Conjugacy classes in maximal subgroups

Question

Let $G$ be a finite group, $H$ a maximal subgroup. If $[G:H] = 2$, it is very well known how to determine the conjugacy classes of elements of $H$: they either stay the same or split depending on whether the representatives are centralized by some element outside of $H$.

Can this argument be generalized to a maximal subgroup which is normal, but not necessarily of index 2? Can it be generalized to any maximal subgroup? The last one might be tricky because the $G$-conjugacy class is not necessarily contained in $H$.

Answer 1

The case where $H$ is normal in $G$ is dealt with in the comments. If $H$ Is not a normal subgroup, things become quite arbitrary. Consider $G=A_4$. $cl_{A_4}(123)=\{(123),(142),(134),(234)\}$ while $cl_{\langle (123)\rangle}(123)=\{ (123)\}$. $cl_{A_4}(12)(34)=\{(12)(34),(13)(24),(14)(23)\}$ while $cl_{\langle (12)(34)\rangle}(12)(34)=\{ (12)(34)\}$

This shows that $|cl_G(x)|/|cl_H(x)|$ is in general not proportional to $[G:H]$ if $H$ is not normal.

Answer 2

In general, the number of conjugates of an element is the index of it's normalizer in the group.

So one would simply compare $[G:N(x)]$ to $[H:N(x)]$. If we want to exclude H being normal I would use the fact that a maximal subgroup is self normalizing.