MathematicalGrouP

The concept of a GrouP is one of the foundations of ModernAlgebra.

The definition of a GrouP is brief.

A GrouP is a non-empty set, say G and a binary operation, say, "*" denoted (G,*) such that:

1) G has closure, that is, if a and b belong to G, then a*b belongs to G.

2) The operation * is associative, that is, if a, b, and b belong to G, (a*b)*c=a*

(b*c).

3) G contains an identity element, say i, that is, if a belongs to G then i*a = a and a*i = a.

4) Every element in G has an inverse, that is, if a belongs to G, there is an

element b in G such that a*b=i.

A GrouP that we are introduced to in elementary school is the integers under addition. Thus, let I be the set of

integers={?-4,-3,-2,-1,0,1,2,3,4?} and let the symbol "+" indicate the operation of addition. Then, (I,+) is a GrouP.

Proof:

1) If a and b are integers then a+b is an integer: Closure.

2) If a, b, and b are integers, then (a+b)+c=a+(b+c). Associativity.

3) 0 is an integer and for any integer a, a+0=a.

(I,+) has an identity element.

4) If a is an integer, then there is an integer b= (-a), such that a+b=0.

Every element of (I,+) has an inverse.

Question: Given the set of integers, I, as above, and the operation multiplication,

denoted by "x" is (I,x) a GrouP?

1) If a and b are integers then axb is an integer. Closure.

2) If a and b are integers, then (axb)xc=ax(bxc). Associativity.

3) 1 is an integer and for any integer a, ax1=a.

(I,x) has an identity element.

4) BUT, if a is an integer, there is not necessarily an integer b =1/a such that

(a)x(1/a)=1.

Then, every element of (I,x) does not have an inverse.

For example, given the integer 4, there is no integer b such that 4xb=1.

Therefore, (I,x) is not a GrouP.

Question: Given the set of rational numbers Z, that is the set of number a/b such that

a and b are integers, but b is not = to 0,

and the operation multiplication, denoted by "x," is (Z,x) a GrouP?