Let \(G\) be a group, let \(G'\) be its commutator subgroup, and let \(N\triangleleft G\) be cyclic. Prove that \[ gn=ng \] for every \(g\in G'\) and \(n\in N\).
Solution
1. Conjugation defines a homomorphism\[ \varphi:G\longrightarrow\operatorname{Aut}(N). \]
Because \(N\triangleleft G\), conjugation by any \(g\in G\) maps \(N\) to itself. Define \[ \varphi(g)(n)=gng^{-1}. \] Each \(\varphi(g)\) is an automorphism of \(N\), and \[ \varphi(gh)=\varphi(g)\varphi(h), \] so \(\varphi\) is a group homomorphism.
2. The group \(\operatorname{Aut}(N)\) is abelian.
Let \(N=\langle a\rangle\). Every automorphism of \(N\) is determined by the image of \(a\), and has the form \[ a\longmapsto a^u \] where, if \(N\) has finite order \(m\), the exponent \(u\) is a unit modulo \(m\), while if \(N\) is infinite, \(u\in\{1,-1\}\). If two automorphisms correspond to exponents \(u\) and \(v\), then their composites send \[ a\longmapsto a^{uv} \] in either order. Hence all automorphisms commute.
3. The commutator subgroup \(G'\) lies in \(\ker\varphi\).
Since the target \(\operatorname{Aut}(N)\) is abelian by step 2, every commutator in \(G\) maps to the identity. Therefore the subgroup generated by all commutators satisfies \[ G'\subseteq\ker\varphi. \]
4. Every element of \(G'\) centralizes every element of \(N\):\[ \boxed{gn=ng\qquad(g\in G',\ n\in N)}. \]
If \(g\in G'\), then step 3 gives \(\varphi(g)=\operatorname{id}_N\). Thus for every \(n\in N\), \[ gng^{-1}=n. \] Multiplying on the right by \(g\) gives \(gn=ng\).
5. Q.E.D.
Step 4 is the required conclusion.