How to define the order parameter of the q-state Potts model?

First one needs to gain a deeper understanding of the order parameter of the Ising model. The magnetization $m$ can be written as $m=(+1)n_1+(-1)n_2$, where $n_p=N_p/N$ ($p=1,2$) is the number density of each spin. So what do the coefficients $(\pm1)$ stand for? They are two possible magnetizations of a single Ising spin, which form the representation of the $\mathbb{Z}_2$ group. So the Ising spin is also known as the $\mathbb{Z}_2$ spin.

Now it is straight forward to extend the formulation to the $q$-state Potts model. Simply replace the number 2 by $q$, i.e. to consider the Potts spin as the $\mathbb{Z}_q$ spin, whose "magnetization" must be taken from the representation of the $\mathbb{Z}_q$ group, which are $q$th roots of unity $e^{2\pi ip/q}$ ($p=1,2,\cdots,q$). Given the representation of the Potts spin, it is easy to write down the order parameter $$m=\sum_{p=1}^qe^{2\pi i\frac{p}{q}}n_p,$$ with $n_p=N_p/N$ (let $N=\sum_pN_p$) being the number density of the $p$th type of the Potts spin. Note that the order parameter is complex in general. In the disordered phase, all types of Potts spin appear with equal probability, i.e. $n_1=n_2=\cdots=n_q$, and in this case, we do have $m=0$ due to the cancellation of the phase, which is consistent with the idea of the order parameter: "a quantity that is zero in the disordered phase and non-zero in the ordered phase".