Associativity for non-binary operators

I have trouble finding a definition for the associativity from operators with arity higher than 2. At first I tried to think how it would be done for an operator with arity=3, and then generalize it, but I am not sure how many elements I need, if just adding one more, or adding n-1 elements. Could someone help me with a definition or explanation? Thak you!

So the reason I am looking into this is because I want to show that a universal algebra (G,p) from type (3), with $$p(x,y,z):= x\cdot y^{-1}\cdot z$$ is a Group. That means I have to express the axioms (associativity, neutral element and inverse elemet) using the operation $$p$$.

• Mathematicians don't usually look for a generalization like this for its own sake. Generalizations happen when there's some interesting context that suggests them. If you have some operation in mind for which you need an idea of associativity please edit the question to tell us about it. Then perhaps we can help. – Ethan Bolker Jun 2 at 19:17
• I was going to guess something like $f(f(a,b,c),d,e)=f(a,f(b,c,d),e)=f(a,b,f(c,d,e))$, but Ethan's point is better. – Rahul Jun 2 at 19:24
• i hope it is understood what i mean? – M-S-R Jun 2 at 19:31
• You're thinking about this all wrong. Your goal should be to just express the group operation in terms of $p$, so then you can write the group axioms in terms of $p$. This has nothing to do with $p$ itself being "associative": rather, you want a certain binary operation you can express in terms of $p$ to be associative. – Eric Wofsey Jun 2 at 19:33
• A universal algebra of type (3) cannot be a group. What you claim to “want to show” is not what you probably actually want to show. Note also that the operation you have is not coherent of itself, because it invokes a multiplication and an inverse function that are not defined. Presumably you are trying to show that if you have a group already, then you can actually express the usual multiplication, inverse, and identity operators in terms of the $3$-ary derived operation $p$. What that tells you is that the clone of the group is the same as the clone generated by $p$. – Arturo Magidin Jun 2 at 19:50