If we look at the cross ratio $(x_0 x_1:x_2 x_3) = \lambda$ of 4 points in projective space, I can see that by looking at all possible permutations (24 of them) of the points we can see that only 6 of them give the same cross ratio.
I.e: $\lambda, \frac{1}{\lambda}, 1-\lambda, \frac{1}{1-\lambda}, \frac{\lambda}{\lambda -1}, \frac{\lambda-1}{\lambda}$
Is there a more efficient way to see this rather than calculating all 24 permutations?
Write the cross product as
$$ \lambda = \frac{(x_1-x_3)(x_2-x_4)}{(x_2-x_3)(x_1-x_4)} $$
As we have the permutation group $S_4$ we only need the generators $(12)$, $(13)$ and $(14)$.
Now
$$(12) \lambda = \frac{(x_2-x_3)(x_1-x_4)}{(x_1-x_3)(x_2-x_4)} = \frac{1}{\lambda}$$ $$(13) \lambda = \frac{(x_3-x_1)(x_2-x_4)}{(x_2-x_1)(x_3-x_4)} = \frac{\lambda}{\lambda - 1} $$ $$(14) \lambda = \frac{(x_4-x_3)(x_2-x_1)}{(x_2-x_3)(x_4-x_1)} = 1 - \lambda $$
What we find is that
$$1 \lambda = 1$$ $$(12)(34) \lambda = \lambda$$ $$(13)(24) \lambda = \lambda$$ $$(14)(23) \lambda = \lambda$$
which is the Klein-Four group $V_4$,
So we can write $S_n = V_4 \otimes S_3$
$V_4$ has 4 elements and $S_3$ has 6 elements. The $S_3$ gives the
$$\lambda, \frac{1}{\lambda}, 1 - \lambda, \frac{1}{1-\lambda}, \frac{\lambda}{1-\lambda}, \frac{1-\lambda}{\lambda}$$
Total we have $24 = 4 \times 6$.
