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I am aware that the octahedron and the cube have the same symmetry group but I was wondering how we show this concretely. I have looked/been thinking about for an answer to this and I have got that as they are the dual of each other then they have the same symmetry group. However whilst I can intuitively understand why this is true I can't find/write a proof of the following statement which the argument seems to rely on:

Every polyhedra has the same symmetry group as its dual

Thanks very much for any help

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up vote 3 down vote accepted

How about: put the octahedron in a cube in the natural way (each vertex of the octahedron is at the center of one of the cube's faces). Now prove that any symmetry of the octahedron corresponds to a symmetry of the cube, and vice-versa.

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Can this approach be done in general for any convex hull of finitely many points? – Dustan Levenstein Mar 18 '12 at 16:04
yes well this is basically what I was asking (just a special case of every polyhedra has the same symmetry group as it dual) but I am not sure how to start doing this – hmmmm Mar 18 '12 at 16:33
At least for the case of a cube it is immediate: after the transformation, each of the octaheadron's vertices defines the center of a face. From this, the cube is easily determined. The inverse is equally simple. – yohBS Mar 18 '12 at 18:47

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