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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.

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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 $a$.

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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.

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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.

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  • $\begingroup$ Ah, of course! I forgot that the way an inverse is declared already states commutativity of the inverse. Now it makes sense :) $\endgroup$ – Erik S Apr 1 '13 at 16:10
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    $\begingroup$ If every element of the group has a right inverse, then the right inverses are also left inverses: Suppose $ab = bc = e$. Then $e = b(ab)c = (ba)bc = ba$. $\endgroup$ – Pedro M. Apr 1 '13 at 16:14
  • $\begingroup$ @Pedro Absolutely correct, thanks for the comment. $\endgroup$ – Lord_Farin Apr 1 '13 at 16:23

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