Problem Statement
Given finite $X \neq \emptyset$ together with an associative binary operation, if left-cancellation and right-cancellation laws hold, is $X$ a group?
Attempt
We know that for all $x, y, z \in X$, left-cancellation $$xy = xz \implies y = z$$ holds. Thus, the function $L_{x}:X \to X$ defined by $$L_{x}(w)=xw$$ is injective. Since $X$ is finite, $L_{x}$ must also be surjective. This means that there exists an element $e$ of $X$ such that $L_{x}(e)=x$, i.e. $x=xe$.
First, we want to show that $e$ is in fact a two-sided identity for $X$. So let $v$ be an arbitrary element of $X$, and consider $xv=(xe)v=x(ev)$ (by associativity). Thus, by left-cancellation, $v=ev$ for all $v$, i.e. $e$ is a left identity. Now consider $vv=v(ev)=(ve)v$ (again, by associativity). By right-cancellation, we conclude that $v=ve$, i.e. $e$ is a right identity.
Next, we need to show the existence of inverses. From before, for every $w \in X$, the function $x \mapsto wx$ is surjective, so there is an element, say $w_R$, such that $w{w_R} = e$. Similarly, there is an element $w_L$ such that ${w_L}w = e$. Since $$w_L = {w_L}e = {w_L}(w{w_R}) = ({w_L}w){w_R} = e{w_R} = w_R,$$ we conclude that $w_L = w_R$ is a two-sided inverse for $w$.
Caveat
But wait! I have not yet established closure of the binary operation!
Additionally, I also need to show that the result in this problem may be false if $X$ is infinite.
Any hints you can provide will be gladly appreciated!
Question
Is my attempt above acceptable for the given problem at hand? Or would there be a more direct approach?
