Wikipedia 10K Redux by Reagle from Starling archive. Bugs abound!!!
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.