Is the set \mathbb{Z} a group under the following operations:a.) a*b = a + b - 1b.) a*b = a - b + ab

Accepted Solution

Answer:a) yesb) noStep-by-step explanation:[tex](\mathbb{Z}, *)[/tex] is a gruop if satisfies the following conditions:1. If a and b are two elements in [tex]\mathbb{Z}[/tex], then the product a*b is also in [tex]\mathbb{Z}[/tex].2. The defined multiplication is associative, i.e., for all a,b,c in [tex]\mathbb{Z}[/tex], (a*b)*c=a*(b*c).3. There is an identity element e such that e*a=a*e=a for every element a in [tex]\mathbb{Z}[/tex].4. There must be an inverse of each element. Therefore, for each element a of [tex]\mathbb{Z}[/tex], the set contains an element b=a^(-1) such that a*a^(-1)=a^(-1)*a=e.Let's see if the conditions are satisfied:a) 1. if x and y are integers then x+y-1=a*y is an integer2. If x,y and z are integers then    (x*y)*z= (x+y-1)*z= (x+y-1) + z - 1= x +y+z-2,    x*(y*z)= x*(y+z-1)= x + (y+z-1) -1 = x+ y + z -2   Then (x*y)*z=x*(y*z), i.e, * is associative.3. Let e=1 and b an integer. Observe that     1*b=1+b-1=b and b*1= b + 1 -1= b.   Then e is an identity element.4. a and integer and b= 2- a. Observe that     b*a= 2-a+a-1= 1 and a*b= a+2-a-1=1, the b= a^(-1) is the inverse of a.We conclude that [tex](\mathbb{Z}, *)[/tex] is a group.b) 1. If x,y and z are integers then    (x*y)*z= (x-y+xy)*z= (x-y+xy) - z + (x-y+xy)z= x -y-z+xy+xz-yz+xyz    x*(y*z)= x*(y-z+yz)= x - (y-z+yz) +x(y-z+yz) = x-y +z + xy -xz -yz+xyz   Then (x*y)*z≠x*(y*z), i.e, * isn't associative.We conclude that [tex](\mathbb{Z}, *)[/tex] isn't a group.