Derived subgroup centralizes normal subgroup whose automorphism group is abelian

Statement
Suppose $$N$$ is a normal subgroup whose automorphism group is abelian (i.e., a normal subgroup that is a  group whose automorphism group is abelian -- it has an abelian automorphism group) of a group $$G$$. Then, the derived subgroup $$[G,G]$$ is contained in the fact about::centralizer $$C_G(N)$$.

Equivalently, since centralizing is a symmetric relation, we can say that $$N$$ is contained in the centralizer of derived subgroup $$C_G([G,G])$$.

Related facts about cyclic normal subgroups

 * Commutator subgroup centralizes cyclic normal subgroup
 * Normal of least prime order implies central
 * Cyclic normal Sylow subgroup for least prime divisor is central

Related facts about descent of action

 * Quotient group acts on abelian normal subgroup

Related facts about containment in the centralizer of commutator subgroup

 * Commutator subgroup centralizes cyclic normal subgroup, so any cyclic normal subgroup is contained in the centralizer of commutator subgroup
 * Abelian-quotient abelian normal subgroup is contained in centralizer of commutator subgroup
 * Abelian subgroup is contained in centralizer of commutator subgroup in generalized dihedral group
 * Abelian subgroup equals centralizer of commutator subgroup in generalized dihedral group unless it is a 2-group of exponent at most four

Proof
Given: A group $$G$$. A cyclic normal subgroup $$N$$.

To prove: $$[G,G] \le C_G(N)$$.

Proof: Consider the homomorphism:

$$\varphi: G \to \operatorname{Aut}(N)$$

given by:

$$\varphi(g) = n \mapsto gng^{-1}$$.

Note that this map is well-defined because $$N$$ is normal in $$G$$, so $$\varphi(g)$$ gives an automorphism of $$N$$ for any $$g \in G$$.


 * 1) The kernel of $$\varphi$$ is $$C_G(N)$$: This is by definition of centralizer: $$C_G(N)$$ is the set of $$g \in G$$ such that $$gng^{-1} = n$$ for all $$n \in N$$, which is equivalent to being in the kernel of $$\varphi$$.
 * 2) The kernel of $$\varphi$$ contains $$[G,G]$$: Since $$\varphi$$ is a homomorphism to an abelian group, $$\varphi([g,h]) = [\varphi(g),\varphi(h)]$$ is the identity. Thus, every commutator lies in the kernel of $$\varphi$$, so $$[G,G]$$ is in the kernel of $$\varphi$$.
 * 3) $$[G,G] \le C_G(N)$$: This follows by combining steps (1) and (2).