Inversion Mapping is Permutation

Theorem

Let $\struct {G, \circ}$ be a group.

Let $\iota: G \to G$ be the inversion mapping on $G$.

Then $\iota$ is a permutation on $G$.

Proof 1

The inversion mapping on $G$ is the mapping $\iota: G \to G$ defined by:

$\forall g \in G: \map \iota g = g^{-1}$

where $g^{-1}$ is the inverse or $g$.

By Inversion Mapping is Involution, $\iota$ is an involution:

$\forall g \in G: \map \iota {\map \iota g} = g$

The result follows from Involution is Permutation.

$\blacksquare$

Proof 2

Proof of Surjection

Let $a \in G$.

By definition of $\iota$:

$\iota(a^{-1}) = \left({a^{-1}}\right)^{-1}$
$\left({a^{-1}}\right)^{-1} = a$

Hence $a$ has a preimage.

Since $a$ was arbitrary, $\iota$ is a surjection.

Proof of Injection

Suppose for some $a, b \in G$ that:

$\iota \left({a}\right) = \iota \left({b}\right)$

Then by the definition of $\iota$:

$a^{-1} = b^{-1}$

It follows from Inverse in Group is Unique that:

$a = b$

Hence $\iota$ is an injection.

$\Box$

Hence by definition $\iota$ is a bijection.

A bijection from a set to itself is by definition a permutation.

$\blacksquare$