Inversion Mapping is Isomorphism to Opposite Group

Definition

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

Let $\struct {G, *}$ be its opposite group.

That is:

$\forall g, h \in G: g \circ h = h * g$

Let $\iota: G \to G$ be the inversion mapping for $\struct {G, \circ}$.

Then $\iota: \struct {G, \circ} \to \struct {G, *}$ is a group isomorphism.

Proof

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

By Involution is Permutation, $\iota$ is a permutation and hence by definition a bijection.

It remains to show that $\iota$ is a group homomorphism.

Let $g, h \in G$.

Then:

\(\ds \map \iota {g \circ h}\)

\(=\)

\(\ds \paren {g \circ h}^{-1}\)

Definition of Inversion Mapping

\(\ds \)

\(=\)

\(\ds h^{-1} \circ g^{-1}\)

Inverse of Group Product

\(\ds \)

\(=\)

\(\ds g^{-1} * h^{-1}\)

Definition of $*$

\(\ds \)

\(=\)

\(\ds \map \iota g * \map \iota h\)

Hence $\iota$ is a group homomorphism.

By definition it follows that $\iota$ is also a group isomorphism.

$\blacksquare$