Showing for a group $G$ with $a,b,x \in G$ that $ax=bx \implies a=b$

Question

Show that for a group $G$, for $a,b,x \in G, ax=bx \implies a=b$ .

I'm not sure why none of the proofs I founds just say something along the lines of:

Assume $ax=bx$ holds. Then $axx^{-1}=bxx^{-1}$ $\implies$ $a=b$. Is there a particular reason?

Answer 1

The reason why is that every element $x \in G$, since $G$ is a group, has an inverse $x^{-1}$ such that

$$x^{-1}x = xx^{-1} = e$$

where $e$ denotes the identity. The identity itself has the property that, for all $g \in G$, $eg=ge=g$. Then, in more steps,

$$ax = bx \implies axx^{-1} = bxx^{-1} \implies ae = be \implies a = b$$