ElementaryGroupTheory

FirstTheorems

Given a Group (G,*) defined as:

G in a NonEmpty SeT and “*” is a BinaryOperation, 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 c belong to (G,*), then (a*b)*c)=(a*(b*c).

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

*4). Every element in (G,*) has an inverse, that is, of a belongs to (G,*), there is an element b in (G,*) such that a*b=b*a=e.

First Theorem:

The identity element of a GrouP (G,*) is unique.

*Proof:

• Suppose there were two elements, e and e’ (G,*), such that for all a in (G,*)

a*e=e*a=e and a*e’=e’*a=a.

• Then e*e’=e’ by the definition of identity element (3).
• And e*e’=e, also by the definition of the identity element (3).
• Then e=e’.
• The identity element in a GrouP (G,*) is unique.

Second Theorem:

Given a group (G,*), and an element x in (G,*), there is only one element y such that y*x=x*y=e. ( The inverse of each element in (G,*) is unique.)

*Proof

• Suppose there are two elements, y and y’ in (G,*) such that for x, an element of (G,*), y*x=x*y=e and y’*x=x*y’=e.
• Then y’=y’*e, by definition of identity element (3).
• Then y’=y’*(x*y), since x*y=e.
• Then y’=(y’*x)*y, by the associativity property of (G,*) (2).
• Then y’=e*y, since y’*x=e
• Then y’=y, since e is the identity element (3).
• Therefore, the inverse of an element x in a GrouP, (G,*) is unique.

Notice the method of proof, which is the same for both theorems and quite common in

MathematicS. It is called, the Indirect Method of Proof.

• First, one assumes that the proposition one is trying to prove if false.
• Then, one tries to get a contradiction.
• If this is successful, then the assumption that the proposition is false, is, itself, false. Hence, the proposition is true.