Theorem-Group is unique and has unique inverse

Question

Let (G; o) be a group. Then the identity of the group is unique and each element of the group has a unique inverse.

In this proof, we will argue completely formally, including all the parentheses and all the occurrences of the group operation o. As we proceed with later work, we will very soon relax our level of formality, omitting avoidable parentheses and uses of the operation symbol.

Answer 1

Proof:

First, we prove uniqueness of the identity. Suppose that e; e' ≡ G both have the property stated in the axiom for the identity; that is,

g o e = e o g = g and g o e' = e' o g = g

for all g ≡ G. For uniqueness, we need to prove that e = e'.

Applying the First equation above to g = e' and the second to g = e, we get

e'o e = e o e' = e' and e o e' = e'o e = e:

Comparing these gives e = e', as required.

Second, we prove that each element of G has a unique inverse. Suppose that for a fixed g ≡ G there are elements h and k which both have the property required of an inverse; that is,

g o h = h o g = e and g o k = k o g = e:

We need to prove that h = k.

Multiplying through the equation k o g = e on the right by h gives

(k o g) o h = e o h;

associativity gives

k o (g o h) = e o h;

and then since g o h = e we have

k o e = e o h:

Using the fact that e is the identity, we nally

get k = h, as required.