Question

Need help checking closure, associativity, Identity and Inverse

1. Let the binary operation on Z be defined by ** Y = (x + 1)(y+1). Determine whether or not Z is a group with respect to *, and justify your answer.

Answer 1

x*y = (x+1)(y+1) on Z

if x and y are in Z then clearly x+1 and y+1 are in Z and so is their product (since Z is closed under multiplication)

hence the closure property holds in Z

clearly * is associative on Z, since, (x+1){(y+1)(z+1)} = {(x+1)(y+1)} (z+1)

hence, x*(y*z) = (x*y)*z

hence * is associative on Z

clearly * is commutative on Z since, x*y=y* x, since product is commutative in Z

now, for y in Z, we need y for all x in Z such that x*y=x that is, (x+1)(y+1)=x for all x in Z

that is, xy+x+y+1=x

i.e. xy+y+1=0 but here the solution of y depends on x

hence there is no y in Z such that y is the identity element of Z w.r.t the operation *

hence Z,* does not form a group