TITLE: The finite subgroups of SU(n)
QUESTION [42 upvotes]: This question is inspired by the recent question "The finite subgroups of SL(2,C)". While reading the answers there I remembered reading once that identifying the finite subgroups of SU(3) is still an open problem. I have tried to check this and it seems it was at least still open in the Eighties.
Can anyone confirm or deny that the finite subgroups of SU(3) are not all known? And if this is true, then what is the source of the difficulty? 
Secondly, what is known of the finite subgroups of SU(n)  for n > 3?
UPDATE: Thanks to those below who have corrected my ignorance! It seems that I may have been tricked by some particularly sensationalised abstracts (or perhaps just misunderstood them.)

REPLY [2 votes]: Just want to point out that an algorithm to answer,
given $n$ and a presentation of a finite group $G$, whether $G$ is isomorphic to a finite subgroup of $SU(n)$,
exists as a consequence of Tarski's Theorem on the decidability of the real field $(\mathbb R,+,\times)$!
Namely, the statement that certain matrices $A_1,\dots,A_k$ exist that satisfy the corresponding equations under matrix multiplication can be written as a statement in the first-order theory of algebraically closed fields of characteristic zero, i.e., the theory of $\mathbb C$. This reduces to $\mathbb R$ as pointed out by Joel elsewhere on MO.
How much faster the Zassenhaus algorithm mentioned by José is, I don't know...