Generating random finite groups I would like a method to efficiently generate a random finite group of a given order $n$.
If there are $g(n)$ non-isomorphic groups of order $n$,
ideally each group would occur with probability $1/g(n)$.
So if $n=64$, each of the $267$ groups would be generated with equal probability.
($g(n)$ is A000001 in OEIS.)
Groups of order $n=2^k$ would be of special interest.
This is far from my expertise, and my searches must be using the wrong terminology,
because I have not found such methods. I'd appreciate pointers—Thanks!
Addendum. The comments indicate that
this appears to be an open problem, with little chance of resolution in the
near future. Now so tagged.
 A: Are you running computer experiments to verify conjectures?  If so, the GAP SmallGroups library will give you exactly what you want up to $n = 1023$.
For example, the GAP commands
  n:=16;; G:=SmallGroup(n, Random(1,NumberSmallGroups(n)));
will return you a group chosen uniformly at random from the groups of order $n=16$.  Similar commands will work up to $n=1023$.  Indeed, it will work for also for all orders up to 2000 except for 1024, and for a considerable number of other orders.
You might also find the SmallGroups library web page to be helpful:
http://www.icm.tu-bs.de/ag_algebra/software/small/
It describes some of the methods involved.  If you're willing to select a group uniformly at random from a subcollection of all the groups of order $2^k$, then there are several papers on groups of order $2^k$ (for varying values of $k$) cited there.  Applying the methods therein might be enough for you, depending on what your specific needs are.
A: Here is a stupid approach: Fill in a $n\times n$ multiplication table randomly, then check whether if satisfies the properties of a group. 
(The hard part is associativity, apart from that we basically want our multiplication table to be a "magic square", maybe there's more efficient ways to uniformly generate magic squares, and then only check associativity).
Repeat until you actually get a group. 
The end result should be uniformly distributed over all group multiplication tables, and since each group $G$ admits $(n-1)!/\left|Aut(G)\right|$ different multiplication tables, this leads to a distribution as described in Michael Zieve's comment.
This is of course far from practical, but illustrates that what you want is, in principle, possible.
