Why do we need to invoke the cancellation laws to prove uniqueness of solution of ax = b in group theory? Quoting Introduction to Analytic Number Theory by Tom M. Apostol:

Theorem 6.1 Cancellationi laws. If elements $a$, $b$, $c$ in $G$ satisfy
$$ ac = bc $$ or $$ ca = cb $$,
then $ a = b $.


Theorem 6.2 (d) For all $a$ and $b$ in $G$ thee equation $ax = b$ has the unique solution $ x = a^{-1}b$; the equation $ya = b$ has the unique solution $y = ba^{-1}$.

Quoting the proof of Theorem 6.2 (d) now:

Again by associativity we have
$$ a(a^{-1}b) = (aa^{-1})b = b\hspace{1.5em} \text{and}\hspace{1.5em} (ba^{-1})a = b(a^{-1}a) = b. $$
The solultions are unique because of the cancellation laws.

Why do we have to invoke the cancellation laws the prove that the solutions are unique?
In other words, which step of the proof will not work if we do not invoke the cancellation laws?
 A: The calculation $a(a^{-1}b)=(aa^{-1})b=b$ shows that the equation $ax=b$ has at least one solution, namely, $x=a^{-1}b$; it does not show that the equation has no other solution, so it does not show that the solution $x=a^{-1}b$ is unique. The cancellation law, however, says that if $c$ and $d$ are both solutions to $ax=b$, so that $ac=b=ad$, then $c=d$. In other words, the cancellation law shows that if there is any solution at all to $ax=b$, that solution is unique. It takes both parts to show both that there is at least one solution and that there is at most one solution, i.e., that there is a unique solution.
A: In fact we don't need to use any additional argument using cancellation to prove uniqueness. Rather it follows immediately from the obvious existence proof, i.e. $\,ax = b\!\iff\! x = a^{-1}b,\,$ via scaling by $\,a^{\pm1}.\,$ Thus if $x_1,x_2\,$ are both roots then $\,x_1 = a^{-1}b = x_2,\,$ so $\,x_1 = x_2\,$
Remark $ $ This point is often misunderstood.  See here and here for much further discussion.
