Let $A$ be an nonempty set, and $f: A^3 \rightarrow A$ mapping which satisfies:
$f(x,y,y)=f(y,y,x)$ for each $x,y \in A$
$f(f(x_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3))=f(f(x_1,y_1,z_1),f(x_2,y_2,z_2),f(x_3,y_3,z_3))$ for each $x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3 \in A$
Prove that for arbitrary fixed $a\in A$ operation $x+y=f(x,a,y)$ defines a structure of commutative group (A,"+").
I've tried to check basic properties related to the definition of a group but I find it very confusing...
This is false. Consider a map $f$ that maps all of $A^3$ to the same element $a_0$ of $A$. Such a map satisfies the conditions. Then $x + y = a_0$ for all $x, y \in A$ (the choice of $a$ is irrelevant). Unless $A$ contains just a single element, this doesn't give a group structure, as you'd get $x + x = a_0 = x + y$, so $x = y$.