Queries of Proof of Wigner-Eckart Theorem With regard to the Wigner-Eckart Theorem the following is stated: The following is an outline of the proof in a text I am using:

"Consider the action of a tensor-operator component on an angular-momentum  state $$T^{(k)}_{q}| \alpha' j' m' \rangle$$ where $\alpha'$ represents other quantum numbers that do not represent angular dependence of the state. $T^{(k)}_{q}$ transforms via the rotation matrix in the same way as the angular-momentum ket $| k q \rangle$. Thus, the state $T_{q}^{(k)} | \alpha' j' m' \rangle$ transforms as the composite state $| k q \rangle |j' m' \rangle$. We can consider the usual angular-momentum addition relation $$|k q;j' m' \rangle = \sum_{k'}\sum_{q'} |k' q' \rangle \langle k' q' | j' m' ; k q \rangle\tag{*}$$ and write in analogy to it the same superposition $$T^{(k)}_{q}| \alpha' j' m' \rangle = \sum_{k' q'} | \tilde{\alpha} k' q ' \rangle \langle k' q' |j' m'; k q \rangle$$ where $\tilde{\alpha}$ is some set of transformed radial quantum numbers since the states in the relations transform equivalently. Now we can operate from the left with $\langle \alpha j m |$, we find the matrix element: $$\langle \alpha j m | T_{q}^{(k)}| \alpha' j' m' \rangle = \sum \langle \alpha j m| \tilde{\alpha} k' q'| j' m' ; k q \rangle = \langle \alpha j m | \tilde{\alpha} j m \rangle \langle j m | j' m'; k q \rangle. "$$

Question: Can anyone see how $T^{(k)}_{q}| \alpha' j' m' \rangle = \sum_{k' q'} | \tilde{\alpha} k' q ' \rangle \langle k' q' |j' m'; k q \rangle$ is in analogy with (*)? Instead of $|k' q' \rangle$ we have $| \tilde{\alpha k' q' \rangle}$ but then we kept $\langle k' q' |j' m'; k q \rangle$? How is this form decided on in analogy with $(*)$?
 A: The quick and dirty argument starts with
$$
T^{(k)}_{q}| \alpha' j' m' \rangle = \sum_{k' q'} | \tilde{\alpha} k' q ' \rangle \langle k' q' |j' m'; k q \rangle\, .
$$
Act on it with - say - $\hat L_+$:
\begin{align}
\hat L_+ T^{(k)}_{q}| \alpha' j' m' \rangle&= 
\left(\hat L_+ T^{(k)}_{q}-T^{(k)}_q \hat L_+ + T^{(k)}_q \hat L_+
\right)
| \alpha' j' m' \rangle\, ,\\
&=[\hat L_+,T^{(k)}_{q}]| \alpha' j' m' \rangle+
 T^{(k)}_q \left[\hat L_+| \alpha' j' m' \rangle\right]\, ,\\
&= \sqrt{(k-q)(k+q+1)}T^{(k)}_{q+1}| \alpha' j' m' \rangle\\
&\quad + T^{(k)}_q\sqrt{(j'-m')(j'+m'+1)}| \alpha' j' m'+1 \rangle
\tag{1}
\end{align}
where the next-to-last line follows because $T^{(k)}_{q}$ is the component of a tensor operator.
Notice this is exactly the same action as 
\begin{align}
\hat L_+ \left(\vert\alpha kq\rangle| \alpha' j' m' \rangle\right)
&=\left[\hat L_+\vert\alpha kq\rangle\right]\vert \alpha' j' m' \rangle + \vert\alpha kq\rangle 
\left[\hat L_+| \alpha' j' m' \rangle\right]\, ,\\
&=\sqrt{(k-q)(k+q+1)}\vert\alpha k,q+1\rangle\vert\alpha'\,j'm'\rangle\\
&\qquad 
+ \sqrt{(j'-m')(j'+m'+1)}\vert \alpha kq\rangle\vert\alpha' j',m'+1\rangle \tag{2}
\end{align} 
which means Eq(1) has the same combination rules as $\vert\alpha kq\rangle| \alpha' j' m' \rangle$, i.e. can reorganized as suggested using CG technology. 
Basically, if you compare (2) and (1), you realize that the action of $\hat L_+$ in (1) is the same as the action of $\hat L_+$ in (2).  Knowing that the left hand side of (2) can be expanded  over $\vert \alpha JM\rangle$ states with resulting overlaps that are CG coefficients, you expand (1) over the same set of $\vert \alpha JM\rangle$ states and the expansion coefficients will be identical to those of (2), i.e. will be a CG.
