Hello everyone,

I'm having some trouble constructing an isomorphism between groups. I have a set of parameters $\displaystyle \left{j_{kl} | 1 \leq k,l \leq 3 \right}$ and a coefficient depending on these parameters. It is generally written as the 3x3 array
$\displaystyle \left{ \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix} \right}$
(usually surrounded by curly brackets, but the tex code doesn't seem to work).

The coefficient (a 9j symbol, Wigner 9j-Symbol -- from Wolfram MathWorld) is (up to a sign, which is irrelevant) unchanged upon any permutation of the rows or columns and under transposition of the array. Thus one can define symmetry operations as such permutations that leave the symbol invariant - a simple example would be the interchanging of rows 1 and 2.

It's not hard to show that these symmetry operations form a group, but I now wish to show that this group is isomorphic to the group $\displaystyle S_3 \times S_3 \times S_2$. The statement seems nearly trivial : both of the $\displaystyle S_3$'s that appear are linked to the permutations of the three rows and columns, and the group $\displaystyle S_2$ corresponds to the transposition operation. However, I can't seem to find a complete, satisfying argument or proof.

What I have found is that labelling the rows $\displaystyle r1,r2,r3$ shows that the row operations on the symbol form a group isomorphic to $\displaystyle S_3$. The same holds for the column operations by labeling the columns $\displaystyle c1,c2,c3$. Furthermore, any row permutation acts as an identity operator on $\displaystyle c1,c2,c3$ and any column permutation does not affect $\displaystyle r1,r2,r3$. Thus, the subgroup of row and column operations form a group isomorphic to $\displaystyle S_3 \times S_3$.
I don't see, however, how I could nicely introduce transpositions and show that the total group is isomorphic to $\displaystyle S_3 \times S_3 \times S_2$.

Can anyone help finding a nice argument? It does not matter if it is entirely different from the semi-proof above.
Thanks in advance!