 here see multiple inverse of two is one by two but one by two does not belong to z. So, therefore, z with multiplication has a binary operation is not a group. Next example five say whether n plus n with addition has a binary operation is a Avalian group. So, here n with a set of all natural numbers clearly satisfies closure and associative property under the usual addition as a binary operation. So, zero is the identity but zero does not belongs to n. Here set n is also called counting number. So, here the zero does not belongs to this set. So, also moreover for A belongs to n we do not have this type of element A minus A is equal to zero. So, zero does not belongs to n. So, inverse of A is equal to minus A does not belongs. There is no negative number. So, hence n plus is not a group. So, next example six. So, that q plus set of positive rational number is an Avalian group where star is defined by this. A star B is equal to E B by 2 for all A B belongs to q. So, we have to show that q plus q plus comma star is an Avalian group and here a binary operation is defined by this relation A star B is equal to A B by 2. So, here is a particular solution of example six. So, here q plus is a set of all positive rational numbers. So, first we see the closure property. So, clearly A star B is equal to A B by 2. So, means any element we consider from does this set set of all positive rational number. So, this relation will always satisfy. If we take any element suppose we consider 3 and 2 belongs to q plus means 3 star 2 means 6 by 2. So, this element will always belongs to q. So, you take any elements from q plus 2. So, this relation is always satisfied. So, this means closure property is satisfied. Next associative property. So, first we define we find this a bracket A star B star C. So, A star B A B by C star C then after again applying this operation defined by this relation we will get A B C by 2 divided 2 is equal to A B C by 4. Similarly, we find this here also we get A B C by 4. So, from this two relation we see that this is equal to this. So, A and B their relationship is same. So, associative property is satisfied. So, here you see this property is satisfied. So, next you see identity we find identity. So, we consider E B the identity element. So, then A star E is equal to A. So, for existence of identity this condition must be satisfied. Therefore, A star E is equal to A. So, A star E is equal to E means A E by 2 is equal to A. So, this implies E is equal to 2. So, here identity elements E is equal to 2 belongs to q. So, this is the identity element. So, next we find the inverse element. So, suppose this is the inverse element A to the power minus beta inverse of A, then for inverse of element we must have this property A star B is equal to E is equal to B star E. So, already we find that 2 is the identity element. So, A star A to the power minus 1 is equal to D is the identity. So, after applying the relation A star A to the power minus 1 we find this relation and after calculation we get A inverse A to the power minus 1 is equal to 4 by A. So, this is the inverse element. So, inverse of A is 4 by A which is belongs to q plus. So, now, we check the commutative property. So, for commutative we have this A star B is equal to B star A. So, A star B is equal to A B by 2. Similarly, B star A is equal to B by A by 2. So, this condition is satisfied. So, therefore, q plus with binary operation star is an abelian group. So, next we discuss example 3. Here, z is the zivita set of all 2 by 2 matrices with non-zero determinant. Then, so that z is an abelian group under matrix multiplication, but it is not an abelian group. So, here we are given that the zivita set of 2 by 2 matrices with non-zero determinant. Then, we have to prove that z is a group under matrix multiplication, but this is not an abelian group. So, first we check for closure property. So, A B belongs to z, any 2 matrices we consider from z, any 2 matrix we consider from z. So, your multiplication will be always belongs to z. So, your closure property is satisfied. Then, associative since matrix multiplication, we know that matrix multiplication is associative, though we have this condition is satisfied. Then, identity. So, identity is a matrix. So, we consider the identity, this identity matrix we consider. So, here we know that A, so this condition is always satisfied in matrix. So, therefore, this will be the identity element. So, now, we find the inverse element. So, inverse of matrix A is this, A to the power minus 1 is equal to 1 by determinant of A, the element of this matrix is D minus C A, this belongs to z. So, all conditions of group is satisfied. So, therefore, z is a group, but matrix multiplication is not commutative. So, this set with matrix multiplication is not an abelian group. Next, we move to the example 8. Here, example 8 is let G be denoted as the set of all matrix of the form. So, G is the set of all matrices of the form x, x, x, x, where x belongs to R, where x not is equal to 0. So, we have to show that G is a group under matrix multiplication. Your closure property. So, we consider any two matrix from the set G, where A is equal to this and B is equal to this. A matrix is x, x, x and B is y, y, y, y, 2 by 2 matrix. So, after multiplication, we will get ab is equal to 2 x y, 2 x y, 2 x y, 2 x y. So, this belongs to G. So, therefore, closure property is satisfied. Then identity. So, we consider E is the identity. So, A is equal to A is equal to EA. For existence of identity, we have these conditions. So, after simplifying this condition, we get 2 x y, 2 x is equal to x. So, identity element is 1 by x. So, for E 1 by x, 1 by x, 1 by x, 1 by x is the identity element of G. Next, we find inverse element. So, let this matrix be the inverse of this matrix. So, we apply the condition of existence of inverse, this AB means we have this condition AB is equal to I is equal to B. Then B is called inverse of A. So, applying this condition, we get this AB into B is equal to identity element. This is the identity matrix. So, after simplifying, we get this. After modulation of this 2 matrix, we get this. So, 2 x y is equal to 1 by 2. So, y is equal to this 1 by 4 x. So, inverse of this matrix is 1 by 4 x, 1 by 4 x, 1 by 4 x and 1 by 4 x. So, this belongs to G. So, all conditions of a group is satisfied by this set. So, hence G is a group. So, next we discuss example 9 and 10. So, here you have to show that G with this matrix is an abelian group under matrix modulation. So, Lana requested to solve this problem yourself. Similarly, example 10, the set of matrices, this set forms a group under matrix modulation. So, this example also you try yourself. So, now, we come to the end of today class. In today class, we have discussed what is a group and properties of group and we solve various type of problem. So, here we discuss a group. A group is an algebraic structure that is a non-empty set equip with a binary operation which satisfies certain postulates and properties of group also discussed in today class. So, next class we will discuss about subgroup, coset and normal subgroup. So, thank you.