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.
 A: Indeed we have to show that the set is closed under addition in each case. For example, consider $2/3$ and $1/3$. Both of them satisfy $|.|<1$ but $|2/3+1/3|$ is not strictly smaller than $1$. So c doesn't make a groupid either. The similar story can be seen for d for example. In fact $|3/2+(-4/2)|<1$ while $|3/2|\ge1,~~|-4/2| \ge1$. 
A: 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. 
A: If $\cdot$ is a binary operation from $G\times G\to G$, then closure is trivially statisfied. In these cases however, you have the binary operation $+$ from $\mathbb{Q}\times\mathbb{Q}\to\mathbb{Q}$, which you are now restricting to some set $G\subset\mathbb{Q}$, considering it as a map from $G\times G\to\mathbb{Q}$. You need to verify the group axioms, as well as showing that the image of this restriction of the binary operation is actually contained in $G$, i.e. that $+$ defines a binary operations from $G\times G\to G$.
