1. Prove that the group of symmetries of the tetrahedron is isomorphic to S₄ in two ways:

Find an explicit isomorphism between the group of symmetries of the tetrahedron and the group Sym(corners) of permutations of the corners of the tetrahedron. Which subgroup of Sym(corners) corresponds to the rigid motions of the tetrahedron?

Call the corners {1,2,3,4}. Rotations about the axis from 4 to the centre of face 123 realise (123) and (132). Similarly rotations about the other three axes realise (124), (142), (134), (143), (234) and (243).

Rotations about the axis joining the midpoint of side 12 to the midpoint of side 34 realise (12)(34) Similar rotations about the other two side axes realise (13)(24) and (14)(23). With the identity map, these rigid motions realise the twelve even permutations A₄.

 Reflection in the plane containing 3 and 4 and the midpoint of the edge 12 realises (12). The products of this reflection and each of the rigid motions realises the rest of the elements of Sym(corners), for example this reflection followed by rotation by 2p/3 about the axis (originally) through 2, i.e. (134), realises (1234).

 Thus we have a 1-1 correspondence F:Sym(corners) -> Symmetries of the tetrahedron. To show that F is an isomorphism, we can either check that F(ab) = F(a)F(b) for each pair a, b in S₄, or we can use the fact that S₄ = <(12), (1234)>. If a in S₄, then a is a product of finitely many copies of these two permutations. Call the length of a the minimum number of factors in such a product. Now prove that F(ab) = F(a)F(b) by induction on n = length a + length b, starting with a and b = (12) or (1234) with n = 2.

 First note that if a is any permutation in S₄ and F(a) the corresponding symmetry of the tetrahedron, then a(12) is the permutation a changed so that the element mapped to 1 is instead mapped to 2 and the element mapped to 2 is instead mapped to 1. Also F(a(12)) is the symmetry F(a) followed by reflection in the plane originally containing the edge 34. Hence F(a(12)) = F(a)F(12).

Similarly, a(1234) = a(12)(134) is the permutation a changed so that the element mapped to 1 is instead mapped to 2, the element mapped to 2 instead goes to 3, the element mapped to 3 instead goes to 4, and the element mapped to 4 instead goes to 1. Also F(a(1234)) is the symmetry F(a) followed by reflection in the plane containing 34 followed by rotation by 2p/3 in the axis originally containing 2. Hence F(a(1234)) = F(a)F(1234).

 It follows that F((12)(12))=F(1) = 1 = F(12)F(12), F((12)(1234)) = F((12)(12)(134)) = F(134) = F(12)F(1234), F((1234)(12)) = F((12)(134)(12)) = F(234) = F(1234)F(12) and F(1234)F(1234) = F(13)(24)=F((1234)(1234)). This completes the proof that F(ab) = F(a)F(b) when the sum length is 2.

 Assume that F(cd) = F(c)F(d) whenever length c + length d is less than n+1, and suppose a and b are such that length a + length b = n+1. Then b = b'(12) or b'(1234) for some b' with length a + length b' = n. Also length b' + length (12) < n+1. By induction, F(ab') = F(a)F(b') and F(b) = F(b')F(12), and similarly for (1234) in place of (12).

 Hence F(ab) = F(ab'(12)) = F(ab')F(12) = F(a)F(b')F(12) = F(a)F(b'(12)) = F(a)F(b), and similarly with (12) replaced by (1234). 

On each face of the cube, draw one diagonal red and the other one blue in such a way that the 6 red diagonals form the edges of a regular tetrahedron and the 6 blue diagonals from the edges of another regular tetrahedron. Show that the group of symmetries of the cube acts on the set {red tetrahedron, blue tetrahedron} and the kernel of this action acts as the group of symmetries of the red tetrahedron. Deduce that the group of symmetries of the tetrahedron is isomorphic to the group of rigid motions of the cube.

The action described is a group action:

1. The identity maps red to red and blue to blue.

2. Call  a  symmetry  a  (RB)-map if it maps red to blue and blue to read, and similarly call a  a (RR)-map if it maps red to red and blue to blue. Then the composition of symmetries shows (RR)(RR) = (RR), (RB)(RR) = (RB) , (RR)(RB) = (RB) and (RB)(RB) = (RR).

Thus the two axioms of group actions are satisfied.

Rotation by p/2 about a face axis is (RB) and the identity maps red to red. Hence the action is onto S₂. Thus the kernel of the action, which is the set of symmetries mapping red to red and blue to blue has index 2 and hence order 24. Each non-identity element of the kernel causes a non-trivial transformation of the red tetrahedron and hence realises all symmetries. But this group also realises all permutations of the long diagonals of the cube and hence is the group of rigid motions of the cube.

(a) Show that G_x is a subgroup of G.

If g and h  1   G_x, then gh and g^-1 also fix x. Hence G_x is a group.

 (b) Let K be the kernel of the action. Prove the following:

 K= 0_{{x 1 X}} G_x

A group element f is in every G_x if and only if f fixes every element of X, if and only if f is in K.

(c) If the action is transitive on X, then for each x in X,
K= 0_{{a 1 G}} a ^-1 G_x a

Hence a ^-1 G_x a  is contained in G_y. Conversely, if f in a^-1 G_x a  then f fixes y so a^-1 G_x a = G_y.

 Hence the two intersections are equal.

3. The circle group T = {z in C : |z|=1 }, where C is the field of complex numbers, and the operation is multiplication of complex numbers.

Show that T is isomorphic to the group of orthogonal matrices of determinant 1 over R, the real numbers, and to the group R/Z, the factor group of the additive group of real numbers by the additive group of integers.

Each element of T is cis (f) for some 0 <= f <= 2 p. Map this to the orthogonal matrix with Row 1 = (cos f, sin f), Row 2 = (-sin f, cos f). Every orthogonal matrix has this form so the map is onto. Since cis (f) cis (a) = cis (f + a) the map is a homomorphism. Only 0 maps to the identity matrix so the map is 1--1. Hence this map is an isomorphism.

 Define F: R --> T by a --> cis (2p a ). Since cis (2p (a + b )) = cis (2p a ) cis (2p b) the map F is a homomorphism which is onto and has kernel Z because a -> 0 = cis(2na) if and only if a is an integer.

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 Last update March 24, 2000

 Author: Phill Schultz, schultz@maths.uwa.edu.au