Derived subgroup centralizes normal subgroup whose automorphism group is abelian

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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

Related facts about descent of action

Related facts about containment in the centralizer of commutator subgroup

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).