Proof of: a commutator subgroup is a normal subgroup The proof of the statement "a commutator subgroup $G'$ is a normal subgroup" goes like this:
1) Show that $g^{-1}[a,b]g \in G'$
2) Show that $g^{-1}[a_1,b_1][a_2,b_2]...[a_n,b_n]g \in G'$
I don't understand why step 1 doesn't suffice itsel

Step $1$ doesn't suffice becaue the commutator subgroup is not the set of all commutators (as this set doesn't usually form a group), rather, it is the subgroup generated by the set of commutators. If you only prove step $1$, it could still be the case that, say, $g^{-1}[a,b][c,d]g \notin G'$ for some $a,b,c,d,g\in G$. 
