My target is to prove that, in order to abelianize a group, it sufficies to take the subgroup generated by the commutators of the generators instead of the commutators of the whole group, to make the factor group.
I checked he answers in here The generator set of the commutator subgroup of a free group, but there's something that doesn't convince me at all in the last answer (the one given by SundayMorning).
He has proved that $K = [G, G]$ is the least normal subgroup that contains $Y$, the set of the commutators of the generators. However I don't see why $K = \langle Y \rangle$, because it's not clear for me that $\langle Y \rangle$ is normal.
The definition I know of $\langle H \rangle$ states that it's the least subgroup containing $H$, not necessarily normal.