Question

prove that the inverse of an element in a group is unique​

Answer 1

Answer:nverse element, that: ∀x∈G:∃x−1∈G:x∘x−1=e=x−1∘x. By Group has Latin Square Property, there exists exactly one a∈G such that a∘x=y.

Step-by-step explanation:

Answer 2

Note :

• Group : An algebraic system (G,*) is said to be a group if the following condition are satisfied :

1. G is closed under *
2. G is associative under *
3. G has a unique identity element
4. Every element of G has a unique inverse in G

• Moreover , if a group (G,*) also holds commutative property , then it is called commutative group or abelian group .

Solution :

To prove :

The inverse element of a group is unique .

Proof :

Let a be any arbitrary element of a group G and let e be the identity element in group G .

Let's assume that , b and c are two inverses of a .

Then , we have

ab = ba = e and ac = ca = e

Now ,

b(ac) = be = b

(ba)c = ec = c

But in a group , the associative property holds .

Hence , we have

b(ac) = (ba)c

→ b = c

Hence ,

The inverse of an element in a group is unique .

Hence proved .