Jordan-Schur theorem on abelian normal subgroups of small index

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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 \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:

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: