Let $\mathcal{G}$ be a finite set and $e: \mathcal{G}\times \mathcal{G} \rightarrow \mathcal{G}$ be a function such that
- For all $a,c\in \mathcal{G}$, there exists a unique $b$ such that $e(a,b) = c$.
- For all $b, c \in \mathcal{G}$, there exists a unique $a$ such that $e(a,b) = c$.
My hunch is that $\mathcal{G}$ is isomorphic to addition modulo $|\mathcal{G}|$, so I started checking if the pair $(G,e)$ forms a group. (The function properties are motivated by encoder-decoder systems.)
Right now, I want to know if $(G,e)$ has an identity. For a fixed $a\in \mathcal{G}$, the definition of $e$ implies that there exist two unique elements $0_L,0_R$ such that
- $e(a, 0_R)=a$
- $e(0_L, a)=a$.
My question is: how do I prove (or disprove) that $0_L=0_R$?
Your hunch is sadly, not correct.
Consider $\cal G = \Bbb Z/n\Bbb Z$ and $e(a, b) := a-b$.
Then, the axioms you've mentioned are satisfied but your results clearly do not follow. (The operation does not even have an identity.)
EDIT: As pointed out correctly, we would need $|n| > 2$ with $n$ being an integer, of course! ($n = 0$ would give an infinite $\cal G$.)