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.
$\blacksquare$
Sources
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): $\S 3.3$: Group actions and coset decompositions