a ), this implies that x C ( a ) = { y G : ay = ya } for all a G and x commutes with all a G , i.e., x Z ( g ). Page67:18 If a and b are distinct group elements, prove that either a 2 6 = b 2 , or a 3 6 = b 3 . This problem requires us to prove that one a 2 6 = b 2 and if not, prove an- 2
other a 3 6 = b 3 . We just need to prove one of the statements at least.(NOT mutually exclusive) Let’s see the a 2 6 = b 2 .When it is true, then nothing to prove. But assume that a 2 = b 2 with distinct elements a 6 = b . It goes like this.

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aab - contact email donsen2 at hotmail.com Contemporary...

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