Problem statement
Let $A$ be a subgroup of $G$. It is known that $A = A^{(1)}$, where $A^{(1)}$ is $A$'s commutator subgroup.
Normal closure of $A$ is defined as $N(A) := \left\langle g A g^{-1} \ | \ g \in G \right\rangle$.
How do I prove that $N(A) = N(A)^{(1)}$?
My thoughts so far
It is evident that $N(A)^{(1)} \subset N(A)$, so it is only needed to prove the reverse inclusion, that is $$\forall z \in N(A) \ \ z = [x_1, y_1]^{\varepsilon_1} \dots [x_n, y_n]^{\varepsilon_n};\ \varepsilon_i \in \{-1, 0, 1\};\ x_i, y_i \in N(A)$$
I have tried to prove this at least for the simplest case, when $z = g a g^{-1};\ a \in A, g \in G$ and $a = [a_1, b_1];\ a_1, b_1 \in A$, but even there I am stuck.
I believe if I could do that, some form of induction could be used to prove the rest, still I am struggling with the "base case".
Is this approach even viable? How do I do this?
You can complete your step by noting that: $[x,y]^g = [x^g,y^g]$ for all $x,y,g\in G$.
A bit easier: you know every element of $A$ is a product of commutators of elements of $A$; hence you know that every element of $gAg^{-1}$ is a product of commutators of elements of $gAg^{-1}$. Hence you know that every element of $A^G$ is...