Abstract algebra notes

ABSTRACT ALGEBRA Contents 1 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Subgroups and Cyclic groups . . . . . . . . . . . . . . 1.3 Cosets and Lagrange’s theorem . . . . . . . . . . . . . 1.4 Normal Subgroups, Quotient group, Homomorphism . . 1.5 Direct Products of Groups . . . . . . . . . . . . . . . . 1.6 Class Equation and Sylow’s Theorem . . . . . . . . . . 1.7 Symmetric Group Sn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 8 12 14 19 20 24 Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 1 2 Preface This is a draft of abstract algebra notes, which is prepared during the lectures of our NET coaching classes and the aim is to supplement the students with their preparation for CSIR-NET exam in Mathematics. Most of the material presented here is not the original work. We are not responsible for any typos in these notes. You are free to modify these notes and correct the mistakes, if any, and distribute among your friends. If you have any suggestions or constructive critisism, which will improve the quality of these notes, please write to : [email protected]. Your suggestions are welcome. 4 Chapter 1 Groups 1.1 Introduction Let G be a non-empty set. A binary operation on G is a map from G × G to Binary operation : G. A non-empty set G with a binary operation ∗ is called a group if it satisfies the following Group : properties. (i) ∗ : G × G → G (closure) (ii) (a ∗ b) ∗ c = a ∗ (b ∗ c), ∀a, b, c ∈ G (associative law) (iii) ∃! e ∈ G such that x ∗ e = e ∗ x = x, ∀x ∈ G (identity law) (iv) for all x ∈ G, ∃! y ∈ G such that x ∗ y = y ∗ x = e Abelian Group : Order: (inverse law) A group G is called abelian group if a ∗ b = b ∗ a, ∀ a, b ∈ G Let G be a group. Then the number of elements in G, as a set, is called the order of G. Order of a group G is usually denoted by o(G) or |G|. NOTATION : A group G with a binary operation ∗ is usually denoted by (G, ∗) Examples 1. (i) (Z, +), o(Z) = ∞, ( = ℵ0 ) (ii) (Q, +), o(Q) = ∞, ( = ℵ0 ) (iii) (R, +), o(R) = ∞, ( = ℵ1 ) 5 6 (iv) (C, +), o(C) = ∞, ( = ℵ1 ) 2. (Z, ×), (Q, ×), (R, ×), (C, ×), does not form group. Because the element 0 does not have inverse in any of the above sets. 3. For every n ∈ N let A be a set with n elements and Sn = {f : A −→ A | f is a bijection }. Then (Sn , ◦) forms a group, where ◦ denotes the composition of mappings. This group is called Symmetric group of n symbols. o(Sn ) = n! (why ?) 4. Let Mn (R) = {n × n real matrices}. Then (Mn , +) is a group. 5. (Mn , ×) is not a group. Since not all matrices in Mn (R) have inverses. 6. Let GLn (R) = {An×n | det(A) 6= 0}, i.e. the set of all n × n invertible real matrices. Then (GLn (R), ×) forms a group. This group is called general linear group of degree n. In general (GLn (F), ×) forms a group, where F can be any field. 7. Let n ∈ Z. Define Zn = {ā | a is the remainder when divided by n }. Then (Zn , ⊕) forms a group, where ⊕ denotes addition modulo n. Here o(Zn ) = n. Let ⊗ denotes multiplication modulo n. Then (Zn \{0}, ⊗) form a group when n = p, a prime number. (What happens when n is not a prime ?) Therefore when n = p, a prime number, (Zn , ⊕, ⊗) becomes a field. (You will learn about fields later.) 8. Let n ∈ N , G = {ai | i = 0, 1, 2, 3, · · · (n − 1)} such that a0 = an = e, ai ∗ aj = ai+j ai ∗ aj = ai+j−n if i + j < n, if i + j > n. Then G is a group and o(G) = n 9. Let G = {e, a, b, c}. Let the binary operation ∗ on G be associative and commutative and e be the identity of G. The elements a, b, c of G satisfies following relations: a ∗ a = b ∗ b = c ∗ c = e, a ∗ b = c, a ∗ c = b, b ∗ c = a. Then (G, ∗) forms an abelian group of order 4. This group is called Klein’s 4-group. Exercise :        a b   Consider GL2 (Z2 ) =   | ad − bc 6= 0, a, b, c, d ∈ Z2 .     c d Prove that |GL2 (Z2 )| = 6. Can you guess a general formula for |GLn (Zp )|, where n ∈ N and p is a 7 prime number. Remark: From now onwards we do not use the sign ∗ to denote the binary operation. Instead we simply write ab to denote a ∗ b. We use the binary operation explicitly, if it is available. Lemma 1.1.1 : If G is a group, then (i) The identity element of the group is unique (ii) For every a ∈ G has a unique inverse in G (iii) For every a ∈ G, (a−1 )−1 = a (iv) For all a, b ∈ G, (ab)−1 = b−1 a−1 , Lemma 1.1.2 If G is a group and a, b ∈ G. Then the equations ax = b and ya = b have unique solutions in G. In particular, the two cancellation laws, au = aw ⇒ u = w and ua = wa ⇒ u = w hold in G. Worked Examples: 1. If G is a group such that (ab)2 = a2 b2 , ∀ a, b ∈ G. Show that G is abelian. Solution: Given (ab)2 = a2 b2 , ∀ a, b ∈ G ⇒ (ab)(ab) = aabb ⇒ aba = aab ⇒ ba = ab (using associative law and cancellation property) (using associative law and cancellation property) ⇒ G is abelian 2. Find the order of GLn (Zp ), where Zp is a finite field of prime order p. Solution: We know that, by definition, GLn (Zp ) contains non-singular matrices, whose entries are chosen from a finite field of order p. ⇒ All the rows of the matrices in GLn (Zp ) are linearly independent. So we can fill the first row in (pn − 1), ways. ( -ve 1 is because we are omitting a row where all the entries are 0) Second row in (pn − p) ways, since we need second row to be linearly independent from first row. Third row in (pn − p2 ) ways, since we need third row to be linearly independent from both first 8 and second row. .. . .. . nth row in (pn − pn−1 ) ways, since we need nth row to be linearly independent from the first (n − 1) rows. Therefore |GLn (Zp )| = (pn − 1)(pn − p)(pn − p2 ) · · · (pn − pn−1 ). e.g. |GL2 (Z2 )| = 6 |GL2 (Z3 )| = 48 |GL3 (Z2 )| = 168 |GL3 (Z3 )| = 11232(check !) Exercise : 1. Consider the group (S3 , ◦). Show that there are 4 elements satisfying x2 = e and 3 elements satisfying x3 = e, Even though the order of this group is 6. 2. If G is a group such that ∀ a ∈ G, a−1 = a . Prove that G is abelian. 3. If G is a group of even order, prove that G has at least one element a 6= e satisfying a2 = e. 1.2 Subgroups and Cyclic groups Definition: A non-empty subset H of a group G is said to be subgroup of G if (i) a, b ∈ H ⇒ ab ∈ H and, (ii) a ∈ H ⇒ a−1 ∈ H The conditions (i) and (ii) can be replaced by a single condition a, b ∈ H ⇒ ab−1 ∈ H. NOTATION : H is a subgroup of G is denoted by H ≤ G. Remark: If H is a subgroup of a group G, then the binary of H is always the binary operation of G. Example: 1. If G = (R, +) and H = Z or Q, then H ≤ G. (check using the definition) 9 2. If G = (Z, +) and H = mZ , where m ∈ Z, then H ≤ G. (check using the definition) 3. G = (R\{0}, ×) and H = Q+ , the set of all positive rationals, then H ≤ G 4. G = (C\{0}, ×) and H = {a + ib | a2 + b2 = 1}, then H ≤ G             a b   a b     5. G =   | ad − bc 6= 0, a, b, c, d ∈ R and H =   | ad 6= 0, a, b, d ∈ R       c d  0 d  Then H ≤ G. (check ) 6. Let G be any group of finite order and a ∈ G. Define H =< a >= {ai |i ∈ Z}, then < a >≤ G. This subgroup is called cyclic subgroup. The element a ∈ G is called the generator of the group H. 7. Let G = (GLn (R), ×) and H = {An×n real matrix | det(A) = 1}. where det(A) denotes the determinant of the matrix A. Then H ≤ G. (check using the definition when n = 2) The subgroup H is called special linear group and is denoted by SLn (R)       a b    Let G = (GL2 (R), ×) and H =   | ad 6= 0, a, b, c, d ∈ R .    c d  Then H 6≤ G. (why ?) Lemma 1.2.1 If H is a non empty finite subset of a group G and H is closed under multiplication(binary operation of G), then H is a subgroup of G. Exercise : Let G be a group. 1. If H ≤ G and K ≤ G, then prove that H ∩ K is a subgroup. What about H ∪ K ? 2. Let a ∈ G. Prove that C(a) = {x ∈ G | xa = ax } is a subgroup of G. The set C(a) is called centralizer of a. 3. Let H ≤ G. Prove that L(H) = { x ∈ G | xh = hx, ∀ h ∈ H } is a subgroup. 4. Prove Z = { a ∈ G | xa = ax, ∀ x ∈ G } is a subgroup. ( This subgroup is called centre of the group. ) 5. Let H ≤ G. Define NG (H) = { a ∈ G | aHa−1 = H}. Prove that NG (H) is a subgroup and H ⊂ NG (H), where NG (H) is called normalizer of H in G. When the group G is clear, we drop the subscript G and simly use N (H) to denote normalizer of H in G. 10 order of an element : Let G be a group and a ∈ G. Then order of a is denoted by o(a) and defined as o(a) = | < a > |, where < a > is the cyclic subgroup of G, generated by a. Also o(a) = least positive integer n such that an = e. Check above two definitions are equivalent. Cyclic Group : A group G is cyclic if ∃ a ∈ G and a 6= e such that G = {e, a, a2 , . . . . . . , an , . . . . . .}, where a is called the generator of group G. Example: 1. G = (Z, +) =< 1 > 2. G = ({i, −1, −i, 1}, ×) =< i >, where i2 = −1. 3. Let n be a positive integer and let S = {cos( 2kπ ) + i sin( 2kπ | k = 0, 1, 2, . . . , (n − 1)} be the 2 2 set of nth roots of unity. Then (S, ×) is a cyclic group and |S| = n. Lemma 1.2.2 If b is a generator of a cyclic group G, then b−1 is also a generator of G. Lemma 1.2.3 Every cyclic group is abelian. Remark: The converse is not true. i.e. Every abelian group is not cyclic. e.g. Klein’s 4-group. Example: The symmetric group Sn is not cyclic, since it is not abelian. Lemma 1.2.4 If G =< a > and o(a) = n, then G = {a, a2 , a3 , · · · · · · an (= e)} Lemma 1.2.5 Let (G, ∗) be a cyclic group generated by a. Then G is infinite if and only if o(a) is infinite. Lemma 1.2.6 A finite group (G, ∗) of order n is cyclic if and only if there exists an element b in G such that o(b) = n. Example: 1. The group (Zn , +) is a finite group of order n. Then 1̄ ∈ Zn and the order of 1̄ in the group is n. Therefore (Zn , +) is a cyclic group with 1̄ as a generator. 11 2. The symmetric group S3 is not cyclic, since the order of S3 is 6 and there exists no element of order 6 in S3 . Lemma 1.2.7 Let G =< a > and o(G) = n > 1. Then for r ∈ N, ar is also a generator of the group iff r < n and g.c.d.(r, n) = 1. Corollary 1.2.8 If G is a finite cyclic group of order n, then the number of generators of G is φ(n), where φ(n) is Euler’s φ function. NOTE : φ(n) = number of positive integers less than n and relatively prime to n. Example: 1. The number of generators of the cyclic group S = ({i, −1, −i, 1}, ×) is 2, since φ(4) = 2. 2. The number of generators of a cyclic group (Z8 , +) is 4, since φ(8) = 4. Lemma 1.2.9 Let G be a group, a ∈ G and let m ∈ Z\{0}. 1. If o(a) = ∞ then o(am ) = ∞ 2. If o(a) = n < ∞ then o(am ) = n g.c.d(n, m) 3. In particular if o(a) = n < ∞ and m is a positive integer dividing n, then o(am ) = n m Lemma 1.2.10 Every subgroup of a cyclic group is cyclic. Lemma 1.2.11 Every non-trivial subgroup of an infinite cyclic group is infinite. Example: In the group G = (Z, +), all subgroups are of the form mZ f or m ∈ N. Lemma 1.2.12 A cyclic group of finite order n has one and only one subgroup of order d, for every positive divisor d of n. Remark: If G is an infinite cyclic group generated by a, then a and a−1 are the only generators of G. e.g. (Z, +) is an infinite cyclic group with 1 and −1 as generators. For a given m ∈ N, (mZ, +) is an infinite cyclic group with m and −m as generators. 12 1.3 Cosets and Lagrange’s theorem Definition: Let G be a group, H ≤ G. For a, b ∈ G we say that “a is congruent to b mod H” , written as “a ≡ b mod H” iff ab−1 ∈ H. Lemma 1.3.1 The relation “a ≡ b mod H” is an equivalence relation. Definition: Let G be a group, H ≤ G and a ∈ G, then Ha = {ha | h ∈ H} is called a right coset of H in G. Similarly we can define left coset. Lemma 1.3.2 For all a ∈ G, Ha = {x ∈ G | a ≡ x mod H} = [a] Remark: Since equivalence relation on a set partitions the set into different equivalence classes, any two right cosets of H in G is either identical or have no elements in common (disjoint). Lemma 1.3.3 There is one-to-one correspondence between any two right cosets of H in G i.e. there exists a bijection between any two right cosets of H in G. Hint: The map f : aH −→ bH given by f (ah) = bh for every h ∈ H Theorem 1.3.4 (Lagrange) If G is a finite group and H is a subgroup of G, then o(H) is a divisor of o(G) If H ≤ G then the index of H in G is the number of distinct right cosets of H in G. We o(G) will denote it by i(H) or IG (H) or |G : H| and |G : H| = o(H) Index : Corollary 1.3.5 If G is a finite group and a ∈ G, then o(a)|o(G) Corollary 1.3.6 Let G be a finite group and a ∈ G. Then ao(G) = e. Example: Let n ∈ N and U (n) = {a < n | g.c.d(a, n) = 1}(a is relatively prime to n). Then U (n) is a group under multiplication mod n. (check) Corollary 1.3.7 (Euler) Let n ∈ N. Let a be a positive integer such that g.c.d(a, n) = 1, then aϕ(n) ≡ 1 mod n. Corollary 1.3.8 (Fermat) If p is a prime number then ap ≡ a mod p . Corollary 1.3.9 If G is a finite group whose order is a prime number p, then G is a cyclic group. 13 Example: 1. Let G = S4 and H = A4 . Then H ≤ G and o(S4 ) = 24 and o(A4 ) = 12. ⇒ order H divides order of G. We will give more details of Sn later. 2. Let G = Klein′ s 4 − group. and H =< a >. Then o(G) = 4 , o(H) = 2 and o(H)|o(G). Remark: 1. Every group of order less than 6 is commutative 2. If G is a non-commutative group of order 2n where n is an odd prime(prime number greater than 2), then G must have a subgroup of order n. 3. The results that are true for “right coset” is also true for “left coset” 4. Converse of the Lagrange’s theorem is not true. i.e. if m is a divisor of o(G), then G need not have a subgroup of order m. e.g. A4 the alternating group of 4-symbols, is a group of order 12 and it has no subgroup of order 6, even though 6|12. But the partial converse of Lagrange’s theorem will be given by Sylow’s theorem, which we will see later. Definition: Let G be a group, H ≤ G and K ≤ G. Define HK = {hk | h ∈ H, k ∈ K}. Similarly we can define KH also. Question : Is HK = KH ?(Think over it) (What if G is abelian ?) Lemma 1.3.10 Let H ≤ G and K ≤ G. Then HK ≤ G iff HK = KH Corollary 1.3.11 Let G be an abelian group and H ≤ G, K ≤ G. Then HK ≤ G. Theorem 1.3.12 If H and K are finite subgroups of G of orders o(H) and o(K) respectively. Then o(H) o(K) o(HK) = o(H ∩ K) Corollary 1.3.13 If G be a finite group, H ≤ G, K ≤ G and o(H) > H ∩ K 6= {e} ( or in other words, o(H ∩ K) > 1). p p o(G), o(K) > o(G) , then Remark: Use the above corollary to prove that, if G is a finite group of order pq, where p and q are primes with p > q, then G can have at most one subgroup of order p. 14 1.4 Normal Subgroups, Quotient group, Homomorphism Normal Subgroup : A subgroup N of a group G is called normal subgroup if gN g −1 = N, ∀g ∈ G. Where gN g −1 = { gng −1 | n ∈ N } NOTATION : If N is a normal subgroup of G, then we denote it by N E G. Example: 1. Let G be a group. Then G and {e} are normal subgroups of G. (These are trivial normal subroups off G). 2. Z(G) = {x ∈ G | xa = ax, ∀ a ∈ G}. Then Z(G) E G 3. If G = (Z, +) and N = (mZ, +), m ∈ N ∪ {0}. Then N E G. 4. For n ∈ N, An is normal in Sn (Verify for n = 3) Lemma 1.4.1 If N E G, then every left coset of N in G, is also a right coset. Remark: Let G be a group and N E G. Let G/N = { N g | g ∈ G }, i.e. the set of all right cosets of N in G. Then G/N forms a group under the operation ‘·’ given by, N x · N y = N xy, where x and y are composed using the group composition.(Verify). This group (G/N, ·) is called quotient group. Lemma 1.4.2 If G is a group of finite order and N E G, then o(G) = o(G/N ) o(N ) Remark: 1. If H is a subgroup of a commutative group G, then the quotient group G/H is commutative. 2. If H is a subgroup of a cyclic group G, then the quotient group G/H is cyclic. Note that the converse of above statements are not true. e.g. G = S3 and H = A3 Theorem 1.4.3 Let G be group and Z(G) be its centre. If G/Z(G) is cyclic, then G is abelian. Theorem 1.4.4 Let N be a subgroup of a group G. Then the following are equivalent. (i) N E G (ii) NG (N ) = N . (Recall NG (N ) is normalizer of N in G). (iii) gN = N g, ∀ g ∈ G 15 (iv) gN g −1 = N ∀ g ∈ G (v) The composition defined by N x · N y = N xy, on the set of all right cosets, makes the set of all right cosets into a group. Theorem 1.4.5 If H is a subgroup of G and K is a normal subgroup of G, then HK is a subgroup of G for any subgroup H of G. Exercise : 1. Prove that every subgroup H of a commutative group G is normal in G. What about the converse ? i.e. If every subgroup H of a group G is normal in G, then is the group G commutative ? 2. If H is a subgroup of index 2 in G, then prove that H E G. 3. Let G be a group and N E G and H ≤ G. Check N H E G or not. 4. If H ≤ G and N E G, then prove that H ∩ N E H 5. Let G = S3 . Prove that Z(G) = {e} Homomorphism : Let (G1 , ∗) and (G2 , ◦) be two groups. A map φ : G1 → G2 is said to be homomorphism, if φ(a ∗ b) = φ(a) ◦ φ(b) for all a, b ∈ G If φ is 1 − 1 then φ is called injective homomorphism. If φ is onto then φ is called surjective homomorphism. If φ is both 1 − 1 and onto then φ is called an isomorphism. If there exists an isomorphism between two groups G1 and G2 , then we say that the two groups G1 and G2 are isomorphic and we write this as G1 ∼ = G2 . Example: 1. Let G1 and G2 be two groups and φ : G1 → G2 defined by φ(x) = e2 (identity of G2 ), for all x ∈ G. Then φ is a homomorphism (Verify). This homomorphism is called trivial homomorphism and this example shows that always, there exists a homomorphism between any two groups. 2. Let G1 = (R, +) and G2 = (R − {0}, ·). Then φ : G1 → G2 defined by φ(x) = 2x is a homomorphism (Verify). 16 3. Let G1 = (Z, +) and G2 = (3Z, +). Then φ : G1 → G2 defined by φ(x) = 3x is a homomorphism (Verify). 4. Let G be a group and N E G. Define a map φ : G → G/N by φ(x) = N x. Then φ is a homomorphism of groups (Verify). This homomorphism is called natural homomorphism of G onto G/N 5. Let G1 = S3 and G2 = ({−1, 1}, ·). Then φ : G1 → G2 defined by φ(σ) = 1 if σ is an even permutation in S3 φ(σ) = −1 if σ is an odd permutation in S3 Then φ is a homomorphism(Verify). 6. Let G1 = R2 (operation vector addition) and G2 = R (operation addition). Let φ : G1 → G2 defined by φ((x, y)) = x. Then φ is a homomorphism (Verify). Remark: Let φ : G1 → G2 be a homomorphism and e1 and e2 be the identity elements of G1 and G2 respectively. Then φ(e1 ) = e2 . Kernel and Image of a homomorphism : 1. Let φ : G1 → G2 be a homomorphism. Then kernel of φ is denoted by ker φ and is defined as ker φ = { x ∈ G1 | φ(x) = e2 } ⊂ G1 . 2. Let φ : G1 → G2 be a homomorphism. Then image of φ is denoted by Im φ or φ(G1 ) and is defined as Im φ = { φ(x) | x ∈ G1 } ⊂ G2 . Note that ker φ and Im φ are both non-empty as e1 ∈ ker φ and e2 ∈ Im φ . Theorem 1.4.6 Let (G1 , ∗) and (G2 , ◦) be two groups and φ : G1 → G2 be a homomorphism. Then (i) φ(e1 ) = e2 (ii) φ(an ) = {φ(a)}n , for all n ∈ Z (iii) If a ∈ G and o(a) is finite, then o(φ(a)) is a divisor of o(a). Theorem 1.4.7 Let (G1 , ∗) and (G2 , ◦) be two groups and φ : G1 → G2 be a homomorphism. Then (i) ker φ is a normal subgroup of G1 17 (ii) Im φ is a subgroup of G2 (iii) φ is one-one if and only if ker φ = {e2 } Lemma 1.4.8 Let (G1 , ∗) and (G2 , ◦) be two groups and φ : G1 → G2 be an homomorphism. Then (i) If G1 is cyclic, then φ(G1 ) is cyclic, and the converse is not true. (ii) If G1 is abelian, then φ(G1 ) is abelian, and the converse is not true. Facts: (Verify) 1. Note that if a is the generator of G1 , then φ(a) is the generator of φ(G1 ) 2. There does not exists and onto homomorphism from (Z4 , +) to Klein’s 4-group. 3. Let G1 and G2 be two finite groups such that gcd(o(G1 ), o(G2 )) = 1. Then trivial homomorphism is the only homomorphism between G1 and G2 Theorem 1.4.9 Let (G1 , ∗) and (G2 , ◦) be two groups and φ : G1 → G2 be an homomorphism. Then (i) If H ≤ G1 then φ(H) ≤ G2 (ii) If H E G1 then φ(H) E G2 (iii) If K ≤ G2 then φ−1 (K) ≤ G1 (iv) If K E G2 then φ−1 (K) E G1 Lemma 1.4.10 A subgroup N of a group G is normal if and only if it is the kernel of some homomorphism. Lemma 1.4.11 If G is a finite group of order n and p is the smallest prime dividing o(G), then any subgroup of index p is normal in G. Example: 1. For n ∈ N, An is normal subgroup of Sn . (Show that An is kernel of some homomorphism). 2. Let G1 = (R, +) and G2 = ({z ∈ C : |z| = 1}, ·) and φ : G1 → G2 be defined by φ(x) = e2πix . Then φ is a homomorphism and ker φ = Z (Verify) 18 Theorem 1.4.12 (First Theorem on homomorphism) Let φ : G1 → G2 be a homomorphism. G ∼ Then = φ(G1 ) kerφ (OR) If φ : G1 → G2 be an onto homomorphism and H = kerφ, then G ∼ = G2 H Exercise : 1. Let G be a group and fix g ∈ G. Let φg : G → G be defined by φg (x) = gxg −1 . Prove that, for every g ∈ G, φg is an isomorphism 2. Let G be a group of finite order n and let k ∈ N be such that gcd(k, n) = 1. Prove that any g ∈ G can be written as xk for some x ∈ G. (Hint: Prove x 7→ xk is an onto homomorphism.) 3. Let G = GLn (Fp ) = {An×n | elements of A are in Fp and det(A) 6= 0} and S = SLn (Fp ) = {An×n | elements of A are in Fp and det(A) = 1} (i) Show that S E G (ii) Compute o(S) Automorphism : Let G be a group. An isomorphism from G onto itself is called an automorphism of G. Remark: The set of all automorphisms of G is denoted by Aut(G) and Aut(G) is a group under composition of mappings (Verify). Example: 1. Let G be a group and {e} be its identity element. Define φ : G → G by φ(x) = e for all x ∈ G. Then φ is an automorphism of G onto G. (Verify) 2. Let G be an abelian group and define φ : G → G by φ(x) = x−1 for all x ∈ G. Then φ is an automorphism of G onto G. (Verify) 3. Let G = (C, +) and φ : G → G is defined by φ(z) = z̄. Then φ is an automorphism of G (Verify) Lemma 1.4.13 Let G be a group and g ∈ G. Then the mapping Tg : G → G defined by Tg (x) = gxg −1 , for all x ∈ G is an automorphism of G. 19 Definition: Let G be a group and g ∈ G. The automorphism Tg of G given by Tg (x) = gxg −1 is called an inner automorphism of G and the subgroup of Aut(G) consisting of all inner automorphisms is denoted by Inn(G). Lemma 1.4.14 Let G be a group. Then Inn(G) E Aut(G). Lemma 1.4.15 Let G be a group and Z(G) be its centre. Then G ∼ = Inn(G) Z(G) (Hint: Define φ : G →Aut(G) by φ(g) = Tg and show kerφ = Z(G), and use 1st isomorphism theorem) Theorem 1.4.16 If G is a finite cyclic group of order n, then Aut(G) ∼ = Un . ( Un = {Set of all positive integers less than n and relatively prime to n}) Lemma 1.4.17 For all n 6= 6, we have Aut(Sn ) = Inn(Sn ) ∼ = Sn . For n = 6, we have |Aut(S6 ) : Inn(S6 )| = 2 Lemma 1.4.18 Let p be a prime and (V, +) be an abelian group with the property that pv = 0 for all v ∈ V . If |V | = pn , then V is an n-dimensional vector space over the field Zp . The automorphisms of V are precisely the non-singular linear transformations from V to itself, i.e. Aut(V ) ∼ = GLn (Zp ). Therefore |Aut(V )| = (pn − 1)(pn − p)(pn − p2 ) · · · (pn − pn−1 ) Remark: 1. If G is a finite cyclic group of order n. Then the number of automorphisms of G is ϕ(n), where ϕ(n) = number of positive integers less than n and relatively prime to n. 2. If G is an infinite cyclic group, then the number of automorphisms of G is 2. 3. The number of automorphisms of S3 is 6 1.5 Direct Products of Groups Definition: Let (G1 , ∗) and (G2 , ◦) be two groups. Then G1 × G2 forms a group under the binary composition ‘·’ defined by (a, b) · (c, d) = (a ∗ c, b ◦ d). This group (G1 × G2 , ·) is called the direct product of groups (G1 , ∗) and (G2 , ◦) Example: G1 = (R, +) and G2 = (Z, +). Then G1 × G2 , G1 × G1 , G2 × G2 are all direct product of groups. 20 Remark: 1. If e1 and e2 are identity elements of G1 and G2 respectively, then (e1 , e2 ) is the identity of G1 × G2 . 2. Let (a, b) ∈ G1 × G2 , then (a−1 , b−1 ) is the inverse of (a, b) in G1 × G2 Lemma 1.5.1 Let A1 = {(g, e2 ) | g ∈ G1 } ⊂ G1 × G2 . Then A1 ≤ G1 × G2 and G1 ∼ = A1 Similarly Let A2 = {(e1 , g) | g ∈ G2 } ⊂ G1 × G2 . Then A2 ≤ G1 × G2 and G2 ∼ = A2 . Remark: 1. Verify that A1 and A2 are normal subgroups of G1 × G2 . i.e. Show that A1 and A2 are kernel of some homomorphism of G1 and G2 respectively. 2. If G is abelian, then G × G is abelian. 3. If G is cyclic, then G × G need not be cyclic. e.g. (Zn , ⊕n ) is a cyclic group for all n ∈ N. But Zm × Zn is cyclic if and only if gcd(m, n) = 1. If gcd(m, n) = 1, then Zm × Zn ∼ = Zmn . Problem: Let G be a group and T = G × G. (i) Prove that D = {(g, g) : g ∈ G} is a subgroup of T isomorphic to G. (ii) Prove that D is normal if and only if G is abelian. Internal Direct Product: Let G be a group and N1 , N2 , N3 , . . . Nk are normal subgroups of G. G is called internal direct product of N1 , N2 , N3 , . . . , Nk if every g ∈ G has a unique representation in the form g = n1 n2 n3 . . . nk ni ∈ Ni for i = 1, 2, 3, . . . k. In this case we write G = N1 , N2 , N3 , . . . , Nk Theorem 1.5.2 Let G be a group and G = N1 , N2 , N3 , . . . , Nk . Let T = N1 × N2 × N3 × . . . × Nk . Then G ∼ =T 1.6 Class Equation and Sylow’s Theorem Conjugacy : Let a, b ∈ G. Then b is said to be conjugate of a, if there exists an x ∈ G such that b = x−1 ax. We write this as a ∼ b 21 Definition: Two subsets S and T in a group G are said to be conjugate in G, if there is some g ∈ G such that T = g −1 Sg. Lemma 1.6.1 Conjugacy is an equivalence relation on G Definition: Let G be a group and a ∈ G. Then the class of a is given by Ca = {x ∈ G | a ∼ x} = {x−1 ax | x ∈ G}. We denote the order of Ca by na . Recall that if a ∈ G then the centralizer of a is given by C(a) = {x ∈ G : ax = xa}. Theorem 1.6.2 Let G be a group and a ∈ G. Then na = |Ca | = |G : C(a)| = |G| . i.e. Number |C(a)| of conjugates of an element a of G is the index of the centralizer of a. Theorem 1.6.3 (The Class Equation) Let G be a finite group and let g1 , g2 , g3 , . . . , gr be representatives of distinct conjugacy classes of G not contained in the center Z(G) of G. Then |G| = X na = |Z(G)| + a∈G r X |G : C(gi )| i=1 Results : 1. Let G be a group of order pn , where p is a prime. Then the center Z(G) of G is non trivial. i.e. Z(G) 6= {e} 2. If o(G) = p2 , where p is a prime, then G is an abelian group. 3. If G is a non abelian group of order p3 , where p is a prime, then o(Z(G)) = p. 4. If o(G) = pn , where p is a prime, then G contains a normal subgroup of order pn−1 5. Let G be a group of order pq, where p and q are primes and p < q. If p 6 |q − 1 then G is cyclic. Theorem 1.6.4 (Cauchy’s Theorem) If p is a prime and p divides order of G, then G contains an element of order p. Definition: Let G be a group and let p be a prime. 1. A group of order pα for some α ≥ 1 is called a p-group.Subgroups of G which are p-groups are called p-subgroups. 2. If G is a group of order pα m, where p 6 | m, then a subgroup of order pα is called a Sylowp − subgroup of G. 22 3. The set of Sylow p-subgroups of G will be denoted by Sylp (G) and the number of Sylow psubgroups of G will be denoted by np (G)( or just np when G is clear from the context) Theorem 1.6.5 (Sylow’s Theorem) Let G be a group of order pα m, where p is a prime not dividing m. 1. Sylow p-subgroups of G exist, i.e. Sylp (G) 6= ∅ 2. If P is a Sylow p-subgroup of G and Q is any p-subgroup of G, then there exists g ∈ G such that Q ≤ g −1 P g, i.e. Q is contained in some conjugate of P . In particular, any two Sylow p-subgroups of G are conjugate in G. 3. The number of Sylow p-subgroups of G is of the form 1 + kp i.e. np ≡ 1(modp) . Further, np divides m, (Since, for any Sylow p-group np is the index of the normalizer NG (P ) of P in G) Corollary 1.6.6 Let P be a Sylow p-subgroup of G. Then the following are equivalent: 1. P is the unique Sylow p-subgroup of G. i.e. np = 1 2. P is normal in G Theorem 1.6.7 If |G| = n and p is the smallest prime that divides n. Then any subgroup of index p is normal Remark: If o(G) = n and p|n then G need not have a sub-group of index p. Theorem 1.6.8 (Cayley’s theorem) Every finite group is isomorphic to some subgroup of Sn Theorem 1.6.9 Let G be a group and H ≤ G, S = {Ha | a ∈ G}. Then there exists homomorphism θ : G → Sn and ker θ is the largest normal subgroup of G containing H. Theorem 1.6.10 Let G, H, S and θ be as in above theorem. If o(G) 6 | i(H)! then ker {θ} = 6 {e} Examples: 1. If |G| = 28, prove that G has a normal subgroup of order 7. Soln : Here 28 = 22 · 7. Since 7|28 and 7 6 | 4 there exists a Syl7 (G). Also n7 = 1 + k7 = 1, 8, 15, . . . and n7 |4 implies n7 = 1 ⇒ Syl7 (G) is normal in G. 2. Let P =< x > , Q =< y >, o(P ) = p and o(Q) = q, where p and q are primes. If xy = yx then |xy| = pq. 23 3. Let G be a group of order pq, where p and q are primes p < q. If p 6 | q − 1, then G is cyclic. Soln : Let P ∈ Sylp (G) and Q ∈ Sylq (G). We show that both P and Q are normal in G. Here nq = 1 + kq and nq |p and p < q ⇒ nq = 1 ⇒ Q is normal in G. Now np = 1 + kp and np |q ⇒ np = 1 or q. Since p 6 |(q − 1) implies np 6= q ⇒ np = 1. Therefore P is a normal subgroup of G. Now P and Q are are normal subgroups of order p and q respectively and hence P Q is a subgroup of G of order pq. Now, from (2), it follows that there exists an element of order pq in G and hence G is cyclic. 4. Let G be a group of order p2 q, where p and q are primes. Then G has a normal subgroup. Soln : Let P ∈ Sylp (G) and Q ∈ Sylq (G). (a) Let p > q. Since np ≡ 1( mod p) and np |q implies np = 1. This implies P is normal in G. (b) Let q > p. We have nq ≡ 1( mod q) or nq = 1 + kq. Since nq |p2 implies nq = 1, p orp2 . Now if nq = 1, then Q is normal in G. If nq 6= 1 then nq ≥ q and since q > p, we have nq 6 |p. Therefore we have nq |p2 . i.e. 1 + kq|p2 . ⇒ q|(p2 − 1) since p2 − 1 = (p + 1)(p − 1) and q > p > p − 1 ⇒ q 6 |(p − 1) we have q|(p + 1). It is possible only if p = 2 and q = 3. Then |G| = p2 q = 12. Remember, there are 5 isomorphic groups of order 12. Now we prove that there exists a normal subgroup in G. Here |G| = 12 = 22 3. Let H ∈ Syl2 (G) and |H| = 4. K ∈ Syl3 (G) and |K| = 3. Claim: Either H or K is normal in G. Suppose K is not normal. But n3 = 1 + 3k|4 implies n3 = 1 or 4. Since K is not normal, n3 6= 1, we have n3 = 4. That means there are 4 distinct normal Sylow 3-subgroups K1 , K2 , K3 , K4 such that |Ki | = 3, i = 1, 2, 3, 4 and Ki ∩ Kj = {e} when i 6= j (why ?). ⇒ K1 ∪ K2 ∪ K3 ∪ K4 contains 8 + 1 = 9 distinct elements of G. Therefore remaining 3 elements has to be in H. Therefore H is a unique Sylow 2 subgroup of G and hence H is normal in G. 5. Groups of Order 30 Let G be a group of order 30. Then 30 = 5 · 3 · 2 = 5 · 6 = 3 · 10 = 2 · 15. Let P ∈ Syl5 (G), Q ∈ Syl3 (G), T ∈ Syl2 (G). If either P or Q are normal, then P Q ≤ G and o(P Q) = 15 and hence P Q is a normal subgroup, since P Q is of index 2 in G. Now we will show that either P or Q is normal in G. Lets assume that, none of them are normal. Then n5 6= 1 and n3 6= 1. But n5 = 1 + k5, n3 = 1 + k3. Therefore, n5 = 1, 6, 11, . . . and n3 = 4, 7, 10, 13, . . .. Since n5 |6 and n3 |10, we have n5 = 6 and n3 = 10. Let P1 , P2 , P3 , P4 , P5 , P6 be 6 distinct Sylow 5-subgroups of G. Then |Pi | = 5, i = 1, 2, 3, 4, 5, 6 and Pi ∩Pj = {e} when i 6= j. Therefore G contains 6 × 4 + 1 = 25 distinct elements of order 5. Similarly Qi , i = 1, 2, . . . 10 are 24 distinct Sylow 3-subgroups of G such that |Qi | = 3 and Qi ∩ Qj = {e} when i 6= j. Therefore there exists 10 × 2 = 20 distinct elements of order 3 in G. Therefore, there are total of 45 distinct elements in G, but this is absurd as o(G) = 30. Therefore, at our assumption is wrong and hence at least P or Q is normal and hence P Q is a subgroup of G of order 15. 6. If |G| = 60 and G has more than one Sylow 5-subgroup, then G is simple. 7. An is simple for n ≥ 5 8. If G is a group of order 60, then G ∼ = A5 9. Let o(G) = 36 H is a subgroup and o(H) = 9. Prove that G is not simple. Soln : Here i(H) = 36 9 = 4, 4! = 24 and 24 6 |36. This implies kerθ 6= {e} ⊃ H. From the above theorems, kerθ is the largest normal subgroup of G containing H. There exists a normal subgroup of G. 1.7 Symmetric Group Sn Definition: Sn is the set of all bijective maps from a set : Ω = {1, 2, 3, . . . n} (containing n elements onto itself). Along with the operation “composition of mappings” (◦), the set Sn becomes a group. Facts : 1. Identity of Sn is the identity map on {1, 2, 3, . . . n}, i.e. Id(i) = i, ∀ i ∈ {1, 2, 3, . . . n} 2. Order of Sn is n! (why ?) 3. Sm is a subgroup of Sn , where m < n 4. S3 =< a, b >, where a = (1 2) and b = (1 2 3), i.e. S3 is generated by a and b Theorem 1.7.1 If σ, τ ∈ Sn are disjoint cycles, then στ = τ σ, (i.e. disjoint cycles commutes) Theorem 1.7.2 Every permutation in Sn is the product of disjoint cycles. Theorem 1.7.3 The order of the k-cycle, as an element of Sn , is k; i.e. τ k = e and τ j 6= e for 0<j<k Theorem 1.7.4 Let σ ∈ Sn have its cycle decomposition into disjoint cycles of lenght m1 , m2 , . . . , mk . Then the order of σ is the least common multiple of m1 , m2 , . . . , mk . 25 Remark: Note that the disjointness of the cycles in the theorem is necessary. For example, (1 2) and (1 3) which are not disjoint, are each of order 2, but their product (1 2)(1 3) = (1 3 2) is of order 3. Examples: 1. Let σ = (1 2)(3 4 5 6)(7 8 9). Then σ contains 3 disjoint cycles of order 2, 4 and 3 respectively. Therefore, by the theorem, o(σ) = lcm(2, 4, 3) = 12 Definition: A 2-cycle is called transposition Example: The element (2 3) is a transposition Theorem 1.7.5 Every permutation is the product of transpositions. Definition: A permutation σ ∈ Sn is an odd permutatoin if σ is the product of an odd number of transpositions, and even permutation if σ is the product of an even number of transpositions. Theorem 1.7.6 A permutation in Sn is either an odd permutation or an even permutation, but can not be both Theorem 1.7.7 The permutation σ is odd if and only if the number of cycles of even length in its cycle decomposition is even. Definition: The set of all even permutation is denoted by An . The product of two even permutations is again even(why?). Therefore An is closed in Sn . Since Sn is a finite group, An is a subgroup of Sn . An is called the alternating group of degree n. Remark: 1. Transpositions are odd permutations
2. Let G = {−1, 1}, · . Then G is a group w.r.t multiplication and o(G) = 2. Consider the map φ : Sn −→ G defined by ( −1 if σ is an odd permutation φ(σ) = 1 if σ is an even permutation Then φ is an onto homomorphism of groups and ker φ is An (Verify). Therefore An is normal subgroup of Sn . |Sn | n! By First Homomorphism theorem, Sn /ker φ ∼ = |G| ⇒ = 2 ⇒ = G. Therefore |An | |An | n! |An | = 2 26 3. For n > 1, Sn contains n! n! even permutations and odd permutations 2 2 4. An is a non-abelian simple group for all n ≥ 5 5. (a) Product of two even permutation is even (b) Product of two odd permutation is even (c) Product of an even permutation and odd permutation is odd 6. If σ is a k-cycle, show that σ is an odd permutation if k is even and an even permutation if k is odd. 7. Any non-abelian group of order 6 is isomorphic to S3 8. Z(Sn ) = {e} for all n ≥ 3 Theorem 1.7.8 Let σ, τ be elements of the symmetric group Sn and suppose σ has a cycle decomposition (a1 a2 a3 . . . ak1 ) (b1 b2 b3 . . . bk2 ) . . .. Then τ στ −1 has cycle decomposition    τ (a1 ) τ (a2 ) τ (a3 ) . . .) τ (ak1 ) τ (b1 ) τ (b2 ) τ (b3 ) . . . τ (bk2 ) That is τ στ −1 is obtained from σ by replacing each entry i in the cycle decomposition for σ by the entry τ (i) Definition: 1. If σ ∈ Sn is the product of disjoint cycles of lengths n1 , n2 , n3 , . . . , nr with n1 ≤ n2 ≤ n3 ≤ · · · ≤ nr (including its 1-cycles) then the integers n1 , n2 , n3 , . . . , nr are called the cycle type of σ 2. If n ∈ N, a partition of n is any non-decreasing sequence of positive integers whose sum is n. Theorem 1.7.9 Two elements of Sn are conjugate in Sn if and only if they have the same cycle type. The number of conjugacy classes of Sn equal to the number partitions of n Examples: 1. Let σ = (1 2)(3 4 5)(6 7 8 9) and let τ = (1 3 5 7)(2 4 6 8). Then τ στ −1 = (3 4)(5 6 7)(8 1 2 9) 2. Let σ1 = (1)(3 5)(8 9)(2 4 7 6) and let σ2 = (3)(4 7)(8 1)(5 2 6 9). Then define τ by τ (1) = 3, τ (3) = 4, τ (5) = 7, τ (8) = 8, τ (9) = 1, τ (2) = 5, τ (4) = 2, τ (7) = 6, τ (6) = 9). Then τ = (1 3 4 2 5 7 6 9)(8) and τ σ1 τ −1 = σ2 . 27 3. The cycle type of a m-cycle in Sn is 1, 1, 1, . . . , 1, m, where m is preceeded by n − m ones. 4. If σ is a m-cycle in Sn , then the number of conjugates of σ is n · (n − 1) · (n − 2) · · · (n − m + 1) = m (n!) m · (n − m)! But number of conjugates of an element σ is equal to the index of centralizer of σ in Sn which |Sn | n! (n!) is given by : ⇒ |CSn (σ)| = m · (n − m)! = = |CSn (σ)| |CSn (σ)| m · (n − m)! i.e. If σ is an m-cycle in Sn , then the number of elements which commutes with sigma is m · (n − m)!. The elements with which σ commutes are 1, σ, σ 2 , σ 3 , σ 4 , . . . , σ m−1 and any per- mutation in Sn which is disjoint from σ(viz. all those elements of Sn which fixes all the integers appearing in m-cycle σ). · · · · · · · · · · · · · · · TO BE CONTINUED :) ··············· 28 Bibliography [1] David Dummit and Richard Foote, Abstract Algebra, Second edition, Wiley Publications, New York, 2010 [2] I. N. Herstein, Topics in Algebra, 2nd edition, John Wiley Publications, Singapore. [3] S. K. Mapa, Higher Algebra : Abstract and Linear, Sarat Book Distributors, Kolkata 2005. 29