 Solution

In the charge-flux composite model, particle $1$ acquires the Aharonov-Bohm effect phase $q_1{\Phi }_2/\hslash$ when it circles the flux of particle $2$. The reciprocal Aharonov-Casher effect contributes $q_2{\Phi }_1/\hslash$. Thus the full braid isThe assumption for $a=(q_0,0)$ and $b=(0,{\Phi }_0)$ saysFor any allowed particle $(n{q}_0,m{\Phi }_0)$, the full braid with $a^2=(2q_0,0)$ is $e^{2im{q}_0{\Phi }_0/\hslash }=1$; similarly every particle braids trivially with $b^2=(0,2{\Phi }_0)$. Under the stated operational identification,For $c=a\times b=(q_0,{\Phi }_0)$, exchanging two identical composites is half their full braid and givesTherefore $a$ and $b$ reproduce the bosonic $e$ and $m$ particles, in either order, and $c$ reproduces their fermionic fusion product $f$.