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At fixed finite $L$, the Helmholtz free energy of a periodic Potts chain receives its first thermal correction from a pair of domain walls. The two-domain-wall degeneracy of a Potts ring divided by the ground state degeneracy factor supplies the coefficient. Expanding the exact canonical partition function proves the formula. The expansion is not uniform in $L$: taking the thermodynamic limit first gives $F/L=-J-{\beta }^{-1}\log [1+(Q-1)e^{-\beta J}]$, whose leading correction is of order $e^{-\beta J}$.

Assume $Q\ge 2$ and $L\ge 2$; the trivial exceptions are described below. In this periodic Potts chain, every satisfied bond contributes $-J$ to the energy, and no bond can contribute less. All $L$ bonds can be satisfied simultaneously by choosing the same state at every site. Conversely, satisfying every bond forces this uniform configuration. Therefore the ground state energy and degeneracy factor areThe $Q$ choices are distinct configurations on the labelled lattice: a global relabelling of the states is a symmetry, not an identification of microstates.

For $Q,L\ge 2$, a periodic Potts chain has $Q$ uniform ground states with energy $-JL$. A single domain wall cannot close around the ring. Its first excited states have two domain walls and energy $-JL+2J$. Choose the wall bonds in $\left(\genfrac{}{}{0ex}{}{L}{2}\right)$ ways and assign distinct states to the two resulting arcs in $Q(Q-1)$ ways. This proves the displayed degeneracy factor without identifying globally relabelled configurations.