= Solution
Use the <canonical anticommutation relations> $\{c_n,c_m^\dagger\}=\delta_{nm}$ and $\{c_n,c_m\}=0$. In particular, $n_m=c_m^\dagger c_m$ is a <projector>, so $(1-2n_m)^2=1$. Define $K_n=\prod_{m<n}(1-2n_m)$. Then $K_n^\dagger=K_n$, $K_n^2=1$, and $K_n$ commutes with $c_n$ and $c_n^\dagger$. Taking the adjoint of the given raising operator gives $S_n^-=c_n^\dagger K_n$. Hence
$$
[S_n^+,S_n^-]
=K_nc_nc_n^\dagger K_n-c_n^\dagger K_n^2c_n
=c_nc_n^\dagger-c_n^\dagger c_n
=1-2n_n=2S_n^z.
$$
Also $[n_n,c_n]=-c_n$ and $[n_n,c_n^\dagger]=c_n^\dagger$ imply $[S_n^z,S_n^\pm]=\pm S_n^\pm$. For different sites, exchanging two <fermion> operators contributes one minus sign and commuting an operator through the other site's parity string contributes another; their product is positive. Thus distinct <spin> algebras commute. \b[The string supplies precisely the signs required for the <Jordan–Wigner transformation>.] The <fermion> vacuum corresponds to <spin> up, while one occupied site corresponds to <spin> down; this convention fixes subsequent signs.
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