Symmetry of Wigner's 3j symbol I'm trying to understand the symmetry of Wigner's 3$j$ symbols.
$3!=6$ permutations, as well as flipping the sign of all magnetic quantum numbers yields 12 operations, which are sometimes called the "12 classical ones".
Wikipedia lists two additional "Regge" symmetries, which would (to my understanding) both double the number of possible operations, yielding 48 symmetry operations. However, the 3$j$ symbols apparently have 72 symmetries.

What am I missing?

 A: The two Regge symmetries both have order $2$, but they don't both double the number of operations, because they don't commute with all of the other operations (or with each other). The Wikipedia article linked in the question actually mentions a nice way of expressing the operations that makes the full structure of the group (and the fact that it has $72$ elements) more obvious. I'll explain it in more detail. Use the abbreviation
$$
\newcommand{\bfv}{\mathbf{v}}
\newcommand{\magic}{{M}}
\newcommand{\threej}{\Omega}
 \bfv \equiv (j_1,\, j_2,\, j_3,\, m_1,\, m_2,\, m_3)
$$
for the list of numbers appearing in the $3$-$j$ symbol
$$
\threej(\bfv) \equiv
\left(\begin{matrix}
   j_1 & j_2 & j_3 
 \\ m_1 & m_2 & m_3
\end{matrix}\right).
$$
Using the same list of numbers, define the matrix
$$
\magic(\bfv) \equiv
\left[\begin{matrix}
   -j_1+j_2+j_3 & j_1-j_2+j_3 & j_1+j_2-j_3 \\
 j_1 + m_1 & j_2 + m_2 & j_3 + m_3 \\
 j_1 - m_1 & j_2 - m_2 & j_3 - m_3
\end{matrix}\right].
$$
Notice that each row sums to $j_1+j_2+j_3$, as does each column. Now, define these linear operations on $\bfv$:

*

*$\bfv\to C(a,b)\bfv$ is the linear transformation whose effect on $\magic$ is to swap columns $a$ and $b$.


*$\bfv\to R(a,b)\bfv$ is the linear transformation whose effect on $\magic$ is to swap rows $a$ and $b$.


*$\bfv\to T\bfv$ is the linear transformation whose effect on $\magic$ is to take the transpose.
Notice that $j_1+j_2+j_3$ is invariant under all of these operations. Using the abbreviation
$$
 s(\bfv) \equiv (-1)^{j_1+j_2+j_3},
$$
the corresponding symmetries of $\threej(\bfv)$ are:

*

*$\threej(\bfv)\to s(\bfv)\threej(C(a,b)\bfv)$. This swaps columns $a$ and $b$ in the $3$-$j$ symbol and multiplies the result by the sign $s(\bfv)$.


*$\threej(\bfv)\to s(\bfv)\threej(R(2,3)\bfv)$. This replaces $(m_1,m_2,m_3)\to(-m_1,-m_2,-m_3)$ in the bottom row of the $3$-$j$ symbol and multiplies the result by the sign $s(\bfv)$.


*$\threej(\bfv)\to \threej(T\bfv)$. This is the first of the Regge symmetries shown in the Wikipedia article. Explicitly:
$$
\threej(T\bfv) = 
\left(\begin{matrix}
   j_1 & \frac{j_2+j_3-m_1}{2} & \frac{j_2+j_3+m_1}{2} 
 \\ j_3-j_2 & \frac{j_2-j_3-m_1}{2}-m_3 
     & \frac{j_2-j_3+m_1}{2}+m_3 
\end{matrix}\right).
$$


*$\threej(\bfv)\to s(\bfv)\threej(R(1,3)\bfv)$. This is the second of the Regge symmetries shown in the Wikipedia article. Explicitly:
$$
s(\bfv)\threej(R(1,3)\bfv) = 
(-1)^{j_1+j_2+j_3}
\left(\begin{matrix}
   \frac{j_2+j_3+m_1}{2} &
   \frac{j_3+j_1+m_2}{2} &
   \frac{j_1+j_2+m_3}{2} \\
 j_1 - \frac{j_2+j_3-m_1}{2} &
 j_2 - \frac{j_3+j_1-m_2}{2} &
 j_3 - \frac{j_1+j_2-m_3}{2}
\end{matrix}\right).
$$
This correspondence shows that the group generated by these symmetries of the $3$-$j$ symbol is identical (as an abstract group) to the group generated by the linear transformations $C,R,T$ of $\bfv$. Because of the way those transformations were defined by their effect on the matrix $\magic(\bfv)$, we can count the number of elements in this group relatively easily. The group generated by the $C$s has $3!=6$ elements, as does the group generated by the $R$s, and the group generated by $T$ has $2$ elements. The $C$s and the $R$s commute with each other and cannot undo each other, so the group generated by the $C$s and $R$s has $6\times 6 = 36$ elements. The operation $T$ doesn't commute with the $C$s and $R$s, but it does satisfy $TG=GT$ where $G$ is the group generated by the $C$s and $R$s, so the group generated by the $C$s, $R$s, and $T$ has $36\times 2=72$ elements.
