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

**(ab)**^{-1}= b^{-1}. a^{-1}**(a**^{-1})^{-1}= a**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^{-1}a^{-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^{-1}a^{-1})(ab) = b^{-1}(a^{-1}a)b [ by associative law]

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

= b^{-1}b [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]

- e = c . e [by inverse law]

b = c [by identity law]

## One thought on “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”

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