# Elementary group theory

Given a group (G,*) defined as:

G is a set and * is a binary operation on 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 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, if 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, e1 and e2 in (G,*), such that for all a in (G,*) i.e.
• a*e1=e1*a=a. (proposition P1)
• a*e2=e2*a=a. (proposition P2)
• Then e1*e2=e2 by the definition of identity element (3) (P1).
• And e1*e2=e1, also by the definition of the identity element (3) (P2).
• Then e1=e2 .
• 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.

Since the inverse of the element x is uniquely defined by x, we may denote it by x^-1.

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 is 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 a 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.