Jordan-Schur theorem on abelian normal subgroups of small index

Statement

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 $d\geq 71$. 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:

$d$	$K(d)$	Proof or example where extreme bound is attained
1	1	Any subgroup of $U_{1}(\mathbb {C} )$ is abelian by definition.
2	60	The case of special linear group:SL(2,5)

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 $d$.

Related facts

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