# symmetry groups of the regular solids

• Nov 25th 2007, 11:43 AM
anncar
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.

• Nov 25th 2007, 04:36 PM
anncar
...................
• Nov 25th 2007, 04:43 PM
Plato
Quote:

Originally Posted by anncar
...................

Is that a BUMP?
Whatever that is.
• Nov 25th 2007, 05:39 PM
anncar
nope. i posted something and decided I wanted to delete it. so i just put dots in place of it.
• Nov 25th 2007, 05:43 PM
ThePerfectHacker
Quote:

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 \$\displaystyle (1)\$ you can swap 1 and 4 leaving 2 and 3 unchanged that is \$\displaystyle (1,4)\$. You can swap 2 and 3 leaving 1 and 4 unchanged that is \$\displaystyle (2,3)\$. And so on ... You will get a group table.
• Nov 25th 2007, 11:46 PM
Opalg
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.
• Nov 26th 2007, 09:33 AM
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?

And if the above question is true is there any nice geometric signifigance to that, i.e. what happens when \$\displaystyle n\geq 5\$ and the group is not solvable? Is this any way related to the 5 Platonic solids?
• Nov 26th 2007, 10:20 AM
Opalg
Quote:

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.