College Algebra : Groups

Question

For each of the following, let the operation ∗ be defined on Z by the given rule.

Determine in each case whether Z is a group with respect to ∗ and whether it is an abelian group. State which, if any, conditions fail to hold.

(a) x ∗ y = x + y + 1

(b) x ∗ y = x − y

(c) x ∗ y = x + xy + y

My thoughts :

In the question 'Z' in here is given by the usual notation of the set of integers(Appears hear as Z due to format changes occurs in the site)

So in this type of a question,

1) can we assume Z denote the set of integers because it's given in the usual notation? And hence can we apply the properties of real numbers to prove that 'Z' here is a group, since the set of integers is a subset of the set of integers?

2) Can we assume the '+' given here denotes the usual operation of addition?

Answer 1

You can assume you are working with integers. Note that in order to check these definitions, you need to know how to add, subtract, and multiply the group elements (x,y). If the set Z in this problem were not the integers, they would need to tell us how to add, subtract, and multiply elements.

Let's walk through part a). To see if the operation * defined in part a) together with Z is a group, we need to verify four axioms.

1) Closure. If x,y in Z, is it true that x*y = x+y+1 in Z? Yes, because integers are closed under addition.

2) associativity. If x,y,z in Z, is it true that x*(y*z) = (x*y)*z? Well,

x*(y*z) = x*(y+z+1) = x+ (y+z+1)+1 and

(x*y)*z = (x+y+1)*z = (x+y+1)+z+1. Due to commutativity of addition, this too is satisfied.

3) identity. Is there an integer e such that x*e = e*x = x? Well, notice that

x*(-1) = x+(-1)+1 = x. So, yes. The identity element is -1.

4) inverse. Is there an integer x' such that x*x' = -1? Well, notice that

x*(-x-2) = x+(-x-2)+1 = -2+1 = -1. So, yes, and part a would be a group.

Is a) Abelian? We need to check commutativity: is it true that x*y = y*x? Yes. Again, this is due to commutativity of addition.