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)