Jordan-Schur theorem on abelian normal subgroups of small index

In terms of existence of a bounding function on index
There exists a function $$K:\mathbb{N} \to \mathbb{N}$$ such that the following holds:

Suppose $$G$$ is a finite group with a faithful linear representation of degree $$d$$ over the field of complex numbers (equivalently, $$G$$ is isomorphic to a subgroup of the unitary group $$U_d(\mathbb{C})$$). Then, $$G$$ has an abelian normal subgroup of index at most $$K(d)$$.

Explicit description of bounding function
The smallest possible function $$K$$ satisfying the above has the property that $$K(d) = (d + 1)!$$ for $$d \ge 71$$. The proof that this works relies on the classification of finite simple groups. For the first few values of $$d$$, $$K(d)$$ is not explicitly known for all of them. However, we do have the first few values:

Corollary for quasirandomness
If $$G$$ is a perfect group and has no proper normal subgroup of index at most $$K(d - 1)$$, then the quasirandom degree of $$G$$ is at least equal to $$d$$.

Converse
These aren't strict converses, but converse-type statements:


 * Degree of irreducible representation is bounded by index of abelian subgroup
 * Degree of irreducible representation divides index of abelian normal subgroup