# Conjugacy classes in maximal subgroups

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 $H$.

• Why not? $|cl_H (x)|=[H:C_H(x)]=[H:C_G(x)\cap H]=[HC_G(x):C_G(x)]$ while $|cl_G (x)|=[G:C_G(x)]$. $cl_H (x)\subseteq cl_G (x)$. Conclude that they're either equal or the latter is of order $[G:H]$ times the former. – hellotinfish Dec 8 '17 at 10:22
• Great! That solves it for $H$ normal in $G$. However, if $H$ is not normal, $H\operatorname{C}_G(H)$ is not necessarily a group, so we might well have that $H\operatorname{C}_G(x)$ is something between $H$ and $G$. It is a nice partial result however, one I can still work with. Thanks! – frafour Dec 8 '17 at 10:43
• @DavidReed well... I don't see the class equation of much use here, unfortunately. If $H$ Is not normal in $G$, things also become quite messed up. – hellotinfish Dec 8 '17 at 10:50
• @DavidReed consider $S_5$ and $A_5$. List out all of their conjugacy classes. You'll see that a conjugacy class of an element $x$ in $A_5$, $cl_{A_5}(x)$, is either equal to the conjugacy class of it in $S_5$, $cl_{S_5}(x)$, or is half the size of its conjugacy class in $S_5$. :D – hellotinfish Dec 8 '17 at 10:59

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.
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.