Unit 2 Group Theory PDF

1 GG GROUP THEORY Notation: ∀: 𝑓𝑜𝑟 𝑎𝑙𝑙 ∃: 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡 ∈: 𝑏𝑒𝑙𝑜𝑛𝑔 𝑡𝑜 ∋: 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 ∅: 𝑛𝑜𝑛 − 𝑒𝑚𝑝𝑡𝑦 ∄: 𝑡ℎ𝑒𝑟𝑒 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 Binary Operation Let 𝐺 be...

1 GG GROUP THEORY Notation: ∀: 𝑓𝑜𝑟 𝑎𝑙𝑙 ∃: 𝑡ℎ𝑒𝑟𝑒 𝑒𝑥𝑖𝑠𝑡 ∈: 𝑏𝑒𝑙𝑜𝑛𝑔 𝑡𝑜 ∋: 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 ∅: 𝑛𝑜𝑛 − 𝑒𝑚𝑝𝑡𝑦 ∄: 𝑡ℎ𝑒𝑟𝑒 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 Binary Operation Let 𝐺 be a non-empty set. Any mapping or function from 𝐺 × 𝐺 to itself is called binary operation on a set 𝐺. i.e., ∗: 𝑮 × 𝑮 → 𝑮. Algebraic structure A non-empty-set equipped with one or more binary operation is called algebraic structure. The algebraic structure consisting of a set 𝐺 and ∗, ∙ denoted by (𝐺,∗,∙). Examples: 1. 𝐺 = 𝑁 (set of all natural numbers). If ∀𝑎, 𝑏 ∈ 𝑁, we define ∗ on 𝑁 as 𝑎 ∗ 𝑏 = 𝑎 + 𝑏, then (𝑁, +) is algebraic structure since sum of two natural numbers is again natural number. 2. Let 𝐺 = 𝑁. If ∀𝑎, 𝑏 ∈ 𝑁, we define ∗ on 𝑁 as 𝑎 ∗ 𝑏 = 𝑎 − 𝑏. Then ∗ is not binary operation on 𝑁 as for 1, 2 ∈ 𝑁 but 1 − 2 = −1 ∉ 𝑁. 3. For 𝐺 = 𝑅 (set of all real numbers). If ∀𝑎, 𝑏 ∈ 𝑅, we define ∗ on 𝑅 as 𝑎 ∗ 𝑏 = min {𝑎, 𝑏} then (𝑅,∗) is a algebraic structure. 4. Let 𝐺 = 𝑅. If ∀𝑎, 𝑏 ∈ 𝑅, we define ∗ on 𝑁 as 𝑎 ∗ 𝑏 = 𝑎𝑏 is not binary operation on 𝑅 as for 1 1 𝑎 = −1, 𝑏 = 2 then 𝑎 ∗ 𝑏 = (−1)2 = 𝑖 ∉ 𝑅. | Group Theory 2 GG 1. Quasi Group/Groupoid A non-empty set equipped with unique binary operation is called groupoid/quasi group. Ex: (𝑁, +), (𝑍, +), (𝑄, ⋅) & (𝑍, −) 2. Semi-Group A quasi group in which the binary operation is associative is called semi-group. Ex: (𝑁, +), (𝑍, ⋅), (𝑄, +), (𝑄, ⋅), (𝑅, +)& (𝑅, ⋅) Identity element (neutral element): Let 𝐺 be a non-empty set and ∗ be a binary operation on 𝐺. Then the element 𝑒 ∈ 𝐺 such that 𝑎 ∗ 𝑒 = 𝑎 = 𝑒 ∗ 𝑎, ∀𝑎 ∈ 𝐺. Then 𝑒 is called identity element with respect to ∗. 3. Monoid A semi-group (𝐺, ∗) is said to be monoid if it has an identity element. Ex: (𝑁, ⋅) is a monoid with identity element 1. (𝑍, +) is a monoid with identity element 0. (𝑅, ⋅) is a monoid with identity element 1. Inverse element: Let 𝐺 be a non-empty set and ∗ be a binary operation on 𝐺. Let 𝑒 ∈ 𝐺 be the identity element in 𝐺 for ∗. The element 𝑎−1 ∈ 𝐺 is said to be an inverse of 𝑎 ∈ 𝐺 if 𝑎 ∗ 𝑎−1 = 𝑒 = 𝑎−1 ∗ 𝑎. Ex: In (𝑁, ∙), the identity element 1 is the only element which has inverse. In (𝑍, +), every element has an inverse and for each 𝑛 ∈ 𝑍 its inverse is (−𝑛). In (𝑍, ∙) only 1 and −1 have inverse elements. 4. Group A Monoid in which each element has its inverse. i.e., A non-empty set 𝐺 with a binary operation ∗ is said to be a group if for every 𝑥, 𝑦, 𝑧 ∈ 𝐺, the following properties hold: 1. Closure law: ∀𝑥, 𝑦 ∈ 𝐺 ⟹ 𝑥 ∗ 𝑦 ∈ 𝐺. 2. Associative law: ∀𝑥, 𝑦, 𝑧 ∈ 𝐺, 𝑥 ∗ (𝑦 ∗ 𝑧) = (𝑥 ∗ 𝑦) ∗ 𝑧. 3. Existence of identity: ∀𝑥 ∈ 𝐺, ∃𝑒 ∈ 𝐺 ∋ 𝑒 ∗ 𝑥 = 𝑥 ∗ 𝑒 = 𝑥. The element 𝑒 is said to be an identity element of G. | Group Theory 3 GG 4. Existence of inverses: ∀𝑥 ∈ 𝐺 ∃𝑦 ∈ 𝐺 ∋ 𝑥 ∗ 𝑦 = 𝑦 ∗ 𝑥 = 𝑒. The element 𝑦 is said to be an inverse of the element 𝑥. Then we say that (𝐺, ∗) is a group, or that 𝐺 is a group under the operation ∗. Definition: If (𝐺, ∗) is a group, the number of elements in the set 𝐺 is said to be the order of 𝐺, and is denoted |𝑮| or 𝑶(𝑮). A finite group is a group whose order is finite, and an infinite group is a group of infinite order. Note: When the group operation is clear from context, we shall write 𝑥𝑦 to mean 𝑥 ∗ 𝑦. Remark: Note that 𝑥𝑦 is different from 𝑦𝑥, as the group operation need not be commutative. If 𝑥 and 𝑦 are two elements such that 𝑥𝑦 = 𝑦𝑥, then we say that 𝑥 and 𝑦 commute with each other. Every element commute with the identity element. Obviously, each element also commutes with itself. If 𝑦 is an inverse of 𝑥, then 𝑥𝑦 = 𝑦𝑥 = 𝑒, which means that 𝑥 and 𝑦 commute with each other. Commutative law: A group in which the law of composition is commutative. i.e., ∀𝑥, 𝑦 ∈ 𝐺 ∋ 𝑥 ∗ 𝑦 = 𝑦 ∗ 𝑥. 5. Abelian Group A group in which binary operation is commutative is called Commutative group or Abelian group. A group that is not Abelian is non-Abelian. Examples: Groups of Numbers 1. (𝒁, +): The set 𝑍 of integers forms a group under the usual addition +. The sum of any two integers is also an integer, thus + is indeed a binary operation on 𝑍 (in other words, 𝑍 is closed under +). The identity element of this group is 0, since 𝑛 + 0 = 0 + 𝑛 = 𝑛 for all 𝑛 ∈ 𝑍. Finally, for any integer 𝑛, we know that 𝑛 + (−𝑛) = (−𝑛) + 𝑛 = 0, so that (−𝑛) which is also an integer, is the inverse of 𝑛. This is an example of infinite group. | Group Theory 4 GG 2. (𝒁, −): is not a group, since (−) is not associative (nor is there an identity element for subtraction, since 𝑛 − 𝑒 = 𝑛 only if 𝑒 = 0, but then 𝑒 − 𝑛 = 0 − 𝑛 = −𝑛 ≠ 𝑛; and without identity element inverses are not defined). 3. Similarly, (𝑄, +), (𝑅, +), (𝐶, +) are infinite groups. 4. The set of non-negative integers 𝑁0 = {0, 1, 2,···} does not form a group under addition. 5. Let 𝐺 = {0}, the singleton set containing only the real number 0. Then (𝐺, +) is a group. Its order is 1, and it is therefore the smallest group. 6. The set of non-zero real numbers forms a group under multiplication (𝑅 − {0}, ×). The 1 identity element is 1 and the inverse of 𝑥 ≠ 0 ∈ 𝑅 is ∈ 𝑅. But, if we include 0, it is 𝑥 no longer a group, since 0 has no (multiplicative) inverse. 7. If 𝑅>0 denotes the set of all positive real numbers, (𝑅>0 , × ) is also a group. 8. Let 𝐺 = {1}, 𝐻 = {1, −1} and 𝐾 = {1, −1, 𝑖, −𝑖} where 𝑖 = √−1. Then (𝐺, ×), (𝐻, ×) and (𝐾, ×) are groups of order 1, 2 and 4 respectively. Groups of Matrices 9. A square matrix 𝐴 is non-singular (or invertible) if its determinant is non-zero (detA ≠ 0). The set of all 𝑛 × 𝑛 non-singular real matrices forms a group under matrix multiplication, denoted 𝐺𝐿𝑛 (𝑅). Matrix multiplication is associative, the identity matrix is the identity element for this multiplication, and every non-singular matrix 𝐴 has an inverse 𝐴−1 (which is also non-singular). Note: 𝐺𝐿𝑛 (𝑅) is certainly closed under multiplication, as 𝑑𝑒𝑡(𝐴𝐵) = (𝑑𝑒𝑡 𝐴)(𝑑𝑒𝑡 𝐵) ≠ 0 whenever 𝑑𝑒𝑡 𝐴, 𝑑𝑒𝑡𝐵 ≠ 0. For 𝑛 > 1, 𝐺𝐿𝑛 (𝑅) is an example of a group where the operation is not commutative – if 𝐴 and 𝐵 are two 𝑛 × 𝑛 matrices 𝐴𝐵 is generally not equal to 𝐵𝐴. 1 1 1 0 Ex: 𝐴 = [ ] &𝐵 =[ ]. 0 1 1 1 | Group Theory 5 GG 𝑎𝑏 Exercise-1: Prove that (𝑅 − {0}, ∗) form a group with 𝑎 ∗ 𝑏 = , where 𝑎, 𝑏 ∈ 𝑅 − {0}. 2 𝑎𝑏 Solution: 1. Closure law: For any 𝑎, 𝑏 ∈ 𝑅 − {0}, the result of the operation 𝑎 ∗ 𝑏 = is 2 still in the set 𝑅 − {0} since dividing by 2 does not change the fact that the result is a non-zero real number. 𝑎𝑏 2. Associativity: Operation ∗ is defined by 𝑎 ∗ 𝑏 = 2 Let us consider another variable 𝑐, the operation is said to be associative if 𝑎 ∗ (𝑏 ∗ 𝑐) = (𝑎 ∗ 𝑏) ∗ 𝑐. Consider 𝑎 ∗ (𝑏 ∗ 𝑐) = 𝑎 ∗ (𝑏𝑐/2) = 𝑎𝑏𝑐/4 𝑎𝑏 𝑎𝑏𝑐 Also, (𝑎 ∗ 𝑏) ∗ 𝑐 = ( 2 ) ∗ 𝑐 = 4 Thus, 𝑎 ∗ (𝑏 ∗ 𝑐) = (𝑎 ∗ 𝑏) ∗ 𝑐 3. Identity Element: An identity element 𝑒 in a group satisfies 𝑎 ∗ 𝑒 = 𝑒 ∗ 𝑎 = 𝑎 for all a ∈ 𝑎𝑒 𝑅−{0}. Consider, 𝑎 ∗ 𝑒 = 𝑎 ⇒ = 𝑎 ⇒ 𝑒 = 2. 2 4. Inverse Element: For each 𝑎 ∈ 𝑅 − {0}, we need to find an inverse element 𝑎⁻¹ such that 𝑎 ∗ 𝑎−1 = 𝑎−1 ∗ 𝑎 = 𝑒. 𝑎 𝑎−1 2𝑒 4 Consider, = 𝑒, then 𝑎−1 = = 𝑎. 2 𝑎 Since all four group axioms are satisfied, we can conclude that the set (𝑅 − {0},∗) forms a 𝑎𝑏 group with the given operation 𝑎 ∗ 𝑏 = , where 𝑎, 𝑏 ∈ 𝑅 − {0}. 2 Exercise-2: Prove that (𝑍,∗) form an abelian group with 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 − 2, where 𝑎, 𝑏 ∈ 𝑍. Solution: 1.Closure law: For any 𝑎, 𝑏 ∈ 𝑍 we see that 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 − 2 ∈ 𝑍. 2. Associativity: For any 𝑎, 𝑏 ∈ 𝑍 we see that (𝑎 ∗ 𝑏) ∗ 𝑐 = (𝑎 + 𝑏 − 2) + 𝑐 − 2 = 𝑎 + (𝑏 + 𝑐 − 2) − 2 = 𝑎 ∗ (𝑏 ∗ 𝑐). 3. Identity element: For any 𝑎, 𝑏 ∈ 𝑍 we need to find 𝑒 ∈ 𝑍 such that 𝑎 ∗ 𝑒 = 𝑎 + 𝑒 − 2 = 𝑎. Solving for 𝑒 yields 𝑒 = 2 which is integer and for any 𝑎 ∈ 𝑍, we have 𝑎 ∗ 2 = 𝑎 + 2 − 2 = 𝑎. 4. Inverse element: For any 𝑎 ∈ 𝑍, we need to find 𝑏 ∈ 𝑍 such that 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 − 2 = 2 the identity element. Solving for 𝑏 yields 𝑏 = 4 − 𝑎 which indeed an integer, and 𝑎 ∗ 𝑏 = 2 is required. | Group Theory 6 GG 5. Commutative law: For any 𝑎, 𝑏 ∈ 𝑍 we see that 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 − 2 = 𝑏 + 𝑎 − 2 = 𝑏 ∗ 𝑎. Hence, it is an abelian group. Note: 𝑍𝑛 is group under addition modulo 𝑛. i.e., 𝑎 ⊕𝑛 𝑏 = 𝑎 + 𝑏 (𝑚𝑜𝑑 𝑛) 𝑍𝑛 = {1, 2, 3, ⋯ , (𝑛 − 1)} Exercise-3: Ex: 𝑍6 = {0, 1, 2, 3, 4, 5}. Cayley table Here, 0 is an identity element. 5 ⊕6 1 = 0 ⇒ 5 and 1 are inverse to each other. 2 ⊕6 4 = 0 ⇒ 2 and 4 are inverse to each other. 3 ⊕6 3 = 0 ⇒ 3 is self-inverse. Note: Similarly, 𝐺 = {3, 6, 9 , 12} is a group under multiplication modulo 5 with the identity element 6. Cayley table | Group Theory 7 GG Basic Properties of Groups 1. Every group has a unique identity element. Proof: For, suppose 𝑒1 and 𝑒2 are both identity elements of a group 𝐺. Then 𝑒1 = 𝑒1 𝑒2 = 𝑒2. The former equality is true because 𝑒2 is an identity element (so 𝑥𝑒2 = 𝑥 for any 𝑥), and the latter equality is true because 𝑒1 is an identity element. Thus, we get 𝑒1 = 𝑒2. 2. Every element of a group has a unique inverse. Proof: Let 𝑥 be an element of a group 𝐺. By inverse law it has an inverse say 𝑦. Suppose that 𝑧 is also an inverse of 𝑥. Let 𝑒 denotes the identity element of 𝐺 then 𝑦 = 𝑦𝑒 = 𝑦(𝑥𝑧) = (𝑦𝑥)𝑧 = 𝑒𝑧 = 𝑧. The second equality follows from 𝑧 being an inverse of 𝑥 the third one from associativity; and the fourth from 𝑦 being an inverse of 𝑥. Since each element 𝑥 is guaranteed to have a unique inverse, we can denote this inverse as 𝑥 −1. Note: If 𝐺 is a group, then for every 𝒂 ∈ 𝑮, (𝒂−𝟏 )−𝟏 = 𝒂. 3. In a group 𝑮, (𝒙𝒚)−𝟏 = 𝒚−𝟏 𝒙−𝟏. Proof: To show that 𝑎−1 = 𝑏, we need to verify that 𝑎𝑏 = 𝑒 and 𝑏𝑎 = 𝑒. We must show (𝑥𝑦)(𝑦 −1 𝑥 −1 ) = 𝑒. Indeed, (𝑥𝑦)(𝑦 −1 𝑥 −1 ) = 𝑥(𝑦𝑦 −1 )𝑥 −1 (Associativity) = 𝑥𝑒𝑥 −1 = 𝑥𝑥 −1 =𝑒 Similarly, (𝑦 −1 𝑥 −1 )(𝑥𝑦) = 𝑒. Thus, (𝑥𝑦)−1 = 𝑦 −1 𝑥 −1. 4. Cancellation Laws Prove that the (left and right) cancellation laws hold in every group. i.e., prove that if 𝑥 and 𝑦 are elements of a group 𝐺, then (1) If ∃𝑎 ∈ 𝐺, 𝑎𝑥 = 𝑎𝑦 then 𝑥 = 𝑦. (2) If ∃𝑏 ∈ 𝐺, 𝑥𝑏 = 𝑦𝑏 then 𝑥 = 𝑦. | Group Theory 8 GG Proof: (1) Let ∃𝑎 ∈ 𝐺 ∋ 𝑎𝑥 = 𝑎𝑦 On left-multiplying by 𝑎 −1 , we get 𝑎−1 (𝑎𝑥) = 𝑎−1 (𝑎𝑦) ⇒ (𝑎−1 𝑎)𝑥 = (𝑎−1 𝑎)𝑦 ⇒ 𝑒𝑥 = 𝑒𝑦 ⇒𝑥=𝑦 (2) Let ∃𝑏 ∈ 𝐺 ∋ 𝑥𝑏 = 𝑦𝑏 On right-multiplying by 𝑏 −1 , we get (𝑥𝑏)𝑏 −1 = (𝑦𝑏)𝑏 −1 ⇒ 𝑥(𝑏𝑏 −1 ) = 𝑦(𝑏𝑏 −1 ) ⇒ 𝑥𝑒 = 𝑦𝑒 ⇒𝑥=𝑦 Exercise: If 𝒙 and 𝒚 are two elements of 𝑮 that commute then show that 𝒙, 𝒚, 𝒙−𝟏 & 𝒚−𝟏 all commute with each other. Solution: Let 𝑥 and 𝑦 are two elements of 𝐺 that commute with each other. i.e., 𝑥𝑦 = 𝑦𝑥. Pre and post multiplying by 𝑥 −1 , the equation 𝑥𝑦 = 𝑦𝑥 becomes, 𝑥 −1 (𝑥𝑦)𝑥 −1 = 𝑥 −1 (𝑦𝑥)𝑥 −1 = 𝑦𝑥 −1 = 𝑥 −1 𝑦 Thus, if 𝑥 and 𝑦 commute, then so do 𝑥 and 𝑦 −1. By symmetry, 𝑥 −1 and 𝑦 commute as well. Applying this result to the commuting pair 𝑥 and 𝑦 −1 , we see that 𝑥 −1 and 𝑦 also commute. Finally, every element commutes with its own inverse. Note: 1. For every 𝑎 ∈ 𝐺, and ∈ 𝑍 + , we define a) 𝑎𝑛 = 𝑎. 𝑎. 𝑎 … 𝑎 (n times) and b) 𝑎−𝑚 = (𝑎−1 )𝑚 = 𝑎−1. 𝑎−1 … 𝑎−1 (m times) c) 𝑎𝑚+𝑛 = 𝑎𝑚 𝑎𝑛 , where 𝑚, 𝑛 ∈ 𝑍. d) (𝑎𝑚 )𝑛 = 𝑎𝑚𝑛 , where 𝑚, 𝑛 ∈ 𝑍. e) 𝑎0 = 𝑒. | Group Theory 9 GG 5. Prove that if every element of 𝑮 is self-inverse, then 𝑮 is Abelian. Is the converse true? Proof: Suppose every element of 𝐺 is self-inverse. That is, for all 𝑥 ∈ 𝐺, 𝑥 −1 = 𝑥. Let 𝑎, 𝑏 ∈ 𝐺. We know that (𝑎𝑏)−1 = 𝑏 −1 𝑎−1. But since all elements are self-inverse, this reduces to 𝑎𝑏 = 𝑏𝑎. Note that the converse is not true. For example, (𝑍, +) is an Abelian group in which not all elements are self-inverse. 6. Prove that G is Abelian if and only if ∀𝒙, 𝒚 ∈ 𝑮, (𝒙𝒚)−𝟏 = 𝒙−𝟏 𝒚−𝟏. Proof: Let 𝐺 be an abelian group. Then (𝑥𝑦)−1 = 𝑦 −1 𝑥 −1 = 𝑥 −1 𝑦 −1. (Since 𝐺 is commutative) Conversely, suppose 𝐺 satisfies the given property (𝑥𝑦)−1 = 𝑥 −1 𝑦 −1 for any two elements 𝑥 and 𝑦. ⇒ (𝑥𝑦) = (𝑥 −1 𝑦 −1 )−1 = (𝑦 −1 )−1 (𝑥 −1 )−1 = 𝑦𝑥 (Since (𝑥 −1 )−1 = 𝑥) 7. Prove that 𝑮 is abelian if and only if ∀𝒂, 𝒃 ∈ 𝑮, (𝒂𝒃)𝟐 = 𝒂𝟐 𝒃𝟐. Proof: Let 𝐺 is abelian group. Then (𝑎𝑏)2 = (𝑎𝑏𝑎𝑏) = (𝑎𝑎𝑏𝑏) = 𝑎2 𝑏 2. (Since 𝐺 is commutative) Conversely, (𝑎𝑏)2 = 𝑎2 𝑏 2 , ∀ 𝑎, 𝑏 ∈ 𝐺. ⇒ 𝑎𝑏𝑎𝑏 = 𝑎𝑎𝑏𝑏 ⇒ 𝑏𝑎 = 𝑎𝑏, ∀ 𝑎, 𝑏 ∈ 𝐺. (By left and right cancellation) Hence, 𝐺 is abelian. Subgroups A subgroup of a group (𝐺,∗) is a subset 𝐻 ⊆ 𝐺 such that (𝐻,∗) is also a group under the same binary operation. Then we write 𝑯 ≤ 𝑮. Ex: We know that 𝑍, 𝑄, 𝑅, and 𝐶 all forms groups under addition. Observe that 𝑍 ⊆ 𝑄 ⊆ 𝑅 ⊆ 𝐶. Similarly, 𝑄 − {0}, 𝑅>0 are subgroups of (𝑅 − {0}, ⋅). Remark: i. It is very important that the operation be the same, for otherwise it is meaningless to call one the subgroup of the other. For example, (𝑄 − {0}, ⋅) is not a subgroup of (𝑅, +). Even though 𝑄 − {0} ⊆ 𝑅. | Group Theory 10 GG ii. Given a subset 𝐻 of a group 𝐺 to check whether 𝐻 is a subgroup, we need to verify that 𝐻 satisfies the group axioms. iii. It is obvious that all elements of 𝐻 will satisfy associativity, since they are elements of 𝐺 as well. iv. The closure of 𝐻 under the group operation of 𝐺 needs to be checked carefully, since the result of applying the operation to two elements of 𝐻 may be an element of 𝐺 that is outside 𝐻. v. To verify the existence of identity and inverses, it is enough to check if the identity element of the group 𝐺, which is already known, is present in the subset 𝐻, and similarly in the case of inverse. These observations are summarised in the following lemma. Lemma: Let (𝐺,∗) be a group and 𝐻 ⊆ 𝐺. Then H is a subgroup of G if and only if all of the following hold. 1. 𝐻 is closed under ∗. That is, ∀ 𝑥, 𝑦 ∈ 𝐻, 𝑥 ∗ 𝑦 ∈ 𝐻. 2. 𝐻 contains the identity element (of 𝐺): 𝑒 ∈ 𝐻. 3. The inverse of every element of 𝐻 is also in 𝐻. That is, ∀𝑥 ∈ 𝐻, 𝑥 −1 ∈ 𝐻. Theorem-1: A non-empty subset 𝑯 of a group 𝑮 is a subgroup of 𝑮 if and only ∀𝒙, 𝒚 ∈ 𝑯, 𝒙𝒚−𝟏 ∈ 𝑯. Proof: Let 𝐻 be a subgroup of group 𝐺. i.e., 𝐻 ≤ 𝐺. Then 𝐻 is obviously non-empty (for 𝑒 ∈ 𝐻) and for any 𝑥, 𝑦 ∈ 𝐻, we have 𝑥, 𝑦 −1 ∈ 𝐻 (since 𝐻 contains the inverses of all its elements) and therefore 𝑥𝑦 −1 ∈ 𝐻. (By closure law) Conversely, suppose that 𝐻 ≠ ∅ and ∀𝑥, 𝑦 ∈ 𝐻, 𝑥𝑦 −1 ∈ 𝐻. Since, 𝐻 ≠ ∅, ∃𝑥 ∈ 𝐻. Then 𝑥𝑥 −1 ∈ 𝐻 ⇒ 𝑒 ∈ 𝐻. If 𝑥 ∈ 𝐻, then 𝑒, 𝑥 ∈ 𝐻 ⇒ 𝑒𝑥 −1 = 𝑥 −1 ∈ 𝐻. Finally, if 𝑥, 𝑦 ∈ 𝐻, then by what we have proved above, 𝑦 −1 ∈ 𝐻. Thus, 𝑥, 𝑦 −1 ∈ 𝐻 ⇒ 𝑥(𝑦 −1 )−1 ∈ 𝐻 ⇒ 𝑥𝑦 ∈ 𝐻. (By assumption on 𝐻) Hence, 𝐻 is a subgroup of 𝐺. Ex: Let 𝐺 be the group of non-zero complex numbers under multiplication, and let 2𝜋𝑖 , 𝜔=𝑒 𝑛 𝑛 − is positive integer so that 𝜔 is a primitive 𝑛𝑡ℎ root of unity. Then 𝐻 = {1, 𝜔, 𝜔2 , … 𝜔𝑛−1 } is a subgroup of 𝐺. | Group Theory 11 GG Clearly, 𝐻 ≠ ∅. The elements of 𝐻 are all the complex numbers of the form 𝜔𝑟 for some integer 𝑟. Thus, if 𝜔𝑟 and 𝜔 𝑠 are any two elements of 𝐻, then 𝜔𝑟 ∙ (𝜔 𝑠 )−1 = 𝜔𝑟−𝑠 is also an element of 𝐻. Hence, H is a subgroup of G. Theorem-2: Let 𝑯 and 𝑲 be any two subgroups of 𝑮. Then 𝟏. 𝑯 ∩ 𝑲 ≤ 𝑮. 𝟐. 𝑯 ∪ 𝑲 is not a subgroup of 𝑮 unless 𝑯 ⊆ 𝑲 or 𝑲 ⊆ 𝑯. Proof: 1. Since 𝑒 ∈ 𝐻 and 𝑒 ∈ 𝐾, 𝑒 ∈ 𝐻 ∩ 𝐾 ≠ ∅. Now if 𝑥, 𝑦 ∈ 𝐻 ∩ 𝐾, then 𝑥, 𝑦 ∈ 𝐻 and 𝑥, 𝑦 ∈ 𝐾. Since, 𝐻 and 𝐾 are subgroups of 𝐺, we have 𝑥𝑦 −1 ∈ 𝐻 and 𝑥𝑦 −1 ∈ 𝐾 ⇒ 𝑥𝑦 −1 ∈ 𝐻 ∩ 𝐾. Hence, 𝐻 ∩ 𝐾 is a subgroup of 𝐺. 2. Suppose that neither one of 𝐻 and 𝐾 is contained in other. Then there exist an element ℎ ∈ 𝐻 such that ℎ ∉ 𝐾 and similarly there exist an element 𝑘 ∈ 𝐾, 𝑘 ∉ 𝐻. Now ℎ𝑘 cannot be an element of 𝐻. (If ℎ𝑘 ∈ 𝐻 ⇒ ℎ−1 ℎ𝑘 = 𝑘 ∈ 𝐻 which is not possible) Similarly, ℎ𝑘 ∉ 𝐾. Therefore, ℎ𝑘 ∉ 𝐻 ∪ 𝐾. Since 𝐻 ∪ 𝐾 is not closed under the operation it is not a subgroup of 𝐺. If 𝐻 ⊆ 𝐾, then 𝐻 ∪ 𝐾 = 𝐾 ≤ 𝐺. The other case is similar. Note: Intersection of any number of subgroups of 𝐺 is again a subgroup of 𝐺. Definition: If 𝐻 and 𝐾 are two subgroups of 𝐺, let 𝑯𝑲 = {𝒙 ∈ 𝑮: 𝒙 = 𝒉𝒌, 𝒉 ∈ 𝑯, 𝒌 ∈ 𝑲}. Note that 𝐻𝐾 and 𝐾𝐻 are subsets of the group G (but not necessarily subgroups), and in general, 𝐻𝐾 ≠ 𝐾𝐻. Every element of 𝐻𝐾 is of the form ℎ𝑘, where ℎ ∈ 𝐻 and 𝑘 ∈ 𝐾. 𝐻𝐾 = 𝐾𝐻 if and only if for every element ℎ𝑘 ∈ 𝐻𝐾, ℎ𝑘 = 𝑘 ′ ℎ′ for some ℎ′ ∈ 𝐻 and 𝑘′ ∈ 𝐾. | Group Theory 12 GG Theorem-3: Let 𝑯 and 𝑲 be any two subgroups of 𝑮. Then 𝑯𝑲 is a subgroup of 𝑮 if and only if 𝑯𝑲 = 𝑲𝑯. Proof: Let 𝐻𝐾 be a subgroup of 𝐺. Let 𝑥 ∈ 𝐻𝐾. ⇒ 𝑥 −1 ∈ 𝐻𝐾. (Since 𝐻𝐾 is a subgroup) ⇒ 𝑥 −1 = ℎ𝑘, ℎ ∈ 𝐻, 𝑘 ∈ 𝐾 Now 𝑥 = (ℎ𝑘)−1 = 𝑘 −1 ℎ−1 which is an element of 𝐾𝐻 since 𝑘 −1 ∈ 𝐾 and ℎ−1 ∈ 𝐻. Thus, 𝑥 ∈ 𝐻𝐾 ⇒ 𝑥 ∈ 𝐾𝐻 which shows that 𝐻𝐾 ⊆ 𝐾𝐻. To see that 𝐾𝐻 ⊆ 𝐻𝐾 as well, consider an arbitrary element 𝑘ℎ ∈ 𝐾𝐻. Now (𝑘ℎ)−1 = ℎ−1 𝑘 −1 ∈ 𝐻𝐾. ⇒ ((𝑘ℎ)−1 )−1 = 𝑘ℎ ∈ 𝐻𝐾. (Since 𝐻𝐾 is a subgroup of 𝐺) Thus, 𝐻𝐾 = 𝐾𝐻. Conversely, suppose that 𝐻𝐾 = 𝐾𝐻. Since 𝐻 and 𝐾 are non-empty being subgroups of 𝐺. So, 𝐻𝐾 ≠ ∅. Let 𝑥, 𝑦 ∈ 𝐻𝐾. We have to show that 𝑥𝑦 −1 ∈ 𝐻𝐾. We know that 𝑥 = ℎ1 𝑘1 , 𝑦 = ℎ2 𝑘2 , for some ℎ1 , ℎ2 ∈ 𝐻, 𝑘1 , 𝑘2 ∈ 𝐾. Then 𝑥𝑦 −1 = (ℎ1 𝑘1 )(ℎ2 𝑘2 )−1 = ℎ1 𝒌𝟏 𝒌−𝟏 −1 𝟐 ℎ2 = ℎ1 𝒌𝟑 ℎ2−1 , where, 𝑘3 = 𝑘1 𝑘2−1 = 𝒉𝟏 𝒉𝟑 𝑘4 , 𝑘3 ℎ2−1 = ℎ3 𝑘4 Since, (𝑘3 ℎ2−1 ∈ 𝐾𝐻 = 𝐻𝐾) = ℎ4 𝑘4 , ℎ4 = ℎ1 ℎ3 ∈ 𝐻 Thus, 𝑥𝑦 −1 ∈ 𝐻𝐾 and hence 𝐻𝐾 ≤ 𝐺. Corollary: If 𝐻, 𝐾 are subgroups of the abelian group 𝐺, then 𝐻𝐾 is a subgroup of 𝐺. Definition: Let 𝐺 be a group. The centre of a group 𝑮 is defined as 𝑍(𝐺) = {𝑧 ∈ 𝐺: 𝑧𝑥 = 𝑥𝑧, ∀𝑥 ∈ 𝐺}. i.e., The center of a group G contains all elements of G that commute with every element of G. Definition: Let 𝑆 ≤ 𝐺. Then the centralizer/normalizer of 𝑆 in 𝐺 is defined as 𝐶𝐺 (𝑆) or 𝑁(𝐺) = {𝑦 ∈ 𝐺: 𝑦𝑥 = 𝑥𝑦, ∀𝑥 ∈ 𝑆}. Note: Centre of a group and centralizer both are subgroups of 𝐺. | Group Theory 13 GG Equivalence relation on 𝑮 Let 𝐻 be a subgroup of 𝐺. Define a relation ~ on 𝐺 as follows: For 𝑎, 𝑏 ∈ 𝐺, 𝑎~𝑏 if and only if 𝑎𝑏 −1 ∈ 𝐻. a) Reflexive relation: As 𝑎𝑎−1 = 𝑒 ∈ 𝐻, we have 𝑎~𝑎. Hence, ~ is reflexive. b) Symmetric relation: Suppose 𝑎~𝑏. Then 𝑎𝑏 −1 ∈ 𝐻. As 𝐻 is a subgroup of 𝐺, (𝑎𝑏 −1 )−1 ∈ 𝐻 i.e., 𝑏𝑎 −1 ∈ 𝐻. This implies that 𝑏~𝑎. Hence, ~ is symmetric. c) Transitive relation: Suppose 𝑎~𝑏 and 𝑏~𝑐. Then 𝑎𝑏 −1 ∈ 𝐻 and 𝑏𝑐 −1 ∈ 𝐻. As 𝐻 is a subgroup of 𝐺, (𝑎𝑏 −1 )(𝑏𝑐 −1 ) ∈ 𝐻. i.e., 𝑎(𝑏 −1 𝑏)𝑐 −1 = 𝑎𝑒𝑐 −1 = 𝑎𝑐 −1 ∈ 𝐻. This implies that 𝑎~𝑐. Thus, ~ is transitive. Therefore ~ is an equivalence relation on 𝐺. Definition: Let 𝐺 be a group, 𝐻 a subgroup of 𝐺; for 𝑎, 𝑏 ∈ 𝐺 we say 𝑎 is congruent to 𝑏 mod 𝐻, written as 𝑎 ≡ 𝑏 𝑚𝑜𝑑 𝐻 if 𝑎𝑏 −1 ∈ 𝐻. Lemma: The relation 𝒂 ≡ 𝒃 𝒎𝒐𝒅 𝑯 is an equivalence relation. Proof: We must prove the following three conditions: 1. 𝑎 ≡ 𝑎 𝑚𝑜𝑑 𝐻. (Reflexive) 2. 𝑎 ≡ 𝑏 𝑚𝑜𝑑 𝐻 implies 𝑏 ≡ 𝑎 𝑚𝑜𝑑 𝐻. (Symmetric) 3. 𝑎 ≡ 𝑏 𝑚𝑜𝑑 𝐻, 𝑏 ≡ 𝑐 𝑚𝑜𝑑 𝐻 implies 𝑎 ≡ 𝑐 𝑚𝑜𝑑 𝐻. (Transitive) 1. Since 𝐻 is a subgroup of 𝐺, 𝑒 ∈ 𝐻, and since 𝑎𝑎−1 = 𝑒, 𝑎𝑎−1 ∈ 𝐻. Hence, 𝑎 ≡ 𝑎 𝑚𝑜𝑑 𝐻. 2. Suppose that 𝑎 ≡ 𝑏 𝑚𝑜𝑑 𝐻. i.e., 𝑎𝑏 −1 ∈ 𝐻. Since 𝑎𝑏 −1 ∈ 𝐻, which is a subgroup of 𝐺, (𝑎𝑏 −1 )−1 ∈ 𝐻. ⇒ 𝑏𝑎 −1 ∈ 𝐻 and hence 𝑏 ≡ 𝑎 𝑚𝑜𝑑 𝐻 3. Finally, suppose 𝑎 ≡ 𝑏 𝑚𝑜𝑑 𝐻 and 𝑏 ≡ 𝑐 𝑚𝑜𝑑 𝐻. These translates into 𝑎𝑏 −1 ∈ 𝐻 and 𝑏𝑐 −1 ∈ 𝐻. Since 𝐻 is a subgroup (𝑎𝑏 −1 )(𝑏𝑐 −1 ) ∈ 𝐻 ⇒ 𝑎(𝑏𝑏 −1 )𝑐 −1 ∈ 𝐻 ⇒ 𝑎𝑒𝑐 −1 ∈ 𝐻 ⇒ 𝑎𝑐 −1 ∈ 𝐻 Hence, it follows that 𝑎 ≡ 𝑐 𝑚𝑜𝑑 𝐻. The relation 𝑎 ≡ 𝑏 𝑚𝑜𝑑 𝐻 is an equivalence relation. | Group Theory 14 GG Equivalence class: For 𝑎 ∈ 𝐺 the equivalence class of 𝑎 is given by [𝑎] = {𝑏 ∈ 𝐺: 𝑏~𝑎} = {𝑏 ∈ 𝐺: 𝑏𝑎−1 ∈ 𝐻} = {𝑏 ∈ 𝐺: 𝑏𝑎−1 = ℎ for some ℎ ∈ 𝐻} = {𝑏 ∈ 𝐺: 𝑏 = ℎ𝑎 for some ℎ ∈ 𝐻} = {ℎ𝑎: ℎ ∈ 𝐻} We denote the set {ℎ𝑎: ℎ ∈ 𝐻} by 𝐻𝑎 and is called as right coset of 𝐻 with respect to 𝑎. Similarly, we can define left coset 𝑎𝐻. Note: Given any set 𝐴 and an equivalence relation ~ on 𝐴, the equivalence classes under ~ form a partition of 𝐴, that is a) 𝐴 = ⋃𝑎∈𝐴[𝑎] b) ∀𝑎, 𝑏 ∈ 𝐴, either [𝑎] = [𝑏] or [𝑎] ∩ [𝑏] = ∅. c) [𝑎] is a non-empty set. Therefore, the left cosets (right cosets) of a subgroup 𝐻 in a group 𝐺 form a partition of G. However, this fact can also be proved independently. Definition: If 𝐻 is a subgroup of 𝐺, 𝑎 ∈ 𝐺, then 𝐻𝑎 = {ℎ𝑎: ℎ ∈ 𝐻}. 𝐻𝑎 is called a right coset of 𝑯 in 𝑮. Definition: If 𝐻 is a subgroup of 𝐺, 𝑎 ∈ 𝐺, then 𝑎𝐻 = {𝑎ℎ: ℎ ∈ 𝐻}. 𝑎𝐻 is called a left coset of 𝑯 in 𝑮. Note: 1. The subgroup 𝐻 itself is a coset of itself. 2. The left cosets of a subgroup partition the whole group into equally sized part. Lemma: For all 𝑎 ∈ 𝐺, 𝐻𝑎 = {𝑥 ∈ 𝐺: 𝑎 ≡ 𝑥 𝑚𝑜𝑑 𝐻}. Definition: If 𝐻 is a subgroup of 𝐺, then the index of 𝑯 in 𝑮 is the number of distinct right or left cosets of 𝐻 in 𝐺 and is denoted by 𝑖𝐺 (𝐻) or (𝐺: 𝐻). | Group Theory 15 GG Lemma-1: Let 𝑮 be a finite group, 𝑯 ≤ 𝑮, and let 𝒙𝟏 𝑯, ⋯ , 𝒙𝒌 𝑯 be all the distinct left cosets of 𝑯 in 𝑮. Then 𝑮 = ⋃𝒌𝒊=𝟏 𝒙𝒊 𝑯. Proof: Let 𝑥1 𝐻, ⋯ , 𝑥𝑘 𝐻 be the distinct left cosets of 𝐻 in 𝐺. Since, 𝑥𝑖 𝐻 ⊆ 𝐺, ∀ 𝑖, we have ⋃𝑘𝑖=1 𝑥𝑖 𝐻 ⊆ 𝐺. Now for each 𝑥 ∈ 𝐺, we have 𝑥𝐻 = 𝑥𝑖 𝐻 for some 𝑖 = 1, 2, … , 𝑘. Also, we have 𝑥 = 𝑥𝑒 ∈ 𝑥𝐻 = 𝑥𝑖 𝐻, for some 𝑖. This implies 𝑥 ⊆ ⋃𝑘𝑖=1 𝑥𝑖 𝐻. Hence, 𝐺 ⊆ ⋃𝑘𝑖=1 𝑥𝑖 𝐻. Therefore, 𝐺 = ⋃𝑘𝑖=1 𝑥𝑖 𝐻. Lemma-2: Distinct left cosets of 𝑯 are disjoint. Proof: Let 𝑥𝐻 and 𝑦𝐻 are two distinct left cosets of 𝐻. On a contrary, let 𝑥𝐻 ∩ 𝑦𝐻 ≠ ∅. Then ∃𝑧 ∈ 𝑥𝐻 ∩ 𝑦𝐻 so, 𝑧 = 𝑥ℎ1 = 𝑦ℎ2 where, ℎ1 , ℎ2 ∈ 𝐻. Then 𝑦 = 𝑥ℎ1 ℎ2−1. [By post multiplying with ℎ2−1 ] Now for any 𝑦ℎ ∈ 𝑦𝐻, 𝑦ℎ = 𝑥ℎ1 ℎ2−1 ℎ ⇒ 𝑦ℎ ∈ 𝑥𝐻. (∵ ℎ1 ℎ2−1 ℎ ∈ 𝐻) Thus, 𝑦𝐻 ⊆ 𝑥𝐻. Similarly, 𝑥𝐻 ⊆ 𝑦𝐻. Hence, 𝑥𝐻 = 𝑦𝐻 which is a contradiction. Therefore, distinct left cosets of 𝑯 are disjoint. Note: Number of elements in a set 𝐴 is called Cardinality and is denoted by |𝐴|. Lemma-3: Every coset (left or right) of a subgroup 𝑯 of a group 𝑮 has the same cardinality as 𝑯 (i.e., same number of elements as H). (Or) Prove that there is a one-to-one correspondence between any two left cosets of 𝑯 in 𝑮. Proof: Let 𝑥𝐻 be any left coset of 𝐻. We need to show that |𝑥𝐻| = |𝐻|, by proving the existence of a bijection between 𝐻 and 𝑥𝐻. Define 𝑓: 𝐻 → 𝑥𝐻, ∀ℎ ∈ 𝐻, 𝑓(ℎ) = 𝑥ℎ. Suppose that 𝑓 (ℎ1 ) = 𝑓 (ℎ2 ) ⇒ 𝑥ℎ1 = 𝑥ℎ2 | Group Theory 16 GG ⇒ ℎ1 = ℎ2 by left cancellation. Thus 𝑓 is injective. It is surjective since every element of 𝑥𝐻 is of the form 𝑥ℎ, where ℎ ∈ 𝐻. ⇒ 𝑥ℎ = 𝑓(ℎ). Thus, 𝑓 is a bijection and |𝑥𝐻| = |𝐻|. A similar one-to-one map of 𝐻 onto the right coset 𝐻𝑥 can be constructed. Hence, Every coset (left or right) of a subgroup 𝐻 of a group 𝐺 has the same cardinality as 𝐻. Lagrange’s Theorem For any finite group 𝑮, the order (number of elements) of every subgroup 𝑯 of 𝑮 divides the order of 𝑮. i.e, 𝑶(𝑯)|𝑶(𝑮). Proof: Let 𝐻 be a subgroup of 𝐺. We know, 𝐺 = ⋃𝑘𝑖=1 𝑥𝑖 𝐻 ⇒ 𝑂(𝐺) = |⋃𝑘𝑖=1 𝑥𝑖 𝐻 | = ∑𝑘𝑖=1 |𝑥𝑖 𝐻| = ∑𝑘𝑖=1 |𝐻| (Since, every coset of 𝐻 has same number of elements as 𝐻) ⇒ 𝑂(𝐺) = 𝑘. 𝑂(𝐻) Thus, 𝑂(𝐻)|𝑂(𝐺). Remark: Note that all the results above could equivalently be written in terms of right cosets. Thus, the number of right cosets of 𝐻 in 𝐺 is also (𝐺: 𝐻). Definition: The order of an element 𝑥 of 𝐺 is defined as the least positive integer 𝑛, if any such that 𝒙𝒏 = 𝒆. If there is no such positive integer then the element is said to have infinite order. The order of 𝑥 is denoted by |𝑥| or 𝑂(𝑥). Corollary: If 𝑮 is a finite group and 𝒂 ∈ 𝑮, then order of 𝒂 divides order of 𝑮 i.e., 𝑶(𝒂)|𝑶(𝑮). Theorem: Let 𝐺 be a group and let 𝑎 ∈ 𝐺. Then 𝐻 = {𝑎𝑛 | 𝑛 ∈ 𝑍} is a subgroup of 𝐺 and is the smallest subgroup of 𝐺 that contains 𝑎, that is, every subgroup containing a contains 𝐻. | Group Theory 17 GG Cyclic Groups A group is cyclic if it is generated by a single element of the group. i.e., 𝐺 = 〈𝑎〉, where, 〈𝑎〉 = {𝑎𝑛 : 𝑛 ∈ 𝑍}. Here, 𝑎 is the generator of 𝐺. Definition: An element 𝑎 of a group 𝐺 generates 𝐺 and is a generator for 𝐺 if < 𝑎 >= 𝐺. Note: A group 𝐺 is cyclic if there is some element 𝑎 in 𝐺 that generates 𝐺. Ex-1: Infinite group (𝑍, +) is a cyclic group generated by 1. Ex-2: Finite cyclic group 𝑍𝑛 under addition modulo 𝑛. Ex-3: 𝑍4 is cyclic and both 1 and 3 are generators. i.e., < 1 >=< 3 >= 𝑍4. Theorem: Every cyclic group is abelian. Proof: Let 𝐺 be a cyclic group and let 𝑎 be a generator of 𝐺 so that 𝐺 = < 𝑎 >= {𝑎𝑛 | 𝑛 ∈ 𝑍}. If 𝑔1 and 𝑔2 are any two elements of 𝐺, there exist integers 𝑟 and 𝑠 such that 𝑔1 = 𝑎𝑟 and 𝑔2 = 𝑎 𝑠. Then 𝑔1 𝑔2 = 𝑎𝑟 𝑎 𝑠 = 𝑎𝑟+𝑠 = 𝑎 𝑠+𝑟 = 𝑎 𝑠 𝑎𝑟 = 𝑔2 𝑔1 , Hence, 𝐺 is abelian. Note: But the converse need not be true. i.e., Abelian group need not be cyclic. Remark: In any group the identity element generates the trivial subgroup: < 𝑒 >= {𝑒}. It is the only one that does (since for all ∀ 𝑥 ∈ 𝐺, 𝑥 ∈< 𝑥 >). Any element generates the same subgroup as its inverse i.e., < 𝑥 >=< 𝑥 −1 >. A group G is said to be cyclic if it is equal to the cyclic subgroup generated by one of its elements. For any group element 𝒂, |𝒂| = | < 𝒂 > |. Ex-4: Let 𝐺 be the group of non-zero complex numbers under multiplication, and let 𝜔 = 2𝜋𝑖 , 𝑒 𝑛 𝑛 − is positive integer so that 𝜔 is a primitive 𝑛𝑡ℎ root of unity. | Group Theory 18 GG Then 𝐺 = {1, 𝜔, 𝜔2 , …. 𝜔𝑛−1 } is a group under (complex) multiplication. 𝐺 is cyclic, since every element of 𝐺 is a power of 𝜔, so that 𝑔 =< 𝜔 >. Ex-5: Show that 𝑍5 = {0,1,2,3,4} forms a cyclic group under addition modulo 5. List all the elements of 𝑍5 that generates it. ⊕5 0 1 2 3 4 0 0 1 2 3 4 1 1 2 3 4 0 2 2 3 4 0 1 3 3 4 0 1 2 4 4 0 1 2 3 < 1 >= {1 , 1 ⊕5 1, 1 ⊕5 1 ⊕5 1, ⋯ } = {1, 2, 3 ,4, 0} < 2 >= {2, 4, 1, 3, 0} Here, < 1 >, < 2 >, < 3 >, < 4 > are generators of 𝑍5. Klein 4-group (Vierergruppe) Let 𝑉4 denote the group formed by {1, 3, 5, 7} under multiplication modulo 8. The group axioms are easily seen to hold from the multiplication table given below: ⨂ 1 3 5 7 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1 ✓ Every element of this group is self-inverse, and therefore the group is Abelian. ✓ This is not a cyclic group. ✓ Each non-identity element generates a subgroup of order 2 and are < 3 >= {1, 3}, < 5 > = {1, 5}, < 7 > = {1, 7}. ✓ This group 𝑉4 is called the Klein 4-group (Vierergruppe) and is the smallest non-cyclic group. ✓ This is an example of abelian group which is not cyclic. Theorem: For any element 𝑥 ∈ 𝐺, 𝑂(𝑥) = 𝑂(< 𝑥 >). i.e., order of an element is the same as the order of the cyclic subgroup generated by the element. | Group Theory 19 GG Corollary: Any group of prime order is cyclic and hence abelian. Proof: Let 𝐺 be a group of order 𝑝, where 𝑝 is a prime number i.e., 𝑂(𝐺) = 𝑝. Since 𝑝 ≥ 2, 𝐺 has at least one non-identity element, say 𝒈. We know that “If 𝐺 is a finite group and 𝑔 ∈ 𝐺, then order of 𝑔 divides order of 𝐺 i.e., 𝑂(𝑔)|𝑂(𝐺)”. Now 𝑂(𝑔)|𝑝 which implies that 𝑂(𝑔) = 1 or 𝑝. But 𝑔 ≠ 𝑒 ⇒ 𝑂(𝑔) ≠ 1. Thus, 𝑂(< 𝑔 >) = 𝑂(𝑔) = 𝑝 = 𝑂(𝐺). Hence, group of prime order is cyclic. Division Algorithm for Z: If 𝑚 is a positive integer and 𝑛 is any integer, then there exist unique integers 𝑞 and 𝑟 such that 𝑛 = 𝑚𝑞 + 𝑟 and 0 ≤ 𝑟 < 𝑚. In the notation of the division algorithm, we regard 𝑞 as the quotient and 𝑟 as the non-negative remainder when 𝑛 is divided by 𝑚. Theorem: A subgroup of a cyclic group is cyclic. Proof: Let 𝐺 be a cyclic group generated by 𝑎 and let 𝐻 be a subgroup of 𝐺. If 𝐻 = {𝑒}, then 𝐻 =< 𝑒 > is cyclic. If 𝐻 ≠ {𝑒}, then 𝑎𝑛 ∈ 𝐻 for some 𝑛 ∈ 𝑍. Let 𝑚 be the smallest positive integer such that 𝑎𝑚 ∈ 𝐻. We claim that 𝑐 = 𝑎𝑚 generates H; that is, 𝐻 =< 𝑎𝑚 >=< 𝑐 >. We must show that every 𝑏 ∈ 𝐻 is a power of 𝑐. Since 𝑏 ∈ 𝐻 and 𝐻 ≤ 𝐺, we have 𝑏 = 𝑎𝑛 for some 𝑛. Find 𝑞 and 𝑟 such that 𝑛 = 𝑚𝑞 + 𝑟 for 0 ≤ 𝑟 < 𝑚 in accord with the division algorithm. Then 𝑎𝑛 = 𝑎𝑚𝑞+𝑟 = (𝑎𝑚 )𝑞 𝑎𝑟 , So 𝑎𝑟 = (𝑎𝑚 )−𝑞 𝑎𝑛. Now since 𝑎𝑛 ∈ 𝐻, 𝑎𝑚 ∈ 𝐻, and 𝐻 is a subgroup, both (𝑎𝑚 )−𝑞 and 𝑎𝑛 are in 𝐻. Thus (𝑎𝑚 )−𝑞 𝑎𝑛 ∈ 𝐻; that is, 𝑎𝑟 ∈ 𝐻. Since 𝑚 was the smallest positive integer such that 𝑎𝑚 ∈ 𝐻 and 0 ≤ 𝑟 < 𝑚, we must have 𝑟 = 0. Thus 𝑛 = 𝑞𝑚 and 𝑏 = 𝑎𝑛 = (𝑎𝑚 )𝑞 = 𝑐 𝑞 , so 𝑏 is a power of 𝑐. Corollary: The subgroups of 𝑍 under addition are precisely the groups 𝑛𝑍 under addition for 𝑛 ∈ 𝑍. | Group Theory 20 GG Definition: Conjugate of an element If 𝑎, 𝑏 ∈ 𝐺, then 𝑏 is said to be a conjugate of 𝑎 in 𝐺 if there exists an element 𝑐 ∈ 𝐺 such that 𝑏 = 𝑐 −1 𝑎𝑐. i.e., 𝐶𝑙(𝑎) = {𝑐 −1 𝑎𝑐| 𝑐 ∈ 𝐺} Normal Subgroup A subgroup 𝑁 of a group 𝐺 is said to be normal subgroup of 𝐺 (𝑵 ⊴ 𝑮) if for every 𝑔 ∈ 𝐺 and 𝑛 ∈ 𝑁, 𝑔𝑛𝑔−1 ∈ 𝑁. Equivalently, if by 𝑔𝑁𝑔−1 we mean the set of all 𝑔𝑛𝑔−1 , 𝑛 ∈ 𝑁, then 𝑁 is a normal subgroup of 𝐺 if and only if 𝑔𝑁𝑔−1 ⊆ 𝑁 for every 𝑔 ∈ 𝐺. Theorem: A subgroup 𝑁 of 𝐺 is normal in 𝐺 if and only if 𝑔𝑁𝑔−1 ⊆ 𝑁 for all 𝑔 in G. Note: For any group, the trivial subgroup and the whole group are always normal subgroups. A non- trivial group that has no normal subgroups other than these two is called a simple group. Theorem: 𝑵 is a normal subgroup of 𝑮 if and only if 𝒈𝑵𝒈−𝟏 = 𝑵 for every 𝒈 ∈ 𝑮. Proof: Suppose that 𝑁 is a normal subgroup of 𝐺. Then for any 𝑔 ∈ 𝐺, 𝑔𝑁𝑔−1 ⊆ 𝑁. We must prove that 𝑁 ⊆ 𝑔𝑁𝑔−1 as well. Let 𝑛 ∈ 𝑁. Clearly, 𝑔−1 ∈ 𝐺. Therefore, 𝑔−1 𝑛𝑔 ∈ 𝑁. ⇒ 𝑔−1 𝑛𝑔 = 𝑛′ , for some 𝑛′ ∈ 𝑁. Then 𝑛 = 𝑔𝑛′ 𝑔−1 ∈ 𝑔𝑁𝑔−1. Thus, ∀𝑛 ∈ 𝑁, 𝑛 ∈ 𝑔𝑁𝑔−1 , which shows that 𝑁 ⊆ 𝑔𝑛𝑔−1. Conversely, if 𝑔𝑁𝑔−1 = 𝑁 for every 𝑔 ∈ 𝐺, certainly 𝑔𝑁𝑔−1 ⊆ 𝑁, so 𝑁 is normal in 𝐺. | Group Theory 21 GG Theorem: The subgroup 𝑵 of 𝑮 is a normal subgroup of 𝑮 if and only if every left coset of 𝑵 in 𝑮 is a right coset of 𝑵 in 𝑮. i.e., 𝒙𝑵 = 𝑵𝒙, ∀𝒙 ∈ 𝑮. Proof: Suppose that 𝑁 is a normal subgroup of 𝐺 (𝑁 ⊴ 𝐺). Then for every 𝑥 ∈ 𝐺, 𝑥𝑁𝑥 −1 = 𝑁, ⇒ (𝑥𝑁𝑥 −1 )𝑥 = 𝑁𝑥 ⇒ 𝑥𝑁 = 𝑁𝑥, and so the left coset 𝑥𝑁 is the right coset 𝑁𝑥. Conversely, suppose that ∀𝑥 ∈ 𝐺, 𝑥𝑁 = 𝑁𝑥. Let 𝑥 ∈ 𝐺, 𝑛 ∈ 𝑁. Then 𝑥𝑛 = 𝑛′ 𝑥, for some 𝑛′ ∈ 𝑁, which implies that 𝑥𝑛𝑥 −1 = 𝑛′ ∈ 𝑁. Thus, 𝑥𝑁𝑥 −1 ⊆ 𝑁 and hence 𝑁 ⊴ 𝐺. Exercise 1. If 𝐺 is an abelian group which subgroup of 𝐺 is normal? 2. Prove that a finite Abelian group is simple if and only if its order is a prime number. Hint: Every element generates a subgroup. 3. Let 𝐻 ≤ 𝐺 (not necessarily) and define 𝑁 = ⋂𝑥∈𝐺 𝑥𝐻𝑥 −1. Show that 𝑁 ⊴ 𝐺. 4. Let 𝑁 be a subgroup of index 2 in 𝐺 (i.e., (𝐺: 𝑁) = 2). Show that 𝑁 ⊴ 𝐺. Hint: If 𝑁 has only two left cosets and only two right cosets and one of them is 𝑁 in each case, what is the other? 5. The centre of a group G is defined by Z(G)= {𝑧 ∈ 𝐺 |𝑧𝑥 = 𝑥𝑧, ∀ 𝑥 ∈ 𝐺}. Prove that 𝑍(𝐺) ≤ 𝐺. Prove that its centre is always a normal subgroup. 6. Let G be an Abelian group. H= {𝑥 ∈ 𝐺 |𝑥 2 = 𝑒}. Show that 𝐻 ≤ 𝐺. 7. Let G be an Abelian group. H= {𝑥 ∈ 𝐺 |𝑥 𝑛 = 𝑒}. Show that 𝐻 ≤ 𝐺. 8. Let G be an Abelian group. H= {𝑥 𝑛 |𝑥 ∈ 𝐺}. Show that 𝐻 ≤ 𝐺. 9. If 𝐻 ≤ 𝐺 then prove that 𝑥𝐻𝑥 −1 ≤ 𝐺. 10. Let 𝐺 be a group and 𝐻 be a cyclic subgroup of 𝐺, and 𝑔 ∈ 𝐺. Let 𝐾 = {𝑔ℎ𝑔−1 |ℎ ∈ 𝐻}. Show that 𝐾 is also cyclic subgroup of 𝐺. Also show that if 𝐻 is of order 𝑛 then 𝐾 is also of order 𝑛. 11. Show that any group with at most 5 elements is abelian. | Group Theory