# 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 $C_G([G,G])$.

## 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).