all1 - Math 4124 Wednesday, January 26 First Homework...

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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 ± = 0, then xy ± = 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 group ( Z / 36 Z ) × : 1, - 1, 5, 13, - 13, 17. Answer: 1,2,6,3,6,2. I won’t 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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Math 4124 Wednesday, February 1 Second Homework Solutions 1. 2.1.10(a) on page 48 Prove that if H and K are subgroups of the group G , then so is H K . Let e be the identity of G . Then e H , K because H and K are subgroups of G , consequently e H K . Next let x , y H K . Then xy H because H G and xy K because K G . Therefore xy H K . Finally x - 1 H and K because H and K are subgroups of G , and we deduce that x - 1 H K . It now follows that H K is a subgroup of G as required. 2. 1.2.3 on page 27 Use the standard generators and relations to show that every ele- ment of D 2 n which is not a power of r has order 2. Deduce that D 2 n is generated by the two elements s and sr , both of which have order 2. Each element of D 2 n can be written uniquely in the form r i or sr i , where 0 i n - 1 (see page 25). The elements r i are of course powers of r , so we need to prove that sr i has order 2. Since sr i ± = e , it will be sufficient to show that ( sr i ) 2 = e ; we shall show that this is true for all i . We have ( sr i ) 2 = sr i sr i . Now the relation rs = sr - 1 applied i times shows that r i s = sr - i , consequently ( sr i ) 2 = ssr - i r i = s 2 r - i + i = ee = e , which is what is required. Finally we need to show that D 2 n is generated by the two elements s and sr , both of which have order two. That they both have order two follows from the previous paragraph. Also every element of D 2 n can be written as a product of r , s , r - 1 , s - 1 , and since r = s - 1 ( sr ) , we see that every element can be written as a product of s , ( sr ) , s - 1 , ( sr ) - 1 . This establishes that D 2 n is generated by s and sr .
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all1 - Math 4124 Wednesday, January 26 First Homework...

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