Let $(G,\cdot)$ be a set with an associative operation. Show that the following two Axioms are equivalent Let $(G,\cdot)$ be a set with an associative operation. Show that the following two Axioms are equivalent:
(a) : there exists a left-hand neutral element $e'$, so that $\forall a \in G: e'a=a$
(b): There exists a neutral element $e$, so that $\forall a\in G:ea=ae=a$

My attempt:
$(a)\Longrightarrow (b) :$
Let $e'$ be the left-hand inverse on $(G,\cdot)$.
Now let's take $a,b \in G$:
$$ab=a(e'b)=(ae')b=ab.$$
So in order for the associativity on $(G,\cdot)$ to hold, $e'$ has to be right-hand neutral as well.
$(b) \Longrightarrow (a):$
Is obvious ?

Is this correct? I mean, its quit obvious, thats why I suspect myself jumping to conclusions..
 A: The two statements are not equivalent. Although (b) implies (a), it is not the case that (a) implies (b).
To verify this, let $G=\{e,a\}$, and define the operation as follows: $ea=a$, $aa=a$, $ae=e$, $ee=e$. That is, the result of multiplying $x$ by $y$ is always $y$.
This is easily seen to be associative, since $x(yz) = yz = z$ and $(xy)z=z$.
It is also clear that both $e$ and $a$ are left inverses, since $ee=e$, $ea=a$ (and also $ae=e$ and $aa=a$). However, neither $e$ nor $a$ are two-sided inverses.
The flaw in your attempt, as has been pointed out, is that associativity does not imply cancellativity. You cannot go from $xy=xz$ to $y=z$, or from $xy=zy$ to $x=z$, from just knowing the operation is associative. But that is what you are attempting to do when claiming that $(ae’)b = ab$ requires $ae’=a$.
A: As Arturo Magidin answer shows $A\not \implies B$.
Perhaps you were supposed to have A) be that there is a left inverse and a (potentially different[1]) right identity:
That is
A) There exists $e', e’’$ so that $e'a = a$ for all $a$ and $be’’ = b$ for all $b$.
If that is true
Claim:  $e'= e"$.
Pf:   $e'e’’ = e'$ by definition of $e'$.  And $e'e’’= e’’$ by definition $e' =e'e’’ = e’’$.
Thus $e'a = a = ae’’ = ae'$ for all $a$ so $B$ is implied.
$A\implies B$.
And as you say $B \implies A$ is obvious (just let $e' = e$ and $e’’ = e$.)
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[1] hypothetically potentially different, but as we prove, not actually possible to be different.
