# Centralisers in the symmetric group

Let $G$ be a $k$-transitive subgroup of the symmetric group $Sym(n)$, $k\geq 2$, $n$ large. (Make $k$ larger if you think it necessary to make the question below non-trivial/interesting.)

Write $C(g)$ for the centraliser of an element and $|C(g)|$ for its number of elements.

What can you say about $\min_{g\in G} |C(g)|$, relative to the size of $G$? Must it be small?

(Note that $k\geq 2$, $n\geq 3$ immediately imply that $G$ is non-abelian (thus making the question possibly non-stupid).)

(I should clarify that I expect (hope?) the answer to be quite a bit smaller than $\ll_{\epsilon} |G|^{\epsilon}$. (Assume $k\geq 3$ or $k\geq 4$ if needed.) In this, a $k$-transitive permutation group feels like a different sort of animal from a linear algebraic group $G$, where $|C(g)|$ is most often no smaller than $|G|^{\dim(T)/\dim(G)}$ ($T$ = a maximal torus of $G$).)

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Also: please, please do not use the classification of finite simple groups in answering my question. – H A Helfgott May 13 '11 at 13:01
A very old theorem of Jordan (which has no doubt been improved) asserts that a primitive (eg 2-transitive) subgroup of $S_n$ with small minimal degree $m$ is either $S_n$ or $A_n$. I'd have to look up the exact meaning of small' here. But the size of the smallest centraliser is clearly bounded below by some function of $m$, so this seems like a start... – HJRW May 13 '11 at 13:06
I suspect the minimum size of a centralizer is bounded above by n-1. So yes, it is very small. The larger k is, the fewer groups to check. If k ≥ 4, then the size of the centralizer is bounded above by n-1. – Jack Schmidt May 13 '11 at 15:01
@HW: Elements of small degree (must) have large centralizers. Elements of large degree (can) have small centralizers. I think your inequality is backwards. – Jack Schmidt May 13 '11 at 15:03
@Jack Schmidt - ah, right. So it is. – HJRW May 13 '11 at 15:08

(Written before clarification at end of question was added). Here is a result which seems to be of a somewhat negative nature in the context of your problem, and your suggested line of attack, I think. I will denote the number of conjugacy classes of $G$ by $k(G)$. The group $G = {\rm SL}(2,2^{n})$ (n>1) is a triply transitive permutation group on $1+2^{n}$ points and has order $(2^{n}+1)2^{n}(2^{n}-1)$. It has $1+2^{n}$ conjugacy classes, so that $k(G) > |G|^{\frac{1}{3}}$. Furthermore, the only orders of centralizers of non-identity elements of $G$ are $2^n$,$2^{n}-1$ and $2^{n}+1$. Hence the minimum centralizer order for an element of $G$ is only marginally smaller than $|G|^{\frac{1}{3}}.$ Admittedly this is a rather special group and a rather rare situation, but it is an infinite family of "bad" examples.

Added later: come to think of it, there are even bad solvable examples. If we take any prime power $p^{n}$ there is a doubly transitive solvable permutation group $G$ of order $p^{n}(p^{n}-1)$ and degree $p^{n}$(this is a Frobenius group which is the semidirect product of a vector space of size $p^n$ acted on by a Singer cycle of order $p^{n}-1$. A point stablizer is cyclic of order $p^n -1)$. In this group, the only centralizer orders for non-identity elements are $p^n$ and $p^{n}-1$. Hence the minimum centralizer order for $G$ is not much less than $\sqrt{|G|}.$ So it seems that you can't expect to get much less than $\sqrt{|G|}$ for the smallest centralizer order for $G$. I suspect that even this would be very difficult to prove without the classification of finite simple groups.

Later remark: It is perhaps worth remarking explicitly here (though I am sure everybody knows it), that $S_n$ contains an element whose centralizer has order $n-1$ (and this is minimal), and that for $n > 4$, $A_n$ contains an element whose centralizer has order $n-2$, which is also minimal. Hence the "generic" highly transitive permutation group of degree $n$ behaves as hoped for large $n$ (and for sufficiently large $k$ there are no others, using the Schreir conjecture for $k >7$ or the disallowed Classification of finite simple groups for $k > 5$).

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Geoff - I am very well aware of how things work out for linear algebraic groups - solvable and simple. I am asking about $k$-transitive subgroups of $Sym(n)$ ($k$ as large as needed) - they seem to be a different sort of animal. – H A Helfgott May 19 '11 at 5:30
OK, but your question did not really say that. I'll leave my answer for the benefit of others, as indicating what can go wrong for $k=2$ or $3$. If you want to make your $k$ really large, it is certainly difficult to cook up examples (since, as you know, in reality there is only $S_n$ and $A_n$ for $k >7$). As to your approach using $k(G)$, A.Maroti has strong bounds for $k(G)$ for primitive $G$ using CFSG, but I would still be interested to see even a bound like $k(G) \leq \sqrt{|G|}$ by elementary methods for this question. – Geoff Robinson May 19 '11 at 5:56
I should point out that the absence of 8-fold transitive groups other than alternating and symmetric groups was proved by Wielandt using the Schreier conjecture (solvability of outer automorphism groups of finite simple groups). This seems milder somehow than the full force of CFSG (though there is presently no proof of Schreier without CFSG). – Geoff Robinson May 19 '11 at 6:08

One possible approach - which would probably give non-optimal but non-trivial bounds - is to look at the number of conjugacy classes $Cl(G)$ in $G$.

Say the number of conjugacy classes $|Cl(G)|$ is $\leq K$. Then, by pigeonhole, there is a conjugacy class $Cl(g)$ with $\geq |G|/K$ elements. By the orbit-stabiliser theorem, we must then have $|C(g)|\leq K$ for some element $g$ of $G$.

Question: what upper bounds can one prove on $|Cl(G)|$ (relative to the size of $G$) without using the classification theorem?

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