Let $2Z$ be the set of even integer numbers. Consider the operation $⊥$ defined by placing, for every $a, b ∈ 2Z$,
$a ⊥ b = a + b - 2$.
• Prove that the algebraic structure $(2Z, ⊥)$ is an abelian group, highlighting in particular what is the neutral element and what is the symmetric of each element $a ∈ 2Z$.
I think the structure isn't an abelian group because subtraction ($-2$ in this case) is not associative. Please tell me what you think.
$$a\perp(b\perp c)=a\perp(b+c-2)=a+(b+c-2)-2=a+b+c-4$$
$$(a\perp b)\perp c=(a+b-2)\perp c=(a+b-2)+c-2=a+b+c-4$$ So yes: the operation is associative. Observe that the operations on the right side of the definition of $\;a\perp b\;$ are the usual operations on the integers.
Now try to check the other axioms (closedness is immediate, though)