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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

FigureItOut
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    What is your difficulty? If $x^{-1}$ is the inverse of $x$ then (directly from the axioms) we have $xx^{-1}=1=x^{-1}x$. But that is also precisely the requirement that $x$ be the inverse of $x^{-1}$. – almagest Jun 18 '16 at 16:07

3 Answers3

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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 $y = (y^{-1})^{-1}$.

goblin GONE
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  • @MorganRodgers: There's nothing in this definition that implies that we need to have an explicit formula for the inverse. It just say that a function taking each element to an inverse must exist (which is completely true), not that this function needs to be given by an explicit formula. As for the names $1$ and (juxtaposition), how would you propose to define a group without using any name for its components? No matter which symbol you use to write down the group axioms, the same criticism could be made. – hmakholm left over Monica Jul 18 '16 at 11:04
  • @MorganRodgers: There is no "full stop" in this definition. It says "an element denoted $1$ such that (properties)". – hmakholm left over Monica Jul 18 '16 at 13:58
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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.

Thyrð Åkyrïkr
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From the definition of inverses, $ x^{-1}x = xx^{-1} = e $, we can obtain, $$ \begin{align} ({x^{-1}})^{-1} x^{-1} &= e \\({x^{-1}})^{-1} x^{-1} x &= ex \\({x^{-1}})^{-1} &= x \end{align} $$.