I'm interested in Jordan's theorem which (after applying the unitary trick) states that any finite subgroup of $U_n(\mathbb{C})$ has an abelian subgroup of index $F(n)$, a function depending only on $n$ and not on the finite group.

The "standard" reference for this theorem is, I think, the book of Curtis and Reiner (Representation theory of finite groups and associative algebras, Chapter 36), and I think the argument there is due to Frobenius and Schur. But on Tao's blog

http://terrytao.wordpress.com/2010/02/18/a-proof-of-gromovs-theorem/

there's a slight variant on this argument which I personally find a little more natural.

Here's a very brief sketch. We work by induction on n. Let A be our finite subgroup of $U_n(\mathbb{C})$. Look at the intersection $A'$ of $A$ with a small ball about the identity; this must constitute a reasonable fraction of $A$. Divide into two cases. Case 1: everything in $A'$ is a multiple of $I_n$. Then we are done simply by looking at the subgroup of $A$ generated by $A'$. Case 2: there is an element $\gamma \in A'$ which is not a multiple of $I_n$. Take the nearest such element to $I_n$. Then one may argue that for any $x \in A'$ the commutator $[\gamma,x]$ is both nearer to $I_n$ than $\gamma$ and not a nontrivial multiple of $I_n$: hence it must *equal* $I_n$ and so $\gamma$ is centralised by the whole of $A'$. However the centraliser of an element such as $\gamma$ is a product of $U_{n_i}(\mathbb{C})$'s with $n_i < n$, and that means we can proceed by induction.

I asked Terry about this and he said someone had sketched the "choose an element closest to the identity" part of the argument and this had seemed the most natural way to conclude.

Anyway, I'd be interested in any comments people have on this or on proofs of Jordan's theorem in general (for example on Jordan's original proof). Can one find the above argument somewhere in the literature? Or was it passed over because it doesn't give especially good bounds on the index of the abelian subgroup?