= Solution
For the <Transverse-field Ising model>, use dimensionless <spin one-half> operators, so $S^x$ has eigenvalues $\pm1/2$. At zero transverse field every aligned bond contributes $-J$. The two <ground states> are
$$
\boxed{|X_\pm\rangle=\bigotimes_{n=1}^N|\!\pm x\rangle_n,\qquad E_0=-JN.}
$$
They span an exactly two-dimensional ground space on the finite periodic ferromagnetic chain. Any linear combination is also a ground state; parity eigenstates are the symmetric and antisymmetric combinations. Choosing one ordered product state in the thermodynamic limit is <spontaneous symmetry breaking>.
For large positive field the leading term is $-2Jg\sum_nS_n^z$. Its unique minimizing state is
$$
\boxed{|Z\rangle=\bigotimes_n|\!+z\rangle_n,\qquad E_0\sim-JgN\quad(g\to\infty).}
$$
The exchange has zero first-order expectation in this state and produces virtual spin-pair fluctuations at finite g. Thus the polarized product is the limiting wavefunction, not an exact finite-field ground state. Bulk perturbation theory gives $E_0/N=-Jg-J/(4g)+O(J/g^3)$. The field-polarized phase has no twofold ordered-state degeneracy.
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