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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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
1 Answers
A better way of completing your argument would be: The group $G$ of rigid motions acts transitively on the set of 4 vertices. Thus from general principles of transitive group actions it follows that $$ |G|=4\cdot Stab_G(v), $$ where $v$ denotes a given vertex, and $Stab_G(v)$ is the subgroup of those rigid motions that stabilize the vertex $v$.
The group $G$ can be written as a union of cosets of $Stab_G(v)$. Each coset corresponds to the set of rigid motions that maps the given vertex $v$ to another vertex $w$. This set is a subgroup only, if $w=v$. If $\rho_{v,w}$ is one rigid motion that maps $v\mapsto w$, then the coset $\rho_{v,w}Stab_G(v)$ contains all such motions. This is a different set of motions from the conjugate subgroup $$ \rho_{v,w}Stab_G(v)\rho_{v,w}^{-1} $$ that is also equal to the subgroup $Stab_G(w)$ consisting of motions sending vertex $w$ to itself.
The 'confusion' (it is, perhaps, a stretch to call it a confusion, because you know how it goes) seems to be that the cosets of $Stab_G(v)$ do form a partition of the whole group, but its conjugates won't. For the purposes of tallying the elements it doesn't matter, because if a finite group $G$ acts transitively on the set $X$, then $$ |G|=|X|\cdot |Stab_G(x)|, $$ for any fixed element $x\in X$. Because conjugate subgroups have the same size, we also always have $$ |G|=\sum_{x\in X}|Stab_G(x)|, $$ even though the subgroups $Stab_G(x)$ do not form a partition of $G$.