All of the maps studied thus far are involutions. The question now arises which group is generated by these involutions. We proceed to answer this question.

For this purpose it is expedient to switch to different couplings. We do this in two steps. First, we set $$\begin{gathered}
\label{lg}
\lambda =g-\zeta/2,\qquad \zeta\equiv (1,1,1,1),\end{gathered}$$ so that $$\begin{gathered}
\label{Hl}
H(g)\to -\frac{d^2}{dx^2}+\sum_t\left(\lambda_t^2-\frac{1}{4}\right)\wp(x+\omega_t).\end{gathered}$$ Also, (\[triv\]) and (\[refl\]) become $$\begin{gathered}
\label{ltr}
\lambda_2\to -\lambda_2,\qquad \lambda_3\to -\lambda_3,
\\
\label{lre}
(\lambda_0,\lambda_1,\lambda_2,\lambda_3)\to (\lambda_1,\lambda_0,\lambda_3,\lambda_2),\end{gathered}$$ and we have $$\begin{gathered}
\lambda = \lambda' \ \Leftrightarrow  \ \lambda_0 +\lambda_2 =\lambda_1 + \lambda_3,\end{gathered}$$ since $\zeta$ is self-dual.

The second step may seem unmotivated at first sight. It consists in defining couplings $$\begin{gathered}
\label{cl}
c_0=\lambda_0 +\lambda_3,\qquad c_3=\lambda_0 -\lambda_3, \qquad c_1=\lambda_1 +\lambda_2,\qquad c_2=\lambda_1 -\lambda_2.\end{gathered}$$ The point of this definition is that the maps (\[ltr\]) become permutations $$\begin{gathered}
\label{ctr}
c_0\leftrightarrow c_3,\qquad c_1\leftrightarrow c_2,\end{gathered}$$ while (\[lre\]) turns into the permutation $$\begin{gathered}
\label{cre}
(c_0,c_1,c_2,c_3)\to (c_1,c_0,c_3,c_2).\end{gathered}$$ Moreover, the $J_N$-action transforms into the permutation $$\begin{gathered}
\label{Jc}
c_2\leftrightarrow c_3,\end{gathered}$$ as is easily checked.

Since the three transpositions (\[ctr\]) and (\[Jc\]) already generate $S_4$, it follows that (\[cre\]) may be viewed as a suitable product of these maps. The symmetries (\[ctr\]), (\[cre\]) generate an 8-element $S_4$-subgroup isomorphic to the dihedral group $I_2(4)$. This subgroup 