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Elementary Properties of a Group

Elementary Properties of a Group

Theorem I: The identity element in a group (G,౦) is unique.

Proof:

Let e and e’ be two identity elements in a group (G,౦), if possible.

Then, 

When e is the identity element,

e ౦ e’ = e’ = e’ ౦ e

When e’ is the identity element,

e ౦ e’ = e = e’ ౦ e


∵ e ౦ e’ = e’ ౦ e

∴ e = e’

Hence, an identity element is unique.


Theorem II:

Each element in a group (G,౦) has a unique inverse.


Proof:

Let ‘a’ be an element of group (G,౦) with an identity element ‘e’.

Let b and c be two inverses of ‘a’, if possible.

Then,

a ౦ b = e = b ౦ a

a ౦ c = e = c ౦ a


Now,

c = c ౦ e                 [ Existence of identity element, c = c ౦ e  ] 

= c ౦ (a ౦ b)        [ Inverse Law, a ౦ b = e ] 

= (c ౦a) ౦ b         [ Associative Law ] 

= e ౦ b                 [ Inverse Law, c ౦ a = e ]

∴ c = b

Hence, this shows that each element in a group has a unique inverse.


Theorem III:

Cancellation Law:

If a, b, c are the elements of a group (G,౦) and if a ౦ b = a ౦ c, then b=c

Also if b౦a = c౦a, them b=c


Proof:

For a ∈ G, there exists an inverse element. 

As a ∈ (G,౦), a has an inverse element a-1 such that,

a ౦ a-1 = a-1 ౦ a = e     [here, e is the identity element]


We have,

a ౦ b = a ౦ c 

Operating both sides by a-1 on the left, we get,

a-1 ౦ (a ౦ b) = a-1 ౦ (a ౦ c) 

or,   (a-1 ౦ a) ౦ b = (a-1 ౦ a) ౦ c       [ Associative Law ]

or,   e ౦ b = e ౦ c                             [ Inverse Law, a-1 ౦ a = e ]

∴     b = c                                          [ Identity Law, e * b = b and e * c = c ]

 

Similarly, we have,

b ౦ a = c ౦ a

Operating both sides by a-1 on the right, we get,

(b ౦ a) ౦ a-1 = (c ౦ a) ౦ a-1

or,  b ౦ (a ౦ a=1) = c ౦ (a ౦ a-1)       [ Associative Law ]

or,   b ౦ e = c ౦ e                             [ Inverse Law,  a ౦ a-1 = e ]

∴     b = c                                          [ Identity Law, b ౦ e = b and c ౦ e = c ]    


Hence, proved.


Theorem IV:

If a, b ∈ (G, ౦), then

(i) (a ౦ b)-1 = b-1 ౦ a-1

(ii) (a-1)-1 = a


Proof:

(i) To prove: (a ౦ b)-1 = b-1 ౦ a-1

If the inverse of (a ౦ b) is equal to  (b-1 ౦ a-1), then (a ౦ b) ౦ (b-1 ౦ a-1) must be equal to  the identity element [ ∵ a ౦ a-1 = e ]. In this case, the identity element is e.

So, to prove: (a ౦ b) ౦ (b-1 ౦ a-1) = e

Taking, (a ౦ b) ౦ (b-1 ౦ a-1)

= a ౦ (b ౦ b-1) ౦ a-1                       [ Associative Law ]

= a ౦ e ౦ a-1                                    [ Inverse Law, b ౦ b-1 = e ]

= (a ౦ e) ౦ a-1  

= a ౦ a-1                                           [ Identity Law, a ౦ e = a ]

=e 

∴ (a ౦ b) ౦ (b-1 ౦ a-1) = e

∴ a ౦ b is the inverse of b-1 ౦ a-1

i.e. (a ౦ b)-1 = b-1 ౦ a-1


(ii) To prove: (a-1)-1 = a

We know that, a ౦ a-1 = e       [Inverse Law, since a-1  is the inverse of a]

or,   a-1 ౦ a = e

Operating both sides by (a-1)-1 on the right, we get,

or,   (a-1)-1 a-1 ౦ a = (a-1)-1 e

or,   {(a-1)-1 a-1} ౦ a = (a-1)-1 e

or,   e ౦ a = (a-1)-1 e         [ Inverse Law, (a-1)-1 a-1 = e ]

∴   a = (a-1)-1                       [ Identity Law, e ౦ a = a and (a-1)-1 e = (a-1)-1 ]                    


NEXT METHOD:

To prove: (a-1)-1 = a

We know that, a ౦ a-1 = e = a-1 ౦ a    [Inverse Law, since a-1  is the inverse of a]

Now, RHS = a

= a ౦ e                           [ Identity Law, a = a ౦ e ]

= a ౦  a-1 ౦ (a-1)-1            [ Inverse Law, a-1 ౦ (a-1)-1 = e ]

= {a ౦  a-1} ౦ (a-1)-1

= e ౦ (a-1)-1                     [ Inverse Law,  a ౦ a-1 = e ]

= (a-1)-1                               [ Identity Law, e ౦ (a-1)-1 = (a-1)-1 ]

= LHS

Hence, proved.


Theorem V:

If a and b are the elements of a group (G,౦), then

a ౦ x = b and x ౦ a = b 

have unique solutions in (G,౦)


Proof:

We have, a ౦ x = b

Operating both sides by a-1 on the left, we get,

(a-1 ౦ a) ౦ x = a-1 ౦ b

or,   e ౦ x = a-1 ౦ b                             [ Inverse Law, a-1 ౦ a = e ]

∴   x = a-1 ౦ b                                    [ Identity Law,  e ౦ x = x ]

This is the required solution.


To show uniqueness,

Let x1 and x2 be two solutions of a ౦ x = b, then

a ౦ x1 = b

a ౦ x2 = b


From results, a ౦ x1 = a ౦ x2 

Operating both sides by a-1 on the left, we get,

a-1 ౦ (a ౦ x1) = a-1 ౦ (a ౦ x2

or,   (a-1 ౦ a) ౦ x1 = (a-1 ౦ a) ౦ x2       [ Associative Law ]

or,   e ౦ x1 = e ౦ x2                             [ Inverse Law, a-1 ౦ a = e ]

∴     x1 = x2                                          [ Identity Law, x1 ౦ e = x1 and  x2 ౦ e =  x2 ]


NEXT METHOD:

For x ౦ a = b

Operating both sides by a-1 on the right, we get,

x ౦ (a ౦ a-1) = b ౦ a-1 

or,   x ౦ e = b ౦ a-1                               [ Inverse Law, a ౦ a-1 = e ]

∴   x = b ౦ a-1                                      [ Identity Law,  x ౦ e  = x ]

This is the required solution.


To show uniqueness,

Let x1 and x2 be two solutions of x ౦ a = b, then

x1 ౦ a = b

x2 ౦ a = b


From results, x1 ౦ a = x2 ౦ a

Operating both sides by a-1 on the right, we get,

(x1 ౦ a) ౦ a-1 = (x2 ౦ a) ౦ a-1

or,  x1 ౦ (a ౦ a=1) = x2 ౦ (a ౦ a-1)       [ Associative Law ]

or,   x1 ౦ e = x2 ౦ e                             [ Inverse Law,  a ౦ a-1 = e ]

∴     x1 = x2                                          [ Identity Law, x1 ౦ e = x1 and x2 ౦ e = x2 ]