If G is a group and a, b are any two elements of G then i) (ab)-1 = b-1. a-1 ii) (a-1)-1 = a iii) If a, b, c are the elements of G then ab = bc ⇒ b = c and ba = ca ⇒ b = c

If G is a group and a, b are any two elements of G then

  1. (ab)-1 = b-1. a-1
  2. (a-1)-1 = a
  3. If a, b, c are the elements of G then ab = bc ⇒ b = c [left cancellation law] and ba = ca ⇒ b = c [right cancellation law]

Proof:

i) Suppose a and b are any two elements of a group G

If a-1  and b-1 are the inverse of a and b respectively then,

a.a-1 = a-1.a = e

b.b-1 = b-1.b = e, where e is the identity element of G.

Consider (ab).(b-1a-1) = a (b.b-1).a-1 [by associative law]

= a. e . a-1 [by inverse law]

= a . a-1 [by identity law]

= e [by inverse law]

Similarly, (b-1a-1)(ab) = b-1(a-1a)b [ by associative law]

= b-1(e)b [ by inverse law]

= b-1b [by identity law]

= e [ by inverse law]

 

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

If e is the identity element of a group G then we have a-1.a = e

Since a-1 ϵ G , (a-1)-1 ϵ G

We have  a-1. a = e ………………….(1)

Multiply left side by (a-1)-1  to equation (1)

(a-1)-1  [a-1. a] = (a-1)-1  . e

[(a-1)-1. a-1] . a = (a-1)-1  [ by associative law and identity law]

⇒ e . a = (a-1)-1  [by inverse law]

⇒ a = (a-1)-1  [by identity law]

⸫ (a-1)-1  = a

 

iii) Since a ϵ G, a-1 ϵ G [since G is a group]

Let ab = ac ……….(2)

Multiply left side of equation (2) by a-1

a-1. ab = a-1 . ac

(a-1. a) b = (a-1. a) c [by associative law]

e . b = e . c [by inverse law]

b = c [ by identity law]

Let ba = ca ………….(3)

Multiply right side of (3) by a-1

Then, ba.a-1 = ca. a-1

b(a.a-1) = c. (a.a-1) [by associative law]

  1. e = c . e [by inverse law]

b = c [by identity law]

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