Your question is indeed somewhat vaguely formulated, but here is an attempt to state some known and fairly up to date facts. If $G$ is a finite subgroup of ${\rm GL}(n,\mathbb{C})$ and $n > 151$, then $G$ has an Abelian normal subgroup $A$ with $[G:A] \leq 60^{n-1}$ UNLESS $G$ has a composition factor $A_{m}$ (alternating group) with $m > 151.$
On the other hand, for large $n$, it is possible to find finite nilpotent subgroups $G$ of ${\rm GL}(n,\mathbb{C})$ with $|G| > c^{n-1}$ for some constant $c > 1$. Also, for any finite nilpotent subgroup $H$ of ${\rm GL}(n,\mathbb{C}),$ there is an Abelian normal subgroup $A$ of $H$ with $[H:A] \leq 2^{n-1}.$
For a finite simple group $G$ of Lie type, it is a (slight extension of a) Theorem of E. Artin that $G$ has a nilpotent subgroup $U$ with $|G| < |U|^{3}.$
From these facts, it is possible to deduce (among other things) that there is a single fixed constant $h$ such that if $G$ is a finite simple group of Lie type which is a subgroup of ${\rm GL}(n,\mathbb{C})$ for some $n$, then $|G| \leq h^{n-1}$. Since there are only finitely many sporadic simple groups, we can enlarge the constant $h$ to cover sporadic simple groups.
This is basically the reason why we need large Alternating composition factors for $G$ if we are to be unable to bound $[G:A]$ (for $G$ a finite subgroup of ${\rm GL}(n,\mathbb{C})$ and $A$ an Abelian normal subgroup of $G$ of maximal order) by a bound of the form $[G:A] < d^{n}$ for a chosen constant $d$.