Right group axioms from left group axioms I was working a question in group theory where we are given that the left axioms hold for a set $G$ together with a binary operation $*$. We would then like to prove that $(G, *)$ is a group. I read a solution, but I came up with something different and wanted to clarify that what I've done is correct for one part of the proof, namely that the left identity $e$ is also a right identity. Here's what I wrote. 
The left axioms that we'll use are $e * x = x$ and $x' * x = e$. 
Let $x \in G$ and $e$ be the left identity in $G$. We need to show that $x * e = x$. 
Now, $x * e = x \iff x * e = e * x  \iff x' * ( x * e) = x' * (e * x) \iff e * e = x' * x \\ \iff e = e$ 
which is true hence the right identity property holds. 
 A: Yes, that is correct. It might be clearer to write it backwards:
$e = e$ (self-evident)
$e \ast e = x' \ast x$ (since $x$ has a left-inverse, and $e$ is a left-identity).
$(x' \ast x) \ast e = x'\ast (e\ast x)$ (same reasoning, but applied to different sides of the equation).
$x'\ast(x \ast e) = x' \ast (e\ast x)$ (by associativity-this is crucial).
$(x')'\ast(x'\ast(x \ast e)) = (x')'\ast(x'\ast(e\ast x))$ (since every element has a left-inverse, including $x'$).
$((x')'\ast x')\ast(x \ast e) = ((x')'\ast x')\ast(e\ast x)$ (associativity, again)
$e\ast(x\ast e) = e\ast(e\ast x)$ (since $(x')'\ast x' = e$)
$x\ast e = e\ast x = x$ (since $e$ is a left-identity (used twice)).

Notice that your FORWARD implications are clear, but the REVERSE implications are not quite as clear, particularly:
$x'\ast(x\ast e) = x'\ast(e\ast x) \implies x\ast e = e\ast x$, in other words you have blithely assumed that such a set $G$ is left-cancellative: this is indeed true, but it needs clarification (which is why I introduce $(x')'$ to "cancel with"). You should also understand that we need associativity for this proof to go through (or else we cannot be certain that $x'\ast(x \ast e) = (x'\ast x)\ast e = e\ast e = e$).
