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

This preview shows pages 1–3. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 24

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online