= Solution
In the <charge-flux composite> model, particle $1$ acquires the <Aharonov-Bohm effect> phase $q_1\Phi_2/\hbar$ when it circles the flux of particle $2$. The reciprocal <Aharonov-Casher effect> contributes $q_2\Phi_1/\hbar$. Thus the full braid is
$$
\exp(2i\theta_{12})
=\exp\left[\frac i\hbar(q_1\Phi_2+q_2\Phi_1)\right].
$$
The assumption for $a=(q_0,0)$ and $b=(0,\Phi_0)$ says
$$
\frac{q_0\Phi_0}{\hbar}=\pi\pmod{2\pi}.
$$
For any allowed particle $(nq_0,m\Phi_0)$, the full braid with $a^2=(2q_0,0)$ is $e^{2imq_0\Phi_0/\hbar}=1$; similarly every particle braids trivially with $b^2=(0,2\Phi_0)$. Under the stated operational identification,
$$
a\times a=b\times b=1.
$$
For $c=a\times b=(q_0,\Phi_0)$, exchanging two identical composites is half their full braid and gives
$$
e^{i\theta_{cc}}=e^{iq_0\Phi_0/\hbar}=-1.
$$
Therefore $a$ and $b$ reproduce the bosonic $e$ and $m$ particles, in either order, and $c$ reproduces their fermionic fusion product $f$.
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