Opposite Group is Group/Proof 2

Theorem

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

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

Then $\struct {G, *}$ is a group.

Proof

Let $e$ denote the identity of $\struct {G, \circ}$:

So: $e \in \struct {G, *}$

Hence:

$\struct {G, *}$ is non-empty.

$\Box$

As $\struct {G, \circ}$ is a group:

$\struct {G, \circ}$ is closed

every element of $\struct {G, \circ}$ has an inverse.

Therefore:

$\quad \forall a, b \in \struct {G, \circ}: b^{-1} \circ a \in \struct {G, \circ}$

Hence, by definition of $*$:

$\quad \forall a, b \in \struct {G, *}: a * b^{-1} \in \struct {G, *}$

$\Box$

The result follows from the One-Step Subgroup Test and Group is Subgroup of Itself.

$\blacksquare$