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

Proof: Consider the homomorphism:

$\phi :G\to \mathrm{A}\mathrm{u}\mathrm{t}(N)$

given by:

$\phi (g)=n\mapsto gn{g}^{-1}$.

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

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