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If I have a natural number $o=2n$ or $4n$ I can create a polyhedron whose group of symmetries has order $o$ by making a polygon like $C_n$ and then dragging it out to make a prism (I believe this is dihedral symmetry in the case $o=4n$; you need colors in the $o=2n$ case, not sure what this is called).

What if $o=3n$? Is the group $C_n \times C_3$ isomorphic to the group of symmetries for any polyhedron? It seems like there should be some clever way to make a pyramid-type structure, but I can't quite figure it out.

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Your terminology seems to be somewhat unusual. By "a polyhedron with symmetries of order $o$", do you mean a polyhedron whose symmetry group is of order $o$? (Usually "symmetries of order $o$" would refer to the individual symmetries themselves having order $o$.) Also, what is the "element" at the beginning? It seems that in the rest of the question you're only interested in $n$ and $o$ and the "element" doesn't occur anymore. Also, it's unusual to speak of groups "representing" polyhedrons. I presume you're asking whether that group is (isomorphic to) the symmetry group of any polyhedron? – joriki Dec 10 '11 at 19:02
@joriki: I apologize for my poor word choice. I edited it a bit; hopefully it makes more sense now. Yes, I'm basically asking about polyhedral symmetry groups isomorphic to $C_n \times C_3$ (or any other group of order $3n$). – Xodarap Dec 10 '11 at 19:14
up vote 2 down vote accepted

If you're looking for a way to get the factor of $3$ in a similar systematic way as you got the factors of $2$ and $4$, there isn't one. The $2$ and $4$ are related to reflections, which are of order $2$. You can read more here and here about why there are sequences of point groups of orders $2n$ and $4n$ but not $3n$.

However, if you're just looking for any polyhedron whose symmetry group is of order $3n$, or in fact of any order $k$ you like, you can use the cyclic groups $C_k$, which exist for all orders. Take a regular $k$-gon and break the reflection symmetries without breaking the rotational symmetry, for instance by cutting off all the corners in the same asymmetric way. Then form a pyramid over the resulting polygon with a point above the centre. The resulting polyhedron has rotational symmetry of order $k$ and nothing else.

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So it is not possible to have 3 fold rotational symmetry about one axis and $k$ fold symmetry about another? Those wikipedia articles seem to indicate that for at least some $k$ (e.g. tetrahedra) this is possible, but I couldn't find a general restriction. – Xodarap Dec 10 '11 at 22:06
@Xodarap: There are two types of points groups in three dimensions. The first type leaves a plane invariant. All the infinite sequences belong to this type. In this case there are only two types of axes: $2$-fold axes within the plane and one $n$-fold axis perpendicular to it. If you want a $3$-fold axis, it has to be the one perpendicular to the plane, and then you can only get orders $3$, $2\cdot3$ and $4\cdot3$. The second type are the polyhedral symmetries. They don't form infinite sequences. They all have three-fold axes, but they only have particular orders: $12, 24, 48, 60$ and $120$. – joriki Dec 11 '11 at 1:03

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