# Normal Subgroups and Factor Groups

Size: px
Start display at page:

Transcription

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 non-zero 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 A-1 = [ ]. 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 well-defined 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 (non-trivial) 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 non-abelian 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: 1-1 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 1-1 -k k j -j j j -j -k k -1 1 i -i -j -j j k -k 1-1 -i i k k -k j -j -i i k -k k -j j i -i 1-1 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 90-1 = 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 1-1 = n ). Also, (n 1h 1) -1 = h 1-1 n 1-1 = nh 1-1 NH. By two-step 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 one-one, 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

### CHAPTER 9. Normal Subgroups and Factor Groups. Normal Subgroups

Normal Subgroups CHAPTER 9 Normal Subgroups and Factor Groups If H apple G, we have seen situations where ah 6= Ha 8 a 2 G. Definition (Normal Subgroup). A subgroup H of a group G is a normal subgroup

### Homework #11 Solutions

Homework #11 Solutions p 166, #18 We start by counting the elements in D m and D n, respectively, of order 2. If x D m and x 2 then either x is a flip or x is a rotation of order 2. The subgroup of rotations

### Introduction to Groups

Introduction to Groups Hong-Jian Lai August 2000 1. Basic Concepts and Facts (1.1) A semigroup is an ordered pair (G, ) where G is a nonempty set and is a binary operation on G satisfying: (G1) a (b c)

### First Semester Abstract Algebra for Undergraduates

First Semester Abstract Algebra for Undergraduates Lecture notes by: Khim R Shrestha, Ph. D. Assistant Professor of Mathematics University of Great Falls Great Falls, Montana Contents 1 Introduction to

### DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3. Contents

DISCRETE MATH (A LITTLE) & BASIC GROUP THEORY - PART 3/3 T.K.SUBRAHMONIAN MOOTHATHU Contents 1. Cayley s Theorem 1 2. The permutation group S n 2 3. Center of a group, and centralizers 4 4. Group actions

### Selected exercises from Abstract Algebra by Dummit and Foote (3rd edition).

Selected exercises from Abstract Algebra by Dummit Foote (3rd edition). Bryan Félix Abril 12, 2017 Section 4.1 Exercise 1. Let G act on the set A. Prove that if a, b A b = ga for some g G, then G b = gg

### 1.1 Definition. A monoid is a set M together with a map. 1.3 Definition. A monoid is commutative if x y = y x for all x, y M.

1 Monoids and groups 1.1 Definition. A monoid is a set M together with a map M M M, (x, y) x y such that (i) (x y) z = x (y z) x, y, z M (associativity); (ii) e M such that x e = e x = x for all x M (e

### 0 Sets and Induction. Sets

0 Sets and Induction Sets A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set

### Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3

Solutions to odd-numbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3 3. (a) Yes; (b) No; (c) No; (d) No; (e) Yes; (f) Yes; (g) Yes; (h) No; (i) Yes. Comments: (a) is the additive group

### Assigment 1. 1 a b. 0 1 c A B = (A B) (B A). 3. In each case, determine whether G is a group with the given operation.

1. Show that the set G = multiplication. Assigment 1 1 a b 0 1 c a, b, c R 0 0 1 is a group under matrix 2. Let U be a set and G = {A A U}. Show that G ia an abelian group under the operation defined by

### Algebra-I, Fall Solutions to Midterm #1

Algebra-I, Fall 2018. Solutions to Midterm #1 1. Let G be a group, H, K subgroups of G and a, b G. (a) (6 pts) Suppose that ah = bk. Prove that H = K. Solution: (a) Multiplying both sides by b 1 on the

### Teddy Einstein Math 4320

Teddy Einstein Math 4320 HW4 Solutions Problem 1: 2.92 An automorphism of a group G is an isomorphism G G. i. Prove that Aut G is a group under composition. Proof. Let f, g Aut G. Then f g is a bijective

### 2) e = e G G such that if a G 0 =0 G G such that if a G e a = a e = a. 0 +a = a+0 = a.

Chapter 2 Groups Groups are the central objects of algebra. In later chapters we will define rings and modules and see that they are special cases of groups. Also ring homomorphisms and module homomorphisms

### School of Mathematics and Statistics. MT5824 Topics in Groups. Problem Sheet I: Revision and Re-Activation

MRQ 2009 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet I: Revision and Re-Activation 1. Let H and K be subgroups of a group G. Define HK = {hk h H, k K }. (a) Show that HK

### MODEL ANSWERS TO THE FIFTH HOMEWORK

MODEL ANSWERS TO THE FIFTH HOMEWORK 1. Chapter 3, Section 5: 1 (a) Yes. Given a and b Z, φ(ab) = [ab] = [a][b] = φ(a)φ(b). This map is clearly surjective but not injective. Indeed the kernel is easily

### Groups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems

Group Theory Groups Subgroups Normal subgroups Quotient groups Homomorphisms Cyclic groups Permutation groups Cayley s theorem Class equations Sylow theorems Groups Definition : A non-empty set ( G,*)

### Solutions of exercise sheet 4

D-MATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 4 The content of the marked exercises (*) should be known for the exam. 1. Prove the following two properties of groups: 1. Every

### Groups. Chapter 1. If ab = ba for all a, b G we call the group commutative.

Chapter 1 Groups A group G is a set of objects { a, b, c, } (not necessarily countable) together with a binary operation which associates with any ordered pair of elements a, b in G a third element ab

### φ(xy) = (xy) n = x n y n = φ(x)φ(y)

Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =

### 120A LECTURE OUTLINES

120A LECTURE OUTLINES RUI WANG CONTENTS 1. Lecture 1. Introduction 1 2 1.1. An algebraic object to study 2 1.2. Group 2 1.3. Isomorphic binary operations 2 2. Lecture 2. Introduction 2 3 2.1. The multiplication

### DEPARTMENT OF MATHEMATIC EDUCATION MATHEMATIC AND NATURAL SCIENCE FACULTY

HANDOUT ABSTRACT ALGEBRA MUSTHOFA DEPARTMENT OF MATHEMATIC EDUCATION MATHEMATIC AND NATURAL SCIENCE FACULTY 2012 BINARY OPERATION We are all familiar with addition and multiplication of two numbers. Both

### Solutions to Assignment 4

1. Let G be a finite, abelian group written additively. Let x = g G g, and let G 2 be the subgroup of G defined by G 2 = {g G 2g = 0}. (a) Show that x = g G 2 g. (b) Show that x = 0 if G 2 = 2. If G 2

### A while back, we saw that hai with hai = 42 and Z 42 had basically the same structure, and said the groups were isomorphic, in some sense the same.

CHAPTER 6 Isomorphisms A while back, we saw that hai with hai = 42 and Z 42 had basically the same structure, and said the groups were isomorphic, in some sense the same. Definition (Group Isomorphism).

### Lecture Note of Week 2

Lecture Note of Week 2 2. Homomorphisms and Subgroups (2.1) Let G and H be groups. A map f : G H is a homomorphism if for all x, y G, f(xy) = f(x)f(y). f is an isomorphism if it is bijective. If f : G

### Section III.15. Factor-Group Computations and Simple Groups

III.15 Factor-Group Computations 1 Section III.15. Factor-Group Computations and Simple Groups Note. In this section, we try to extract information about a group G by considering properties of the factor

### Elements of solution for Homework 5

Elements of solution for Homework 5 General remarks How to use the First Isomorphism Theorem A standard way to prove statements of the form G/H is isomorphic to Γ is to construct a homomorphism ϕ : G Γ

### Basic Definitions: Group, subgroup, order of a group, order of an element, Abelian, center, centralizer, identity, inverse, closed.

Math 546 Review Exam 2 NOTE: An (*) at the end of a line indicates that you will not be asked for the proof of that specific item on the exam But you should still understand the idea and be able to apply

### Algebra I: Final 2012 June 22, 2012

1 Algebra I: Final 2012 June 22, 2012 Quote the following when necessary. A. Subgroup H of a group G: H G = H G, xy H and x 1 H for all x, y H. B. Order of an Element: Let g be an element of a group G.

### D-MATH Algebra I HS 2013 Prof. Brent Doran. Solution 3. Modular arithmetic, quotients, product groups

D-MATH Algebra I HS 2013 Prof. Brent Doran Solution 3 Modular arithmetic, quotients, product groups 1. Show that the functions f = 1/x, g = (x 1)/x generate a group of functions, the law of composition

### Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.

Glossary 1 Supplement. Dr. Bob s Modern Algebra Glossary Based on Fraleigh s A First Course on Abstract Algebra, 7th Edition, Sections 0 through IV.23 Abelian Group. A group G, (or just G for short) is

### Math 31 Take-home Midterm Solutions

Math 31 Take-home Midterm Solutions Due July 26, 2013 Name: Instructions: You may use your textbook (Saracino), the reserve text (Gallian), your notes from class (including the online lecture notes), and

### Exercises on chapter 1

Exercises on chapter 1 1. Let G be a group and H and K be subgroups. Let HK = {hk h H, k K}. (i) Prove that HK is a subgroup of G if and only if HK = KH. (ii) If either H or K is a normal subgroup of G

### MATH 436 Notes: Cyclic groups and Invariant Subgroups.

MATH 436 Notes: Cyclic groups and Invariant Subgroups. Jonathan Pakianathan September 30, 2003 1 Cyclic Groups Now that we have enough basic tools, let us go back and study the structure of cyclic groups.

### A. (Groups of order 8.) (a) Which of the five groups G (as specified in the question) have the following property: G has a normal subgroup N such that

MATH 402A - Solutions for the suggested problems. A. (Groups of order 8. (a Which of the five groups G (as specified in the question have the following property: G has a normal subgroup N such that N =

### 2MA105 Algebraic Structures I

2MA105 Algebraic Structures I Per-Anders Svensson http://homepage.lnu.se/staff/psvmsi/2ma105.html Lecture 7 Cosets once again Factor Groups Some Properties of Factor Groups Homomorphisms November 28, 2011

### MATH HL OPTION - REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis

MATH HL OPTION - REVISION SETS, RELATIONS AND GROUPS Compiled by: Christos Nikolaidis PART B: GROUPS GROUPS 1. ab The binary operation a * b is defined by a * b = a+ b +. (a) Prove that * is associative.

### Physics 251 Solution Set 1 Spring 2017

Physics 5 Solution Set Spring 07. Consider the set R consisting of pairs of real numbers. For (x,y) R, define scalar multiplication by: c(x,y) (cx,cy) for any real number c, and define vector addition

### 7 Semidirect product. Notes 7 Autumn Definition and properties

MTHM024/MTH74U Group Theory Notes 7 Autumn 20 7 Semidirect product 7. Definition and properties Let A be a normal subgroup of the group G. A complement for A in G is a subgroup H of G satisfying HA = G;

### MATH 101: ALGEBRA I WORKSHEET, DAY #3. Fill in the blanks as we finish our first pass on prerequisites of group theory.

MATH 101: ALGEBRA I WORKSHEET, DAY #3 Fill in the blanks as we finish our first pass on prerequisites of group theory 1 Subgroups, cosets Let G be a group Recall that a subgroup H G is a subset that is

### SPRING BREAK PRACTICE PROBLEMS - WORKED SOLUTIONS

Math 330 - Abstract Algebra I Spring 2009 SPRING BREAK PRACTICE PROBLEMS - WORKED SOLUTIONS (1) Suppose that G is a group, H G is a subgroup and K G is a normal subgroup. Prove that H K H. Solution: We

### Definitions, Theorems and Exercises. Abstract Algebra Math 332. Ethan D. Bloch

Definitions, Theorems and Exercises Abstract Algebra Math 332 Ethan D. Bloch December 26, 2013 ii Contents 1 Binary Operations 3 1.1 Binary Operations............................... 4 1.2 Isomorphic Binary

### Block Introduction. Some self study examples and proof of the theorems are left to the readers to check their progress.

Abstract Algebra The word Algebra is derived from the Arabic word al-jabr. Classically, algebra involves the study of equations and a number of problems that devoted out of the theory of equations. Then

### Fall /29/18 Time Limit: 75 Minutes

Math 411: Abstract Algebra Fall 2018 Midterm 10/29/18 Time Limit: 75 Minutes Name (Print): Solutions JHU-ID: This exam contains 8 pages (including this cover page) and 6 problems. Check to see if any pages

### The Outer Automorphism of S 6

Meena Jagadeesan 1 Karthik Karnik 2 Mentor: Akhil Mathew 1 Phillips Exeter Academy 2 Massachusetts Academy of Math and Science PRIMES Conference, May 2016 What is a Group? A group G is a set of elements

### Algebra I: Final 2018 June 20, 2018

1 Algebra I: Final 2018 June 20, 2018 ID#: Quote the following when necessary. A. Subgroup H of a group G: Name: H G = H G, xy H and x 1 H for all x, y H. B. Order of an Element: Let g be an element of

### its image and kernel. A subgroup of a group G is a non-empty subset K of G such that k 1 k 1

10 Chapter 1 Groups 1.1 Isomorphism theorems Throughout the chapter, we ll be studying the category of groups. Let G, H be groups. Recall that a homomorphism f : G H means a function such that f(g 1 g

### Theorems and Definitions in Group Theory

Theorems and Definitions in Group Theory Shunan Zhao Contents 1 Basics of a group 3 1.1 Basic Properties of Groups.......................... 3 1.2 Properties of Inverses............................. 3

### Algebra I Notes. Clayton J. Lungstrum. July 18, Based on the textbook Algebra by Serge Lang

Algebra I Notes Based on the textbook Algebra by Serge Lang Clayton J. Lungstrum July 18, 2013 Contents Contents 1 1 Group Theory 2 1.1 Basic Definitions and Examples......................... 2 1.2 Subgroups.....................................

### SUMMARY OF GROUPS AND RINGS GROUPS AND RINGS III Week 1 Lecture 1 Tuesday 3 March.

SUMMARY OF GROUPS AND RINGS GROUPS AND RINGS III 2009 Week 1 Lecture 1 Tuesday 3 March. 1. Introduction (Background from Algebra II) 1.1. Groups and Subgroups. Definition 1.1. A binary operation on a set

### Homework Problems, Math 200, Fall 2011 (Robert Boltje)

Homework Problems, Math 200, Fall 2011 (Robert Boltje) Due Friday, September 30: ( ) 0 a 1. Let S be the set of all matrices with entries a, b Z. Show 0 b that S is a semigroup under matrix multiplication

### Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018

Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The

### Chapter I: Groups. 1 Semigroups and Monoids

Chapter I: Groups 1 Semigroups and Monoids 1.1 Definition Let S be a set. (a) A binary operation on S is a map b : S S S. Usually, b(x, y) is abbreviated by xy, x y, x y, x y, x y, x + y, etc. (b) Let

### MODEL ANSWERS TO HWK #4. ϕ(ab) = [ab] = [a][b]

MODEL ANSWERS TO HWK #4 1. (i) Yes. Given a and b Z, ϕ(ab) = [ab] = [a][b] = ϕ(a)ϕ(b). This map is clearly surjective but not injective. Indeed the kernel is easily seen to be nz. (ii) No. Suppose that

### Written Homework # 2 Solution

Math 516 Fall 2006 Radford Written Homework # 2 Solution 10/09/06 Let G be a non-empty set with binary operation. For non-empty subsets S, T G we define the product of the sets S and T by If S = {s} is

### CONSEQUENCES OF THE SYLOW THEOREMS

CONSEQUENCES OF THE SYLOW THEOREMS KEITH CONRAD For a group theorist, Sylow s Theorem is such a basic tool, and so fundamental, that it is used almost without thinking, like breathing. Geoff Robinson 1.

### Automorphism Groups Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G.

Automorphism Groups 9-9-2012 Definition. An automorphism of a group G is an isomorphism G G. The set of automorphisms of G is denoted Aut G. Example. The identity map id : G G is an automorphism. Example.

### Extra exercises for algebra

Extra exercises for algebra These are extra exercises for the course algebra. They are meant for those students who tend to have already solved all the exercises at the beginning of the exercise session

### Algebra Review. Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor. June 15, 2001

Algebra Review Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor June 15, 2001 1 Groups Definition 1.1 A semigroup (G, ) is a set G with a binary operation such that: Axiom 1 ( a,

### Group Theory. Hwan Yup Jung. Department of Mathematics Education, Chungbuk National University

Group Theory Hwan Yup Jung Department of Mathematics Education, Chungbuk National University Hwan Yup Jung (CBNU) Group Theory March 1, 2013 1 / 111 Groups Definition A group is a set G with a binary operation

### Algebra. Travis Dirle. December 4, 2016

Abstract Algebra 2 Algebra Travis Dirle December 4, 2016 2 Contents 1 Groups 1 1.1 Semigroups, Monoids and Groups................ 1 1.2 Homomorphisms and Subgroups................. 2 1.3 Cyclic Groups...........................

### ABSTRACT ALGEBRA: REVIEW PROBLEMS ON GROUPS AND GALOIS THEORY

ABSTRACT ALGEBRA: REVIEW PROBLEMS ON GROUPS AND GALOIS THEORY John A. Beachy Northern Illinois University 2000 ii J.A.Beachy This is a supplement to Abstract Algebra, Second Edition by John A. Beachy and

### BASIC GROUP THEORY : G G G,

BASIC GROUP THEORY 18.904 1. Definitions Definition 1.1. A group (G, ) is a set G with a binary operation : G G G, and a unit e G, possessing the following properties. (1) Unital: for g G, we have g e

### Homomorphisms. The kernel of the homomorphism ϕ:g G, denoted Ker(ϕ), is the set of elements in G that are mapped to the identity in G.

10. Homomorphisms 1 Homomorphisms Isomorphisms are important in the study of groups because, being bijections, they ensure that the domain and codomain groups are of the same order, and being operation-preserving,

### MATH 1530 ABSTRACT ALGEBRA Selected solutions to problems. a + b = a + b,

MATH 1530 ABSTRACT ALGEBRA Selected solutions to problems Problem Set 2 2. Define a relation on R given by a b if a b Z. (a) Prove that is an equivalence relation. (b) Let R/Z denote the set of equivalence

### Course 311: Abstract Algebra Academic year

Course 311: Abstract Algebra Academic year 2007-08 D. R. Wilkins Copyright c David R. Wilkins 1997 2007 Contents 1 Topics in Group Theory 1 1.1 Groups............................... 1 1.2 Examples of Groups.......................

### Exercises MAT2200 spring 2013 Ark 4 Homomorphisms and factor groups

Exercises MAT2200 spring 2013 Ark 4 Homomorphisms and factor groups This Ark concerns the weeks No. (Mar ) and No. (Mar ). Plans until Eastern vacations: In the book the group theory included in the curriculum

### Math 210A: Algebra, Homework 5

Math 210A: Algebra, Homework 5 Ian Coley November 5, 2013 Problem 1. Prove that two elements σ and τ in S n are conjugate if and only if type σ = type τ. Suppose first that σ and τ are cycles. Suppose

### Math 451, 01, Exam #2 Answer Key

Math 451, 01, Exam #2 Answer Key 1. (25 points): If the statement is always true, circle True and prove it. If the statement is never true, circle False and prove that it can never be true. If the statement

### GROUP THEORY EXERCISES AND SOLUTIONS

GROUP THEORY EXERCISES AND SOLUTIONS Mahmut Kuzucuoğlu Middle East Technical University matmah@metu.edu.tr Ankara, TURKEY November 10, 2014 ii iii TABLE OF CONTENTS CHAPTERS 0. PREFACE..................................................

### Examples: The (left or right) cosets of the subgroup H = 11 in U(30) = {1, 7, 11, 13, 17, 19, 23, 29} are

Cosets Let H be a subset of the group G. (Usually, H is chosen to be a subgroup of G.) If a G, then we denote by ah the subset {ah h H}, the left coset of H containing a. Similarly, Ha = {ha h H} is the

### Cosets and Normal Subgroups

Cosets and Normal Subgroups (Last Updated: November 3, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from

### Groups and Symmetries

Groups and Symmetries Definition: Symmetry A symmetry of a shape is a rigid motion that takes vertices to vertices, edges to edges. Note: A rigid motion preserves angles and distances. Definition: Group

### S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES

S11MTH 3175 Group Theory (Prof.Todorov) Final (Practice Some Solutions) 2 BASIC PROPERTIES 1 Some Definitions For your convenience, we recall some of the definitions: A group G is called simple if it has

### M3P10: GROUP THEORY LECTURES BY DR. JOHN BRITNELL; NOTES BY ALEKSANDER HORAWA

M3P10: GROUP THEORY LECTURES BY DR. JOHN BRITNELL; NOTES BY ALEKSANDER HORAWA These are notes from the course M3P10: Group Theory taught by Dr. John Britnell, in Fall 2015 at Imperial College London. They

### Math 430 Final Exam, Fall 2008

IIT Dept. Applied Mathematics, December 9, 2008 1 PRINT Last name: Signature: First name: Student ID: Math 430 Final Exam, Fall 2008 Grades should be posted Friday 12/12. Have a good break, and don t forget

### Class Equation & Conjugacy in Groups

Subject: ALEBRA - V Lesson: Class Equation & Conjugacy in roups Lesson Developer: Shweta andhi Department / College: Department of Mathematics, Miranda House, University of Delhi Institute of Lifelong

### PROBLEMS FROM GROUP THEORY

PROBLEMS FROM GROUP THEORY Page 1 of 12 In the problems below, G, H, K, and N generally denote groups. We use p to stand for a positive prime integer. Aut( G ) denotes the group of automorphisms of G.

### Cover Page. The handle holds various files of this Leiden University dissertation

Cover Page The handle http://hdl.handle.net/1887/54851 holds various files of this Leiden University dissertation Author: Stanojkovski, M. Title: Intense automorphisms of finite groups Issue Date: 2017-09-05

### book 2005/1/23 20:41 page 132 #146

book 2005/1/23 20:41 page 132 #146 132 2. BASIC THEORY OF GROUPS Definition 2.6.16. Let a and b be elements of a group G. We say that b is conjugate to a if there is a g G such that b = gag 1. You are

Answers to Final Exam MA441: Algebraic Structures I 20 December 2003 1) Definitions (20 points) 1. Given a subgroup H G, define the quotient group G/H. (Describe the set and the group operation.) The quotient

### FROM GROUPS TO GALOIS Amin Witno

WON Series in Discrete Mathematics and Modern Algebra Volume 6 FROM GROUPS TO GALOIS Amin Witno These notes 1 have been prepared for the students at Philadelphia University (Jordan) who are taking the

### The Gordon game. EWU Digital Commons. Eastern Washington University. Anthony Frenk Eastern Washington University

Eastern Washington University EWU Digital Commons EWU Masters Thesis Collection Student Research and Creative Works 2013 The Gordon game Anthony Frenk Eastern Washington University Follow this and additional

### Elementary Algebra Chinese Remainder Theorem Euclidean Algorithm

Elementary Algebra Chinese Remainder Theorem Euclidean Algorithm April 11, 2010 1 Algebra We start by discussing algebraic structures and their properties. This is presented in more depth than what we

### HOMEWORK Graduate Abstract Algebra I May 2, 2004

Math 5331 Sec 121 Spring 2004, UT Arlington HOMEWORK Graduate Abstract Algebra I May 2, 2004 The required text is Algebra, by Thomas W. Hungerford, Graduate Texts in Mathematics, Vol 73, Springer. (it

### On The Number Of Distinct Cyclic Subgroups Of A Given Finite Group

BearWorks Institutional Repository MSU Graduate Theses Spring 2016 On The Number Of Distinct Cyclic Subgroups Of A Given Finite Group Joseph Dillstrom As with any intellectual project, the content and

### 1.5 Applications Of The Sylow Theorems

14 CHAPTER1. GROUP THEORY 8. The Sylow theorems are about subgroups whose order is a power of a prime p. Here is a result about subgroups of index p. Let H be a subgroup of the finite group G, and assume

### INTRODUCTION TO THE GROUP THEORY

Lecture Notes on Structure of Algebra INTRODUCTION TO THE GROUP THEORY By : Drs. Antonius Cahya Prihandoko, M.App.Sc e-mail: antoniuscp.fkip@unej.ac.id Mathematics Education Study Program Faculty of Teacher

### Math 2070BC Term 2 Weeks 1 13 Lecture Notes

Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic

7. Let K = 15 be the subgroup of G = Z generated by 15. (a) List the elements of K = 15. Answer: K = 15 = {15k k Z} (b) Prove that K is normal subgroup of G. Proof: (Z +) is Abelian group and any subgroup

### Modern Algebra (MA 521) Synopsis of lectures July-Nov 2015 semester, IIT Guwahati

Modern Algebra (MA 521) Synopsis of lectures July-Nov 2015 semester, IIT Guwahati Shyamashree Upadhyay Contents 1 Lecture 1 4 1.1 Properties of Integers....................... 4 1.2 Sets, relations and

### Chapter 5 Groups of permutations (bijections) Basic notation and ideas We study the most general type of groups - groups of permutations

Chapter 5 Groups of permutations (bijections) Basic notation and ideas We study the most general type of groups - groups of permutations (bijections). Definition A bijection from a set A to itself is also

### Two subgroups and semi-direct products

Two subgroups and semi-direct products 1 First remarks Throughout, we shall keep the following notation: G is a group, written multiplicatively, and H and K are two subgroups of G. We define the subset

### Problems in Abstract Algebra

Problems in Abstract Algebra Omid Hatami [Version 0.3, 25 November 2008] 2 Introduction The heart of Mathematics is its problems. Paul Halmos The purpose of this book is to present a collection of interesting

### Algebra homework 6 Homomorphisms, isomorphisms

MATH-UA.343.005 T.A. Louis Guigo Algebra homework 6 Homomorphisms, isomorphisms Exercise 1. Show that the following maps are group homomorphisms and compute their kernels. (a f : (R, (GL 2 (R, given by

### Normal Subgroups and Quotient Groups

Normal Subgroups and Quotient Groups 3-20-2014 A subgroup H < G is normal if ghg 1 H for all g G. Notation: H G. Every subgroup of an abelian group is normal. Every subgroup of index 2 is normal. If H

### Algebraic structures I

MTH5100 Assignment 1-10 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one

### (1) Let G be a finite group and let P be a normal p-subgroup of G. Show that P is contained in every Sylow p-subgroup of G.

(1) Let G be a finite group and let P be a normal p-subgroup of G. Show that P is contained in every Sylow p-subgroup of G. (2) Determine all groups of order 21 up to isomorphism. (3) Let P be s Sylow

### MA441: Algebraic Structures I. Lecture 26

MA441: Algebraic Structures I Lecture 26 10 December 2003 1 (page 179) Example 13: A 4 has no subgroup of order 6. BWOC, suppose H < A 4 has order 6. Then H A 4, since it has index 2. Thus A 4 /H has order