= Solution
For a finite-range isotropic <Heisenberg model>, the <Mermin-Wagner theorem> excludes spontaneous magnetic order in one dimension at every nonzero <temperature>. Long-wavelength rotations cost very little <energy>: a smooth orientational fluctuation has $\delta F=(\rho_s/2)\int dx\,(\partial_x\theta)^2$, so its thermal variance contains $(k_{\rm B}T/\rho_s)\int dq/q^2$, which diverges in the infrared. In a <Heisenberg ferromagnet>, the same obstruction appears as a divergent thermal occupation of quadratic long-wave <spin waves>. This explains why a finite magnetization cannot be supported, although correlations can be strong over a long finite distance. The rigorous statement requires short-range interactions and the stated continuous symmetry; interchain coupling and anisotropy change the problem. Symmetry forcing the mean to zero in a finite chain alone would not prove absence of thermodynamic order. <Absence of Ferromagnetism or Antiferromagnetism in One- or Two-Dimensional Isotropic Heisenberg Models by N. D. Mermin and H. Wagner>[1966 abstract and theorem: finite-range isotropic spin models at nonzero temperature]{ref}
For the <Ising model>, use dimensionless $s_i=2S_i^z=\pm1$, $h=g\mu_{\rm B}B$, $K=\beta J/2$ and $b=\beta h/2$. Its pair <Boltzmann factor> is
$$
P_{s,s'}=\exp\left[Kss'+\frac b2(s+s')\right].
$$
Multiplying N such factors around the periodic ring counts each bond once and each field term twice with weight one half. Summing the closed spin-index sequence therefore gives the <transfer matrix for the one-dimensional Ising model>
$$
\boxed{P=\begin{pmatrix}e^{K+b}&e^{-K}\\e^{-K}&e^{K-b}\end{pmatrix},\qquad Z=\operatorname{Tr}P^N.}
$$
The two <eigenvalues> are
$$
\lambda_\pm=e^K\left[\cosh b\pm\sqrt{\sinh^2b+e^{-4K}}\right],\qquad
Z=\lambda_+^N+\lambda_-^N.
$$
At finite temperature every entry of P is positive, so the <Perron–Frobenius theorem> gives $\lambda_+>|\lambda_-|$. Consequently
$$
\boxed{F=-\frac{NJ}{2}-Nk_{\rm B}T\ln\left[\cosh b+\sqrt{\sinh^2b+e^{-2\beta J}}\right]+o(N).}
$$
The exact finite-N correction to the extensive expression is $-k_{\rm B}T\ln[1+(\lambda_-/\lambda_+)^N]$. No assumption that $\lambda_-$ is positive is needed.
Differentiate at fixed temperature. For $D=\sqrt{\sinh^2b+e^{-2\beta J}}$, $\partial_b\ln(\cosh b+D)=\sinh b/D$. The <Ising-chain magnetization with spin-half normalization> gives total <magnetic moment>
$$
\boxed{M=-\frac{\partial F}{\partial B}=\frac{Ng\mu_{\rm B}}2\frac{\sinh(\beta g\mu_{\rm B}B/2)}{\sqrt{\sinh^2(\beta g\mu_{\rm B}B/2)+e^{-2\beta J}}}.}
$$
This is the thermodynamic-limit result; the moment per site is M/N. For every finite $T>0$, \b[$\lim_{B\to0}M/N=0$], even with the thermodynamic limit taken first. The zero-field <magnetic susceptibility> per site is $(g\mu_{\rm B})^2e^{\beta J}/(4k_{\rm B}T)$, large but finite.
The <Ising-chain domain-wall energy convention> explains the result for $J>0$: changing a parallel bond into an antiparallel bond costs J in this spin-half normalization. Domain walls, occurring in pairs on a ring, acquire a nonzero density at any nonzero temperature. Indeed the zero-field <correlation length> is $\xi=-a/\ln[\tanh(\beta J/2)]\sim(a/2)e^{\beta J}$ at low temperature. The Ising symmetry is discrete, so this is a separate one-dimensional domain-wall mechanism rather than an application of the continuous-symmetry theorem. At zero temperature a ferromagnetic chain can select a fully polarized state; the $B\to0$ and $T\to0$ limits do not commute.
\Image[/past-exam-of-the-advanced-quantum-condensed-matter-physics-course-of-the-university-of-cambridge/2012/paper-1-aqc2012-ising-magnetization.png]
{title=Ising magnetization}
{source=paper-1-aqc2012-ising-magnetization.py}
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