= Solution
Use <spin one-half> states $|\uparrow\rangle,|\downarrow\rangle$ with $S^z=\pm1/2$, so $\sigma^i=2S^i$ and $H=-J\sum_n(\sigma_n^x\sigma_{n+1}^x+g\sigma_n^z)$. At $g=0$, every ferromagnetic bond has its lowest energy $-J$ when adjacent $x$ spins agree. On the connected periodic chain the normalized <ground states> are
$$
\boxed{|{+x}\rangle^{\otimes N},\quad|{-x}\rangle^{\otimes N},\qquad E_0=-NJ,}
$$
where $|\pm x\rangle=(|\uparrow\rangle\pm|\downarrow\rangle)/\sqrt2$. Their span is the two-dimensional <ground-state subspace>; any normalized superposition is another <ground state>. They are exchanged by the global spin-flip <discrete symmetry> generated by $\prod_n\sigma_n^z$.
As $g\to+\infty$, the field dominates and selects
$$
\boxed{|\uparrow\rangle^{\otimes N},\qquad E_0\sim-NJg.}
$$
The limiting polarized state is unique. At large finite $g$, exchange admixes virtual spin flips, so the product state is the limiting <wavefunction> rather than an exact finite-field <eigenstate>. The bulk expansion below gives $E_0/N=-J[g+1/(4g)+O(g^{-3})]$. The exact twofold degeneracy stated at $g=0$ should be distinguished from thermodynamic <spontaneous symmetry breaking> in the ordered phase: a finite chain at nonzero field can have a split pair of symmetry <eigenstates>.
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