= Solution
Let $[a,a^\dagger]=1$, $n=a^\dagger a$, and take $S^+=(S^-)^\dagger=\sqrt{2S-n}\,a$. The physical <Holstein–Primakoff occupation constraint> limits the <Hilbert space>, which contains only $n=0,1,\ldots,2S$. Acting in this number basis gives
$$
S^-|n\rangle=\sqrt{(n+1)(2S-n)}|n+1\rangle,\qquad S^+|n\rangle=\sqrt{n(2S-n+1)}|n-1\rangle.
$$
The upper and lower endpoints are annihilated as required. Consequently
$$
[S^+,S^-]|n\rangle=\{(n+1)(2S-n)-n(2S-n+1)\}|n\rangle=2(S-n)|n\rangle=\boxed{2S^z|n\rangle}.
$$
The change in $n$ also gives $[S^z,S^\pm]=\pm S^\pm$. These <matrix elements> prove the <spin> algebra on the full physical spin-$S$ space. Extending the oscillator space to arbitrary occupation would not give a physical finite <spin>: the occupation constraint and operator ordering in the square root matter. <bosons> on distinct sites have $[a_m,a_n^\dagger]=\delta_{mn}$ and all other site <commutators> zero.
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