# Polya theorem, isometries

I decided to use free time to refresh my knowledge about Polya enumeration theorem. Unfortunately I can't move forward, because I completely don't understand a few things.

I'm confused, how to understand the term isometry in tasks on this theorem? From the definition isometry is a transformation that doesn't change the distance among objects.

But for example one day I was told, on my Discrete Mathematics course, that the cycle index of all isometries of $n$-polygon is: $I_G(x_1,...,x_n)=\frac{1}{2n}\left( nx_1x_2^{(n-1)/2} +\sum_{d|n}\phi(d) x_d^{n/d} \right)$ for an odd $n$. So we consider here rotations and symmetries as isometries. So it is quite easy, after thinking a while (maybe a little longer while).

But on the same course I was told that the cycle index of all isometries of regular tetrahedron is: $I_G(x_1,x_2,x_3,x_4)=\frac{1}{24}(x_1^4+6x_1^2x_2+8x_1x_3+3x_2^2+6x_4)$ because $S_4$ (group of $4$-permutations) is affecting this solid's vertices. But what we consider here as isometries? I can't find all transformations. Are we simply permutating all vertices? But it is in conflict with "isometries" of $n$-polygon above (that is: symmetries and rotations). I can see that permutations: $1\times x_1^4, \ 3\times x_2^2, \ 8\times x_1x_3$ are rotations, but I don't know what are the rest transformations responsible for.

In one task that I found, I am asked to find the cycle index of all isometries of cuboid with length of edges: $1$ and $2$. The hint is that the result is $16$. I'm confused, I don't have any Rubik's cube in that shape to help. And I don't see this in my mind, can't count all isometries. Is there any way to do it for a person that has poor imagination?

I know that I want a lot of explanations, but I just can't handle on my own. I would be very, very grateful for any help.

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You have 12 isometries for the tetrahedron or 24 if you consider those ones with negative determinant (sometimes called non proper). –  Sigur Aug 20 '12 at 23:01
So these non proper isometries have nothing to do with symmetries? They are something else? –  ray Aug 21 '12 at 8:03
They are, but symmetries around a plane, for example. Consider the reflexion that preserves an edge and swap the other two vertices (a plane like a mirror). You can not do this without self intersection in 3-dimensional space, but you can in 4-dimensional. –  Sigur Aug 21 '12 at 12:33
Please answer your own question and accept it to close the issue. –  vonbrand Feb 10 at 3:12