For a group $G$, let $C = \{aba^{-1}b^{-1}: a, b \in G\}$, and $H = \langle C\rangle$. (Recall that an element of the form $aba^{-1}b^{-1}$ is a commutator; $C$ is not in general a group.) Let $\phi_g$ be any inner automorphism of $G$.
(a) Show that $\phi_g$ gives a bijection from $C$ to $C$. In other words, show that for every $a, b \in G$ there exists (unique) $c, d \in G$ such that $\phi_g(aba^{-1}b^{-1}) = cdc^{-1}d^{-1}$.
(b) Using the previous, or otherwise, show that $\phi_g(H) = H$. Thus, $\langle C\rangle$ is a normal subgroup of $G$.
I was able to do part A. showing that it was a bijection, however i do not not how to use what i proved in part A to prove part B. I dont really see the relation between part a and part b. however this is how i proved part b:
If $g\in G$ and $h\in H$, then $ghg^{-1}h^{-1}=h'$, for some $h'\in H$ (since $H$ contains the commutator subgroup). But then $ghg^{-1}=h'h\in H$. Therefore, $gHg^{-1}\subset H$, which is equivalent to saying $\phi_g(H) = gHg^{-1} = H$
I was wondering how I would prove part B using part A. because in the proof i didnt use that $\phi_g$ was a bijection form $C$ to $C$
Hint: Part (a) tells you that $\phi_g$ is a bijection from the commutator subgroup to itself, in particular $\phi_g([a,b]) = g[a,b]g^{-1} = [c,d]$ for some $c,d \in G$. So if $h \in H$ then $$\phi_g(h) = g([a_1,b_1]\dots[a_n,b_n])g^{-1} = (g[a_1,b_1]g^{-1})\dots(g[a_nb_n]g^{-1})= [c_1,d_1]\dots[c_n,d_n], $$ i.e $\phi_g(h) \in H$.