Statistics of bound states of anyons with order pq Anyons with fractional statistics are possible in 2 spatial dimensions, as shown by Wilczek. Suppose we have two identical anyons of spin 1/pq, where p and q are integers more than 1. Then, interchanging both of them will pick up a phase factor of $e^{-2\pi i/pq}$, right? Suppose there is a bound state of p such anyons. Then, they've got to have a spin of 1/q. However, interchanging two such identical bound states will pick up a phase factor of $e^{-2\pi i p/q}$ instead of $e^{-2\pi i/q}$?
 A: The OP is missing something very important. The case where p=q is the most interesting. Consider the specific example where p=q=2. We have two spin-1/4 anyons. Exchanging them counterclockwise (orientation matters!) picks up a phase factor of $i$ or $-i$ depending upon the specifics. If you're not careful, you'd say a bound state of two such anyons has spin-1/2, and this makes it fermionic. However, exchanging two such bound states picks up a phase factor of $i^4$ or $(-i)^4$, which is one. 
Actually, the spin of the bound state is the sum of the spins of its constituents and also their relative orbital angular momentum. The orbital angular momentum can't be integral. Look at what happens when one anyon encircles the other counterclockwise. That's two counterclockwise exchanges. Its wavefunction picks up a phase factor of $i^2$ or $(-i)^2$, which is -1. So, the orbital angular momentum modulo 1 is 1/2. So, the spin of the bound state is actually integral, and not half odd integral. This makes the bound state bosonic after all. It's no surprise that exchanging two bosons doesn't pick up any phase factor.
Back to the original question in its full generality. If you have a bound state of p anyons of spin 1/pq, the spin of the bound state is the sum of the spins of its constituent anyons, which is $1/q$, with the sum of the relative orbital angular momenta, which is $p(p-1)/pq=(p-1)/q \mod 1$. So, the overall spin of the bound state is $p/q \mod 1$, and not $1/q$.
