TITLE: If $G$ is a group, then $a \ast x = b$ has a unique solution.
QUESTION [6 upvotes]: From "A First Course In Abstract Algebra" by John B. Fraleigh

TheoremIf $G$ is a group with binary operation $\ast$, and if $a$ and $b$ are any elements of $G$, then the linear equations $a \ast x = b$ and $y \ast a = b$ have unique solutions $x$ and $y$ in $G$.
ProofFirst we show the existence of at least one solution by just computing that $a' \ast b$ is a solution of $a \ast x = b$.
Note that
$$\begin{align*}
a \ast (a' \ast b) &= (a \ast a') \ast b & &\text{associative law} \\ 
&= e \ast b & &\text{definition of $a'$} \\ 
&= b & &\text{property of $e$} \\
\end{align*}$$
Thus $x = a' \ast b$ is a solution of $a \ast x = b$. In a similar fashion, $y = b \ast a'$ is a solution of $y \ast a = b$.
To show uniqueness of $y$, we use the standard method of assuming that we have two solutions, $y_1$ and $y_2$, so that $y_1 \ast a = b$ and $y_2 \ast a = b$. Then $y_1 \ast a = y_2 \ast a$, and by Theorem 4.15, $y_1 = y_2$. The uniqueness of $x$ follows similarly. $\blacksquare$

My question.
Why didn't Fraleigh simply solve the equation to show that a solution exists?
I mean: given the equation $a \ast x = b$ we can do as follows
$$\begin{align*}a \ast x &= b \\
a' \ast (a \ast x) &= a' \ast b \\
(a' \ast a) \ast x &= a' \ast b \\
e \ast x &= a' \ast b \\
x &= a' \ast b\end{align*}$$ and this shows two things: that a solution exists and also that the solution is unique.
Is it a matter of style or in this context my way is wrong?

REPLY [5 votes]: The difference between "a sequence of equations" and a proof are: clearly expressed logical relations between steps; and clearly expressed justification of each step.
Regarding your comment in reponse to @MatthewTowers, you may have "simply used the properties of operation $*$ in a group", but you did not justify your use of those properties, nor did you explain the logical relations amongst your equations.
If you include those justifications and implications, then your list of equations can be converted into a true proof of uniqueness, something like this (which I did by simply cutting and pasting your align* environment and adding some stuff):
$$\begin{align*}
a \ast x &= b \\
\implies a' \ast (a \ast x) &= a' \ast b \quad\text{(well-definedness of $(*)$)} \\
\iff (a' \ast a) \ast x &= a' \ast b \quad\text{(associative law applied on the LHS of the equation)}\\
\iff e \ast x &= a' \ast b \quad\text{(inverse law applied on the LHS of the equation)} \\
\iff x &= a' \ast b \quad\text{(identity law applied on the LHS of the equation)}
\end{align*}
$$
and therefore, following all the implication arrows forwards, we have $a*x=b \implies x = a'*b$, which proves uniqueness.
But, has existence been proved??!?
Look carefully at the implications that I have written out. The first one is only $\implies$; the remainder are $\iff$. In each $\iff$ step, the group law I am applying allows me to rewrite one side of the equation by substitution, leaving the other side unchanged. But the first implication is different: well-definedness of $(*)$ gives only the forward direction of implication $\implies$.
Notice that if the converse of the first implication is true then you can follow all of the implication arrows backwards to prove existence.
So, is the converse of the first implication true?
Yes, of course it is, and one can prove it like this:
\begin{align*}
a' * (a * x) &= a' * b \\
\implies a * (a' * (a * x)) &= a * (a' * b) \quad\text{(well-definedness of $(*)$)}\\
\end{align*}
and so on applying the associative law and inverse law and identity law...
However, that leads to a rather gloppy proof of existence. Fraleigh's proof of existence is much shorter and better.