Jordan–Schur theorem

Remarks

Jordan proved the theorem for finite groups. Schur extended it to torsion groups of matrices: every torsion subgroup of $\mathrm{GL}_n(\C)$ has an abelian normal subgroup of finite index.

Idea of the proof: a finite group $G$ preserves a Hermitian inner product, so after a change of basis $G$ lies in the unitary group $U(n)$. If two unitary matrices are close enough to the identity, their commutator is even closer to the identity. In a finite group this forces the elements in a small ball around the identity to generate an abelian subgroup $A$. This $A$ is normal, since conjugation by unitary matrices preserves distances to the identity. Its index is at most the number of small balls needed to cover $U(n)$, which depends only on $n$.

Example: for $n = 2$, the dihedral group of order $2m$ acts on $\C^2$ by rotations and reflections of the plane. It is not abelian for $m \ge 3$, but its rotations form a cyclic normal subgroup of index $2$.

The finiteness or torsion hypothesis is necessary. The group $\mathrm{SL}_2(\Z) \subseteq \mathrm{GL}_2(\C)$ contains a nonabelian free group of finite index, and a nonabelian free group has no abelian subgroup of finite index.

Going deeper

For a student

The bound $J(n)$ does not depend on the order of $G$, and this is the whole point. The same commutator argument in the Euclidean group is behind the first of the Bieberbach theorems on crystallographic groups. The theorem also shows that a finite group with no abelian normal subgroup of small index, such as a large alternating group, has no faithful complex representation of small degree.