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.



Generating subgroups of large index by a large chunk of a conjugacy class
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 $m \to \infty$.


