It is known that for a finite non-abelian simple group $G$ we have $|G|<|C\_G(x)|^3$ for some involution $x$. Is there a better bound for the order of centralizer of a nontrivial element of $G$ (not necessarily involution)? For instance, is there always a nontrivial element $x\in G$ such that $|G|<|C\_G(x)|^2$?



On the size of centralizers in a non-abelian finite simple group
(Later edit: Note that in general, the maximal order of the centralizer of a non-identity element in $G\_{n} = {\rm SL}(2,2^{n})$ is $2^{n}+1,$ while the group order is $(2^{n}-1)2^{n}(2^{n}+1).$ Hence as $n \to \infty$, the limit of $\frac{|G\_{n}|^{\frac{1}{3}}}{{\rm max}\_{x \neq 1}(|C\_{G\_{n}}(x)|)}$ is $1 ).$


I do not know the reference for the claim about $|C\_{G}(x)|^{3}$ for $x$ an involution when $G$ is non-Abelian simple. Here are some comments on related results. It is certainly true ( and a consequence of results of Brauer and Fowler) that $|G| <|C\_{G}(x)|^{3}$ for some non-central element $x$ when $G$ is simple. There may be later results with which I am not familiar which show that $x$ may be chosen to be an involution. In general, Brauer and Fowler that a finite group $G$ of even order greater than $2$ has a proper subgroup $H$ with $|H|^{3} >|G|.$ I think it is a consequence of results of Burness, Liebeck and Saxl that the only non-trivial finite groups $G$ which do not have a proper subgroup $H$ with $|H|^{2} > |G|$ have order $p$ or $p^{2}$ for some prime $p$ (though this result may be older- it certainly requires CFSG).


Here is an outline: there is clearly such an $H$ if $G$ is a $p$-group of order at least $p^{3}$ for some prime $p.$ There is clearly such an $H$ ( which may be taken to be a suitable Hall subgroup) when $G$ is solvable, but not a $p$-group for any prime $p.$
Suppose then that $G$ is not solvable, and let $M$ be the terminal member of the derived series of $G.$ Suppose that $M < G.$ If $[G:M] = p$ for some prime $p,$ then we can take $H=M$ unless $|M| <p.$ But in the latter case, we can take $H$ to be a subgroup of order $p.$ If $[G:M] = p^{2}$ for some prime $p,$ then we can take $H$ to be a maximal normal subgroup of $G$ containing $M.$ Otherwise, we can choose $H$ to be a proper subgroup of $G$, containing $M$, of maximal order.


Suppose then that $G$ is perfect. The results cited above deal with the case that $G$ is non-Abelian simple, so suppose otherwise, and let $M$ be a maximal normal subgroup of $G.$ Then $G/M$ is non-Abelian simple, and there is a proper subgroup $H$, containing $M,$ of the required order.


