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Mathematics 541Homework

October 25, 2000

Chapter 6, Problem 4. Show that U(8) is not isomorphic to U(10).

Answer: We know that U(8) ={1, 3, 5, 7} and U(10) ={1, 3, 7, 9}. In U(8), some compu-tation shows that 3, 5, and 7 all have order 2, while in U(10), 3 and 7 have order 4 and9 has order 2. Since there are differing numbers of elements of order 2, we see that U(8)cannot be isomorphic to U(10).

If you prefer, you can observe that U(10) is cyclic (generated either by 3 or 7), whileU(8) is not cyclic, so the groups cannot be isomorphic.

Chapter 6, Problem 5. Show that U(8) is isomorphic to U(12).

Answer: There doesnt seem to be any easy way to do this problem other than by trial anderror. In fact, any 1-1 map fromU(8) toU(12) that sends 1 to 1 will be an isomorphism. Totake a specific example, let : U(8) U(12) be defined by (1) = 1, (3) = 5, (5) = 7,and (7) = 11. Then writing out the Cayley tables shows that this is an isomorphism:

1 3 5 71 1 3 5 73 3 1 7 55 5 7 1 37 7 5 3 1

1 5 7 111 1 5 7 115 5 1 11 77 7 11 1 5

11 11 7 5 1

Chapter 6, Problem 11. Find two groups G and H such that G H, but Aut(G) Aut(H).

Answer: The simplest answer here is to take G = Z3 and H = Z4. Then we know thatGH (because Z3 and Z4 have different numbers of elements), but Aut(Z3) U(3) andAut(Z4) U(4). SinceU(3) andU(4) both have two elements, both groups are isomorphicto a cyclic group with 2 elements.

In view of the answer to problem 6.5 above, another answer would be Z8 and Z12,which are clearly not isomorphic. We know that Aut(Z8) U(8), and Aut(Z12) U(12).

Chapter 6, Problem 21. Prove or disprove thatU(20) and U(24) are isomorphic.

Answer: Both U(20) and U(24) contain 8 elements, so there is no obvious reason whythey cant be isomorphic. But if we compute orders of elements, we find that the orders

of elements in U(20) looks like this:

n 1 3 7 9 11 13 17 19o(n) 1 4 4 2 2 4 4 2

The orders of elements in U(24) looks like this:

n 1 5 7 11 13 17 19 23o(n) 1 2 2 2 2 2 2 2

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Therefore, the groups cannot be isomorphic.

Chapter 6, Problem 27. Let C be the complex numbers, and

M=

a bb a

a, b R

.

Prove that C and M are isomorphic under addition and that C and M, the nonzeroelements ofM, are isomorphic under multiplication.

Answer: Define a function (a+ bi) =

a bb a

. You can check that this map is a

bijection. To check that we have have an isomorphism under addition, we compute

((a + bi) + (c + di)) =((a + c) + (b + d)i) =

a + c (b + d)b + d a + c

= a bb a

+ c dd c

= (a + bi) + (c + di).To check that we have an isomorphism under multiplication, we compute

((a + bi)(c + di)) =(ac bd + (ad + bc)i) =

ac bd (ad + bc)ad + bc ac bd

=

a bb a

c dd c

= (a + bi)(c + di).

Chapter 6, Problem 30. Suppose thatG is a finite abelian group and G has no elementof order 2. Show that the mapping g g2 is an automorphism ofG. Show, by example,that ifG is infinite, the mapping need not be an automorphism.

Answer: Define (g) = g2. We see trivially that (ab) = (ab)2 = a2b2 = (a)(b), usingthe fact that Gis an abelian group.

To show that is 1-1, suppose that (a) = (b). Then a2 = b2, or a2(b1)2 = e, or(ab1)2 = e. Since G contains no elements of order 2, we can conclude that ab1 = e, ora= b.

Because Gis a finite group, we know that any 1-1 map from Gto Gis automaticallya surjection, and therefore is a bijection. That makes an automorphism.

IfGis infinite, the picture changes. The function is still 1-1, but need not be onto.One example is the group Q+, the positive rational numbers with the group operation of

multiplication. The function(x) =x2 satisfies(xy) =(x)(y), and the function is 1-1,but clearly not onto.

Chapter 6, Problem 35. Let a belong to a group G and let o(a) be finite. Let a be theautomorphism of G given by a(x) = axa

1. Show that o(a) divides o(a). Exhibit anelementa from a group for which 1< o(a)< o(a).

Answer: Because we know that gh = gh, we can conclude by induction that ak =(a)

k. Suppose now that o(a) = n. Then (a)n = an = e, which is the identity

permutation. Therefore, o(a)|n.

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In the groupD4, consider the permutationR90 . We compute (R90)2 =R2

90

=R180 ,

which is the identity permutation (as shown previously). Since R90 is not the identitypermutation, we know that o(R90) = 2. On the other hand, o(R90) = 4.

Chapter 7, Problem 1. Let H={(1), (12)(34), (13)(24), (14)(23)}. Find the left cosets of

H in A4.Answer: We have

eH={(1), (12)(34), (13)(24), (14)(23)}= (12)(34)H= (13)(24)H= (14)(23)H

(123)H={(123), (134), (243), (142)}= (134)H= (243)H= (142)H

(132)H={(132), (234), (124), (143)}= (234)H= (124)H= (143)H

Chapter 7, Problem 7. Find all of the left cosets of{1, 11} in U(30).

Answer: The elements ofU(30) are {1, 7, 11, 13, 17, 19, 23, 29}. Let H={1, 11}. Then we

have 1H={1, 11}= 11H

7H={7, 17}= 17H

13H={13, 23}= 23H

19H={19, 29}= 29H

Chapter 7, Problem 18. Use Corollary 1 of Lagranges Theorem to prove that the orderofU(n) is even when n >2.

Answer: The elementn 1 is always in U(n), and when n >2, we can see that n 1= 1.On the other hand, (n 1)2 =n2 2n + 1, which is equal to 1 in U(n). Therefore, n 1

has order 2 in U(n). Since U(n) has an element of order 2, we can conclude that it musthave even order.