Given $a, b$ in the group $G$, then the equations $a\cdot x = b$ and $y\cdot a = b$ have unique solutions for $x$ and $y$ in $G$. Given $a, b$ in the group $G$, then the equations $a \cdot x = b$ and $y \cdot a = b$ have unique solutions for $x$ and $y$ in $G.$
This is a very common property of groups. Maybe, there are similar threads in the forum, but I want to verify a proof. My solution goes like this:

Given $a\cdot x=b$, we see that, if $x=a^{-1}\cdot b$, then $a\cdot x= (a\cdot a^{-1})\cdot b=e\cdot b=b.$ Also, in the case of $y\cdot a=b$, we see that $y=b\cdot a^{-1}$, so then, $y\cdot a=b\cdot (a^{-1}\cdot a)=b$.

Is the proof alright? And $x$ has no other solutions apart from $a^{-1}\cdot b$, as it is a linear equation, right? Is this sufficient to prove the uniqueness of the solution?
 A: You gave A solution to $ax=b$, but you did not show that it is the only solution.
Let's show that the equation $ax=b$ has at most one solution. We do this by showing the following: If $x$ and $x'$ are distinct elements in $G$ then $ax$ and $ax'$ are also distinct. Suppose otherwise: that $x$ and $x'$ are distinct elements in $G$ satisfying $ax=ax'$. Then let $a^{-1}$ be $a$'s inverse in $G$.

*

*Then on the one hand
$a^{-1}ax = (a^{-1}a)x = x$.


*But then also $a^{-1}ax = a^{-1}(ax)$ $= a^{-1}(ax')$ [because $ax=ax'$] $=(a^{-1}a)x' =x'$.
Do you see the contradiction between 1. and 2.?
And then can you use the above to show that the equation $ya=b$ has at most $1$ solution?
A: You have not demonstrated uniqueness in either case.
To this end, suppose $as=b$ and $at=b$. Then $as=at$, which implies
$$\begin{align}
s&=es\\
&=(a^{-1}a)s\\
&=a^{-1}(as)\\
&=a^{-1}(at)\\
&=(a^{-1}a)t\\
&=et\\
&=t.
\end{align}$$
This establishes uniqueness. Then if $ax=b$, we have $x=a^{-1}b$, which will be the only solution by uniqueness.
The case $ya=b$ is entirely analogous.
