Group theory - prove that $\forall x((x^{-1})^{-1}=x)$ so I got this question for homework:
Prove that this property can be deduced from group theory:
The inverse of an inverse is the identity: $\forall x((x^{-1})^{-1}=x)$
I tried building this statement from the group's axioms but was having difficulties. 
Some help/hints?
Thanks
 A: With questions as basic as this, you really need to give your definition of terms like 'group.'
My preferred definition is:

Definition. A group is a set $G$ together with:
  
  
*
  
*an element of $G$ denoted $1$
  
*a function $G \times G \rightarrow G$ denoted $x,y \mapsto xy$.
  
*a function $G \rightarrow G$ denoted $x \mapsto x^{-1}$
  
  
  such that the following identities hold.
  
  
*
  
*$(xy)z =x(yz)$
  
*$x1=x$
  
*$1x = x$
  
*$xx^{-1} = 1$
  
*$x^{-1}x = 1$
  

The following result is easy to prove, but extremely useful all throughout group theory:

Proposition 0. Let $G$ denote a group. Then for all $x,y \in G$, we have: $$xy = 1 \rightarrow y = x^{-1}$$

Proof. Since $xy=1$, hence $x^{-1}xy= x^{-1}$, hence $y = x^{-1}$.
Let us now turn out attention to your problem:

Proposition 1. Let $G$ denote a group. Then for all $y \in G$, we have: $$y = (y^{-1})^{-1}$$

Proof. Taking $x$ to be $y^{-1}$ in Proposition 0, we obtain: $$y^{-1}y = 1 \rightarrow y = (y^{-1})^{-1}.$$ Thus $$1 = 1 \rightarrow y = (y^{-1})^{-1}.$$
So $y = (y^{-1})^{-1}$.
A: The inverse of $x$ is unique, because suppose that $a^{-1}$ and $b^{-1}$ are inverse to $x$ then
$$
a^{-1} = a^{-1}\cdot e = a^{-1} \cdot x\cdot b^{-1} = e \cdot b^{-1} = b^{-1}.
$$
If $x^{-1}$ the inverse of $x$, your proposition follows by the uniqueness of the inverse.
A: From the definition of inverses, $ x^{-1}x = xx^{-1} = e $, we can obtain, $$ \begin{align} ({x^{-1}})^{-1} x^{-1} &= e \\({x^{-1}})^{-1} x^{-1} x &= ex \\({x^{-1}})^{-1} &= x \end{align} $$.
