Product Inverse Operation Properties induce Group

Theorem
Let $\struct {G, \oplus}$ be a closed algebraic structure on which the following properties hold:

Let $\circ$ be the operation on $G$ defined as:
 * $\forall x, y \in G: x \circ y = x \oplus \paren {e \oplus y}$

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

Proof
Taking the group axioms in turn:

$\struct {G, \oplus}$ is closed by definition.

Hence $\struct {G, \circ}$ is likewise closed.

and:

Thus $e$ is the identity element of $\struct {G, \circ}$.

We have that $e$ is the identity element of $\struct {G, \circ}$.

Hence:

and:

Thus every element $x$ of $\struct {G, \circ}$ has inverse $e \oplus x$.

We have:

Let $x \oplus y = e$.

Then we have:

Then:

Hence we have:
 * $x = y$

That is:


 * $(3): \quad x \oplus y = e \implies x = y$

Let $x \oplus z = y \oplus z$.

Then we have:

Then:

Thus we have:
 * $x \oplus y = e$

as both are equal to $\paren {x \oplus z} \oplus \paren {y \oplus z}$.

But then from $(3)$:
 * $x = y$

It follows that:


 * $(4): \quad x \oplus z = y \oplus z \implies x = y$

Then we have:

Thus $\circ$ is associative.

All the group axioms are thus seen to be fulfilled, and so $\struct {G, \circ}$ is a group.