3.3 Constructing Examples

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1 33 JABeachy 1 33 Constructing Examples from A Study Guide for Beginner s by JABeachy, a supplement to Abstract Algebra by Beachy / Blair From this point on, if the modulus n is fixed throughout the problem, we will write a rather than [a n for elements of Z n 19 Show that Z 5 Z 3 is a cyclic group, and list all of the generators of the group Solution: By Proposition 334 (b), the order of an element ([a 5,[b 3 ) in Z 5 Z 3 is the least common multiple of the orders of the components Since [1 5,[2 5,[3 5,[4 5 have order 5 in Z 5 and [1 3,[2 3 have order 3 in Z 3, the element ([a 5,[b 3 ) is a generator if and only if [a 5 [0 5 and [b 3 [0 3 There are 8 such elements, which can easily be listed Comment: The other 7 elements in the group will have at least one component equal to zero There are 4 elements of order 5 (with [0 3 as the second component) and 2 elements of order 3 (with [0 5 as the first component) Adding the identity element to the list accounts for all 15 elements of Z 5 Z 3 20 Find the order of the element ([9 12,[15 18 ) in the group Z 12 Z 18 Solution: Since gcd(9,12) 3, we have o([9 12 ) o([3 12 ) 4 Similarly, o([15 18 ) o([3 18 ) 6 Thus the order of ([9 12,[15 18 ) is lcm[4, Find two groups G 1 and G 2 whose direct product G 1 G 2 has a subgroup that is not of the form H 1 H 2, for subgroups H 1 G 1 and H 2 G 2 Solution: In Z 2 Z 2, the element (1,1) has order 2, so it generates a cyclic subgroup that does not have the required form 22 In the group G Z 36, let H {[x x 1 (mod 4)} and K {[y y 1 (mod 9)} Show that H and K are subgroups of G, and find the subgroup HK Solution: It can be shown(as in Problem 3235) that the given subsets are subgroups A short computation shows that H {1,5,13,17,25,29} and K {1,19} Since x 1 x 19 for x G, the set HK must contain 12 elements, and so HK G since G Z 1 36 has ϕ(36) elements 23 Let F be a field, and let H be the subset of GL 2 (F) consisting of all invertible upper triangular matrices Show that H is a subgroup of GL 2 (F) [ [ [ a11 a Solution: Since 12 b11 b 12 a11 b 11 a 11 b 12 +a 12 b 22, and 0 a 22 0 b 22 0 a 22 b 22 a 11 a 22 0 and b 11 b 22 0 together imply that (a 11 b 11 )(a 22 b 22 ) 0, it follows that H is closed under multiplication It is also closed under formation of inverses, since [ 1 [ a11 a 12 a 1 11 a 1 11 a 12a a 22 0 a 1 The identity matrix is certainly in H 22

2 33 JABeachy 2 Comment: This comment is directed to the readers who remember some linear algebra We will prove a more general result Proposition: For a field F, the set H of upper triangular matrices in GL n (F) is a subgroup of GL n (F) Proof: Of course, the identity matrix is upper triangular, so it belongs to H Suppose that A [a ij and B [b ij belong to H This condition is expressed by that fact that a ij 0 if i > j and b ij 0 if i > j The entries of the product matrix [c ij [a ij [b ij are given by the formula c ij n k1 a ikb kj Now look at c ij when i > j Since a ik 0 for all k < i and b kj 0 for all k > j, we must have a ik b kj 0 for all 1 k n because i > j This argument shows that the set of upper triangular matrices in GL n (F) is closed under multiplication Finally, we need to show that the inverse of an upper triangular matrix is again upper triangular Recall that A 1 1 adj(a), where adj(a) is the adjoint of A, and det(a) that adj(a) is the transpose of the matrix of cofactors of A It can then be checked that the adjoint of an upper triangular matrix is again upper triangular Second proof: We will use induction (which [ doesn t require using the adjoint) If an A B invertible matrix has the block form, then A and C must be invertible 0 C [ 1 [ A B A 1 A matrices, and a direct calculation shows that 1 BC 1 0 C 0 C 1 The induction begins with the 2 2 case proved above [ Now suppose that [a ij is A X an n n upper triangular matrix We can write [a ij, where A is an 0 a nn (n 1) (n 1) upper triangular matrix, and X is an (n 1) 1 column Given the induction hypothesis that the inverse of an (n 1) (n 1) [ upper triangular matrix is A again upper triangular, it is clear that [a ij A 1 Xa 1 nn 0 a 1 is also upper nn triangular 24 Let p be a prime number (a) Show that the order of the general linear group GL 2 (Z p ) is (p 2 1)(p 2 p) Hint: Count the number of ways to construct two linearly independent rows Solution: We need to count the number of ways in which an invertible 2 2 matrix can be constructed with entries in Z p This is done by noting that we need 2 linearly independent rows The firstrow can be any nonzero vector, so there are p 2 1 choices There are p 2 possibilities for the second row, but to be linearly independent of the first row, it cannot be a scalar multiple of that row Since we have p possible scalars, we need to omit the p multiples of the first row Therefore the total number of ways to construct a second row independent of the first is p 2 p (b) Show that the subgroup of GL 2 (Z p ) consisting of all invertible upper triangular matrices has order (p 1) 2 p

3 33 JABeachy 3 Solution: An upper triangular 2 2 matrix has nonzero determinant if and only if the elements on the main diagonal are nonzero There are (p 1) 2 choices for these entries Since the third entry can be any element of Z p, there are p choices for this entry 25 Find the order of the element A i i Solution: For any diagonal 3 3 matrix we have n a 0 0 a n b 0 0 b n c 0 0 c n in the group GL 3 (C) It follows immediately that the order of A is the least common multiple of the orders of the diagonal entries i, 1, and i Thus o(a) 4 26 Let G be the subgroup of GL 2 (R) defined by {[ m b G m 0} [ [ 1 0 Let A and B Find the centralizers C(A) and C(B), and show that C(A) C(B) Z(G), where Z(G) is the center of G [ m b Solution: Suppose that X belongs to C(A) in G Then we must have XA AX, and doing this calculation shows that [ [ [ [ [ [ m m+b m b m b m b+1 Equating corresponding entries shows that we must have m+b b+1, and so m 1 On the other hand, any matrix of this form commutes with A, and so {[ 1 b C(A) b R} Now suppose that X [ m b [ m b [ m b belongs to C(B) Then XB BX, and so [ [ [ m b [ m b Equating corresponding entries shows that we must have b 0, and so {[ m 0 C(B) 0 m R}

4 33 JABeachy 4 This shows that C(A) C(B) is the identity matrix, and since any element in the center of G must belong to C(A) C(B), our calculations show that the center of G is the trivial subgroup, containing only the identity element [ 27 Compute the centralizer in GL 2 (Z 3 ) of the matrix [ [ a b Solution: Let A, and suppose that X belongs to the centralizer of A in GL 2 (Z 3 ) Then XAAX, and so c d [ [ [ a a+b a b c c+d c d [ [ [ a b a c b d Equating corresponding entries shows that c d c d we must have a a c, a + b b d, and c + d d The first equation implies that c 0, while the second equation [ implies that a d It follows that the centralizer in GL 2 (Z 3 ) of the matrix is the subgroup H {[ a b 0 a a,b Z 3 and a 0} Note that the modulus 3 has played no role here Comment: The centralizer contains 6 elements, while it follows from Problem 24 that GL 2 (Z 3 ) has (3 2 1)(3 2 3) 48 elements H {[ [ [ [ [ [ } ,a,a 2,b,ab,a 2 b 0 1 Note that the multiplication table for H will look like that of S 3, since it has order 6 and is not abelian (ab ba) (See Exercises in the text) [ 28 Compute the centralizer in GL 2 (Z 3 ) of the matrix [ [ a b Solution: Let A, and suppose that X belongs to the c d [ a b a centralizer of A in GL 2 (Z 3 ) Then XA AX, and so c d c [ [ [ [ [ a b a b a c b d Equating corresponding entries shows that we must have a b a c, a b d, c d a, and c d c d a b c b The first equation implies that c b, while the second equation [ implies that d a+b It follows that the centralizer in GL 2 (Z 3 ) of the matrix is the subgroup {[ a b b a+b a,b Z 3 and a 0 or b 0} Comment: In this case the centralizer contains 8 of the 48 elements in GL 2 (Z 3 ) H {[ [ [ [ [ [ [ [ } ,,,,,,, Note that H is cyclic, as given above

5 33 JABeachy 5 29 Let H be the following subset of the group G GL 2 (Z 5 ) {[ } m b H GL 2 (Z 5 ) m,b Z 5, m ±1 (a) Show that H is a subgroup of G with 10 elements [ m b Solution: Since in the matrix there are two choices for m and 5 choices for [ b, we will [ have a total[ of 10 elements The set is closed under multiplication since ±1 b ±1 c ±1 b±c, and it is certainly nonempty, and so it is a subgroup since the group is finite [ [ 1 0 (b) Show that if we let A and B, then BA A 1 B [ [ [ 1 1 Solution: We have BA and [ [ [ A 1 B (c) Show that every element of H can be written uniquely in the form A i B j, where 0 i < 5 and 0 j < 2 [ [ [ 1 b 1 c 1 b+c Solution: Since, the cyclic subgroup generated by [ 1 b A consists of all matrices of the form Multiplying on the right by B will create 5 additional elements, giving all of the elements in H 30 Let H and K be subgroups of the group G Prove that HK is a subgroup of G if and only if KH HK Note: This result strengthens Proposition 332 Solution: First assume that HK is a subgroup of G If k K and h H, then we have k ek HK and h he HK, so the product kh must belong to HK This shows that KH HK Conversely, supposethat KH HK It is clear that e ee HK since H and K are subgroups To show closure, let g 1,g 2 HK Then there exist elements h 1,h 2 H and k 1,k 2 K with g 1 h 1 k 1 and g 2 h 2 k 2 Since k 1 h 2 KH, by assumption there exist k 1 K and h 2 H with k 1h 2 h 2 k 1 Thus g 1g 2 h 1 k 1 h 2 k 2 h 1 h 2 k 1 k 2 HK since both H and K are closed under multiplication Finally, (g 1 ) 1 (h 1 k 1 ) 1 k h 1 1 HK since k K and h 1 1 H and KH HK ANSWERS AND HINTS 31 What is the order of ([15 24,[25 30 ) in the group Z 24 Z 30? What is the largest possible order of an element in Z 24 Z 30?

6 33 JABeachy 6 Answer: The element ([15 24,[25 30 ) has order lcm[8,6 24 The largest possible order of an element in Z 24 Z 30 is lcm[24,30 120, 32 Find the order of each element of the group Z 4 Z 4 Answer: (0,1) has order one; (2,1),(0,2),(2,2) have order two; the remaining 4 elements have order four 33 Check the order of each element of the quaternion group in Example 337 by using the matrix form [ of the element [ [ [ i 0 0 i Answer: Order 2: Order 4: ±, ±, ± 0 i i 0 35 Let G Z 10 Z 10 (a) If H (3,3) and K (3,7), list the elements of HK Answer: HK {(1,1),(3,3),(9,9),(7,7),(3,7),(9,1),(7,3),(1,9)} (b) If H (3,3) and K (1,3), list the elements of HK Answer: HK G 37 In G Z 15, find subgroups H and K with H 4, K 2, HK G, and H K {1} Answer: Answer: Let H 2 and K 1 [ [ Find the orders of and in GL (Z 5 ) Answer: The first matrix has order 8; the second matrix is not in GL 2 (Z 5 ) {[ } m b 42 Let K GL 2 (Z 5 ) m,b Z 5, m 0 (b) Show, by finding the order of each element in K, that K has elements of order 2 and 5, but no element of order 10 [ 1 b Answer: The 4 elements of the form, with b 0, each have order 5; the 5 [ [ [ 4 b 2 b 3 b elements each have order 2; the 10 elements or each have order Find the cyclic subgroup of GL 4 (Z 2 ) generated by Answer: , List the orthogonal matrices in GL 2 (Z 3 ) and find the order of each one

7 33 JABeachy 7 [ 1 0 Answer: Case 1: ± [ 1 0 Case 2: ±, ± [, ± have determinant 1 [ have determinant The symmetric orthogonal matrices have order 2 (except for I 2 ), and the remaining two matrices have order 4