THEOREM

*******

 

Most presentations of the axioms of group theory seem to give both right and left identities and inverses, which turns out to be somewhat redundant. Here, given the axioms for group (g,*) with only right identity and inverse, we formally prove that the right identity and inverses also a left identity and inverse respectively.

 

 

Dan Christensen

http://www.dcproof.com

2023-07-05

2023-07-09

 

 

GROUP (g,*) AXIOMS

******************

 

1     Set(g)

      Axiom

 

G1: * is closed on g

 

2     ALL(a):ALL(b):[a e g & b e g => a*b e g]

      Axiom

 

G2: * is associative

 

3     ALL(a):ALL(b):ALL(c):[a e g & b e g & c e g => a*b*c=a*(b*c)]

      Axiom

 

G3: e is a right identity

 

4     e e g

      Axiom

 

5     ALL(a):[a e g => a*e=a]

      Axiom

 

G4: Right inverses

 

6     ALL(a):[a e g => EXIST(b):[b e g & a*b=e]]

      Axiom

 

   

    Lemma

   

    Prove: ALL(a):[a e g => [a*a=a => a=e]]

   

    Suppose...

 

      7     x e g

            Premise

 

         Prove: x*x=x => x=e

        

         Suppose...

 

            8     x*x=x

                  Premise

 

            9     x e g => x*e=x

                  U Spec, 5

 

            10   x*e=x

                  Detach, 9, 7

 

            11   x=x*e

                  Sym, 10

 

            12   x e g => EXIST(b):[b e g & x*b=e]

                  U Spec, 6

 

            13   EXIST(b):[b e g & x*b=e]

                  Detach, 12, 7

 

         Define: x'  (inverse of x)

 

            14   x' e g & x*x'=e

                  E Spec, 13

 

            15   x' e g

                  Split, 14

 

            16   x*x'=e

                  Split, 14

 

            17   x=x*(x*x')

                  Substitute, 16, 11

 

         Apply associativity to obtain x*x*x' on RHS

 

            18   ALL(b):ALL(c):[x e g & b e g & c e g => x*b*c=x*(b*c)]

                  U Spec, 3

 

            19   ALL(c):[x e g & x e g & c e g => x*x*c=x*(x*c)]

                  U Spec, 18

 

            20   x e g & x e g & x' e g => x*x*x'=x*(x*x')

                  U Spec, 19

 

            21   x e g & x e g

                  Join, 7, 7

 

            22   x e g & x e g & x' e g

                  Join, 21, 15

 

            23   x*x*x'=x*(x*x')

                  Detach, 20, 22

 

            24   x=x*x*x'

                  Substitute, 23, 17

 

            25   x=x*x'

                  Substitute, 8, 24

 

         As Required:

 

            26   x=e

                  Substitute, 16, 25

 

    As Required:

 

      27   x*x=x => x=e

            4 Conclusion, 8

 

Lemma

 

As Required:

 

28   ALL(a):[a e g => [a*a=a => a=e]]

      4 Conclusion, 7

 

   

    Prove: ALL(a):ALL(b):[a e g & b e g => [a*b=e => b*a=e]]

   

    Suppose...

 

      29   x e g & y e g

            Premise

 

      30   x e g

            Split, 29

 

      31   y e g

            Split, 29

 

        

        Prove: x*y=e => y*x=e

        

         Suppose...

 

            32   x*y=e

                  Premise

 

            33   y e g => y*e=y

                  U Spec, 5

 

            34   y*e=y

                  Detach, 33, 31

 

            35   y*(x*y)=y

                  Substitute, 32, 34

 

         * both sides by x

 

            36   ALL(b):[y e g & b e g => y*b e g]

                  U Spec, 2

 

            37   y e g & x e g => y*x e g

                  U Spec, 36

 

            38   y e g & x e g

                  Join, 31, 30

 

            39   y*x e g

                  Detach, 37, 38

 

            40   ALL(b):[y e g & b e g => y*b e g]

                  U Spec, 2

 

            41   y e g & x e g => y*x e g

                  U Spec, 40

 

            42   y e g & x e g

                  Join, 31, 30

 

            43   y*x e g

                  Detach, 41, 42

 

            44   y*x=y*x

                  Reflex, 43

 

            45   y*(x*y)*x=y*x

                  Substitute, 35, 44

 

         Apply associativity to obtain y*x*y*x on LHS

 

            46   ALL(b):ALL(c):[y e g & b e g & c e g => y*b*c=y*(b*c)]

                  U Spec, 3

 

            47   ALL(c):[y e g & x e g & c e g => y*x*c=y*(x*c)]

                  U Spec, 46

 

            48   y e g & x e g & y e g => y*x*y=y*(x*y)

                  U Spec, 47

 

            49   y e g & x e g

                  Join, 31, 30

 

            50   y e g & x e g & y e g

                  Join, 49, 31

 

            51   y*x*y=y*(x*y)

                  Detach, 48, 50

 

         As Required:

 

            52   y*x*y*x=y*x

                  Substitute, 51, 45

 

         Apply associativity to obtain y*x*(y*x) on LHS

 

            53   ALL(b):ALL(c):[y*x e g & b e g & c e g => y*x*b*c=y*x*(b*c)]

                  U Spec, 3, 43

 

            54   ALL(c):[y*x e g & y e g & c e g => y*x*y*c=y*x*(y*c)]

                  U Spec, 53

 

            55   y*x e g & y e g & x e g => y*x*y*x=y*x*(y*x)

                  U Spec, 54

 

            56   y*x e g & y e g

                  Join, 39, 31

 

            57   y*x e g & y e g & x e g

                  Join, 56, 30

 

            58   y*x*y*x=y*x*(y*x)

                  Detach, 55, 57

 

         As Required:

 

            59   y*x*(y*x)=y*x

                  Substitute, 58, 52

 

         Apply lemma

 

            60   y*x e g => [y*x*(y*x)=y*x => y*x=e]

                  U Spec, 28, 43

 

            61   y*x*(y*x)=y*x => y*x=e

                  Detach, 60, 39

 

            62   y*x=e

                  Detach, 61, 59

 

    As Required:

 

      63   x*y=e => y*x=e

            4 Conclusion, 32

 

As Required:

 

64   ALL(a):ALL(b):[a e g & b e g => [a*b=e => b*a=e]]

      4 Conclusion, 29

 

   

    Prove: ALL(a):[a e g => e*a=a]

   

    Suppose...

 

      65   x e g

            Premise

 

      66   x e g => x*e=x

            U Spec, 5

 

      67   x*e=x

            Detach, 66, 65

 

      68   x e g => EXIST(b):[b e g & x*b=e]

            U Spec, 6

 

      69   EXIST(b):[b e g & x*b=e]

            Detach, 68, 65

 

    Define: x'  (right inverse of x)

 

      70   x' e g & x*x'=e

            E Spec, 69

 

      71   x' e g

            Split, 70

 

      72   x*x'=e

            Split, 70

 

      73   ALL(b):[e e g & b e g => e*b e g]

            U Spec, 2

 

      74   e e g & x e g => e*x e g

            U Spec, 73

 

      75   e e g & x e g

            Join, 4, 65

 

      76   e*x e g

            Detach, 74, 75

 

      77   e*x=e*x

            Reflex, 76

 

      78   e*x=x*x'*x

            Substitute, 72, 77

 

      79   e*x=x*x'*(x*e)

            Substitute, 67, 78

 

    Apply associativity to obtain x*x'*x*e on RHS

 

      80   x*x' e g

            Substitute, 72, 4

 

      81   ALL(b):ALL(c):[x*x' e g & b e g & c e g => x*x'*b*c=x*x'*(b*c)]

            U Spec, 3, 80

 

      82   ALL(c):[x*x' e g & x e g & c e g => x*x'*x*c=x*x'*(x*c)]

            U Spec, 81

 

      83   x*x' e g & x e g & e e g => x*x'*x*e=x*x'*(x*e)

            U Spec, 82

 

      84   x*x' e g & x e g

            Join, 80, 65

 

      85   x*x' e g & x e g & e e g

            Join, 84, 4

 

      86   x*x'*x*e=x*x'*(x*e)

            Detach, 83, 85

 

    As Required:

 

      87   e*x=x*x'*x*e

            Substitute, 86, 79

 

    Apply associativity to obtain x*(x'*x)*e on RHS

 

      88   ALL(b):ALL(c):[x e g & b e g & c e g => x*b*c=x*(b*c)]

            U Spec, 3

 

      89   ALL(c):[x e g & x' e g & c e g => x*x'*c=x*(x'*c)]

            U Spec, 88

 

      90   x e g & x' e g & x e g => x*x'*x=x*(x'*x)

            U Spec, 89

 

      91   x e g & x' e g

            Join, 65, 71

 

      92   x e g & x' e g & x e g

            Join, 91, 65

 

      93   x*x'*x=x*(x'*x)

            Detach, 90, 92

 

    As Required:

 

      94   e*x=x*(x'*x)*e

            Substitute, 93, 87

 

    Apply previous result

 

      95   ALL(b):[x e g & b e g => [x*b=e => b*x=e]]

            U Spec, 64

 

      96   x e g & x' e g => [x*x'=e => x'*x=e]

      97   x e g & x' e g

 

            U Spec, 95

            Join, 65, 71

 

      98   x*x'=e => x'*x=e

            Detach, 96, 97

 

      99   x'*x=e

            Detach, 98, 72

 

      100  e*x=x*e*e

            Substitute, 99, 94

 

      101  e*x=x*e

            Substitute, 67, 100

 

As Required:

 

102  ALL(a):[a e g => e*a=a*e]

      4 Conclusion, 65

 

 

As Required:

 

103  ALL(a):ALL(b):[a e g & b e g => [a*b=e => b*a=e]]

    & ALL(a):[a e g => e*a=a*e]

      Join, 64, 102