Group Product Identity therefore Inverses

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

Let $x, y \in \struct {G, \circ}$.

Then if either $x \circ y = e$ or $y \circ x = e$, it follows that $x = y^{-1}$ and $y = x^{-1}$.

Proof
From the Division Laws for Groups:
 * $x \circ y = e \implies x = e \circ y^{-1} = y^{-1}$

Also by the Division Laws for Groups:
 * $x \circ y = e \implies y = x^{-1} \circ e = x^{-1}$

The same results are obtained by exchanging $x$ and $y$ in the above.