Why is the Symmetry Group of the 9j Symbol Isomorphic to S_3 x S_3 x S_2?

[Yoran91]

Hello everyone,

I read in Edmond's 'Angular momentum in Quantum Mechanics' that the symmetry group of the 9j symbol is isomorphic to the group $S_3 \times S_3 \times S_2$.

Why is this? Can anyone shed some light on this?

 

[Bill_K]

Edmonds says,

we may permute the rows or columns in the matrix forming the 9-j symbol, or transpose the matrix itself...The symmetry group [is] the product of the three permutation groups of three, three and two objects respectively.

Likewise, the Wikipedia page on 9-j symbols describes it this way.

 

[Yoran91]

I don't see the full picture, yet.

Labelling the rows $r_1,r_2,r_3$ and the columns $c_1,c_2,c_3$, it's easy to show that the subgroups of the row and column operations are both isomorphic to $S_3$. Since any row permutation does not affect the order of the $c_i$, its an element of $S_3 \times e$ and any column permutation is in $e \times S_3$ in the same way.

So the subgroup of all symmetry operations not containing a transposition of the array is $S_3 \times S_3$. But how do you take the transpositions into account?

I see that relevant subgroup is $S_2$, but I don't see exactly how you go from $S_3 \times S_3$ to $S_3 \times S_3 \times S_2$ by taking the transpositions into account.