I am working on a homework problem, and although I do not want a solution, I want to ask what is wrong with a certain approach I have.
If $H\subseteq G$ is a normal subgroup in $G$, and $[g]_G$ is the conjugacy class of some $g\in G$, then either $[g]_G \subseteq H$ or $[g]_G\cap H =\varnothing$.
I wanted to show that if there is any element $h\in G$ such that $hgh^{-1}\in H$, then for any other $k\in G$, then $kgk^{-1}\in H$. I assumed that there was some $h\in G$ such that $hgh^{-1}\in H$. Since $H$ normal in $G$ we have that $$h^{-1}Hh = H $$ So we would have \begin{align*}h^{-1}hgh^{-1}h&\in h^{-1}Hh\\ g&\in h^{-1}Hh\\ g &\in H\end{align*} However, it is not necessarily true that $g$ is in $H$. So I think this is approach is incorrect. I have found another solution which does lead to this problem, but I would like to know why I reach this conclusion which is not necessarily true.
You showed that $[g]_G\cap H\not=\emptyset\implies g\in H$. But clearly $g\in H$ and $H\trianglelefteq G\implies [g]_G\subset H$.
So you were successful.
It's just that the use of $h$ for an arbitrary element of $G$ is a little confusing. (One might think that $h\in H$.)