Proving that a set is a group under addition 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.

Consider the set $G=$ {-1,0,1} under addition.  
It satisfies 3 properties mentioned by you,
namely
1.) $\exists$  0∈G such that a+0=0+a=a , $\forall$ a∈G.
2.) $\forall$ a∈G, $\exists$ -a∈G such that a+(-a)=(-a)+a=0.
3.) the operation + is associative.
But it doesn't form a group under addition since it is not closed.
i.e. 1+1=2 $\notin G$.
Hence G serves as a counter example. 
