This was a qualifying exam question. I'm not sure I even understood it, so I want to see if my reasoning is valid. The question is:
Let H be a subgroup of G. Suppose that for any two elements a,b that are conjugate in G, we can choose an element h $\in H$ so that a = $hbh^{-1}$. Show that the commutator G' is contained in H.
I proceed as follows: First I show H is normal, then I show G/H is abelian. Let h $\in H$. Then $ghg^{-1}$ is conjugate to h for all g, so $ghg^{-1} = khk^{-1}$ for some k $\in H$. So $ghg^{-1}$ is in H, so H is normal in G.
Next, let a $\in G$. Then again, for all g $\in G$, $gag^{-1} = hah^{-1}$ for some h $\in H$. Also, $a^{-1}hah^{-1} = khk^{-1}h^{-1}$ for some k $\in H$. So $aH = hah^{-1}H = gag^{-1}H$. So G/H is commutative, and so G' $\leq$ H.
Is there an error in this logic?