# Centralizer of a $p$-element modulo the $p'$-core and conjugacy class sizes in quotient groups

Does $[ G : C_G(x) ] = [ G/K : C_{G/K}(x) ] [ K : C_K(x) ]$ hold for all finite groups $G$ and $p$-elements $x$, where $K = O_{p'}(G)$ is the largest normal subgroup of $G$ with order coprime to $p$?

A related statement is true: Let $G = H \ltimes K$ be a semidirect product of $H$ with the (arbitrary) normal subgroup $K$, and let $x$ in $H$. Suppose $hk$ centralizes $x$. Then considering the group mod $K$, one gets that $h$ centralizes $x$. Hence $k$ centralizes $x$, and $C_G(x) = C_H(x) \cdot C_K(x)$. Importantly, we have the equality:

$$[ G : C_G(x) ] = [ G/K : C_{G/K}(x) ] [ K : C_K(x) ] .$$

If $G$ splits over $K = O_{p'}(G)$ so that $G = H \ltimes K$, then every $p$-element is conjugate to some element of $H$, and the previous implies the conjugacy class sizes behave as I expect. However, $G$ often does not split over $K$ and I am not sure what happens then.

So suppose $G$ is a finite group with normal subgroup $K$. One still has that every $h$ centralizing $x$ gives rise to an $hK$ in $C_{G/K}(x)$ and of course $C_G(x)$ is a union of cosets of $C_K(x)$. However, it is not clear that all elements $hK$ of $C_{G/K}(x)$ give rise to elements $h$ in $C_G(x)$. In other words, I can only prove that:

$$[ G : C_G(x) ] \geq [ G/K : C_{G/K}(x) ] [ K : C_K(x) ] .$$

I checked for counterexamples (small groups up to order $1000$ except $768$, primitive groups up to degree $500$ except $343$, and perfect groups up to order $10^6$ available in GAP), but found no counterexamples to equality. I am not sure if the coprime hypothesis is truly relevant, but it is the case of interest and I could not think of any other way to say "$x$ in $H$" when $G$ did not split. With absolutely no hypotheses on $x$ or $K$, then $G$ dihedral of order 8 with $K = Z(G)$ and $x$ a non-central involution gives a counterexample, but this is not at all similar to the case I am interested in.

If $H$ is a p-subgroup of $G$, then it is true that: $$[G:C_G(H)]=[G/K:C_{G/K}(H)]\ [K:C_K(H)].$$
$C_K(H) = C_G(H)\cap K$; this is clear.
$C_{G/K}(H)=C_G(H)K/K$. To prove this, note first that $N_{G/K}(H)=N_G(H)K/K$. This is equivalent to $N_G(HK)=N_G(H)K$, and this follows from the Frattini argument applied to $HK$ in $N_G(HK)$. Now the centralizer case follows because $N_G(H)$ is mapped to $N_{G/K}(H)$, $H$ is mapped to an isomorphic copy of itself, and the quotient map respects the action of $N_G(H)$ on $H$ (so, in particular, respects the kernel of this action, $C_G(H)$).