We can do this by considering the "universal" definition of $N$: $N$ has the following properties:
- $H\subseteq N$;
- $N\triangleleft G$; and
- If $K$ is any subgroup such that $H\subseteq K$ and $K\triangleleft G$, then $N\subseteq K$.
Since $\mathrm{ker}(f)\triangleleft G$, and we are assuming that $H\subseteq \mathrm{ker}(f)$, property 3 implies immediately that $N\subseteq \mathrm{ker}(f)$.
We can also do this by considering the top-down construction of $N$: $N$ is the intersection of all normal subgroups of $G$ that contains $H$:
$$N = \bigcap_{H\subseteq K\triangleleft G} K.$$
Then since $H\subseteq \mathrm{ker}(f)\triangleleft G$, it follows that $N\subseteq\mathrm{ker}(f)$.
Or we can do this by considering the bottom-up construction of $N$; but here you are somewhat incorrect above. You write that "every element of $N$ can be written as a product containing elements of $H$", and this is not quite good enough. The correct statement is that $N$ consists exactly of products of conjugates of elements of $H$:
$$N = \Bigl\{ (g_1h_1g_1^{-1})\cdots(g_mh_mg_m^{-1})\Bigm| g_i\in G, h_i\in H, m\in\mathbb{N}\Bigr\}.$$
Given such an element, if we apply $f$ to it we get:
$$\begin{align*}
f\Bigl((g_1h_1g_1^{-1})\cdots(g_mh_mg_m^{-1})\Bigr) &= f(g_1)f(h_1)f(g_1)^{-1}\cdots f(g_m)f(h_m)f(g_m)^{-1}\\
&= f(g_1)1f(g_1)^{-1}\cdots f(g_m)1f(g_m)^{-1} \quad\text{(since }H\subseteq \mathrm{ker}(f)\text{)}\\
&= 1,
\end{align*}$$
so for every $n\in N$ we have $f(n)=1$; hence, $N\subseteq \mathrm{ker}(f)$.
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
$$G/C_G(N)\,\,\,\text{is abelian}\,\,\Longleftrightarrow G'\leq C_G(N)\;\;\;\;\;\;\square$$
Best Answer
Denote the commutator of $a$ and $b$ by $a^{-1}b^{-1}ab = [a,b]$.
If $u$ is an element from the commutator subgroup, then $g^{-1}ug = u(u^{-1}g^{-1}ug) = u[u, g]$ .
Another approach: the commutator subgroup is defined to be the subgroup generated by the commutators, so every element of the commutator subgroup is of the form $$[a_1, b_1][a_2,b_2]\ldots[a_n, b_n].$$ It is enough to show that $g^{-1}[a,b]g$ is always in the commutator subgroup, because then
$$g^{-1}[a_1, b_1][a_2,b_2]\ldots[a_n, b_n]g = (g^{-1}[a_1, b_1]g)(g^{-1}[a_2,b_2]g)(g^{-1}\ldots g)(g^{-1}[a_n, b_n]g)$$
is a product of elements from the commutator subgroup. When $\phi$ is any homomorphism, we have $\phi([a,b]) = [\phi(a), \phi(b)]$. Since for any $g \in G$ the map $\phi$ defined by $\phi(x) = g^{-1}xg$ is a homomorphism, the result follows.