ahw1 - group ( Z / 36 Z ) : 1,-1, 5, 13,-13, 17. Answer:...

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Math 4124 Wednesday, January 26 First Homework Solutions 1. 1.1.9(b) on page 22 Let G = { a + b 2 R | a , b Q } . Prove that the nonzero elements of G form a group under multiplication. If x , y G then we may write x = a + b 2 and y = c + d 2 where a , b , c , d Q , and then xy = ( ac + 2 bd )+( ad + bc ) 2. Also if x , y 6 = 0, then xy 6 = 0. It follows that multiplication is a binary operation on G \ 0. Also multiplication is associative and the identity is 1. Finally we need to check for inverses. If a + b 2 G \ 0, then the inverse will be 1 / ( a + b 2 ) ; the only problem we might have is that this is not obviously in G \ 0. However by multiplying top and bottom by a - b 2, this is a a 2 - 2 b 2 - b 2 a 2 - 2 b 2 . Since a 2 - 2 b 2 is a nonzero rational number, it is now clear that the inverse is in G \ 0 (if a 2 - 2 b 2 = 0, then 2 Q ) and the result follows. 2. 1.1.14 on page 22 Find the orders of the following elements of the multiplicative
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Unformatted text preview: group ( Z / 36 Z ) : 1,-1, 5, 13,-13, 17. Answer: 1,2,6,3,6,2. I wont give explanations for all the answers here, just for 5. We have modulo 36, 5 1 = 5, 5 2 = 25, 5 3 = 17, 5 4 = 13, 5 5 = 29, 5 6 = 1. Therefore the least positive power of 5 which is the identity is 6, consequently the order of 5 is 6. 3. 1.1.25 on page 22 Prove that if G is a group and x 2 = 1 for all x G , then G is abelian. Let x , y G . Then x 2 = y 2 = ( xy ) 2 = 1. Therefore x 2 y 2 = 1 = ( xy ) 2 = xyxy . Multiplying by x-1 on the left and y-1 on the right, we obtain xy = yx as required....
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