SOLUTIONS FOR THE SECOND PROBLEM SET

Transcription

1 SOLUTIONS FOR THE SECOND PROBLEM SET Page 22, problem 11: The elements of Z/12Z are of the form a + 12Z where 0 a 11. For any element a + 12Z in Z/12Z, it is clear that 12(a + 12Z) = 12a + 12Z = Z which is the identity element in Z/12Z. Proposition 8 (on the Basic Propositions Handout) implies that the order of a + 12Z divides 12. Only one element has order 1, namely the identity element Z. One can consider 2(a + 12Z) = 2a + 12Z for 0 a 11 and one finds that 2(a + 12Z) = Z only for a = 0 and a = 6. But 0+12Z has order 1. The only element of order 2 in Z/12Z is therefore Z. Similarly, 3(a + 12Z) = Z means that 3a is divisible by 12. For 0 a 11, this occurs for a = 0, 4,and 8. But Z has order 1. The two elements Z and Z don t have order 1 or 2, and hence have order 3. Similarly, one finds that 4(a + 12Z) = Z means that 4a is divisible by 12. For 0 a 11, this occurs for a = 0, 3, 6, or 12. The orders of the corresponding four elements of Z/12Z must divide 4 according to Basic Proposition 8. Excluding the elements of order 1 or 2 found above, we are left with 3+12Z and 9+12Z as the elements of Z/12Z of order 4. Similarly, one finds the following elements of order 6 in Z/12Z: Z, and Z. These are the elements which satisfy 6(a + 12Z) = Z, excluding the elements of orders 1, 2, or 3 found above. All remaining elements in Z/12Z have order dividing 12, but not equal to 1, 2, 3, 4, or 6. They must have order 12. Those remaining elements are: Z, Z, Z, Z. Page 22, problem 12: For brevity, we will write a to denote a + 12Z, as in the text. We will write down the results of simple calculations in (Z/12Z). The identity element in (Z/12Z) is 1. That element has order 1. As for the other elements, notice that 1 2 = 1, 5 2 = 1, 7 2 = 1, 7 2 = 1, 1

2 but 1 1 1, 5 1 1, 7 1 1, Therefore, the elements 1, 5, 7, and 7 in (Z/12Z) all have order 2. Finally, we have 13 = 1, the identity element in (Z/12Z), which we know has order 1. Page 22, problem 25: Suppose that G is a group and that x 2 = e for all x G, where e is the identity element of G. This means that if x G, then xx = e. Equivalently, we have x = x 1 for all x G. Suppose that a and b are elements of G. We will apply the above property to x = a, x = b, and also to x = ab. We will also use Basic Proposition 4. We then obtain: ab = (ab) 1 = b 1 a 1 = ba for all a,b G. We have used Basic Proposition 4 to obtain the second equality. We have proved that ab = ba for all a,b G. Therefore, G is an abelian group. Page 33, problem 2: This is a tedious calculation. We will just give the results. σ = ( )(2)(3 15 8)( )(6), τ = (1 14)( )(3 10)(5 12 7)(6)(8 11), σ 2 = (1 5)(10 13)(2)(3 8 15)( )(14 7 9), στ = (1 11 3)(2 4)( )(6)(12)(13 14), τσ = (1 4)(2 9)( )(6)(7)( ), τ 2 σ = τ(τσ) = ( )(6)(9). Page 33, problem 3: The cycle decomposition allows us to compute the order of an element of S n easily. The order is the least common multiple of the cycles lengths. Using that fact, we see that the orders of σ, τ,σ 2, στ, τσ, and τ 2 σ are 12, 30, 6, 6, 6, and 13, respectively. Page 33, problem 4(a): The elements of S 3 are ε, (1 2), (1 3), (2 3), (1 2 3), (1 3 2). 2

3 The first is the identity element and has order 1. The next three elements are 2-cycles and have order 2. The last two elements are 3-cycles and have order 3. Page 33, problem 6: The cycle decomposition of an element of order 4 must consist of 1-cycles, 2-cycles, and 4-cycles, with at least one 4-cycle. Since we are considering elements in S 4, an element of order 4 must be a 4-cycle. Thus, we just have to write out a list of all the 4-cycles in S 4. We can always start with a 1. Here is the list: ( ), ( ), ( ), ( ), ( ), ( ). Note that we have found exactly six elements of order 4 in S 4. Page 33, problem 7: The cycle decomposition of an element of order 2 must consist of 1-cycles and 2-cycles, with at least one 2-cycle. In S 4, such an element is either a 2-cycle or a product of two disjoint 2-cycles. Here is the list: (1 2), (1 3), (1 4), (2 3), (2 4), (3 4), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3). Note that we have found nine elements of order 2 in S 4. Page 36, problem 1: The identity element in Q 8 is 1. It has order 1. Also ( 1) 2 = 1, but 1 1, and so 1 has order 2. Now note that i 4 = (i 2 ) 2 = ( 1) 2 = 1, but i 2 = 1 1. By Basic Proposition 8, the order of i divides 4, but does not divide 2. Therefore, i has order 4. The same calculation works for i,j, j,k, and k. Hence all of those elements also have order 4. Page 40, problem 4: We proved in class that if two groups are isomorphic, then, for any m Z +, the two groups must have the same number of elements of order m. We can apply this to the two groups R {0} and C {0} (under the operation of multiplication). The identity element in both groups is 1. Take m = 3. In R {0}, the equation x 3 = 1 has only one solution, namely x = 1. Of course, 1 has order 1. Hence the group R {0} has no elements of order 3. However, note that x 3 = 1 x 3 1 = 0 (x 1)(x 2 +x+1) = 0 and this equation has three distinct roots in C, namely 1, ,

4 These are the elements of C {0} of order dividing 3 since they satisfy the equation x 3 = 1 and no other elements of C {0} satisfy that equation. The first of the above numbers has order 1, but the other two must have order 3. Therefore, we have shown that C {0} has exactly two elements of order 3. Since R {0} has no elements of order 3 and C {0} has two elements of order 3, those two groups cannot be ismorphic. Page 40, problem 6: Suppose that G is a group. Consider the following property: Property: If a G, then there exists an element b G such that b 2 = a. We first prove that this property is preserved by isomorphisms. Suppose that G and G are two groups and that ϕ : G G is an isomorphism from G to G. Assume that G satisfies the above property. We will prove that G also satisfies the above property. To prove this, suppose that a G. Since ϕ is surjective, there exists an element a G such that ϕ(a) = a. Since G satisfies the above property, there exists an element b G such that b 2 = a. Let b = ϕ(b). We then have (b ) 2 = ϕ(b) 2 = ϕ(b 2 ) = ϕ(a) = a. We have proved that for every element a G, there exists an element b G such that (b ) 2 = a. Therefore, we have proved that G also satisfies the above property. For this problem, let us take G = Q and G = Z under the operations of addition. Now the group Q does satisfy the above property. For if a Q, then b = a/2 Q and satisfies the equation 2b = a. However, the group Z does not satisfy the above property. For if one takes a = 1, then a Z, but there does not exist an element b Z satisfying 2b = a. Therefore, we can conclude that Q and Z are not isomorphic. Page 40, problem 7: Note that D 8 has five elements of order 2, namely the four reflections and the rotation by 180 degrees. However, in problem 1 on page 36, it was pointed out that Q 8 only has one element of order 2. Therefore, we can conclude that D 8 and Q 8 cannot be isomorphic. Page 40, problem 8: If G and G are finite groups and if ϕ : G G is an isomorphism, then it follows that G = G. This is because a bijection between two finite sets can exist if and only if the two sets have the same cardinality. Now S n = n! and S m = m!. If n m, then n! m!. Therefore, S n and S m cannot be isomorphic. 4

5 A: Suppose that G is a group and that G = 4. We start with some general remarks. Let G be a group and let e denote the identity element of G. We have ea = a and ae = a for all a G. Thus, ea = ae for all a G. Furthermore, suppose that a,b G and that ab = e. We have ab = e = ab = aa 1 = b = a 1 = ba = a 1 a = e. We have used the cancellation law to derive the second implication. Therefore, if ab = e, then it follows that ba = e and hence that ab = ba. Now suppose that G is a group of order 4. For all a G, we have ea = ae = a. Hence e commutes with every element of G. Now suppose that the other three elements of G are denoted by a,b and c. Thus, e,a,b, and c are all distinct and G = {e,a,b,c}. Obviously, aa = aa. Hence a commutes with itself. Now consider ab. We have ab {e,a,b,c}. We cannot have ab = b or ab = a. To explain this, notice that ab = b = ab = eb = a = e, ab = a = ab = ae = b = e. However, a e and b e. It therefore follows that ab b and ab a. This leaves two possibilities: either ab = e or ab = c. medskip If we reverse the role of a and b in the previous paragraph, then we find that there are two possibilities for ba, namely either ba = e or ba = c. If ab = e, then we showed above that ba = e and hence ab = ba. Similarly, reversing the role of a and b, if we have ba = e, then it follows that ab = e and hence that ab = ba. There is only one case not yet covered, the case where ab and ba are both equal to c. But in that remaining case, we have ab = c and ba = c and so we have ab = ba. Hence a and b commute with each other in that case too. The above argument can be applied to the pair of elements a and c. It shows that ac = ca. The argument applies to the pair b and c, showing that bc = cb. It follows that G is indeed an abelian group. 5