Clebsch-Gordan coefficients let you treat n spins (or generaly - any n particles with arbitrary angular momentum) as a single composite system. The coefficients are simply the matrix element of basis transformation from seperated to composite system.
For 2 particles with total angular momentum eigenvalues $ l_1,l_2 $, such that for example $ m_1=-l_1,-l_1+1,...,+l_1 $ the composite system is -
$$ (l_1) \otimes (l_2) \sim (l_1+l_2) \oplus ...\oplus(|l_1-l_2|) $$
2 spin half particles reduce to -
$$ (\frac{1}{2})\otimes(\frac{1}{2})\sim (1)\oplus(0) $$
the left size decribes 2 particles, each has spin 1/2 ,and you can decribe the sany system configuration by combination of those spins -
$$ |++\rangle , |+-\rangle, |-+\rangle, |--\rangle $$
the right side describes the system as composite system, which can be described by it's total angular momentum and it's total $z$ component -
$$ |S_{tot}=1,m_s=1\rangle ,|S_{tot}=1,m_s=0\rangle , |S_{tot}=1,m_s=-1\rangle , |S_{tot}=0,m_s=0\rangle ,$$
As I said, the Clebsch-Gordan coefficients are the basis transformation matrix elements, for example, one of them is -
$$ \langle++|S_{tot}=1,m_s=1\rangle $$
The way to add 3 spins is first to add 2 of the and then to add the third one to the composite system -
$$ [(\frac{1}{2})\otimes (\frac{1}{2})]\otimes (\frac{1}{2})\sim [(1)\oplus(0)]\otimes(\frac{1}{2})\sim [(1)\otimes(\frac{1}{2})]\oplus[(0)\otimes(\frac{1}{2})]\sim (\frac{3}{2})\oplus (\frac{1}{2})\oplus(\frac{1}{2})$$
(notice the first addition. the later one is "conditional" - if S1+S2=1 then ... if S1+S2=0 then)
so the new basis, in order of $ |S_{tot},S_{1+2},m_{tot}\rangle $ is
$ |\frac{3}{2},1,-\frac{3}{2}\rangle,|\frac{3}{2},1,-\frac{1}{2}\rangle,|\frac{3}{2},1,\frac{1}{2}\rangle,|\frac{3}{2},1,\frac{3}{2}\rangle,|\frac{1}{2},1,\frac{1}{2}\rangle,|\frac{1}{2},1,-\frac{1}{2}\rangle,|\frac{1}{2},0,-\frac{1}{2}\rangle,|\frac{1}{2},0,\frac{1}{2}\rangle $
this is enough for most applications (finding energy spectrum and such)
Best Answer
The Clebsch-Gordan decomposition is for the underlying vector spaces
$$V_{j_1}\otimes V_{j_2}\cong\bigoplus_{j=|j_1-j_2|}^{j_1+j_2}V_j. \tag{1'}$$
Since the Lie group $G=SU(2)$ is simply connected, all its Lie algebra representations are also Lie group representations. [The opposite is trivial.] One can hence act$^1$ with both Lie algebra elements and Lie group elements on the underlying vector spaces (1').
$^1$ Be aware that a Lie algebra (group) element acts via Leibniz rule (simultaneously) on tensor product factors in (1'), respectively.
They both act simultaneously on direct summands in (1').
The Lie group and Lie algebra action are in this way compatible on (1') via the exponential map.