Let $R = \{ a + b \sqrt[3]{2} + c\sqrt[3]{4} \mid a,b,c\in\mathbb Z\}$ is R an abelian group I asked this question here, but I am still confused with the answers I got. I am not asking to do the problem again as I've already got that part. My question is: Is it necessary to go over all the properties of a group to show something is an abelian or just showing for commutativity would suffice. I mean I can prove all those properties, but that'd be time consuming on a test if not necessary. Thanks. 
 A: No, generally it doesn't suffice and the following is true:

To show that a set $H$ with operation "$+$" is an abelian group, you have to verify all group axioms and show that $H$ is commutative regarding "$+$".

Now, often you can take a shortcut. But even in this case it doesn't suffice just to show commutativity of $H$ regarding "$+$".
Say $H \subset G$ and $H$ not empty, and you know that $G$ is a group, then to show that $H$ is a group is much easier: you just have to prove that $H$ is closed under "$+$" and contains inverses of all its elements. Or more concretely,
$$a, b \in H \Rightarrow a+b \in H \text{ and } -a \in H\,.$$
You automatically know it must have a neutral element, because for any $a\in H$ just use $a - a = 0$, also $H$ inherits the property "associativity" from $G$.
So in this case, you don't have to prove all group axioms for $H$.
And then you show that "$+$" is commutative:
$$\text{ For all } a, b \in H: a+b = b+a\, .$$
If in addition you know that $G$ is not just some group, but an abelian group, it gets even easier. Then you can skip the proof for the commutativity of "$+$", because if "$+$" is already commutative for the bigger group $G$ then the same must clearly be true for $H$, which is just a subset of $G$.
