Why every finite non-abelian simple group of order $n$ has a proper subgroup of index at most $kn^{\frac{3}{7}}$?

Question

Can anyone please clarify me on why every finite non-abelian simple group of order $n$ has a proper subgroup of index at most $kn^{\frac{3}{7}}$ for some positive constant $k$? Thanks

Answer 1

$A_d$ has $A_{d-1}$ has a subgroup, so it's true with a much better bound for the alternating groups. That leaves the finite simple groups of Lie type.

At this point I'm only going to talk about $\text{PSL}_d(\mathbb{F}_q)$. I'll also only work up to a multiplicative constant or maybe worse.

Finding subgroups of small index is equivalent to finding small sets on which a group acts transitively. $\text{PSL}_d(\mathbb{F}_q)$ naturally acts transitively on $\mathbb{P}^{d-1}(\mathbb{F}_q)$, which has size $\sim q^{d-1}$. Now, $\text{PSL}_d(\mathbb{F}_q)$ itself has size $n \sim q^{d^2 - 1}$, so we've found a subgroup of index $\sim n^{ \frac{1}{d+1} }$. Since $d \ge 2$, we have $\frac{1}{d + 1} \le \frac{1}{3} < \frac{3}{7}$, so the bound is satisfied here.

The $\frac{3}{7}$ bound might be tight for one of the other infinite families of groups of Lie type, but I don't know enough about them to say. In any case they should all have transitive actions that look roughly like the above so I imagine the argument isn't difficult, although it might be tedious. For example the orthogonalish groups act on unit spheres.