9More on regular polyhedra
IA Groups
9.1 Symmetries of the cube
Rotations
Recall that there are
G
+

= 24 rotations of the group by the orbitstabilizer
theorem.
Proposition. G
+
∼
=
S
4
, where G
+
is the group of all rotations of the cube.
Proof.
Consider
G
+
acting on the 4 diagonals of the cube. This gives a group
homomorphism
ϕ
:
G
+
→ S
4
. We have (1 2 3 4)
∈ im ϕ
by rotation around
the axis through the top and bottom face. We also (1 2)
∈ im ϕ
by rotation
around the axis through the midpoint of the edge connect 1 and 2. Since (1 2)
and (1 2 3 4) generate
S
4
(Sheet 2 Q. 5d),
im ϕ
=
S
4
, i.e.
φ
is surjective. Since
S
4
 = G
+
, ϕ must be an isomorphism.
All sym metries
Consider the reflection in the midpoint of the cube
τ
, sending every point to
its opposite. We can view this as
−I
in
R
3
. So it commutes with all other
symmetries of the cube.
Proposition. G
∼
=
S
4
× C
2
, where
G
is the group of all symmetries of the cube.
Proof.
Let
τ
be “reflection in midpoint” as shown above. This commutes with
everything. (Actually it is enough to check that it commutes with rotations
only)
We have to show that
G
=
G
+
hτi
. This can be deduced using sizes: since
G
+
and
hτi
intersect at
e
only, (i) and (ii) of the Direct Product Theorem gives
an injective group homomorphism
G
+
× hτi → G
. Since both sides have the
same size, the homomorphism must be surjective as well. So
G
∼
=
G
+
× hτi
∼
=
S
4
× C
2
.
In fact, we have also proved that the group of symmetries of an octahedron
is
S
4
× C
2
since the octahedron is the dual of the cube. (if you join the centers
of each face of the cube, you get an octahedron)