Group has Latin Square Property

Theorem
Let $\left({G, \circ}\right)$ be a group.

Then $G$ satisfies the Latin square property.

That is, for all $a, b \in G$, there exists a unique $g \in G$ such that $a \circ g = b$.

Similarly, there exists a unique $h \in G$ such that $h \circ a = b$.

Proof
Thus, such a $g$ exists.

Suppose $x \in G$ where $a x = b$.

Then:

Thus, $x$ is uniquely of the form $a^{-1} b$.

To prove the second part of the theorem, let $h = b a^{-1}$.

The remainder of the proof follows a similar procedure to the above.