Why is the set of commutators called "commutator subgroup"? From wikipedia and this question Why is the set of commutators not a subgroup?, I know that the set of commutators $G'=\{a^{-1}b^{-1}ab|a,b\in G\}$ of a group $G$ is not a group, because $G'$ is not closed.  That is to say, the product of two commutators is not necessarily a commutator.  Since $G'$ is not a group, it is not the subgroup of $G$. Then WHY $G'$ is called "commutator subgroup" or "derived subgroup"? 
 A: $G'$ is a subgroup of $G$. It can be defined in different ways. One way (which aptly explains and justifies its name) is that $G'$ is the smallest normal subgroup of $G$ such that $G/G'$ is commutative. Thus, $G'$, the commutator subgroup, is the smallest part of $G$ that needs to be killed in order to turn $G$ into a commutative group. Hence $G'$ 'commutates' $G$. (Proving this subgroup exists, without using any commutators, is a nice little exercise). 
This same subgroup $G'$ can be defined as the subgroup of $G$ generated by all the commutators $[x,y]=xyx^{-1}y^{-1}$. Notice that groups exist (though it's not easy to find an example) where the set of all commutators is strictly smaller than the commutator subgroup. So, to answer your question, the set of all commutators is generally not called the commutator subgroup.
A: The names "conmutator subgroup" or "derived subgroup" refer to the subgroup generated by the commutator set of the group and generally $G'$ or $[G,G]$ denote this subgroup generated by the set of commutators and not the set itself.
A: We all have to admit that the name "commutator subgroup" is badly chosen as you already mentioned commutators themselves do not serve a group in general. (I can also support this idea with +1 of one of my professors.)
In addition, "commutator subgroup" measures how much non-commutative a group is.
