9More on regular polyhedra

IA Groups



9.2 Symmetries of the tetrahedron
Rotations
Let 1
,
2
,
3
,
4 be the vertices (in any order).
G
+
is just the rotations. Let it act
on the vertices. Then orb(1) = {1, 2, 3, 4} and stab(1) = { rotations in the axis
through 1 and the center of the opposite face } = {e,
2π
3
,
4π
3
}
So |G
+
| = 4 · 3 = 12 by the orbit-stabilizer theorem.
The action gives a group homomorphism
ϕ
:
G
+
S
4
. Clearly
ker ϕ
=
{e}
.
So
G
+
S
4
and
G
+
has size 12. We “guess” it is
A
4
(actually it must be
A
4
since that is the only subgroup of
S
4
of order 12, but it’s nice to see why that’s
the case).
If we rotate in an axis through 1, we get (2 3 4)
,
(2 4 3). Similarly, rotating
through other axes through vertices gives all 3-cycles.
If we rotate through an axis that passes through two opposite edges, e.g.
through 1-2 edge and 3-4 edge, then we have (1 2)(3 4) and similarly we obtain
all double transpositions. So
G
+
=
A
4
. This shows that there is no rotation
that fixes two vertices and swaps the other two.
All symmetries
Now consider the plane that goes through 1, 2 and the mid-point of 3 and 4.
Reflection through this plane swaps 3 and 4, but doesn’t change 1
,
2. So now
stab
(1) =
h
(2 3 4)
,
(3
,
4)
i
=
D
6
(alternatively, if we want to fix 1, we just move
2, 3, 4 around which is the symmetries of the triangular base)
So
|G|
= 4
·
6 = 24 and
G
=
S
4
(which makes sense since we can move
any of its vertices around in any way and still be a tetrahedron, so we have all
permutations of vertices as the symmetry group)