# Up to what level can associativity be guaranteed?

My question is generated from the following question:

It turns out that the inverse of product with an assumption of inverse existence is a necessary condition of associative. Then is there any set with a binary composition rule satisfies $(ab)^{-1}=b^{-1}a^{-1}\,$ but fails associate? The answer is yes, see the following Cayley table:

$$\begin{array}{c|lcr} & 1 & 2 & 3 \\ \hline 1 & 1 & 2 & 3 \\ 2 & 2 & 1 & 2 \\ 3 & 3 & 2 & 1 \end{array}$$ The above set has unique inverse and unique identity but has overlapping number in the same column or row. Then I think what if the Cayley table keeps all row and column entries being different which also has a unique identity and unique inverse, can it be a group? The answer is no, see the following example: $$\begin{array}{c|lcr} & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 1 & 2 & 3 & 4 & 5 \\ 2 & 2 & 1 & 4 & 5 & 3 \\ 3 & 3 & 5 & 1 & 2 & 4 \\ 4 & 4 & 3 & 5 & 1 & 2 \\ 5 & 5 & 4 & 2 & 3 & 1 \end{array}$$ which is not commute and has order 5. This indicates the set cannot be a group since group with order 5 only can be isomorphic to $\mathbb{Z}_5$. However, this Cayley table fails the inverse of product property, e.g. $(2\cdot 3)^{-1}=5^{-1}=5$ while $3^{-1}2^{-1}=3\cdot 2=4$.

Then I am thinking can a set with a binary composition law, a unique identity, unique inverse property, Cayley table having no overlapping element in any row and column and the inverse of product be a group?

Concisely, Cayley table having no overlapping element indicates $\forall a\in S,\,(a\cdot\bullet):S\to S$, and $(\bullet\cdot a):S\to S$ are two bijection. Therefore, the set is a Quasi-group, together with a unit which is called a loop.

Then the question is reduced to ask whether a loop with inverse property and inverse product rule, i.e. $(ab)^{-1}=b^{-1}a^{-1}$ form a group?

Could you find a counterexample? Or just prove it.

On the other hand, what can be an equivalent assumption of the associativity rule.

Thanks.