Group has Latin Square Property

Theorem
For any elements $$a$$ and $$b$$ in a group $$G$$, there exists a unique $$g \in G$$ such that $$ag=b$$.

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

Proof
$$ $$ $$ $$ $$

Thus, such a $$g$$ exists.

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