# Textbook proof for uniqueness of inverse in a group

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?

• Ah, thank you! That made me recheck the definitions and this indeed is the case. :) – dotslash Apr 4 '14 at 16:46

## 2 Answers

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.$aa^{−1}=e=>a=e∗a$ and $a^{−1}a=e=>a=a∗e$

• I don't understand the last two relations, though. How does $a*a^{-1}=e$ imply that $a = e*a$?, etc. – dotslash Apr 4 '14 at 17:02
• Start with $a*a^{-1} = e$, multiply both sides on the right by $a$, and then associate to get $a*e = e*a$. – user7530 Apr 4 '14 at 17:40
• Btw, you can use \implies to get $\implies$. – Ben West Apr 4 '14 at 18:17
• @BenWest Oh, cool! Thanks a lot for the tip. :) I always used \Rightarrow – dotslash Apr 5 '14 at 3:25

If you have already proven that the identity is unique in a group, then you can use $$e*a=a*e=a$$ for all $a\in G$.