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I am in Intro to Algebra, and have a question regarding the commutator subgroup. I am a bit confused with the premise, though, with how the set is a subgroup in the first place.

Let $C$ be the set of commutators of $G$. Then two arbitrary elements of $C$ would be $aba^{-1}b^{-1}$ and $cdc^{-1}d^{-1}$. I don't see how $C$ is closed under multiplication. That is, I don't see how $$aba^{-1}b^{-1}cdc^{-1}d^{-1}\in C.$$ Am I making a wrong assumption in assuming that the binary relation is multiplication? Any help would be appreciated. Again, this is my first semester of algebra, so try to keep it basic.

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    $\begingroup$ The commutator subgroup is generated by elements of the form $aba^{-1}b^{-1}$, it's not just those elements. $\endgroup$ – Zavosh Jul 5 '15 at 4:24
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Indeed, the set of commutators need not be a subgroup (take a look at this thread). But that's not a problem, because the commutator subgroup of a group $G$ is defined to be the subgroup generated by the set of commutators of $G$ (Wikipedia link).

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  • $\begingroup$ Perfect; thanks a lot. $\endgroup$ – Elliot G Jul 5 '15 at 15:44

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