Solutions I.N. Herstein- Second Edition

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1 Solutions I.N. Herstein- Second Edition Sadiah Zahoor 25, July 2016 Please me if any corrections at Problem 0.1. In the following determine whether the systems described are groups. If they are not, point out which of the group axioms fail to hold. (a. G = set of integers, a.b a b. (b. G = set of all positive integers, a.b = ab, the usual product of integers. (c. G = a 0, a 1,..., a 6 where, a i.a j = a i+j if i + j < 7, a i.a j = a i+j 7 if i + j 7 (for instance, a 5.a 4 = a = a 2, since = 9 > 7. (d. G = set of all rational numbers with odd denominators, a.b a + b, the usual addition of rational numbers. Proof. (a. G = set of integers, a.b a b. We note G is not a group as there exists no identity in G. (b. G = set of all positive integers, a.b = ab, the usual product of integers. We note G is not a group as the inverses do not belong in G for all elements of G. For example, inverse of 2 is 1 2 Z+. (c. G = a 0, a 1,..., a 6 where, a i.a j = a i+j if i + j < 7, a i.a j = a i+j 7 if i + j 7 (for instance, a 5.a 4 = a = a 2, since = 9 > 7. We note G has closure by definition. We claim a 0 is the identity of G. This is because a 0 a i = a i a 0 = a 0+i = a i i = 1, 2, 3, 4, 5, 6. Next we claim the associative law holds in G. i + j < 7, j + k < 7, i + j + k < 7 Then (a i a j a k = a i+j a k = a i+j+k and a i (a j a k = a i a j+k = a i+j+k. i + j < 7, j + k < 7, i + j + k > 7 Then (a i a j a k = a i+j a k = a i+j+k 7 and a i (a j a k = a i a j+k = a i+j+k 7. 1

2 Solutions by Sadiah Zahoor 2 i + j > 7, j + k < 7, i + j + k > 7, i + j + k 7 < 7 Then (a i a j a k = a i+j 7 a k = a i+j+k 7 and a i (a j a k = a i a j+k = a i+j+k 7. i + j > 7, j + k < 7, i + j + k > 7, i + j + k 7 < 7 because i + j + k > 14 i + 7 > 14 i > 7 which is a contradiction. Then (a i a j a k = a i+j 7 a k = a i+j+k 7 and a i (a j a k = a i a j+k = a i+j+k 7. i + j < 7, j + k > 7, i + j + k > 7, i + j + k 7 < 7 because i+j +k > 14 7+j > 14 j > 7 which is a contradiction. Then (a i a j a k = a i+j a k = a i+j+k 7 and a i (a j a k = a i a j+k 7 = a i+j+k 7. i + j > 7, j + k > 7, i + j + k > 7, i + j + k 7 < 7 because i + j + k > 14 i + 7 > 14 i > 7 which is a contradiction. Then (a i a j a k = a i+j 7 a k = a i+j+k 7 and a i (a j a k = a i a j+k 7 = a i+j+k 7. Finally we see existence of inverses. a 0.a 0 = a 0.a 0 = a 0 a 1.a 6 = a 6.a 1 = a 7 7 = a 0 a 2.a 5 = a 5.a 2 = a 7 7 = a 0 a 3.a 4 = a 4.a 3 = a 7 7 = a 0 Thus, all group axioms hold and G is a group. (d. G = set of all rational numbers with odd denominators, a.b a + b, the usual addition of rational numbers. We see that G has the identity as 0 G can have any denominator including odd denominator. = ap + bq bp Next we see closure a b, p q G, we have a b + p q where bp is odd as b, p are both odd. Now if bp ap + bq, then we have an integer with denominator 1 and thus belongs to G or if bp ap + bq, then we have an odd denominator. In the latter case, observe that ap+bq < bp, then obviously the denominator is bp. But if ap + bq > bp, then any cancellation would yield an odd denominator because we can t have an even number that divides an odd number bp. Next, associativity is inherited from the group of rational numbers under addition. Lastly, we see existence of inverses. For each a Q with b odd, we have b a with b odd such that a b b + ( a b = ( a b + a b = 0. Thus, all group axioms hold and G is a group. Problem 0.2. Prove that if G is an abelian group, then for all a, b G and all integers n, (a.b n = a n.b n.

3 Solutions by Sadiah Zahoor 3 Proof. Let a, b G. Then (a.b 1 = (a.b. Also (a.b 2 = (a.b.(a.b. Then by associativity in G, we have (a.b 2 = a.(b.a.b and by commutativity in G, we have (a.b 2 = a.(a.b.b. Using associativity again, we get (a.b 2 = (a.a.(b.b = a 2.b 2. Let us assume (a.b k = a k.b k for any k < n. Then (a.b k+1 = (a.b k.(a.b By our assumption, we have (a.b k+1 = (a k.b k.(a.b. Following the same argument as in the case n = 2, (a k.b k.(a.b = a k.(a.b k.b = (a k.a.(b k.b = a k+1.b k+1 because a k, b k G. Thus by mathematical induction, (a.b n = a n.b n for all n Z +. Now let n = m for m Z + We intend to show that (a.b m = a m.b m m Z +. For m = 1, by using commutativity, we get (a.b 1 = b 1.a 1 = a 1.b 1. Also, m = 2, by using associativity and commutativity, we get (a.b 2 = (a.b 1+( 1 = (a.b 1.(a.b 1 = (b 1.a 1.(b 1.a 1 = b 1.(b 1.a 1.a 1 = (b 1.b 1.(a 1.a 1 = b 2.a 2 = a 2.b 2. Let us assume (a.b k = a k.b k for any k < m. Then (a.b (k+1 = (a.b k+( 1 = (a.b k.(a.b 1 By our assumption, we have (a.b k+( 1 = (a k.b k.(b 1.a 1. Following the same argument as in the case n = 2, (a k.b k.(b 1.a 1 = a k.(b k.b 1.a 1 = a k.(b (k+1.a 1 = a k.a 1.b (k+1 = a (k+1.b (k+1 because a k, b k G. Thus by mathematical induction, (a.b m = a m.b m for all m Z +. This further implies (a.b n = a n.b n for all n Z For n = 0, (a.b 0 = e = e.e = a 0.b 0. Thus (a.b n = a n.b n for all n Z. Problem 0.3. If G is a group such that (a.b 2 = a 2.b 2 for all a, b G, show that G must be abelian. Proof. Let a, b G. By associativity in G, we get (a.b 2 = (a.b.(a.b = a.(b.a.b By associativity in G, we also get a 2.b 2 = (a.a.(b.b = a.(a.b.b From the given condition (a.b 2 = a 2.b 2 a.(b.a.b = a.(a.b.b By using the left hand cancellation and right hand cancellation law in a group, we get b.a = a.b Since a, b were arbitrary, we get a.b = b.a a.b G. Thus by definition, G is abelian. Problem 0.4. If G is a group in which (a.b i = a i.b i for three consecutive integers i for all a, b G, show that G is abelian. Proof. Let a, b G. Let i, i + 1, i + 2 be three consecutive integers. From the given condition and associativity in G, we get, (a.b i+1 = a i+1.b i+1 (a.b i.(a.b = a i.a.b.b i a i.b i.a.b = a i.a.b.b i

4 Solutions by Sadiah Zahoor 4 By the left hand cancellation law in a group and operating by b i from the left, we get, a.b = b i (a.bb i Again, from the given condition and associativity in G, we get, (a.b i+2 = a i+2.b i+2 (a.b i.(a.b 2 = a i.a 2.b 2.b i a i.b i.(a.b 2 = a i.a 2.b 2.b i By the left hand cancellation law in a group and operating by b i from the left, we get, (a.b 2 = b i (a 2.b 2 b i (a.b.(a.b = b i (a 2.b 2 b i From the evaluated value of (a.b, we get, (b i (a.bb i.(b i (a.bb i = b i (a 2.b 2 b i b i (a.b 2 b i = b i (a 2.b 2 b i By the left and right hand cancellation law in a group, we get (a.b 2 = a 2.b 2 Since a, b were arbitrary, from the proof of previous problem, we conclude G is abelian. Problem 0.5. Show that the conclusion of Problem 4 does not follow if we assume the relation (a.b i = a i.b i for just two consecutive integers. Proof. Problem 0.6. In S 3, give an example of two elements x, y such that (x.y 2 x 2.y 2. Proof. We recall S 3 is the set of all permutations of the set {1, 2, 3} which forms a group under function composition. Let φ : ψ : Clearly, φ and ψ belong to S 3 and φ ψ. It is also clear that φ 2 = e and ψ 3 = e φ 2 : ψ 3 : We also note that ψ 2 S 3 and ψ 2 ψ and ψ 2 φ is given by: ψ 2 :

5 Solutions by Sadiah Zahoor 5 φ.ψ : ψ.φ : This implies φ.ψ and ψ.φ are other two elements of S 3 distinct from φ, ψ, ψ 2 and e. Thus, we have found our six elements of S 3. S 3 = {e, φ, ψ, ψ 2, φ.ψ, ψ.φ} For simplification we also observe, ψ 3 = e ψ 2 = ψ 1, we observe ψ.φ : = φ.ψ 1 : Now let us take φ, ψ S 3 and consider (φ.ψ 2 = (φ.ψ.(φ.ψ = φ.(ψ.φ.ψ = φ.(φ.ψ 1.ψ = φ 2.e = e.e = e φ 2.ψ 2 = e.ψ 2 = ψ 2 Thus, φ, ψ are two elements of S 3 such that (φ.ψ 2 φ 2.ψ 2. Problem 0.7. In S 3, show that there are four elements satisfying x 2 = e and three elements satisfying y 3 = e. Proof. As seen in the previous problem, S 3 = {e, φ, ψ, ψ 2, φ.ψ, ψ.φ} We have e 2 = e, φ 2 = e, (φ.ψ 2 = e from the previous problem, (ψ.φ 2 = (ψ.φ.(ψ.φ = (φ.ψ 1.(ψ.φ = φ.(ψ 1.ψ.φ = φ.e.φ = φ 2 = e. Thus we have four elements whose square is identity in S 3. Also, e 3 = e, ψ 3 = e, and (ψ 2 3 = ψ 6 = (ψ 3 2. Hence, we have found three elements whose cube is identity in S 3. Problem 0.8. If G is a finite group, show that there exists a positive integer N such that a N = e for all a G. Proof. Let G be a finite group with G = n. Now consider a G, a e. Consider the set generated by a, given by, {a, a 2, a 3,..., a n,...}. Suppose, if possible a N e for every positive integer N. Then by the cancellation law in the group G, we get a i = a j, i < j a j i = e which is a contradiction to our assumption. Thus every element in < a > is distinct from others. This implies there are infinitely many distinct elements in < a >. Also, < a > G. Thus we get a contradiction to the fact that G is finite. Hence, our assumption is wrong. This means N Z + such that a N = e. Since a was arbitrary, we get for each a i : i {1, 2,..., n} G, N i Z + such that a Ni i = e. Take N = lcm{n i : i {1, 2,..., n}}, then for this N, a N i = e for all i {1, 2,..., n}.

6 Solutions by Sadiah Zahoor 6 Problem 0.9. (a. If the group G has three elements, show it must be abelian. (b. Do part (a if G has four elements. (c. Do part (a if G has five elements. Proof. Multiplication Table: Let G = {a 1, a 2,..., a n } be a finite group with a 1 = e. The multiplication table or group table of G is the n n matrix whose i, j entry is the group element a i a j. The multiplication table can also easily show whether the given finite group is abelian iff it is symmetric along the diagonal joining left upper corner and right lower corner. This is precisely the definition of being abelian, ab = ba ab G. (a. Let G = e, a, b. Then either ab = ba = e or a 2 = b 2 = e. In the former case, e a b e e a b a a b e b b e a Clearly, we see the multiplication table is symmetric along diagonal, we conclude by that G is abelian by definition. In the latter case, ab = a or ab = b, which implies a = e or b = e which reduces the group to the trivial group. So we have just one case which is given above and is abelian. (b. Let G = {e, a, b, c}. In the first case, we have a 2 = e, b 2 = e, c 2 = e, that is, all elements are inverses of themselves. Since a, b, c, e are all distinct and each is the inverse of itself, we have ba = ab = c ac = ca = b bc = cb = a e a b c e e a b c a a e c b b b c e a c c b a e From the multiplication table, it s clear that G is abelian. In the second case, let just a and c have each other as inverses and b be the inverse of itself. Let ac = ca = e and b = b 1. Then b 2 = ac = ca = e. Also, because e, a, b, c are distinct, we have ba = ab = c bc = cb = a a 2 = cb.bc = c 2 a 2 = cb.cb = cab = e.b = b e a b c e e a b c a a b c e b b c e a c c e a b

7 Solutions by Sadiah Zahoor 7 From the multiplication table, symmetry along the diagonal implies that the group is abelian. In the third case, each of the three elements has a different inverse,which is not possible cause we have three odd number of non-identity elements. (c. Let G = {e, a, b, c, d}. In the first case, each of the three elements is the inverse of itself. a 2 = b 2 = c 2 = d 2 = e. Let ab = c Then ab = c a 2 b = ac b = ac. Now ad = a d = e which is a contradiction. and ad = b a 2 d = ab d = c which is a contradiction. and ad = c a 2 d = ac d = b which is a contradiction. and ad = d a = e which is a contradiction. and ad = e a = d which is a contradiction. Thus, this case is not possible. In the second case, let two elements be inverses of each other and the other two be inverses of themselves. ab = ba = c 2 = d 2 = e. Then let cd = a, we have cd = a c 2 d = ca d = ca ca = d cab = db c = db c = db dc = d 2 b dc = b dc = b dc 2 = bc d = bc d = bc ad = abc ad = c d = ca bd = bca bd = da = c because otherwise we get a contradiction. Then da = c d 2 a = dc a = b which is a contradiction. Hence this case also doesn t work out. In the last case, let every element in the group have an inverse other than itself. Suppose ab = ba = cd = dc = e. Let ac = b. Then we have ac = b a 2 c = ab = e a 2 cd = d a 2 = d ac = b bac = b 2 c = b 2 ac = b acd = bd a = bd a 2 = d a 2 b = db a = db b 2 = c b 2 a = ca b = ca a = db ad = dbd ad = da If ad = b, then ac = ab a = c which is a contradiction. Thus ad = da = c. ad = c adc = c 2 a = c 2 ad = c bad = bc d = bc bc = d bcd = d 2 b = d 2 ad = c adb = cb a 2 = cb d = cb e a b e e a b a a d e b c b b e a c c b d a e d d c a e b Since the multiplication table is symmetrical along the diagonal, we get

8 Solutions by Sadiah Zahoor 8 that G is abelian. Conclusion: Every group of order 3, 4, 5 is always abelian. Problem Show that if every element of the group G is it s own inverse, then G is abelian. Proof. Let a, b G. Then a.b G. From the given condition, we have a = a 1, b = b 1, (a.b = (a.b 1. Thus, we have b.a = b 1.a 1 = (a.b 1 = a.b Since a, b were arbitrary, we have G is abelian. Problem If G is a group of even order, prove that it has an element a e such that a 2 = e. Proof. Let G be a group of even order, say, 2m for some m N. Then o(g > 1. Thus, a G such that a e. Now we consider the case G = 2. Then G = {e, a} where a e. We know every element has a unique inverse. Now e = e 1. Thus, a 1 is either e or a. If a 1 = e, then a = e which is a contradiction, thus a 1 = a a 2 = e. Now let G = 2m for m N, m > 1. Then G = {e, a 1, a 2,..., a 2m 1 } where a i e and a i a j for i, j {1, 2,.., 2m 1}. Now every element in G has a unique inverse by definition of a group. Now let us assume a i a 1 i for all i {1, 2,.., 2m 1} because if a i = a 1 i for some i {1, 2,.., 2m 1}, then a 2 i = e for that particular i and we are done. So a 1 i = a j for i j; i, j {1, 2,.., 2m 1}. Also by the uniqueness of inverses, we can t have the same a j as the inverse of two elements because then a 1 j = a i = a k, i k will have two different inverses in G. Thus, we group each pair of an element and it s inverse (different from itself and unique and conclude that the set S = {a i G : a i a 1 i } has even number of elements due to existence of pairs. Clearly e / S. Thus S = 2m and can at most be 2m 2. This implies existence of a non-identity element in G S = {a i G : a i = a 1 i }. Let this element be a 2m 1 = a G. Then a G; a e such that a = a 1 a 2 = e. Problem Let G be a non-empty set closed under an associative product, which in addition satisfies: There exists an e G such that a.e = a for all a G. Give a G, there exists an element y(a G such that a.y(a = e. Prove that G must be a group under this product. Proof. We are already given that the non-empty set G is closed under the given product and is also associative. We are given the existence of right identity and right inverse and we need to show that the existence of left identity and left inverse. By the existence of right inverse, given a G, y(a G such that a.y(a = e. Again by the existence of right inverse, given y(a G, z(y(a G such that y(a.z(y(a = e. By the existence of right identity for and associativity in G, we get y(a.a = (y(a.a.e

9 Solutions by Sadiah Zahoor 9 y(a.a = (y(a.a.(y(a.z(y(a y(a.a = y(a.(a.y(a.z(y(a y(a.a = (y(a.e.z(y(a y(a.a = y(a.z(y(a y(a.a = e Since a G was arbitrary, we have given a G, there exists y(a G such that a.y(a = y(a.a = e. Now e.a = (a.y(a.a = a.(y(a.a = a.e = a. Since a was arbitrary, we have shown a.e = e.a = a a G. Thus all the group axioms hold for G under the given product. Problem Prove, by an example, that the conclusion of Problem 12 is false if we assume instead: There exists an e G such that a.e = a for all a G. Give a G, there exists an element y(a G such that y(a.a = e. Proof. Consider a product on the given non-empty set G such that a b = a a, b G. This product is closed by definition and we can check for a, b, c G, this product is associative, The following properties hold for G: a (b c = a = a b = (a b c There exists an e G, such that a e = a for all a G. Infact by our definition all elements satisfy the property of e. Give a G, there exists an element y(a G such that y(a a = e. But G is not a group. This is because e a = e a G. Thus, a e e a a G unless a = e. So we assume that the given group is non-trivial. Then a e and we fail to have an identity for G and G is not a group necessarily. Problem Suppose a finite set G is closed under an associative product and that both cancellation laws hold in G. Prove that G must be a group. Proof. Let G = {a 1, a 2,..., a n } be a finite group which is closed under an associative product in which both cancellation laws hold. We intend to show that G is a group. We are given that G is closed and associative under. Let a G, then consider the set a G = {a a 1, a a 2,..., a a n }. Now if a a i = a a j for i j {1, 2,..., n} in a G, then by the left cancellation law in G, we have a i = a j which is not true. Thus a G has n distinct elements each of which lies in G. This means a G = G. We get a a i = b has a unique solution a i G; i {1, 2,..., n} for each b G. In particular for b = a, we have a unique solution, say e G to a e = a. Now consider the set G a = {a 1 a, a 2 a,..., a n a}. Now if a i a = a j a for i j {1, 2,..., n} in G a, then by the right cancellation law in G, we have a i = a j which is not true. Thus G a has n distinct elements each of which lies in G. This means G a = G. We get a i a = b has a unique solution a i G; i {1, 2,..., n} for each b G. Now consider b e = (a i a e = a i (a e = a i a = b for every b G. Thus e G is a

10 Solutions by Sadiah Zahoor 10 right identity. Also, a a i = e has a unique solution. Since a was arbitrary, we have a unique solution a a i = e for each a G. This implies the existence of right inverse for each a G. By Problem 12, we get G is a group under. Problem (a. Using the result of Problem 14, prove that the non-zero integers modulo p, p a prime number, forms a group under multiplication modulo p. (b. Do part (a for the non-zero integers relatively prime to n under multiplication mod n. Proof. (a. Here G = {1, 2,..., p 1} because we are not considering integers which give remainder 0 on division by p. Clearly, G is finite. By the division algorithm, any non-zero integer which doesn t give a zero remainder modulo p, say, n = qp + r gives a remainder 0 < r < p on division by p. Thus (G, p is closed. Next we show p is associative. Let a, b, c be non-zero integers such that a = q 1 p + r 1 ; b = q 2 p + r 2 ; c = q 3 p + r 3 where r 1, r 2, r 3 G. Now we have a p b = ab mod p (q 1 p + r 1 (q 2 p + r 2 mod p r 1 r 2 mod p (a p b p c = r 1 r 2 c mod p r 1 r 2 (q 3 p + r 3 mod p r 1 r 2 r 3 mod p Similarly, we have b p c = bc mod p (q 2 p + r 2 (q 3 p + r 3 mod p r 2 r 3 mod p a p (b p c = ar 2 r 3 mod p (q 1 p + r 1 r 2 r 3 mod p r 1 r 2 r 3 mod p Now ab ac mod p p a(b c p a or p b c. If p a, then a is a multiple of p. Then a 0 mod p a / G which is a contradiction. Thus p a, then p b c b c mod p. This implies that the left cancellation law holds in G. Similarly the right cancellation law holds too. Thus, by Problem 14, we have (G, p is a group. (b. Here G = {1, 2,..., n 1} because we are not considering integers which give remainder 0 on division by n. Clearly, G is finite. By the division algorithm, any non-zero integer which doesn t give a zero remainder modulo n, say, m = qp + r gives a remainder 0 < r < n on division by n. Thus (G, n is closed. Next we show n is associative. Let a, b, c be non-zero integers such that a = q 1 n + r 1 ; b = q 2 n + r 2 ; c = q 3 n + r 3 where r 1, r 2, r 3 G. Now we have a n b = ab mod n (q 1 n + r 1 (q 2 n + r 2 mod n r 1 r 2 mod n (a n b n c = r 1 r 2 c mod n r 1 r 2 (q 3 n + r 3 mod n r 1 r 2 r 3 mod n Similarly, we have b n c = bc mod n (q 2 n + r 2 (q 3 n + r 3 mod n r 2 r 3 mod n a n (b n c = ar 2 r 3 mod n (q 1 n + r 1 r 2 r 3 mod n r 1 r 2 r 3 mod n

11 Solutions by Sadiah Zahoor 11 Now ab ac mod n n a(b c. Since a is not congruent to 0 mod p, a and n are relatively prime. Then n (b c b c mod n. This implies that the left cancellation law holds in G. Similarly the right cancellation law holds too. Thus, by Problem 14, we have (G, p is a group. Problem In Problem 14 show by an example that if one of just assumed one of the cancellation laws, then the conclusion need not follow. Proof. We consider the same example as before (G, where G is a finite semigroup under defined as a b = a a, b G. Here we have by the definition of our product that a c = b c a = b a, b, c G. But c a = c b does not imply a = b. Thus only one cancellation law holds in this semi-group. We have shown earlier that G is not a group. Problem Prove that in Problem 14 infinite examples exist, satisfying the conditions, which are not groups. Proof. Let G = Z {0}, and be usual multiplication in Z {0}. Now we know (Z {0}, is an infinite set closed and associative under. Also, a c = b c a = b a, b, c Z {0} and a b = a c b = a a, b, c Z {0}. This means both cancellation laws hold in (Z {0},. But (Z {0}, is not a group because the inverses do not exist in Z {0} for elements other than 1, 1. Problem For any n > 2 construct a non-abelian group of order 2n. (Hint: imitate the relation in S 3. Proof. A symmetry is any rigid body motion (isometry of the n gon which can be effected by taking a copy of the n gon, moving this copy in any fashion in 3 space and then placing the copy back on the original n gon so it exactly covers it. If we look at the symmetries of the n gon with labelled vertices 1, 2,..., n, we can notice that we can rotate the n gon in the same plane in a clockwise direction such that n rotations returns us to the original position. This rotation just replaces a vertex i by the vertex i + 1 for i = 1, 2,..., n 1 and the vertex n by vertex 1. But since we are talking about symmetries on an n gon in 3 space, we can also flip the copy of the n gon and then again continue with similar n rotations as before. The flipping of the n gon rotates the n gon by π radians along the line joining two vertices which remain fixed while doing so. We can flip it using any vertex but for convenience and uniformity, we shall take it to be the rotation along the line joining the vertex 1 to the center of the n gon. So including the flip, we have a total 2n distinct symmetries of the n gon from observation. We prove it below. Let n gon have labelled n vertices as 1, 2,..., n. Now, we can send vertex 1 to any of the remaining n vertices by a permutation including the permutation which keeps vertices of n gon unaltered. This gives us n choices for the vertex 1. Suppose 1 is sent to vertex i where i {1, 2,..., n}. Now the vertex 2 has to be adjacent to vertex 1. Thus 2 has two choices. It can either take the place of i 1 vertex or the i + 1 vertex. Now after we choose the vertex for 2, the vertex 3 has just 1 choice to lie adjacent to 2 from the side other than 1. Similarly, all other vertices have just 1 choice. Thus the total number of choices is 2n.

12 Solutions by Sadiah Zahoor 12 Hence, there are 2n distinct symmetries of a regular n gon. We next construct a group of symmetries of a regular n gon and show it s nonabelian. Label the vertices of the square 1, 2,..., n clockwise. We know that there are 2n distinct symmetries of an n gon from the earlier proposition. Let e be the identity permutation of the vertices which keeps all vertices unaltered. e : i i ; i = 1, 2,..., n Let r be the rotation clockwise through the origin by 2π n plane. So basically r is just a permutation given by: radians in the same r : i i + 1 ; i = 1, 2,..., n 1 n 1 Let s be the rotation by π radians along the line joining the vertex 1 to the origin. Then s is again just a permutation given by: n = odd s : n = even s : 1 1 i n (i 2 ; i = 2, 3,..., n i n (i 2 ; i = 2, 3,..., n 2 n n We can see that r n = e and s 2 = e. The other n 3 elements of the set are r 2,...r n 1, sr, sr 2,..., sr n 1. Now we make this set S = {e, r, r 2,..., r n 1, s, sr, sr 2,..., sr n 1 } into a group by defining the product to be function composition of f g f, g S which is obtained by applying the symmetry g first and then f. This again gives a symmetry of a n gon, because the first permutation returns us a symmetry which is again an n gon and then applying another permutation is just like taking permuted n gon as the original object and working with it. Thus, we have closure in our set. Associativity follows from the associativity of function composition. We already have defined our identity for the set as the permutation which leaves all vertices unaltered which by definition gives f e = e f = e f S. Lastly, we define the inverse of each symmetry as the symmetry which reverses it and returns the identity. Thus, all group axioms hold and S is a group. Now we try to show rs sr and thus the group is not abelian. rs(1 = r(1 = 2 and sr(1 = s(2 = n. Thus rs sr. The symmetries of an n gon form a group under function composition called the Dihedral group of order 2n, denoted by, D 2n. Moreover, it s non-abelian. Problem If S is a set closed under the associative operation, prove that no matter how you bracket a 1 a 2...a n, retaining the order of the elements, you get the same element in S (e.g., (a 1.a 2.(a 3, a 4 = a 1.(a 2.(a 3.a 4 ; use induction on n.

13 Solutions by Sadiah Zahoor 13 Proof. For n = 1, 2, there is nothing to prove. For n = 3, we have by the definition of being associative that a 1.a 2.a 3 = a 1.(a 2.a 3 = (a 1.a 2.a 3. For n = 4, we have by using the case for n = 3 (a 1.a 2.(a 3.a 4 = a 1.(a 2.(a 3.a 4 a 1.((a 2.a 3.a 4 = a 1.(a 2.(a 3.a 4 (a 1.(a 2.a 3.a 4 = a 1.(a 2.(a 3.a 4 ((a 1.a 2.a 3.a 4 = a 1.(a 2.(a 3.a 4 Thus any product of four elements a 1.a 2.a 3.a 4 can be eventually written as a 1.(a 2.(a 3.a 4. Now let us assume for k < n that the product a 1.a 2...a k can be written eventually as a 1.(a 2.(...(a k 1.a k... irrespective of the way the brackets are placed. This means associative law will hold for k < n elements. Now let us look at the product a 1.a 2...a k+1. Then the product can be broken down into two products followed by the use of induction hypothesis. No matter how we take the product, we have to break it into two parts and we can use the induction hypothesis on any of the parts being less than or equal to k elements. Then we use the associative law for three grouped elements repeatedly since it is given to be closed under the associative operation. For example, a 1.a 2...a k+1 = (a 1.a 2.a 3 (a 4...a k.a k+1 a 1.a 2...a k+1 = (a 1.(a 2.a 3.(a 4 (...(a k.a k+1... a 1.a 2...a k+1 = a 1.((a 2.a 3.(a 4 (...(a k.a k+1... a 1.a 2...a k+1 = a 1.(a 2.(a 3 (a 4 (...(a k.a k+1... Thus, by induction a 1.a 2...a n can be eventually written as a 1.(a 2.(a 3 (...(a n 1.a n... irrespective of how the brackets are placed. ( a b Problem Let G be the set of all real 2 2 matrices, where ad bc 0 is a rational number. Prove that G forms a group under matrix multiplication. ( 1 0 Proof. We first note G as ad bc = 1 0 is a rational number. 0 1 Thus G and identity belongs ( to G. ( a b p q Next we show closure. Let, G. r s Then ad bc, ps rq Q {0}. ( ( ( a b p q ap + br aq + bs And. = such that r s cp + dr cq + ds apcq+apds+brcq+brds cpaq cpbs draq drbs = apds+brcq cpbs draq = ad(ps rq +bc(rq ps = (ps rq(ad bc Q {0} by closure in (Q {0},. Associativity is inherited from the group of real 2 2 matrices with non-zero determinant. Next we show the existence of inverses.

14 Solutions by Sadiah Zahoor 14 ( ( a b 1 d b For each G with ad bc Q {0}, G ad bc c a ad with (ad bc 2 bc (ad bc 2 = 1 Q {0} such that ad bc ( ( ( ( ( a b 1 d b 1 d b a b 1 0. = = ad bc c a ad bc c a 0 1 Thus, all group axioms hold and G is a group under matrix multiplication. ( a b Problem Let G be the set of all real 2 2 matrices, where ad 0 d 0 is a rational number. Prove that G forms a group under matrix multiplication. Is G abelian? ( 1 0 Proof. We first note G as ad = 1 0. Thus G and identity 0 1 belongs to G. ( ( a b p q Next we show closure. Let, G. 0 d 0 s Then ad, ps 0. ( ( a b p q And. = 0 d 0 s ( ap aq + bs 0 ds such that apds 0 because of properties of real numbers. Associativity is inherited from the group of real 2 2 matrices with non-zero determinant. Next we show the existence of inverses. ( a b For each G with ad 0, 1 0 d ad with ad (ad 2 bc (ad 2 = 1 0 such that ad ( a b 0 d. 1 ad ( d b 0 a = 1 ad ( d b 0 a ( d b 0 a ( a b 0 d G = ( Thus, all group axioms hold and G is a group under matrix multiplication. No, G is not abelian because ( ( ( =, ( ( ( while. = ( ( Thus there exists A =, B = G such that AB BA ( a 0 Problem Let G be the set of all real 2 2 matrices 0 a 1, where a 0. Prove that G is an abelian group under matrix multiplication. ( 1 0 Proof. We first note G as a = 1 0. Thus G and identity 0 1..

15 Solutions by Sadiah Zahoor 15 belongs to G. Next we show closure. Let Then a, p 0. ( ( a 0 p 0 And 0 a 1. 0 p 1 = that ( ( a 0 p 0 0 a 1, 0 p 1 G. ( ( ap 0 ap 0 0 p 1 a 1 = 0 (ap 1 such ap 0 because of properties of real numbers. Associativity is inherited from the group of real 2 2 matrices with non-zero determinant. Next we show the existence of inverses. For each ( a 0 0 a 1 G with a 0, with a 1 0 as a 0 such that ( ( ( a 0 a 1 0 a a 1. = 0 a 0 a ( a a ( a 0 0 a 1 G = ( Thus, all group axioms hold and G is a group under matrix multiplication. We show G is abelian. ( ( a 0 p 0 Let 0 a 1, 0 p 1 G, such that ( ( ( ( a 0 p 0 ap 0 ap 0 0 a 1. 0 p 1 = 0 p 1.a 1 = 0 (ap 1 ( ( ( ( p 0 a 0 pa 0 pa 0 and 0 p 1. 0 a 1 = 0 a 1 p 1 = 0 (pa 1. But ap = pa because R is abelian. Thus, G is abelian. Problem Construct in the G of Problem 21 a subgroup of order 4. {( ( ( ( } Proof. Let H =,,, Clearly every element of H is in G as determinant is non-zero. In fact determinant is 1. Firstly we note that identity belongs to H. Next we show closure. ( ( a 0 p 0 Let, H such that ab = 1, pq = 1 0 b 0 q ( ( ( a 0 p 0 ap 0 Then. = 0 b 0 q 0 bq where apbq = abpq = 1.1 = 1 and thus we have closure. The associativity is inherited from G. The existence of inverse is clear from the fact that for each.

16 Solutions by Sadiah Zahoor 16 ( a 0 A = H with ab = 1, 0 b B = 1 ( b 0 H with ba ab 0 a ab = 1 such that AB = BA = I where I is the identity. Thus all group axioms hold and H is a subgroup of G of order 4. ( a b Problem Let G be the set of all real 2 2 matrices, where a, b, c, d are integers modulo 2, such that ad bc 0. Using matrix multiplication as the operation in G, prove that G is a group of order 6. Proof. Z 2 = {0, 1}. ( a b Number of possible real 2 2 matrices, where a, b, c, d are integers modulo 2 are 2 4 = 16. Now we need only those elements whose determinant is non-zero. We note that since G is a group, each element in G occurs once in every row and once in every column in the group table, thus, we can express it in G distinct products. Let us count the elements with determinant 0. Case 1: ad bc = 0 0 = 0, This implies ad = 0 and bc = 0. For ad = 0, we have 3 choices for the product and for bc = 0, we have 3 choices for the product. Therefore we have 9 choices. Case 2: ad bc = 1 1 = 1, This implies that ad = 1 and bc = 1. For ad = 1, we have 1 choices for the product and for bc = 1, we have 1 choices for the product. Therefore we have 1 choices. Thus we have 10 choices with determinant 0. Thus the elements which have non-zero determinant are = 6 in number and o(g = 6. In fact we can write the elements explicitly and prove it is a group. {( ( ( ( ( ( } G =,,,,, ( 1 0 We first observe the identity G. 0 1 ( ( a b p q Next we show closure. Let, G with ad bc 0 and r s pq rs 0. Since we are working in Z 2, then ad bc = pq rs = 1. ( ( ( a b p q ap + br aq + bs. = r s cp + dr cq + ds such that apcq+apds+brcq+brds cpaq cpbs draq drbs = apds+brcq cpbs draq =

17 Solutions by Sadiah Zahoor 17 ad(ps rq + bc(rq ps = (ps rq(ad bc = ( 1.1 = 1 a b Thus, the determinant is not zero. And we have G is closed. ( p q. r s G and The associativity is inherited from the associativity of group of real 2 2 matrices with non-zero determinant. Finally we need to show existence of inverses in G. ( ( a b d b For each ( a b such that G with ad bc = 1, G with ad bc = 1 c a ( ( ( ( d b d b a b 1 0 = =. c a c a 0 1. Thus, all group axioms hold and G is a group of order 6 under multiplication modulo 2. ( a b Problem (a. Let G be the group of all 2 2 matrices where ad bc 0 and a, b, c, d are integers modulo 3, relative to matrix multiplication. Show that o(g = 48. (b. If we modify the example of G in part (a by insisting that ad bc = 1, then what is o(g? Proof. (a. Z 3 = {0, 1, 2}. ( a b Number of possible real 2 2 matrices, where a, b, c, d are integers modulo 3 are 3 4 = 81. Now we need only those elements whose determinant is non-zero. We note that since G is a group, each element in G occurs once in every row and once in every column in the group table, thus, we can express it in G distinct products. Let us count the elements with determinant 0. Case 1: ad bc = 0 0 = 0, This implies ad = 0 and bc = 0. For ad = 0, we have 5 choices for the product and for bc = 0, we have 5 choices for the product. Therefore we have 25 choices. Case 2: ad bc = 1 1 = 0, This implies that ad = 1 and bc = 1. For ad = 1, we have 2 choices for the product and for bc = 1, we have 2 choices for the product. Therefore we have 4 choices. Case 3: ad bc = 2 2 = 0, This implies ad = 2 and bc = 2. For ad = 2, we have 2 choices for the product and for bc = 2, we have 2 choices for the product. Therefore we have 4 choices. Thus we have 33 choices with determinant 0. Thus the elements which have non-zero determinant are = 48 in number and o(g = 48.

18 Solutions by Sadiah Zahoor 18 (b. If ad bc = 1 ad bc 2. Since we are working in Z 3, we can have the following possibilities for ad bc: ad bc = 2 2 = 0, ad bc = 1 1 = 0, ad bc = 0 0 = 0 which has 33 possibilities as shown in previous part. ad bc = 2 0 = 2, ad bc = 0 1 = 2, ad bc = 1 2 = 1 = 2 which are the cases we want to exclude. ad bc = 1 0 = 1, ad bc = 2 1 = 1, ad bc = 0 2 = 2 = 1 which are the cases we want to include. We note that since G is a group, each element in G occurs once in every row and once in every column in the group table, thus, we can express it in G distinct products. Let us count the elements with determinant 1. Case 1: ad bc = 1 0 = 1, This implies ad = 1 and bc = 0. For ad = 1, we have 2 choices for the product and for bc = 0, we have 5 choices for the product. Therefore we have 10 choices. Case 2: ad bc = 2 1 = 1, This implies that ad = 2 and bc = 1. For ad = 2, we have 2 choices for the product and for bc = 1, we have 2 choices for the product. Therefore we have 10 choices. Case 3: ad bc = 0 2 = 2 = 1, This implies ad = 1 and bc = 2. For ad = 0, we have 5 choices for the product and for bc = 2, we have 2 choices for the product. Therefore we have 4 choices. Thus we have 24 choices with determinant 1. So o(g = 24. ( a b Problem (a. Let G be the group of all 2 2 matrices where a, b, c, d are integers modulo p, p a prime number, such that ad bc 0. G forms a group relative to matrix multiplication. What is o(g? (b. Let H be the subgroup of the G of part (a defined by {( } a b H = G ad bc = 1. What is o(h? Proof. (a. Z p = {0, 1,..., p 1}. ( a b Number of possible real 2 2 matrices, where a, b, c, d are integers modulo p are p 4. Now we need only those elements whose determinant is non-zero. We note that since G is a group, each element in G occurs once in every row and once in every column in the group table, thus, we can express it in G distinct products. Thus x x Z p, we can write x as G = p 1 distinct products. Let us count the elements with determinant 0.

19 Solutions by Sadiah Zahoor 19 Case 1: ad bc = 0 0 = 0, This implies ad = 0 and bc = 0. For ad = 0, when a = 0, we have p choices for d and when d = 0, we have p choices for a, counting a = 0 and d = 0 twice, we have p + p 1 = 2p 1 choices for the product and for bc = 0, we have 2p 1 choices for the product. Therefore we have (2p 1 2 choices. Case 2: ad bc = x x = 0; x Z p, This implies that ad = x and bc = x. For ad = x, we have p 1 choices for the product and for bc = x, we have p 1 choices for the product. Therefore we have (p 1 2 choices for each x and p 1 choices for value of x. Thus we have (p 1 3 choices. Thus we have (2p 1 2 +(p 1 3 = 4p p+p 3 3p 2 +3p 1 = p 3 +p 2 p choices with determinant 0. Thus the elements which have non-zero determinant are o(g = p 4 p 3 p 2 + p in number. Z p = {0, 1,..., p 1}. ( a b Number of possible real 2 2 matrices, where a, b, c, d are integers modulo p are p 4. Now we need only those elements whose determinant is 1. We note that since G is a group, each element in G occurs once in every row and once in every column in the group table, thus, we can express it in G distinct products. Thus x x Z p, we can write x as G = p 1 distinct products. Let us count the elements with determinant 1. Case 1: ad bc = 1 0 = 0, This implies ad = 1 and bc = 0. For ad = 1, we have p 1 choices for writing the product and for bc = 0, when b = 0, we have p choices for c and when c = 0, we have p choices for b, counting b = 0 and c = 0 twice, we have p+p 1 = 2p 1 choices for the product. Therefore we have (p 1(2p 1 = 2p 2 3p + 1 choices. Case 2: ad bc = (x + 1 x = 1; x Z p {p 1}, This implies that ad = x + 1 and bc = x. For ad = x + 1, we have p 1 choices for the product and for bc = x, we have p 1 choices for the product. Therefore we have (p 1 2 choices for each x and p 2 choices for value of x. Thus we have (p 1 2 (p 2 = p 3 4p 2 +5p 2 choices. Case 3: ad bc = 0 (p 1 = 1, This implies ad = 0 and bc = p 1. For ad = 0, when a = 0, we have p choices for d and when d = 0, we have p choices for a, counting a = 0 and d = 0 twice, we have p+p 1 = 2p 1 choices for the product and for bc = (p 1, we have (p 1 choices for the product. Therefore we have (p 1(2p 1 = 2p 2 3p + 1 choices. Thus we have 2(2p 2 3p p 3 4p 2 + 5p 2 = 4p 2 6p p 3

20 Solutions by Sadiah Zahoor 20 4p 2 + 5p 2 = p 3 p choices with determinant 1. Thus the elements which have determinant 1 are o(h = p 3 p in number.