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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.

Zev Chonoles
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pancini
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    The commutator subgroup is generated by elements of the form $aba^{-1}b^{-1}$, it's not just those elements. – Zavosh Jul 05 '15 at 04:24

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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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Zev Chonoles
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