The inverse of the inverse in a group In one of my math exercises, I'm being asked to prove that 
for all $a, b \in G | (a^{-1})^{-1} = a$
with G a group. However, nowhere is stated that it is a commutative group. My first thought was that, because inverses are unique, the inverse of $a^{-1}$ can only be $a$; This thought is correct according to the answers. However, if the group is not commutative, is it not possible that there is another unique inverse for $a$?
The exact words of the answer are:
The result follow from Theorem 16.1(b) because both $(a^{-1})^{-1}$ and $a$ are inverses of $a^{-1}$.
Where Theorem 16.1(b) states The inverse of every element of a group G is unique.
 A: If we wanted the group axioms to say

For each $g\in G$ there exists a $g'\in G$ such that $gg'=1$

then that'd be all we need. Indeed for any $g\in G$, we have that $gg'=1$, $g'\in G$. Then we also have some other $g''$ for which $g'g''=1$. But then, by associativity
$$g(g'g'')=g$$
$$(gg')g''=g$$
$$1g''=g$$
$$g''=g$$
So left and right inverses (which exist axiomatically) coincide.
This said, $a^{-1}$ is just a convenient symbol for the element $g\in G$ for which $ag=ga=1$. Thus, $(a^{-1})^{-1}$ is a convenient symbol for the element $g'\in G$ for which $a^{-1}g'=g'a^{-1}=1$; which is just $a$.
A: How does your book define inverse? Usually it is defined as follows:

If $a$ is an element of your group, then $b$ is said to be an inverse if $ab=ba=1$. 

If $b$ and $c$ are both inverses of $a$, then $ba=ab=1$ and $ca=ac=1$. Then we can write:$$b=b\cdot 1=b(ac) = (ba)c=1\cdot c = c$$
So any two inverses of $a$ are equal.
A: The uniqueness of the inverse follows from associativity and it being both a left- and a right-inverse, i.e. $a^{-1}$ is a $b$ satisfying:
$$ab = ba = e$$
It now follows that for $b, b'$ inverses of $a$:
$$b = be = b(ab') = (ba)b' = eb' = b'$$
i.e. $b = b'$. On the other hand, a right- or left-inverse (only required to satisfy $ab = e$, $ba = e$, respectively) may not be unique. If you would like some examples, tell me and I'll dig some up.
