= Solution
Let $n_n=c_n^\dagger c_n$ and $P_n=\prod_{m<n}(1-2n_m)$. The <Jordan–Wigner transformation> gives $2S_n^x=P_n(c_n+c_n^\dagger)$ and $P_{n+1}=P_n(1-2n_n)$. The <canonical anticommutation relations> imply
$$
(c_n+c_n^\dagger)(1-2n_n)=c_n^\dagger-c_n.
$$
Consequently the <bondwise cancellation of a Jordan-Wigner string> yields
$$
4S_n^xS_{n+1}^x=(c_n^\dagger-c_n)(c_{n+1}^\dagger+c_{n+1}),\qquad
2S_n^z=1-2n_n.
$$
Substitution therefore gives
$$
\boxed{H=-J\sum_n\left[(c_n^\dagger-c_n)(c_{n+1}^\dagger+c_{n+1})+g(1-2n_n)\right].}
$$
The <finite-ring parity qualification of an Ising quasiparticle gap> matters here. For an exact periodic spin chain, the closing bond retains total <fermion parity>. The <parity-dependent boundary condition of the Jordan-Wigner chain> is $c_{N+1}=-Pc_1$, with $P=(-1)^{N_f}$. Fixing that parity determines the allowed Fourier grid, and the resulting many-body states must still have the chosen parity. Dropping this boundary issue preserves the thermodynamic energy density, but can lose a degeneracy or the correct finite-ring excitation gap. <Physics 215C: Quantum Field Theory, Part 3 by John McGreevy>[§5.1, spin-chain fermionization and boundary sectors]{ref}
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