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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"?

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$G'$ is the group generated by commutators. Not the set of all commutators. – Prahlad Vaidyanathan Sep 22 '13 at 10:09
As it happens, in many cases the set of commutators is closed under multiplication, and is a group. If I remember right, the smallest example of a group in which the commutators don't form a subgroup is a group of 96 elements. I know there's a question about it, here or on MathOverflow. – Gerry Myerson Sep 22 '13 at 10:18
up vote 6 down vote accepted

$G'$ 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.

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Like the phrase ' the smallest part of G that needs to be killed in order to turn G into a commutative grooup' as a visual explanation of the quotient group. – Robert Mar 23 at 6:57

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.

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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.

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I don't think commutator subgroup is such a bad name. What would you call it? – Ittay Weiss Sep 22 '13 at 11:02
I do not have any idea for an alternative now. But I suppose and observe that this name is sometimes being misleading for beginners. In general, naming is a really hard thing especially in mathematics as it requires a very broad perspective, foreseeing and knowledge. – Metin Y. Sep 22 '13 at 11:12
I disagree that the name is bad. I find it completely logical that the smallest subgroup containing all commutators is called the commutator subgroup. – Tobias Kildetoft Sep 22 '13 at 11:28

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