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EXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS Mahmut Kuzucuo˘glu Middle East Technical University matmah@metu.edu.tr Ankara, TURKEY April 18, 2012
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iii TABLE OF CONTENTS CHAPTERS 0. PREFACE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v 1. SETS, INTEGERS, FUNCTIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2. GROUPS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3. RINGS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .55 4. FIELDS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 5. INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
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v Preface These notes are prepared in 1991 when we gave the abstract al- gebra course. Our intention was to help the students by giving them some exercises and get them familiar with some solutions. Some of the solutions here are very short and in the form of a hint. I would like to thank B¨ulent B¨uy¨ukbozkırlı for his help during the preparation of these notes. I would like to thank also Prof. Ismail S˙ ¸. G¨ulo˘glu for checking some of the solutions. Of course the remaining errors belongs to me. If you find any errors, I should be grateful to hear from you. Finally I would like to thank Aynur Bora and G¨uldane G¨um¨u¸s for their typing the manuscript in LATEX. Mahmut Kuzucuo˘glu I would like to thank our graduate students Tu˘gba Aslan, B¨u¸sra C¸ ınar, Fuat Erdem and Irfan Kadık¨˙ oyl¨u for reading the old version and pointing out some misprints. With their encouragement I have made the changes in the shape, namely I put the answers right after the questions. 20, December 2011
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M. Kuzucuo˘glu 1. SETS, INTEGERS, FUNCTIONS 1.1. If A is a finite set having n elements, prove that A has exactly 2n distinct subsets. Solution: Apply Induction on n. If |A| = 1, then A has exactly two subsets namely φ and A. So the claim is true for n = 1. Induction hypothesis: For any set having exactly n−1 elements, the number of subsets is 2n−1. Let now A = {a , a ,· · · , a } be a set with 1 2 n |A| = n. Any subset X of A is either contained in B = {a1,· · · , an−1} or a ∈ X. By induction hypothesis, there are exactly 2n−1 subsets of n A contained in B. Any other subset X of A which is not contained in B is of the form. X = {an}∪Y where Y is a subset of B. Their number is therefore equal to the number of subsets of B, i.e. 2n−1. Then the number of all subsets of A is 2n−1 + 2n−1 = 2n. 1.2. For the given set and relations below, determine which define equivalence relations. (a) S is the set of all people in the world today, a ∼ b if a and b have an ancestor in common. (b) S is the set of all people in the world today, a ∼ b if a and b have the same father. 1
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2 M. KUZUCUO GLU˘ (c) S is the set of real numbers a ∼ b if a = ±b. (d) S is the set of all straight lines in the plane, a ∼ b if a is parallel to b. Solution: ′′ b, c and d ′′ are equivalence relations, but ′′ a ′′ is not. 1.3. Let a and b be two integers. If a|b and b|a, then show that a = ±b. Solution: If a|b, then b = ka for some integer k. If b|a, then a = `b for some integer l. Hence b = ka = k`b, then we obtain b − k`b = 0. This implies b(1 − k`) = 0 so either b = 0 or k` = 1. If b = 0, then a = 0 and hence a = ±b and we are done. If k` = 1, then either k = −1 and ` = −1 or k = 1 and ` = 1. In the first case b = −a, in the second case b = a. Hence b = ±a. 1.4. Let p1, p2,· · · , pn be distinct positive primes. Show that (p1p2 · · ·pn) + 1 is divisible by none of these primes. Solution: Assume that there exists a prime say pi where i ≤ n such that pi divides p1p2 · · ·pn + 1. Then clearly pi|p1p2 · · ·pn and pi|p1p2,· · ·pn + 1 implies that pi|1 = (p1 · · ·pn + 1) − (p1 · · ·pn). Which is impossible as pi ≥ 2. Hence none of the pi’s divides p1 · · ·pn + 1. 1.5. Prove that there are infinitely many primes. (Hint: Use the previous exercise.) Solution: Assume that there exists only finitely many primes say the list of all primes {p1, p2, . . . pn}. Then consider the integer p1p2 · · ·pn + 1. Then by previous Question 1.4 none of the primes pi i = 1, . . . , n divides p1p2 · · ·pn + 1. Hence either p1p2 · · ·pn + 1 is a prime which is not in our list or when we write p1p2 · · ·pn + 1 as a product of primes we get a new prime q which does not appear in {p1, p2,· · ·pn}. Hence in both ways we obtain a new prime which is not in our list. Hence we obtain a contradiction with the assumption that the number of primes is n. This implies that the number of primes is infinite.
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EXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS 3 1.6. If there are integers a, b, s, and t such that, the sum at+bs = 1, show that gcd(a, b) = 1. Solution: We have at + bs = 1 Assume that gcd(a, b) = n. Then by definition n|a and n|b and if there exists m|a and m|b, then m|n. Since n|a we have n|at and n|bs. Hence n|at + bs. This implies n|1. i.e. n = 1 1.7. Show that if a and b are positive integers, then ab = lcm(a, b) · gcd(a, b). Solution: Let gcd(a, b) = k and lcm(a, b) = l. Then a = ka and b = kb where gcd(a , b ) = 1 and ab = k2a b . 1 1 1 1 1 1 By definition a|l and b|l, moreover if there exists an integer s such that a|s and b|s, then l|s. Claim. l = ka1b1 = ab1 = a1b. Indeed we have a|ka1b1 and b|ka1b1. Assume that there exists an integer t such that a|t and b|t. Then t = ak1 and t = bk2. We have t = ak1 = bk2 = ka1k1 = kb1k2. It follows that a1k1 = b1k2. Since a1 and b1 are relatively prime we have a1|k2 and b1|k1. Then k2 = a1.c and k1 = b1.u. Then we have a1k1 = a1b1u = b1a1c it follows that u = c and t = ak1 = ka1b1c hence l = ka1b1|t.
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4 M. KUZUCUO GLU˘ 2. GROUPS 2.1. Let S be any set. Prove that the law of multiplication defined by ab = a is associative. Solution: Let x, y, z ∈ S. We want to show that x(yz) = (xy)z. Indeed x(yz) = xy = x by the law of multiplication in S. And (xy)z = xz = x, by the same law so x(yz) = x = (xy)z. 2.2. Assume that the equation xyz = 1 holds in a group G. Does it follow that yzx = 1? That yxz = 1? Justify your answer. Solution: xyz = 1 implies that x(yz) = 1. Let yz = a . Then we have xa = 1 and so ax = 1 since a is invertible and a−1 = x. (See solution 6) It follows that (yz)x = 1. Hence yzx = 1. On the other hand, if xyz = 1, it is not always true that yxz( = 1. To) 1 2 see this, let G be the group of 2×2 real matrices and let x = 0 2 ( ) ( ) ( ) 0 1 −1/2 3/4 1 0 y = and z = . Then xyz = = 1 2 1 1 −1 0 1 ( ) 2 −2 in G. But yxz = 6= 1. 5 −9/2 2.3. Let G be a nonempty set closed under an associative product, which in addition satisfies: (a) There exists an e ∈ G such that ae = a for all a ∈ G. (b) Given a ∈ G, there exists an element y(a) ∈ G such that ay(a) = e. Prove that G must be a group under this product. Solution: Given a ∈ G. Since right inverse exists, there exists y(a) ∈ G such that ay(a) = e. Then, y(a) = y(a)e = y(a)(ay(a)) = (y(a)a)y(a). Also, there exists t ∈ G such that y(a)t = e. This implies
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EXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS 5 that (y(a)a)y(a)t = e then (y(a)a)e = e Hence y(a)a = e. So every right inverse is also a left inverse. Now for any a ∈ G we have ea = (ay(a))a = a(y(a)a) = ae = a as e is a right identity. Hence e is a left identity. 2.4. If G is a group of even order, prove that it has an element a 6= e satisfying a2 = e. Solution: Define a relation on G by g ∼ h if and only if g = h or g = h−1 for all g, h ∈ G. It is easy to see that this is an equivalence relation. The equivalence class containing g is {g, g−1} and contains exactly 2 elements if and only if g2 6= e. Let C , C ,· · · , C be the equivalence classes of G with 1 2 k respect to ∼ . Then |G| = |C1| + |C2| + · · · + |Ck| Since each |Ci| ∈ {1,2} and |G| is even the number of equivalence classes Ci, with |Ci| = 1 is even. Since the equivalence class containing {e} has just one element, there must exist another equivalence class with exactly one element say {a}. Then e 6= a and a−1 = a. i.e. a2 = e. 2.5. If G is a finite group, show that there exists a positive integer m such that am = e for all a ∈ G. Solution: Let G be finite group and 1 6= a ∈ G. Consider the set a, a2, a3,· · · , ak · · · It is clear that ai 6= ai+1 for some integers from the beginning . Since G is a finite group there exists i and j such that ai = aj implies ai−j = 1.Therefore every element has finite order . That is the smallest positive integer k satisfying ak = 1 (One may assume without loss of generality that i > j). One can do this for each a ∈ G. The least common multiple m of the order of all elements of G satisfies am = 1 for all a ∈ G. 2.6. If G is a group in which (ab)i = aibi for three consecutive integers i for all a, b ∈ G, show that G is abelian.
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6 M. KUZUCUO GLU˘ Solution: Observe that if there exist two consecutive integers n, n+ 1 such that (ab)n = anbn and (ab)n+1 = an+1bn+1 for all a, b ∈ G, then an+1bn+1 = (ab)n+1 = (ab)nab = anbnab. Then we obtain an+1bn+1 = anbnab.Now by multiplying this equation from left by an and from right by b−1 we obtain abn = bna. In our case taking n = i and n = i + 1, we have abi = bia and by taking n = i + 1 and i + 2 we have abi+1 = bi+1a. This shows that abi+1 = bi+1a = bbia = babi and now multiplying from right by bi we obtain ab = ba. Hence G is abelian. 2.7. If G is a group such that (ab)2 = a2b2 for all a, b ∈ G, then show that G must be abelian. Solution: abab = a2b2 apply a−1 from left and b−1 from right. We obtain ba = ab for all a, b ∈ G. Hence G is abelian. 2.8. Let a, b be elements of a group G. Assume that a has order 5 and a3b = ba3. Prove that ab = ba. Solution: We have a5 = e and a3b = ba3. Applying a3 to the second equation we obtain a3(a3b) = a3(ba3) = (a3b)a3 = (ba3)a3 and hence a6b = ba6 As a6 = a5 · a = ea = a we obtain ab = ba. 2.9. Let a and b be integers. (a) Prove that the subset aZ + bZ = {ak + bl | l, k ∈ Z } is a subgroup of Z. (b) Prove that a and b + 7a generate the subgroup aZ + bZ. Solution: a) Clearly aZ + bZ 6= φ. Let ak1 + b`1, ak2 + b`2 be two elements in aZ+bZ where k1, k2, `1, `2 ∈ Z.We have (ak1 +b`1)− (ak2 + b`2) = a(k1 − k2) + b(`1 − `2) ∈ aZ + bZ as k1 − k2, `1 − `2 ∈ Z This implies aZ + bZ is a subgroup of Z. b) Firstly a, b+7a ∈ aZ+bZ. Secondly, given any ak+b` ∈ aZ+bZ we can write ak + b` = a(k − 7`) + (b + 7a)` this implies a and b + 7a generate aZ + bZ.
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EXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS 7 2.10. Let H be the subgroup generated by two elements a, b of a group G. Prove that if ab = ba, then H is an abelian group. Solution: The elements of H are of the form: ai1bi2ai3 · · ·aik−1bik where i1,· · ·ik ∈ Z, for some k. So let x, y ∈ H Then we can write x = ai1bi2 · · ·aik−1bik and y = aj1bj2 · · ·aj`−1bj`. Then xy = (ai1 · · ·bik )(aji · · ·bj`) since ab = ba we can interchange each term in this multiplication, and obtain: xy = (aj1 · · ·bj`)(ai1 · · ·bik ) = yx. This implies H is abelian. 2.11. (a) Assume that an element x of a group has order rs. Find the order of xr. (b) Assuming that x has arbitrary order n, what is the order of xr? Solution: (a) Since xrs = 1, we have (xr)s = 1. This implies that the order of xr ≤ s. Let us assume that order of xr is k. Since (xr)k = 1, xrk = 1 = xrs This implies that xrs−rk = 1 It follows that rs−rk = 0 since rs is the order of x and rs−rk < rs.This implies s = k.that is to say the order of xr is s. (b) Let k be the order of xr. Then (xrk) = 1 ⇒ xrk = 1. We also have xn = 1 which implies rk is a multiple of n. Since it should be the smallest such number and it also is a multiple of r, we conclude that rk is the least common multiple of r and n.Then k = `cm(r,n). r 2.12. Prove that in any group the orders of ab and of ba are equal. Solution: Let (ab)k = 1. Then abab· · ·ab = 1 This implies a(ba)(ba)· · ·(ba)b = 1 We have a(ba)k−1b = 1 ⇒ (ba)k−1 = a−1b−1 = (ba)−1 We obtain (ba)k−1(ba) = 1which implies (ba)k = 1. Similarly if (ba)` = 1, then (ab)` = 1. Hence orders of ab and ba are the same.
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8 M. KUZUCUO GLU˘ 2.13. Let G be a group such that the intersection of all its subgroups which are different from {e} is a subgroup different from identity. Prove that every element in G has finite order. Solution: Let ∩{e}6=H≤GH = K 6= {e}. Let e 6= a ∈ G. Then consider the subgroups H = 〈ai〉 for i = 1,2,3· · · are sub- i groups of G. For each i ∈ N such that Hi 6= {e} we have K ≤ Hi. Hence K is cyclic as every subgroup of a cyclic group is cyclic. Hence K = 〈an〉 for some fixed positive integer n. Since 〈an〉 ≤ 〈ai〉 if and only if an = aik if and only if n = ik if and only if i|n, we obtain i|n for each i = 1,2,3, . . . with Hi 6= {e}. As n is a fixed given positive integer it has only finitely many divisors. Since K 6= e we have only finitely many H 6= e i.e. So there exists j, such that H = 〈aj〉 = e i.e. aj = e. i j 2.14. Show that if every element of the group G is its own inverse, then G is abelian. Solution: For all x, y ∈ G we have (xy)−1 = xy and x−1 = x and y−1 = y. Then (xy)−1 = xy. This implies y−1x−1 = xy. Hence yx = xy . ( ) a b 2.15. Let G be the set of all 2×2 matrices where a, b, c, d c d are integers modulo 2, such that ad − bc 6= 0. Using matrix multiplica- tions as the operation in G prove that G is a group of order 6. Solution: In the first row of any matrix belonging to G, each entry could be 0 or 1 in Z2, but (0,0) should be extracted since ad − bc 6= 0. Hence we have 22 − 1 different choices for the first row. The second row is not a multiple of the first row. Hence G has (22 − 1)2 elements, namely 6. It can be checked that this set is closed under multiplication, and also obviously each element has its inverse in this set. Since matrix( ) 1 0 multiplication is associative, G is a group with identity . 0 1 {( ) ( ) ( ) ( ) ( ) ( )} 1 0 1 1 1 0 0 1 1 1 0 1 G = , , , , , 0 1 0 1 1 1 1 0 1 0 1 1
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EXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS 9 2.16. Let G be the group of all non-zero complex numbers a + bi (a, b real, but not both zero) under multiplication, and let H = { a + bi ∈ G | a2 + b2 = 1 } Verify that H is a subgroup of G. Solution: If a = 1, b = 0, then a+bi ∈ H so H is non-empty. Let a + bi, c + di ∈ H. Then (a + bi)(c + di) = (ac − bd) + (ad + bc)i (ac − bd)2 + (ad + bc)2 = a2c2 − 2acbd + b2d2 + a2d2 + b2c2 + 2abcd. = a2(c2 + d2) + b2(c2 + d2) = (a2 + b2)(c2 + d2) = 1. Hence H is closed with respect to multiplication. Now for the in- verse 1 a − bi 2 2 = = a − bi as a + b = 1 a + bi a2 + b2 Hence 1 = (a + bi)−1 ∈ H and H is a subgroup of G. a+bi 2.17. Let G be a finite group whose order is not divisible by 3. Suppose that (ab)3 = a3b3 for all a, b ∈ G. Prove that G must be abelian. Solution: Let |G| = n. Since (3, n) = 1 there exist x, y ∈ Z such that 3x + ny = 1. Here x can be taken positive. Now consider ab = (ab)3x+ny = (ab)3x(ab)ny. Since |G| = n and ab ∈ G, we have (ab)n = e. Hence ab = (ab)3x = (a3b3)x = a3b3a3b3 · · ·a3b3 ︸ ︷︷ ︸ x−times = a3 (b3a3)(b3a3)· · ·(b3a3)b3 ︸ ︷︷ ︸ (x−1) times
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10 M. KUZUCUO GLU˘ = a3(b3a3)x−1b3 = a3(ba)3x−3b3 = a3(ba)3x(ba)−3b3 = a3(ba)3x(ba)ny(ba)−3b3 = a3(ba)3x+ny(ba)−3b3 = a3(ba)(ba)−3b3 = a3(ba)−3(ba)b3 = a3a−3b−3(ba)b3 = b−2ab3 = b−2a−2a3b3 = (b−2a−2)(ab)3 Hence we get ab = b−2a−2(ab)3. Now multiplying from right by (ab)−1 and from left by b2 and a2 we obtain (ab)2 = a2b2 i.e. abab = a2b2 Multiply from left by a−1 and from right by b−1 we obtain ba = ab for all a, b ∈ G. ( ) a b 2.18. Let G be the group of all 2 × 2 matrices where c d ad − bc 6= 0 and a, b, c, d are integers module 3 relative to matrix mul- tiplication. Show that |G| = 48. b) If we modify the example of G in part (a) by insisting that ad − bc = 1, then what is |G|? Solution: For the first row (a, b) of a matrix in G a and b could be anything in Z3, but we must exclude the case a = 0 and b = 0. Hence (3 × 3) − 1 possibilities for the first row. The second row should be not a multiple of the first row. Hence for the second row (3 × 3) − 3 possibilities. Hence the number of elements in G is 8 × 6 = 48. ( ) a b (b) The set of matrices such that ad − bc = 1 forms a c d ( ) a b subgroup H of G. Moreover for any g = ∈ G either ad−bc = c d det (g) = 1 or det(g) = −1(≡ 2(mod 3)). On the other hand for any Hg1, Hg2 ∈ G, Hg1 = Hg2 if and only if det(g1) = det(g2). (Why?) Hence the above subgroup has index 2 in G, i.e. H has order 24.
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EXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS 11 (By using determinant homomorphism; this can be seen more eas- ily.) 2.19. (a) If H is a subgroup of G, and a ∈ G let aHa−1 = { aha−1 | h ∈ H } Show that aHa−1 is a subgroup of G. (b) If H is finite, what is the order o(aHa−1). Solution: (a) Since H is a subgroup clearly aHa−1 is non-empty. Let ah a−1 and ah a−1 ∈ aHa−1. Then 1 2 (ah a−1)(ah a−1)−1 = (ah a−1)(ah−1a−1) 1 2 1 2 = ah h−1a−1 ∈ aHa−1 1 2 as h h−1 ∈ H. 1 2 (b) Define a map α : H → aHa−1 by α(h) = aha−1 for all h ∈ H. The map α is 1 − 1 and onto. Hence if H is a finite group then o(H) = o(aHa−1). 2.20. The center Z of a group G is defined by Z = { z ∈ G | zx = xz for all x ∈ G }. Prove that Z is a subgroup of G. Solution: Let z, w ∈ Z. Hence xw = wx for all x ∈ G. Then we have w−1x = xw−1 for all x ∈ G, i.e. w−1 ∈ Z. Moreover (zw)x = z(wx) = z(xw) = (zx)w = x(zw) Hence zw ∈ Z and Z is a subgroup of G. 2.21. If H is a subgroup of G, then by the centralizer CG(H) of H we mean the set { x ∈ G | xh = hx for all x ∈ H }. Prove that CG(H) is a subgroup of G. Solution: Clearly 1 ∈ CG(H), so CG(H) 6= ∅. Let x, y ∈ CG(H). By assumption xh = hx and yh = hy for all h ∈ H. Hence hy−1 = y−1h for all h ∈ H. Hence y−1 ∈ C (H). Now consider G
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12 M. KUZUCUO GLU˘ (xy)(h) = x(yh) = x(hy) as y ∈ CG(H) = (xh)y = h(xy). Hence xy ∈ CG(H) 2.22. If a ∈ G define CG(a) = { x ∈ G | xa = ax }. Show that CG(a) is a subgroup of G. The group CG(a) is called the centralizer of a in G. Solution: Let x, y ∈ C (a). If ya = ay, then ay−1 = y−1a hence G y−1 ∈ C (a) and moreover (xy)a = x(ya) = x(ay) = a(xy). Hence G xy ∈ CG(a). (Observe that CG(a) = CG(〈a〉).) 2.23. If N is a normal subgroup of G and H is any subgroup of G prove that N H is a subgroup of G. Solution: Let n1h1 and n2h2 be two elements in N H, where ni ∈ N,and hi ∈ H, i = 1,2. Then n h (n h )−1 = n h h−1n−1 1 1 2 2 1 1 2 2 = n h h−1n−1(h h−1)(h h−1) 1 1 2 2 2 1 1 2 = n h h−1n−1(h h−1)−1(h h−1) 1 1 2 2 1 2 1 2 = n n h h−1 ∈ N H as n n ∈ N 1 3 1 2 1 3 where n = (h h−1)n−1(h h−1)−1 ∈ N 3 1 2 2 1 2 and h h−1 ∈ H . 1 2 2.24. Suppose that H is a subgroup of G such that whenever Ha 6= Hb, then aH 6= bH. Prove that gHg−1 ⊆ H. Solution: Assume (if possible) that gHg−1 6⊆ H for some g in G. Then there exists 1 6= h ∈ H such that ghg−1 ∈/ H. Hence ghg−1H 6= H. This implies hg−1H 6= g−1H (otherwise ghg−1H = H). Then by as- sumption and by the above observation we have, Hhg−1 = Hg−1 6= Hg−1. This is a contradiction. Hence gHg−1 must lie in H for any
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EXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS 13 g ∈ G. 2.25. Suppose that N and M are two normal subgroups of G and that N ∩ M = {e}. Show that for any n ∈ N, m ∈ M, nm = mn. Solution: Consider nmn−1m−1 in two ways. Firstly (nmn−1)m−1 ∈ M as nmn−1 ∈ M and m−1 ∈ M, secondly n(mn−1m−1) ∈ N as n ∈ N and mn−1m−1 ∈ N. Hence nmn−1m−1 ∈ N ∩ M = {e}. We obtain nmn−1m−1 = e i.e. nm = mn for all n ∈ N and m ∈ M. 2.26. If G is a group and H is a subgroup of index 2 in G, then prove that H is a normal subgroup of G. Solution: Let H and aH be the left cosets of H in G and H and Hb be right cosets of H in G. Since there are only two cosets aH = G \ H and Hb = G \ H, then we have aH = Hb. In order to show that H is normal in G, we want to see that xH = Hx for any x ∈ G. Let x ∈ G. If x ∈ H, then certainly xH = H = Hx i.e. xH = Hx. If x ∈ G \ H = aH = Hb, then there exist h1, h2 ∈ H such that x = ah1 = h2b. Then xH = (ah1)H = aH = Hb = H(h2b) = Hx, i.e. xH = Hx for any x ∈ G. 2.27. Show that the intersection of two normal subgroups of G is a normal subgroup of G. Solution: It is clear that the intersection of two subgroups of G is a subgroup of G. Let N and M be normal subgroups of G. Then for any g ∈ G and any n ∈ N and any m ∈ M we have gng−1 ∈ N and gmg−1 ∈ M. Let w ∈ N ∩M. Since w ∈ N we have gwg−1 ∈ N and since w ∈ M we have gwg−1 ∈ M. Hence gwg−1 ∈ N ∩ M for any g ∈ G i.e. N ∩ M is a normal subgroup of G. 2.28. If N and M are normal subgroups of G, prove that N M is also a normal subgroup of G.
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14 M. KUZUCUO GLU˘ Solution: By Question 2.23 N M is a subgroup of G. We need to show that for any g ∈ G any nm ∈ N M where n ∈ N, m ∈ M, we have gnmg−1 ∈ N M. But gnmg−1 = gng−1gmg−1 ∈ N M since gmg−1 ∈ M and gng−1 ∈ N. 2.29. If a cyclic group T of G is normal in G, then show that every subgroup of T is a normal subgroup in G. Solution: Let T = 〈t〉 be a cyclic group. Then for any g ∈ G we have gtg−1 = ti for some i ∈ Z. Let H be a subgroup of T. Hence H is cyclic and generated by tn for some n. Now consider gtng−1 = (gtg−1)n = tin = (tn)i ∈ H. Hence H is a normal sub- group of G. 2.30. If N is a normal subgroup in the finite group such that (|G : N|,|N|) = 1. Show that any element x ∈ G satisfying x|N| = e must be in N. Solution: Let x be an element in G such that x|N| = e. Since (|G : N|,|N|) = 1 there exist m, n ∈ Z such that m|G : N| + n|N| = 1. Then x = xm|G:N|+n|N| = xm|G:N|xn|N|. Since x|N| = e we have x = xm|G:N|. Consider the element xN of G/N. Since (xN)|G/N| = (x|G/N| N) we have (xN)|G/N| = N the identity element of G/N that means x|G/N| ∈ N and so x = (x|G:N|)m ∈ N . 2.31. Let G be a group in which for some integer n > 1,(ab)n = anbn for all a, b ∈ G. Show that a) Gn = { xn | x ∈ G } is a normal subgroup of G. b) Gn−1 = { xn−1 | x ∈ G } is a normal subgroup of G. Solution: (a) First we show that Gn is a subgroup of G. Let xn, yn ∈ Gn. Then xnyn = (xy)n ∈ Gn and (x−1)n = (xn)−1 ∈ Gn. Hence Gn is a subgroup of G. Let g ∈ G. Then gxng−1 = (gxg−1)n ∈ Gn hence Gn is a normal subgroup of G.
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