Normal Subgroups and Factor Groups


 Garry Moody
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1 Normal Subgroups and Factor Groups Subject: Mathematics Course Developer: Harshdeep Singh Department/ College: Assistant Professor, Department of Mathematics, Sri Venkateswara College, University of Delhi INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 1
2 Introduction If G is a group, and H is a subgroup of G, and g is an element of G, then gh = {gh : h an element of H} is the left coset of H in G with respect to g, and Hg = {hg : h an element of H} is the right coset of H in G with respect to g. For example, consider a group G = S 3 and the subgroup H = { (1), (12) }. Then for g = (13) G, gh = (13)H = { (13), (123) }, and Hg = H(13) = { (13), (132) }. In this chapter we will study the algebraic properties of the subgroups whose left coset and right coset are same, and for those subgroups, their cosets form a group called the quotient or factor group. Then we will study characteristic, commutator subgroups. We will look into problems for better understanding of the text. Definition: Normal Subgroup A subgroup H of a group G, is called a normal subgroup if gh = Hg, for all g in G. Notation: A subgroup H is normal subgroup of G is denoted by H G. The property gh = Hg, for g G, means that gh 1 = h 2g for some h 1, h 2 H. Theorem 1: Normal Subgroup Test A subgroup H of a group G, is normal in G, if and only if ghg 1 H, for all g G. Proof: ( ) If H G, then for any g G and h H, gh = h 1g for some h 1 H (by definition); i.e., ghg 1 = h 1. Hence, ghg 1 H. ( ) If ghg 1 H, for all g G, then for some a G, aha 1 H, or ah Ha. Also, if g = a 1 (as a 1 is also a member of G), then a 1 Ha H, or Ha ah. Hence, ah = Ha, for all a G. Remarks: INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 2
3 We have just learned that, in order to show a subgroup H of G normal, we have to show, either ah = Ha for all a G, or aha 1 H for all a G; and, In order to show a subgroup H of G, not normal, i.e., H G, we have to find at least one element g G, and h H, such that ghg 1 H. Example 1: Every subgroup of an Abelian group is normal (Abelian group: Group in which xy = yx x, y G). Solution: Let G be an Abelian group, and H be its subgroup. For a G and h H, ah = ha (as h H G, h G). Hence, aha 1 = h, which implies, a 1 Ha H. Therefore, H is normal subgroup of G. Example 2: The center of a group is always normal subgroup of the group. Solution: Let G be a group and Z(G) be the center of group G. We know that Z(G) is a subgroup of G. In order to show Z(G) G, we need to show gz(g)g 1 Z(G), g G. So, let x Z(G), or gxg 1 gz(g)g 1. Since, gx = xg, as x Z(G), Therefore, gxg 1 = x, and hence, gxg 1 Z(G). gz(g)g 1 Z(G). Hence, Z(G) G. Example 3: Let S n be the group of permutations on n elements {1, 2, 3,..., n}. Let A n be the subgroup of S n, consisting of all even permutations. Then A n S n. Solution: For σ S n, we have to show that, σa nσ 1 A n. Let h A n. Then h is an even permutation (by definition of A n). Case 1: If σ is an even permutation, then σhσ 1 is also an even permutation for all σ S n (product of even permutations). Case 2: If σ is an odd permutation, then σh is also an odd permutation, also, σ 1 is an odd permutation, which implies σhσ 1 is an even permutation (product of two odd permutations is an even permutation). Therefore, σhσ 1 A n. Hence, σa nσ 1 A n. Therefore A n S n. Example 4: The subgroup of rotations in D n is a normal subgroup of D n. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 3
4 Solution: Let subgroup of rotations be denoted by R. We need to show that grg 1 R, g D n. Case 1: If g D n, be rotation, then for any rotation r, grg 1 is also a rotation. Case 2: If g D n, be reflection, then for any rotation r, grg 1 = r 1, hence is a rotation. (as, frf 1 = r 1 ) grg 1 R. Hence, subgroup of rotations D n. Example 5: The subgroup SL(n, R) of n x n matrices with determinant 1, is a normal subgroup of GL(n, R), the group of n x n matrices with nonzero determinant. Solution: Let A SL(n, R), B GL(n, R). Since, det(bab 1 ) = det(b)det(a)det(b 1 ), and det(b 1 )= 1/ det(b). det(bab 1 ) = det(b)det(a)/det(b) = det(a) =1(as A SL(n, R). Hence, B SL(n, R) B 1 SL(n, R), B GL(n, R). SL(n, R) GL(n, R). Example 6: Let H = {(1), (12)} be a subgroup of S 3. Then, H S 3. Solution: Let (13) S 3. Since (13) 1 = (13), and (12) H. Since, (13)(12) (13) 1 = (13)(12)(13) = (23) H. Hence, H S 3. Example 7: Let H = {[ a b ] a, b, d R, ad 0}. Then, H GL(2, R). 0 d Solution: Let A = [ ] GL(2, R). Then A1 = [ ]. Let B = [1 ] H. Since, ABA 1 = [ ] [ ] [0 1 1 ] = [2 ] H. Hence, H GL(2, R) Example 8: Let G = GL(2, R), and let K be a subgroup of R* i.e., (R\{0},*). Then H = { A G det(a) K } G. Solution: Let B G, and A H. Then det(bab 1 ) = det(b)det(a)/det(b) = det(a) K (as A H). Therefore, BAB 1 H, and hence, B H B 1 H, which implies, H G. Example 9: If a subgroup H of G has index 2 in G, then H G. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 4
5 Solution: Subgroup H of G has index 2 in G means, there are exactly 2 distinct left(or, right) cosets of H in G. Let x G. Case 1: If x H, then xh = H = Hx. Case 2: If x H, then xh H = φ, and. Also, Hx H = φ. Since, there are exactly 2 distinct left(or, right) cosets of H in G, G = H Hx, and G = H xh. Hence, xh = Hx. Hence, for x G, xh = Hx. H G. Example 10: Let H = { (1), (12)(34) } be a subgroup of A 4. Then, H A 4. Solution: Consider the even permutation (123) A 4. Since, (123)(12)(34)(123) 1 = (123)(12)(34)(132) = (14)(23) H. Hence, H A 4. Factor Groups Let N be a normal subgroup of a group (G,*). We define the set G/N to be the set of all left (or, right) cosets of N in G, i.e., G/N = { an : a G }. We define the binary operation on G/N as follows: For each an and bn G/N, the binary operation of an and bn is written as (an)(bn), which is equal to (a*b)n, where * is the binary operation of the group G. In view of this G/N is a group, which is defined and proved in the succeeding topics. Definition: Factor Groups For any group G, let N be a normal subgroup of G. Then, the set of all left (or right) cosets of N in G, is also a group itself, known as factor group of G by N (or the quotient group of G by N). Notation: The factor group of G by H is denoted by G/H. And, G/H = { ah a G }. Theorem 2: Factor Groups INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 5
6 Let G be a group and let H be a normal subgroup of G. Then the set G/H = { ah a G } is a group under the operation (ah)(bh) = abh. Solution: Firstly, we have to show that the operation is well defined; for this, we must show that the correspondence defined above from G/H x G/H into G/H is actually a function. To do this we assume that for some elements a, a, b, and b from G, we have ah = a H and bh = b H and verify that (ah)(bh) = (a H)(b H). That is, verify that abh = a b H. (This shows that the definition of multiplication depends only on the cosets and not on the coset representatives). From, ah = a H and bh = b H, we have a = ah 1 and b = bh 2 for some h 1, h 2 H, and therefore a b H = ah 1bh 2H = ah 1bH = ah 1Hb = ahb = abh (using H is normal subgroup of G). Now, eh = H is the identity; a 1 H is the inverse of ah; and (ah)(bh)(ch) = (abh)ch = (ab)(ch) = a(bc)h = ah(bc)h = ah(bh)(ch). This proves that G/H is a group. Remark: The group G/H is known as Factor Group G by H. Example 11: Since Z is an Abelian group, 4Z is normal subgroup of Z. Find the factor group Z/4Z. Solution: Z/4Z = { a + 4Z a Z } = { 0 + 4Z, 1 +4Z, 2 +4Z, 3 +4Z } (using the fact that, 0 + 4Z = 4 + 4Z = 8 + 4Z =..., as 0 4 4Z; 1 + 4Z = 5 + 4Z = 9 + 4Z =...; so on). Example 12: Let G = Z 18, and let H = <6> = { 0, 6, 12 }. Find G/H. Solution: G/H = { 0 + H, 1 + H, 2 + H, 3 + H, 4 + H, 5 + H }, (as 0 + H = 6 + H = 12 + H =...; and so on). Example 13: Consider K = { R 0, R 180 } a subgroup of D 4. Write the elements of D 4/K. Solution: Since D 4 = { R 0, R 90, R 180, R 270, H, V, D 1, D 2 }, therefore, D 4/K = { R 0K, R 90K, R 180K, R 270K, HK, VK, D 1K, D 2K }. Since, R 0K = K = R 180K; R 270K = R 90K; HK = VK; D 1K = D 2K, therefore, D 4/K = { K, R 90K, HK, D 1K }. Result used: ah = bh, if and only if, a 1 b H. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 6
7 The cayley table for this group D 4/K is given as: K R 90K HK D 1K K K R 90K HK D 1K R 90K R 90K K D 1K HK HK HK D 1K K R 90K D 1K D 1K HK R 90K K Example 14: Let G = U(32), and let H = U 16(32). a) Find the order of G/H, and is it Abelian? b) Which of three Abelian groups of order 8 is it isomorphic to: Z 8, Z 4 Z 2, or Z 2 Z 2 Z 2? Solution: Since G = U(32) = { a Z 32 gcd(a,32) = 1 } = { 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31 }, and H = U 16(32) = { a U(32) a = 1mod(16) } = {1, 17 }. a) Order of G/H = (order of G)/(order of H). Hence, G/H = 16/2 = 8. Since G is Abelian, so is G/H. b) G/H = { 1H, 3H, 5H, 7H, 9H, 11H, 13H, 15H }(since, 17H = H, 19H = 3H, 21H = 5H, 23H = 7H, 25H = 9H, 27H = 11H, 29H = 13H, 31H = 15H ) Result: ah = least n N a n H Hence, 1H = 1; 3H = 4 (as, 3 2 = 9 H, 3 3 = 27 H, 3 4 = 81 = 17 H); similarly, 7H = 9H = 2. Since, Z 2 Z 2 Z 2 doesn t have any element of order 2, G/H cannot be isomorphic to it. Also, since, only 1 element of order 4 in Z 8, and G/H has more than 1 element of order 4, hence G/H cannot be isomorphic to it. Therefore, without loss of generality, G/H is isomorphic to Z 4 Z 2. Example 15: Let G = U(32) and K = { 1, 15 }. Then, find G/K and, G/K is isomorphic to Z 8, Z 4 Z 2, or Z 2 Z 2 Z 2? INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 7
8 Solution: G/K = G / K = 16/2 = 8. Since, G is Abelian, G/K is also Abelian group. Further, since, 3K G/K, and 3K = 8, and since any element of order 8 in Z 4 Z 2, and Z 2 Z 2 Z 2, therefore G/H is isomorphic to Z 8. Example 16: Show that A 4 has no subgroup of order 6. Solution: If possible, let H be a subgroup of A 4, such that H = 6. Since, A 4 = 12, H has index 2 in G, as G / H = 2, hence H G. Thus the factor group A 4/H exists, and A 4/H = 2. Since, order of an element divides the order of group, for ah A 4/H, ah = 1, or 2. If ah = 1, ah = H, hence, a H. If ah = 2, then a 2 H, for all a A 4. Since there are nine different elements in { a 2 a A 4 }, hence we have a contradiction to order of the subgroup H. Hence, A 4 has no subgroup of order 6. Theorem 3: The G/Z(G) Theorem Let G be a group and consider Z(G) to be the center of G. If G/Z(G) is cyclic, then G is Abelian. Proof: G/Z(G) is a cyclic factor group, so a generator of G/Z(G) say gz(g). Let a, b G, there exist integers p and q respectively such that az(g) = (gz(g)) p = g p Z(G), and bz(g) =(gz(g)) q = g q Z(G) Thus, a = g p y for some y Z(G) ( az(g) = g p Z(G) a 1 g p Z(G) a 1 g p = x for some x in Z(G). g p = ax g p x 1 = a, x 1 = y in Z(G) ); Similarly b = g q t for some t in Z(G). Therefore, ab = (g p y)(g q t) = g p (yg q )t = g p (g q y)t = (g p g q )yt = (g p g q )ty = (g p+q )ty = (g q+p )ty = (g q g p )ty = g q (g p t)y = g q (tg p )y = (g q t)(g p y) = ba. Thus, G is Abelian. Theorem 4: G/Z(G) Inn(G) For any group G, G/Z(G) is isomorphic to Inn(G). INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 8
9 Remark: Inn(G) = { ϕ g ϕ g: G G, ϕ g is an isomorphism, which is defined as, ϕ g(x) = g 1 xg, for x G } Proof: Define a mapping F: G/Z(G) Inn(G) by F(gZ(G)) = ϕ g. Firstly, we have to show that F is a welldefined mapping. Consider gz(g) = hz(g) g 1 h Z(G) g 1 hx = x g 1 h, x G. hxh 1 = gxg 1 x in G ϕ h(x) = ϕ g(x) x in G ϕ h = ϕ g F(gZ(G)) = F(hZ(G)). Now, to show F injective, Consider F(gZ(G)) = F(hZ(G)) ϕ g = ϕ h ϕ g(x) = ϕ h(x) x in G gxg 1 = hxh 1 x in G x g 1 h = g 1 hx x in G g 1 h Z(G) gz(g) = hz(g). F is onto because for any element ϕ g Inn(G) an element g G such that F(gZ(G)) = ϕ g. F is an operation preserving map (homomorphism) as F(gZ(G)hZ(G)) = F(g(Z(G)h)Z(G)) = F(g(hZ(G))Z(G)) = F(ghZ(G)) = ϕ gh = ϕ g ϕ h (by property of ϕ) = F(gZ(G))F(hZ(G)), Thus G/Z(G) is isomorphic to Inn(G). Example 17: Prove that Inn(D 6) D 3. Solution: We know Inn(D 6) D 6/Z(D 6); Since Z(D 6) = { R 0, R 180 }, Z(D 6) = 2, hence, Inn(D 6) = D 6/Z(D 6) = D 6 / Z(D 6) = 12/2 = 6. Since, groups of order 6, up to isomorphism are: D 3, or Z 6. Hence, Inn(D 6) D 3, or Z 6. If Inn(D 6) Z 6, then, Inn(D 6) is cyclic, and thus, by G/Z(G) theorem, D 6 is Abelian, which is a contradiction. Hence, Inn(D 6) D 3. Theorem 5: Existence of Element of Prime Order Let G be a finite Abelian group and let p be a prime number that divides the order of group G. Then G contain an element of order p. Proof: Firstly, suppose that G has order 2 and only prime number which divide the order of Group G is 2 itself; as group of order 2 has one element of order 1(identity) and other of order 2 so result holds trivially. Theorem will be proved by using Second Principle of Mathematical Induction on order of group G, according to which the theorem is assumed to be applicable for all those Abelian groups that have their order less than the order of group G and use this assumption to prove that theorem holds true for G also. Let x G, and x = n (G has finite order so INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 9
10 does x) and let n = mq where q is some prime number (any natural number can be written as such), Thus, x m = y (say) has order q (as q is the least positive integer such that x mq = x n = e, as n is the least), y G has order q, so if q = p then we are done. So assume p q. Since every subgroup of an Abelian group is normal, we can therefore construct factor group G = G/<y> (<y> is subgroup of Abelian group G hence normal). Then G is Abelian as G is Abelian and p divides G since G = G / <y> = G /q and p, q divides G (subgroup <y> order divides order of group). So G is Abelian group with order less than G and p/ G, Thus by Second Principle of Mathematical Induction, an element y<x> of G of order p, Thus (y<x>) p = <x> (identity in G ) y p <x> = <x> y p <x>. So y p = e or y p has order q (y p <x> and <x> has prime order). So if, y p = e y has order p then this will be the required element of order p, if y p has order q, so y q has order p thus y q = y G will be the element of order p. Thus, existence of element of order p in G is ensured. Some problems from Seventh Edition, Contemporary Abstract Algebra, by Joseph A. Gallian Question 1. Prove that a factor group of an Abelian group is Abelian. Solution: Recall: A group G is Abelian if ab = ba for all a, b G. Suppose G is Abelian. Let G/H be any factor group of G. We have to show that G/H is Abelian. Let ah, bh be any elements of G/H. Then (ah)(bh) = (ab)h by definition of multiplication in factor groups. Then (ab)h = (ba)h since G is Abelian. (ba)h = (bh)(ah) again definition of multiplication in factor group. Therefore (ah)(bh) = (bh)(ah). Therefore G/H is Abelian. Question 2. What is the order of the factor group (Z 10 U(10))/<(2, 9)>? Solution: We know that Z 10 U(10) = Z 10 U(10) = 10 x 4 = 40; and (2, 9) = l.c.m.( 2 in Z 10, 9 in U(10)) = l.c.m.(5, 2) = 10; Since, (2, 9) = <(2, 9)>, <(2, 9)> = 10. Hence, (Z 10 U(10))/<(2, 9)> = 40/10 = 4. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 10
11 Question 3. Show, by example, that in a factor group G/H it can happen that ah = bh, but a b. (Do not use a = e or b = e.) Solution: Let G = Z 6, and H = {0, 3}; We know that Z 6 is Abelian, and H is a subgroup of it, therefore, H is normal subgroup of Z 6. Hence, Z 6 /H is a factor group. Let a = 1, and b = 4. Here, ah = 1H = {1, 4}; bh = 4H = {4, 1}. Hence, ah = bh, but, a = 1 = 6, b = 4 = 3; i.e., a b. Question 4. The only subgroup of A 4 of order 4. Why does this imply that this subgroup must be normal in A 4? Generalize this to arbitrary finite groups. Solution: Let H be that only subgroup of A 4 which has order 4. For g A 4, gh is also a subgroup of A 4. Also, gh = H = 4. Since, a unique subgroup of order 4 of A 4, gh = H, similarly, Hg = H. Hence, gh = Hg for all g A 4, which implies H is normal in A 4. For generalization, let G be any group of finite order (say n), then for any m, s.t., m n. If a unique subgroup of order m, then that subgroup is normal (using the fact ah = H ). Question 5. Let p be a prime. Show that if H is a subgroup of a group of order 2p that is not normal, then H has order 2. Solution: Let G be a group, and G = 2p. Then possible orders of (nontrivial) subgroups of G are: 2, or p. If the subgroup has order p, then [G:H] = G / H = 2p/p = 2, i.e., H has index 2 in G, then H is normal in G, which is a contradiction. Hence, H = 2. Question 6. Show that D 13 is isomorphic to Inn(D 13). Solution: Result: For any prime p, Z(D p) = { R 0 }, i.e., Z(D p) = 1. We know G/Z(G) Inn(G), therefore, Inn(D 13) D 13/Z(D 13). Since, Z(D 13) is trivial subgroup of D 13, D 13/Z(D 13) = D 13. Hence, Inn(D 13) D 13. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 11
12 Question 7. If G is a group and G : Z(G) = 4, prove that G/Z(G) Z 2 Z 2. Solution: Since G : Z(G) = 4, G/Z(G) is isomorphic to either Z 4, or Z 2 Z 2. If G/Z(G) Z 4, then G/Z(G) is cyclic, and therefore, G is Abelian, which implies, Z(G) = G, i.e., G : Z(G) = 1, which is a contradiction to hypothesis given. Hence, G/Z(G) Z 2 Z 2. Question 8. Suppose that G is a nonabelian group of order p 3, where p is a prime, and Z(G) {e}. Prove that Z(G) = p. Solution: Since, Z(G) {e}, Z(G) can be p, p 2, or p 3 ; Since, if Z(G) = p 3 then Z(G) = G, and G will be Abelian, which is a contradiction. So, Z(G) can have order either p, or p 2. If Z(G) = p 2, then by G/Z(G) will have order p, hence G/Z(G) will be cyclic, implying G to be Abelian, which is again a contradiction. Hence, Z(G) = p. Question 9. If G = pq, where p and q are primes that are not necessarily distinct, prove that Z(G) = 1 or pq. Solution: Let G = pq, then possibilities for Z(G) can be 1, p, q, or pq (using Lagrange s theorem). Our task is to eliminate the possibilities of Z(G) = p and q. Suppose Z(G) = p, then G/Z(G) will have order q, which implies G/Z(G) is cyclic, and hence, G is Abelian, implying Z(G) = G, or Z(G) = pq, which is a contradiction in itself. Hence, Z(G) p. Similarly, Z(G) q. Hence, Z(G) can be 1 or pq only. Question 10. Let G be an Abelian group and let H be the subgroup consisting of all elements of G that have finite order. Prove that every nonidentity element in G/H has infinite order. Solution: If for ah G/H, where a H, i.e., ah H, and ah has finite order. Then (ah) n = H, for some n N. a n H = H for some n, a n H for some n N. (a n ) m = e, for some m N, since, H contains elements of finite order. Thus, a nm = e, hence a is finite, which is a contradiction. Hence, every nonidentity element in G/H has infinite order. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 12
13 Question 11. Determine all subgroups of R* that have finite index. Solution: Let H be a subgroup of G that has index n. G/H = n, hence for any ah G/H, (ah) n = H, i.e., a n H. Then (R*) n = { x n x R*} H. If n is odd, then (R*) n = R*; if n is even, then (R*) n = R +. So, H = R* or H = R +. Question 12. Let G = {±1, ±i, ±j, ±k}, where i 2 = j 2 = k 2 = 1, i = (1)i, 1 2 = ( 1) 2 = 1, ij = ji = k, jk = kj = i, and ki = ik = j. a) Construct the Cayley table for G. b) Show that H = {1, 1} G. c) Construct the Cayley table for G/H. Is G/H isomorphic to Z 4 or Z 2 Z 2? Solution: a) The cayley table for G is given as: 11 i i j j k k i i j j k k i i j j k k i i i 1 1 k k j j i i i 11 k k j j j j j k k 1 1 i i j j j k k 11 i i k k k j j i i k k k j j i i 11 b) In order to show H = {1, 1} G, we have to show ghg 1 H, for all g G. Let g G be any arbitrary element of G, then g1g 1 = gg 1 = 1 H. Also, g( 1)g 1 = (1)gg 1 = (1)1 = 1 H. Hence, ghg 1 H, and H = {1, 1} G. c) Here G/H = { H, ih, jh, kh } (as ih = (i)h, jh = (j)h, kh = (k)h). Since, (ah) 2 = H for a = 1, i, j, or k, G/H is isomorphic to Z 2 Z 2. Cayley table for G/H is given as: H ih jh kh INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 13
14 H H ih jh kh ih ih H kh jh jh jh kh H ih kh kh jh ih H Question 13. In D 4, let K = { R 0, D } and let L = { R 0, D, D, R 180 }. Show that K L D 4, but that K is not normal in D 4. Solution: Index of K in L is 2 ( L / K = 4/2 = 2 ). Therefore, K L. Also, since, index of L in D 4 is 2 ( as D4 / L = 2 ), L D 4. Now, we have to show that K is not normal in D 4. Since, R 90DR 901 = R 90DR 270 = D K, K is not normal in D 4. Question 14. Show that the intersection of two normal subgroups of G is a normal subgroup of G. Solution: Let H and K be normal subgroups of G. Let x H K. Then x H and x K. For any element g G, gxg 1 H (since H is normal) and gxg 1 K (since K is normal). So gxg 1 H K. Thus H K is normal subgroup of G. Question 15. Let N be a normal subgroup of G and let H be any subgroup of G. Prove that NH is a subgroup of G. Give an example to show that NH need not be a subgroup of G if neither N nor H is normal. Solution: Suppose n 1h and n 2h 2 NH. Then n 1h 1n 2h 2 = n 1n h 1h 2 NH (as h 1n 2h 11 = n ). Also, (n 1h 1) 1 = h 11 n 11 = nh 11 NH. By twostep subgroup test, NH is a subgroup of G. Take G = S 3, H = <(12)> and K = <(23)>. Then H and K are subgroups of G (each containing two elements) and HK = { (1), (12), (23), (132) } a set of size 4. Therefore HK is not a subgroup of G (by Lagrange s Theorem) since 4 does not divide 6. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 14
15 Question 16. Suppose that a group G has a subgroup of order n. Prove that the intersection of all subgroups of G of order n is a normal subgroup of G. Solution: Let D = { S S is a subgroup of G, and S = n }. Let g G. Case 1: If g 1 Dg is a subset of each subgroup of G of order n, then g 1 Dg is a subset of the intersection of all subgroups of G of order n. Hence, g 1 Dg D for each g G, and therefore D is normal in G. Case 2: If g 1 Dg is not contained in a subgroup, say H, of G of order n for some g G, thus D is not contained in ghg 1, for if D is contained in ghg 1, then g 1 Dg is contained in H which is a contradiction. But ghg 1 = n, and hence, D ghg 1, a contradiction. Thus, g 1 Dg = D for each g G. Hence D is normal in G. Question 17. Suppose that H is a normal subgroup of a finite group G. If G/H has an element of order n, show that G has an element of order n. Show, by example, that the assumption that G is finite is necessary. Solution: Let gh = n. Then, g n H for some n N. Hence, (g n ) t = e for some t N; Hence, g = nt and g t = n. Hence, G has an element of order n. Consider G = Z, and H = 2Z, then 1 + 2Z G/H has order 2, but any element in Z of order 2. Hence, for above result, the assumption that G is finite is necessary. Automorphism of a group Let G be a group. An automorphism of G is an isomorphism G G. We write Aut(G) for the set of all automorphisms of G. Characteristic Subgroup A subgroup N of a group G is called a characteristic subgroup if f(n) = N for all automorphisms f of G. Remark: It is sufficient to show f(n) N, in order to prove a subgroup characteristic, as f is bijective, and number of elements in f(n) =number of elements in N. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 15
16 Question 18. Prove that every subgroup of cyclic group is characteristic. Solution: Let G be a cyclic group. So, G = <g> for some g in G. Now, let H be a subgroup of G. So, H is cyclic as well, H = <g k > for some integer k. Next, any automorphism f: G G is of the form f(g) = g n for some integer n, where G = <g n > = <g> (automorphism takes generator to generator) So, f(h) = <f(g k )> (subgroup of cyclic group is cyclic) = <g nk >. Claim: <g nk > = <g k >. (Then f(h) = H, which shows that H is characteristic.) (i) <g nk > <g k >: This is true, because g nk = (g k ) n. (ii) <g k > <g nk >: Since <g n > = G, there exists an integer s such that g ns = g. Hence, g k = (g ns ) k = (g s ) nk, and we are done. Generating set of a group Let G be a cyclic group, we say G is generated by { a }, i.e., G = <a>, or if every element of G can be written as a n, for different values of n Z. Now, if generating set for a group G = { a 1, a 2, a 3, a 4,..., a m }, it means, every n1 n2 n3 nm element of G can be expressed as in the form of a 1 a 2 a 3...a m, for n1, n2, n3,..., nm Z. Question 19. Prove that the center of a group is characteristic. Solution: Let G be a group, and Z(G) be the center of G. We have to show that f(z(g)) = Z(G) for every automorphism f of G. So, let f be some arbitrary automorphism of G. Let b Z(G), Then, f(b) f(z(g)). Let g G, then g = f(u) for some u G (as, f is automorphism). Then, gf(b) = f(u)f(b) = f(ub) = f(bu) (as b Z(G)) = f(b)f(u) = f(b)g. Hence, f(z(g)) Z(G). Next, Let h Z(G), G. So h = f(y) for some y G. We want to show that y Z(G). Consider yg for some g G. Since, g G, g = f(x) for some x G. So yg = f 1 (h)f  1 (x) = f 1 (hx) = f 1 (xh) (as h Z(G)) = f 1 (x)f 1 (h) = gy. Hence, y Z(G) or h = f(y) f(z(g)). Hence, Z(G) f(z(g)), for f an automorphism of G. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 16
17 Hence, the center of a group is a characteristic subgroup. Commutator Subgroup Let G be a group, and G be a subgroup of G. Then G is called commutator subgroup of G if G is generated by the set { x 1 y 1 xy x, y G }. Question 20. Prove that commutator subgroup G of a group G is characteristic subgroup of G. Solution: Let G be a group, G be commutator subgroup of G. Let G is generated by the set { x 1 y 1 xy x, y G }. Let f Aut(G). Then f(g ) is generated by the set { f(x) 1 f(y) 1 f(x)f(y) f(x), f(y) f(g) }. Since f is oneone, and onto homomorphism, the set { f(x) 1 f(y) 1 f(x)f(y) f(x), f(y) f(g) } = { x 1 y 1 xy x, y G }. Hence, f(g ) = G, or G is characteristic subgroup of G. Question 21. Let G be a group and let G be the subgroup of G generated by the set S = {x 1 y 1 xy x, y G}. a) Prove that G is normal in G. b) Prove that G/G is abelian. c) If G/N is abelian, prove that G N. Solution: a) Let x, y, g G be given. Then g(x 1 y 1 xy)g 1 = g(x 1 g 1 gy 1 g 1 gxg 1 gy)g 1 = (gx 1 g 1 )(gy 1 g 1 )(gxg 1 )(gyg 1 ) = (gxg 1 ) 1 (gyg 1 ) 1 (gxg 1 )(gyg 1 ) G. Now let a G be given. Then a = a 1a 2 a n, 1 1 where each a i = x i y i x iy i, for some x i, y i G. By the above argument, ga ig 1 G for every i. Therefore, gag 1 = g(a 1a 2 a n)g 1 = (ga 1g 1 )(ga 2g 1 ) (ga ng 1 ) G. Thus, G is normal in G. b) Let x, y G be given. Then (x 1 y 1 xy)g = G, since x 1 y 1 xy G. Therefore, xy G = yx G (since, a 1 b H, if and only if, ah = bh) c) It suffices to show that x 1 y 1 xy N for every x, y G. Let x, y G be given. Since G/N is abelian, (x 1 y 1 xy)n = N, which implies that x 1 y 1 xy N. INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 17
18 Question 22. Let H be a normal subgroup of a finite group G and let x G. If gcd( x, G/H ) = 1, show that x H. Solution: Let x = m, and G/H = n; Since, xh G/H, (xh) n = H. Hence, x n H = H, which implies, x n H. Since gcd(m, n) = 1, there exist integers t, and q s.t., mt + nq = 1. Hence, x mt + nq = x 1. x mt x nq = x. x nq = x (since x = m). Now, since x n H, and H is a subgroup of G, x nq H; Hence, x H. References: 1. Dummit, David S., Foote, Richard M. (2004), Abstract Algebra (3 rd edition) 2. Joseph A. Gallian (2013), Contemporary Abstract Algebra (8 th edition) INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 18
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