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 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

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: $[G,G]\leq C_{G}(N)$.

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. $[G,G]\leq C_{G}(N)$: This follows by combining steps (1) and (2).