How to prove that $x = a^{-1} b$ is unique? ($a^{-1}$ is already proved unique.)

Question

In the text I read, it proved that $a^{-1}$ unique. But then it goes by saying uniqueness of $x = a^{-1} b$ follows from the fact that $a^{-1}$ is unique.

---

Update: Here is the complete text (from p.20 of Abstract Algebra by Dummit and Foote)

Answer 1

The claim that "the uniqueness of $x$ follows because $\,a^{-1}$ is unique" is incorrect and misleading. The proof uses only the existence of $\,a^{-1}$ to deduce the uniqueness of solutions of $\,ax = b.\,$ Let's examine the proof. If $\,a,b,x\in G\,$ a $\rm\color{#0af}{monoid}$, and $\,a\,$ has a $\rm\color{#c00}{left}$ inverse $\,a'\in G,\,$ i.e. $\,\color{#c00}{a'a = 1}\,$ then

$$\qquad\begin{align} b \,&=\, ax\\ \Rightarrow\ a'b \,&=\, a'(ax)\\ &=\, (\color{#c00}{a'a})x \ \ \ {\rm by\ multiplication\ is\ \color{#0af}{associative}}\\ &=\, \color{#c00}1x\\ &=\, x\end{align}$$

Thus every root $\,x\,$ of $\,ax=b\,$ is equal to $\,a'b,\,$ so roots are unique - at most one root exists (since if $\,r_2,r_1$ are roots then $\,r_2 = a'b = r_1\Rightarrow\, r_2 = r_1).\,$ The proof did not use uniqueness of the (left) inverse $\,a'.\,$ Rather it used only its existence (to cancel $\,a).\,$ Even if - hypothetically - there were multiple inverses the proof would still work fine because it only needs to choose one of them.

---

Remark $ $ Further, the existence of a root follows if $\,a'\,$ is also a $\rm\color{#0a0}{right}$ inverse: $\,\color{#0a0}{aa'=1},\,$ since then the above arrow reverses by scaling by $\,a\,$ (or we can directly verify $\,x = a'b\,$ is a root: $\, ax = a(a'b) = (\color{#0a0}{aa'})b = b)$.

In the special case $\,b = 1\,$ the proof actually yields uniqueness of two-sided inverses, since it shows that if there exists a left inverse $\,a'$ of $\,a,\,$ i.e. $\,a'a = 1\,$ and there also exists a right inverse $\,x,\,$ i.e. $\,ax = 1,\,$ then $\,a' = x,\,$ i.e. left = right inverse, a result which is well-known. Thus - as above - if $\,a',a''$ are left inverses then $a'' = x = a'$ so $\,a'' = a'\,$ Similarly for uniqueness of right inverses.

More generally, to show that a set $S\:\!$ is a singleton it suffices to prove $\,s\in S\!\iff s = a.\,$ Above is the special case when $\,S\,$ is a set of roots of an equation.

There is much further discussion of this subtlety in an old sci.math thread linked here (in the more concrete case $G =$ additive group of reals). As is clear from that thread, this point often proves puzzling to beginners - even in very concrete groups.