FirstTheorems
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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, among other things, the Indirect Method of Proof and Proof by Contradiction.
 First, one assumes that the proposition one is trying to prove is 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.
Four More Elementary Group Theorems
I. For all a. b belonging to a GrouP (G,*), if a*b=e, then a=b^1 and b=a^1.
 Proof
 Part A.
 Let a*b=e.
 Then (a*b)*b^1=e*b^1.
 Then a*(b*b^1)= b^1.
 Then a*e= b^1.
 Therefore, a=b^1.

 Part B.
 Let a*b=e.
 Then a^1*(a*b)= a^1*e.
 Then (a^1*a)*b=a^1.
 Then e*b=a^1.
 Therefore, b=a^1

II. For all a,b belonging to a
GrouP (G,*), (a*b)^1=b^1*a^1.
 Proof
 This theorem says that the inverse of a*b= b^1*a^1.
 So we have to prove that (a*b)*(b^1*a^1)= e.
 Then (a*b)*(b^1*a^1)=(a*(b*(*b^1*a^1))= a*((b*b^1)*a^1).
 Then (a*b)*(b^1*a^1)= a*(e*a^1)= a*a^1=e.
 Therefore, by Theorem I, b^1*a^1 is the inverse of a*b.

III. For all a belonging to a
GrouP (G,*), (a^1)^1=a.
 This theorem states that the inverse of the inverse of an element in a Group, is itself.
 Proof
 Since a*a^1=e, by Theorem I, the inverse of a^1=a.
 Therefore, (a^1)^1=a.

IV. For all a,x,y, belonging to
GrouP (G,*), if a*x=a*y, then x=y, and if x*a=y*a, then x=y.
 Proof
 Part A.
 If a*x=a*y, then a^1*(a*x)=a^1*(a*y).
 Then (a^1*a)*x= (a^1*a)*y.
 Then e*x=e*y.
 Therefore, x=y.

 Part B.
 If x*a=y*a, then (x*a)*a^1=(y*a)*a^1.
 Then x*(a*a^1)= y*(a*a^1).
 Then x*e=y*e.
 Therefore, x=y.
 The results of Theorem IV are often called the cancellation rules for a Group.