2
$\begingroup$

Let $G$ be a finite simple group and let $C$ be a (non-trivial) conjugacy class of $G$. Let $H$ be a subgroup of $G$ such that $$|H\cap C| \geq \epsilon |C|.$$ Can one conclude that the index of $H$ in $G$ is bounded by a constant that depends on $\epsilon$, but not on $G$ or $C$?

This is not a particularly urgent question (it is related to something in my thesis from twenty years back), but I feel that it is simple enough to formulate and is probably known to the experts.

$\endgroup$

1 Answer 1

4
$\begingroup$

It looks to me as if this will not work with $3$-cycles in $G = A_{n}.$ Suppose, for example, that $n =2m$ and take $H=A_{m}.$ Then $G$ contains $\frac{n(n-1)(n-2)}{6}$ $3$-cycles, and $H$ contains $\frac{m(m-1)(m-2)}{6}$ $3$-cycles, so about $\frac{1}{8}$ of the $3$-cycles in $G.$ However, $[G:H] \to \infty$ as $m \to \infty.$

(Later edit in response to question in comment. I think it also fails for transvections in $G = {\rm GL}(2m,2)$ and $H = {\rm GL}(2m-1,2).$ The number of transvections in $G$ is ( I think) $2^{4m-1} - 2^{2m-1}$ and the number of transvections in $H$ is $2^{4m-3} - 2^{2m-2}.$ For large $m,$ the proportion of tranvections from $G$ which lie in $H$ is around $\frac{1}{4}$ but $[G:H] \to \infty$ as $m \to \infty$.

$\endgroup$
2
  • $\begingroup$ Thanks, this makes perfect sense. What do you think will happen if $G$ is a Chevalley group with a fixed type of group but arbitrary finite field? $\endgroup$ Dec 25, 2014 at 2:07
  • $\begingroup$ In response to your edit, I am interested in the cases like $GL(n,q)$ where $n$ is fixed but the field size goes to infinity. $\endgroup$ Dec 25, 2014 at 13:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.