Let us take the group $S_4$. It has $24$ elements. Its group $A_4$ has $12$ elements. It is well known that the group of rigid motions of the tetrahedron has $12$ elements ($A_4$). Why? Because if we fix one vertex, we can move the rest of the vertices $3$ different ways ($120$ degrees, $240$ degrees and $360$ degrees). So it is with each of $4$ vertexes; and $4\times3=12$. Now let's be a bit more careful. One of the $3$ permutations when $1$ vertex is fixed is constant (each of the vertices goes to itself). But it makes $1$ of the permutations of $4$ vertexes the same for each of $4$ fixed vertexes, so reduces the number of different permutations. $|A_4|=12$ is a well known result and I would like to understand where I am wrong.
Group of rigid motion of tetrahedron
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group-theory
polyhedra
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2There are rigid motions on the tetrahedron that don't fix any vertices. – 2011-12-17
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3Not all motions of the tetrahedron leave a vertex fixed. Construct one and play with it: you will see! – 2011-12-17
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1A hint: it might be helpful for you to visualize [the tetrahedron as being embedded within a cube](http://i.stack.imgur.com/stZDK.gif). – 2011-12-17
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2As mentioned in the other comments, you're missing the motions corresponding to the permutations: $(12)(34), (13)(24), (14)(23)$ (the ones interchanging pairs of vertices). – 2011-12-17
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3[Here](http://www.youtube.com/watch?v=5QgIJOy7T7Y)'s a fun, tasty and inexpensive way to create models of Platonic solids. As others have pointed out, you are missing some of the symmetries. – 2011-12-18
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0Thank you for editing my stuff and your suggestions. I suspect I wasn’t clear enough in my question. I know where to get three more permutation. But the computation I presented works (it is giving the correct answer and offered in Benedict Gross lectures (lecture 18) together with additional permutations switching all vertexes). So is it coincident? The correct answer doesn’t guarantee correct thinking or it is something dipper? – 2011-12-18