Thread: symmetry groups of the regular solids

The two paragraphs just explain the problem. You are supposed to prove that

The group of rotational symmetries of a tetrahedron is

A(4).

The full group of symmetries of a tetrahedron is

S(4).

The group of rotational symmetries of a cube or of an octahedron
is S(4).

The group of rotational symmetries of an icosahedron or of a dodecahedron is A(5).

based on:

A regular solid is a 3-
dimensional polyhedron in which each face is a regular polygon. Any two
faces as well as any two vertices can be matched by an isometry (a rigid
motion) of the 3-dimensional space. It is convenient to use a symbol

{p, q}

for a regular solid whose faces are regular

p-gons with q of them situated
around each vertex. We have proved that there are FIVE regular solids:

•

TETRAHEDRON {3, 3},

•

CUBE {4, 3},

•

OCTAHEDRON {3, 4},

•

DODECAHEDRON {5, 3},

•

ICOSAHEDRON {3, 5}.

Any regular solid may be inscribed in a sphere, and then any symmetry of
any regular solid will leave the center of the sphere fixed and will transform
the surface of the sphere onto itself. We call a rotational symmetry of a
regular solid any rotation of the sphere (with respect to an axis passing
through its center) mapping the regular solid into inself. A symmetry of a
regular solid is either a rotational symmetry or a reflection with respect to
a plane (also passing through the center of the sphere), mapping the regular
solid onto itself.

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Originally Posted by anncar

...................

Is that a BUMP?
Whatever that is.

nope. i posted something and decided I wanted to delete it. so i just put dots in place of it.

Originally Posted by anncar

The group of rotational symmetries of a tetrahedron is

A[FONT=CMR10][SIZE=3](4).

[LEFT]The full group of symmetries of a tetrahedron is S[FONT=CMR10][SIZE=3](4).

Look at the picture below and note all the ways you can rotate it. So you can have an identity rotation represented by the cycle you can swap 1 and 4 leaving 2 and 3 unchanged that is . You can swap 2 and 3 leaving 1 and 4 unchanged that is . And so on ... You will get a group table.

For the cube, look at the four long diagonals connecting a vertex to the opposite vertex on the opposite face. You can get any permutation of this set of four diagonals by a suitable rotation.

I am too lazy to check this but is it in general true that the group of rotations on a solid is an alternating group?

And if the above question is true is there any nice geometric signifigance to that, i.e. what happens when and the group is not solvable? Is this any way related to the 5 Platonic solids?

Originally Posted by ThePerfectHacker

I am too lazy to check this but is it in general true that the group of rotations on a solid is an alternating group?

This is not true for the cube or the octahedron, both of which have the full symmetric group S_4 as their symmetry group. There is a good explanation here.