I have the following table and I don't know how to determine if an operation is associative based on the table. Is there an easy way to do it? Or it's just brute force
\begin{array}{|c|c|c|c|c|c|} \hline *& a & b & c &d &e \\ \hline a& a&b &c&b&d\\ \hline b& b&c &a&e&c\\ \hline c& c &a &b&b&a\\ \hline d&b&e&b&e&d\\ \hline e&d&b&a&d&c\\ \hline \end{array}
We can see that it's not commutative because $b*e \neq e*b$, but how do we check if it's associative?
In general, it is not possible to check for associativity simply by glancing at the Cayley table. This is, in part, because associativity is determined from a three termed equation $a(bc) = (ab)c$ whilst the Cayley table shows two-term products only.
However, you don't quite need brute force. You can use Light's associativity test.
Light's associativity test is based on the following Lemma. Let $*$ be a binary operation on the set $S$ (called product).
Definition: A subset $G$ of $S$ generates $S$ if every element of $S$ can be generated as product of elements of $G$.
Lemma: If G generates S then * is associative on S if and only if $$\forall (x \in S) \forall (g \in G) \forall (z \in S): x*(g*z)=(x*g)*z$$
In your example $\{e\}$ generates $\{a,b,c,d,e\}$, because
$$e=e$$ $$c=e^2$$ $$a=e*c=e*(e^2)$$ $$d=a*e=(e*(e^2))*e$$ $$b=a*d=(e*(e^2))*((e*(e^2))*e)$$
so you have to check
$$\forall (x \in S) \forall (z \in S): x*(e*z)=(x*e)*z$$
But we have $$c*(e*e)=c*c=b$$ $$(c*e)*e=a*e=a$$ and so $$c*(e*e) \ne (c*e)*e$$ So $*$ is not associative.
Do two "outer products" using the multiplication table (but the dimensions in different orders). Then you compare the resulting outer products element-wise.
