Let $S$ be a non-empty set and $*$ be a binary operation on $S$ such that the following axioms hold:
for every $x,y,z∈S$ , $x*(y*z)=(x*z)*y$;
$(S,*)$ has a right identity i.e. there exists $e∈S$ such that $a*e=a$, for every $a∈S$, then is it consistent that $(S,*)$ has no left identity?
If $(S,*)$ satisfies 1. and has a left identity, then I can show that $(S,*)$ is commutative and hence has a right identity, but if $(S,*)$ satisfies 1. and has a right identity, then I am not being able to show the existence of a left identity, that's why I'm asking the question.
Isn't any left zero semigroup with more than 1 element a counter-example to this question? A left zero semigroup is a set $S$ with an associative binary operation $*$ such that $x*y=x$ for all $x,y\in S$. In this case:
for every $x,y,z\in S$, $x=x*y=x*(y*z)=(x*z)*y=x*y=x$; and
if $e\in S$ is any element, then $x*e=x$ for all $x\in S$;
But $S$ has no left identity. If $f\in S$ is arbitrary, then $f*x=f$ for all $x\in S$. Since $|S|>2$, it follows that there is $x\in S$ such that $x\not=f$, and so $f*x\not=x$. Thus $f$ is not a left identity, and since it was arbitrary, $S$ doesn't have a left identity.