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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.

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There are rigid motions on the tetrahedron that don't fix any vertices. –  Brandon Carter Dec 17 '11 at 17:31
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Not all motions of the tetrahedron leave a vertex fixed. Construct one and play with it: you will see! –  Mariano Suárez-Alvarez Dec 17 '11 at 17:31
    
A hint: it might be helpful for you to visualize the tetrahedron as being embedded within a cube. –  J. M. Dec 17 '11 at 17:45
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As 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). –  Bill Cook Dec 17 '11 at 20:08
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Here'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. –  Michael Joyce Dec 18 '11 at 4:29
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1 Answer

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 $G$.

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