How many elements have to verify the associativity property in a group? If this is a duplicate please mark it down.
We know that if $(G,\ast)$ is a group then it must verify the associative property, that is, $$\forall x,y,z\in G:\quad x\ast(y\ast z)\quad=\quad(x\ast y)\ast z\,.$$
My question is how many elements have to verify the associativity in a group?
I suspect that it must be $$\frac{n!}{3!},$$ where $n$ is the order of $G$.
Is that right?
Thank you!
EDIT: as you have opined I would like to know the worst case, that is, in those where we have not realized the inheritance that can have an operation within the group or any other factor that reduces the number of check rows (yes, "silly" mode activated!). If you want, you can propose the best level if certain restrictions occur (be Abelian, etc.) :)!

In “Verification of Identities” (1997),  Rajagopalan and Schulman  discuss associativity testing for binary operations on a finite set of $n$ elements.  For certain types of operations, a random algorithm works with high probability.  However:

This random sampling approach does not work in general.  For every $n≥3$, there exists an operation with just one nonassociative triple.

(Page 3.)
So the answer to your question, unfortunately, is that in the general case one must check all $n^3$ triples of elements.  (Not $\frac{n!}6$ as you suggested.)
The Rajagopalan and Schulman paper has more to say that may be useful to you anyway.

Since R&S don't say, here's a very simple example of an operation with only one nonassociative triple.  Consider the set $\{0, 1, 2, \ldots n-1\}$ (for $n\ge 3$) with the following operation: 

$a\ast b = 0$ in all cases, except $2\ast 1=2$.

Then it is easy to show that $a\ast(b\ast c) = (a\ast b)\ast c$ with only one exception: Since  $b\ast c\ne 1$,  the left side must be equal to $0$.  And, except in the one case $(2\ast 1)\ast 1 = 2\ast 1 = 2$, the right side will be $0$ also.)
