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math 435 homework5

# math 435 homework5 - 1 1 x π n 1 n 1 π i ∈ Z and thus...

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Homework 5 for MATH 435 Solutions Problem 1 Book, p. 169, Exercise 2.66 Problem 2 Book, p. 169, Exercise 2.70 Problem 3 Book, p. 170, Exercise 2.78 Problem 4 Book, p. 190, Exercise 2.98 Solution. Let x , y G . Since G/Z ( G ) is cyclic, there exists g G such that G/Z ( G ) = { g m Z ( G ): m Z . Choose m , n Z and z 1 , z 2 Z ( G ) such that x = g m z 1 and y = g n z 2 . Then, using the fact that z 1 , z 2 commute with every element of G and the laws of exponentiation, xy = g m z 1 g n z 2 = g m g n z 1 z 2 = g n g m z 2 z 1 = g n z 2 g m z 1 = yx . Problem 5 Book, p. 190, Exercise 2.100 Solution. If G is generated by a single element, then G is cyclic and so is every subgroup. Now let H 6 G , and suppose G is generated by { x 1 , . . . , x n + 1 } . We claim that G/ < x n + 1 > is generated by n elements, namely the cosets x 1 < x n + 1 > , . . . , x n < x n + 1 > . Let g < x n + 1 > be a coset in G/ < x n + 1 > . Since < x 1 , . . . , x n + 1 > = G , and since G is abelian, we can write g = x
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Unformatted text preview: 1 ) 1 . . . x π ( n + 1 ) n + 1 , π ( i ) ∈ Z , and thus g < x n + 1 > = x π ( 1 ) 1 . . . x π ( n + 1 ) n + 1 < x n + 1 > = x π ( 1 ) 1 . . . x π ( n ) n < x n + 1 > . Now consider the set H < x n + 1 > = { h < x n + 1 > : h ∈ H } . It is not hard to see that this is a sub-group of G/ < x n + 1 > (using the fact that H is a subgroup of G ). By inductive hypothesis H < x n + 1 > is ﬁnitely generated, say by { y 1 , . . . y k } . Given h ∈ H , we have that h ∈ h < x n + 1 > . Thus h can be written as y π ( 1 ) 1 . . . y π ( k ) k x m n + 1 , with π ( i ) , m ∈ Z . Hence H is ﬁnitely generated. ±...
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