A Textbook of Abstract Algebra by Pinter gives the following proof of the property that an element in a group can have only one inverse (consider $*$ to be the operation and $e$ to be the identity element):
Suppose an element $a$ has two inverses, $a_1$ and $a_2$.
Then, $a_1*(a*a_2) = a_1*e = a_1$, and
($a_1*a)*a_2 = e*a_2 = a_2$
The book then says by associativity, the two-left hand sides are equal, and hence so are $a_1$ and $a_2$.
I have an objection to this proof. The identity element is defined as $a*e=a$, and since commutativity is not necessary in a group, the argument that $e*a_2 = a_2$ doesn't look watertight to me.
Any comments?
Best Answer
Irrespective of whether the group is commutative or not, $aa^{−1}=a^{−1}a=e$ and that's why the relation which you mentioned commutes.
$$aa^{−1}=e \implies a=e\cdot a $$ $$\text{ and }$$ $$ a^{−1}a=e \implies a=a\cdot e$$