Commutator subgroup and cyclic normal subgroup Given a finite group $G$, its commutator subgroup $H$ of $G$ and a cyclic normal subgroup $N$ of $G$, I'm trying to show that $hn = nh$ for all $n \in N$ and $h\in H$.
That basically means that every commutator fixes any $n$. My progress so far seems hardly any progress at all; if anything I feel like I'm making things messier. But I'm just going to write them anyway. 
My first thought was that since $hnh^{-1} \in N$, $hnh^{-1} = n^r$ for some $r$ and we can show $r$ must be 1, but this went nowhere. 
My second approach was to write, without lost of generality, $h=aba^{-1}b^{-1}$. Then the result becomes $a^{-1}b^{-1}nba = b^{-1}a^{-1}nab$ for any two elements $a$, $b$ of $G$. I feel like this should either be trivial somehow or additionally complicated. 
Also, I'm having difficulty figuring out how the cyclic-normality of N comes into play here.
Any help would be greatly appreciated!
 A: This is a nice exercise...Hints
1) $\,\operatorname{Aut}(N)\,$ is abelian
2) Every inner automorphism of $\,G\,$ is, when restricted to $\,N\,$ , is an element of $\,\operatorname{Aut}(N)\,$
3) $\,\forall x,y\in G\,\,,\,[x,y]^{-1}=[y,x]\,$
Try now to do something with this and, if after thinking it over for a while you're still stuck, write back below as a comment.
Added on request: As noted, $\,\operatorname{Aut}(N)\,$ is abelian and if $\,\phi_g\,$ denotes the inner automorphism determined by $\,g\,$ , then $\,\forall\,g\in G\,\,\,,\,\,\text{then}\;\; \left.\phi_g\right|_N\,\in\operatorname{Aut}(N)$  . We show now that any basic commutator $\,[x,y]\in H\,$ centralizes any element $\,n\in N\,$ :
$$[x,y]n[x,y]^{-1}=[x,y]n[y,x]=x^{-1}y^{-1}xyny^{-1}x^{-1}yx=\left(\phi_{x^{-1}}\phi_{y^{-1}}\phi_x\phi_y\right)(n)=$$
$$\stackrel{\text{Aut}(N)\,\,\text{is abelian!}}=\left(\phi_{x^{-1}}\phi_x\phi_{y^{-1}}\phi_y\right)(n)=Id_N\circ Id_N(n)=n$$
and since the above is true for any generator of $\,H=G'=[G,G]\,$ then it is true for the whole group.
Second solution: Perhaps easier: for any subgroup $\,K\leq G\,$ , the map $$f:N_G(K)\to\operatorname{Aut}(K)\,\,,\,\,f(k):=\phi_k=\,\text{conjugation by}\,\,k$$
is a group homomorphism (with $\,\phi_k(x):=kxk^{-1}\,$), whose kernel is precisely $\,C_G(K)\,$ , and from here 
$$N_G(K)/C_G(K)\cong T\leq\operatorname{Aut}(K)$$
In our case, we have $\,N\triangleleft G\Longleftrightarrow N_G(N)=G\,$ , so that we get $\,G/C_G(N)\cong T\leq\operatorname{Aut}(N)\,$ .
But $\,\operatorname{Aut}(N)\,$ is abelian, so that 
$$G/C_G(N)\,\,\,\text{is abelian}\,\,\Longleftrightarrow G'\leq C_G(N)\;\;\;\;\;\;\square$$
A: Since $N$ is normal, "conjugation by $g$" gives a group homomorphism $\phi$ from $G$ to $Aut(N)$, the group of automorphisms of $N$. Since $N$ is cyclic, this group of automorphisms is abelian. But you know that the commutator subgroup is the "universal" subgroup that quotients $G$ into something abelian. That is, if there's any group homomorphism from $G$ to an abelian group $A$, it always factors through the group $G/H$. Now I aks ya, what's the kernel of $\phi$?
