To show that a set $G$ is a group under addition, do we first need to show that $G$ is closed under addition, or is that implied by proving the three properties of a group, namely
there exists an identity element $e \in G$ such that $a \cdot e = e \cdot a = a$ for every $a \in G$
for every $a \in G$, there exists an $a^{-1} \in G$ such that $aa^{-1} = a^{-1}a = e$
the operation $\cdot$ is associative
(Here, $\cdot$ is some binary operation.)
For context, I'm doing the following exercise:
and want to know if I should first prove that sets are closed under addition for those that turn out to be groups under addition.