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.

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 $|3/2|\ge1,~~|-4/2| \ge1$. 
