. Its center is the origin (0, 0, 0). Let SO

(1)

Rotations preserving the cube Algebra 2, 2020 Consider the cube C with vertices (±1, ±1, ±1) in R

³

. Its center is the origin (0, 0, 0). Let SO

₃

(R) be the orthogonal group of matrices with determinant 1. Each of its elements is a rotation with respect to a line passing through (0, 0, 0). Let G ⊂ SO

3

(R) be the subgroup of matrices that preserve the cube C.

Proposition. The group G is isomorphic to the symmetric group S

4

.

Proof. Let X be the set of diagonals of C. They are the lines passing through the origin with directions

(1, 1, 1), (1, −1, −1), (−1, 1, −1), and (−1, −1, 1).

The group G permutes the four diagonals in X. This leads to a group homomorphism φ : G −→ S

4

.

If σ ∈ G is in the kernel of φ, then σ fixes the four diagonals. This means that its action on the projective plane P

₂

(R) has four fixed points, no three of which are on a line. By the lemma below, σ is a scalar matrix. Since the only scalar matrix in SO

3

(R) is the identity, σ is the identity. This shows that φ is injective.

To show that φ is surjective, observe that a rotation of 90 degrees around any of the x-, y- or z-axes is an element of G of order 4. Moreover, a rotation of 120 degrees around any of the diagonals has order 3. Therefore #G = #φ(G) is divisible by 12. Since #S

4

= 24, its subgroup φ(G) is normal and S

₄

/φ(G) is abelian. This implies that the commutator subgroup A

4

is contained in φ(G). It follows that A

4

⊂ φ(G) ⊂ S

4

, so that φ(G) is either A

4

or S

4

. Since A

4

does not contain any elements of order 4, we must have φ(G) = S

4

, as required.

Lemma. Let k be a field and let σ ∈ GL

3

(k). Suppose that σ fixes four points in the projective plane P

2

(k), no three of which are on a line. Then σ is a scalar matrix.

Proof. Lift the four fixed points to vectors in k

³

. Since no three points are on a line, any three vectors form basis. Pick any three: e

1

, e

2

, e

3

∈ k

³

. Then the fourth vector e

4

is equal to a

₁

e

₁

+ a

₂

e

₂

+ a

₃

e

₃

with a

₁

, a

₂

, a

₃

6= 0. Since σ fixes the four points in P

₂

(k), it preserves the lines spanned by the vectors e

i

. This means that for each i = 1, 2, 3, 4 there is a λ

_i

∈ k

^∗

for which σ(e

_i

) = λ

_i

e

_i

. It follows that λ

₄

a

_i

= λ

_i

a

_i

for i = 1, 2, 3 and hence λ

1

= λ

2

= λ

3

= λ

4

. We conclude that σ is a scalar matrix, as required.

Exercise. Generalize the lemma to matrices A ∈ GL

_n

. Its center is the origin (0, 0, 0). Let SO

Rotations preserving the cube Algebra 2, 2020 Consider the cube C with vertices (±1, ±1, ±1) in R

. Its center is the origin (0, 0, 0). Let SO

(R) be the orthogonal group of matrices with determinant 1. Each of its elements is a rotation with respect to a line passing through (0, 0, 0). Let G ⊂ SO

(R) be the subgroup of matrices that preserve the cube C.

Proposition. The group G is isomorphic to the symmetric group S

.

Proof. Let X be the set of diagonals of C. They are the lines passing through the origin with directions

(1, 1, 1), (1, −1, −1), (−1, 1, −1), and (−1, −1, 1).

The group G permutes the four diagonals in X. This leads to a group homomorphism φ : G −→ S

.

If σ ∈ G is in the kernel of φ, then σ fixes the four diagonals. This means that its action on the projective plane P

(R) has four fixed points, no three of which are on a line. By the lemma below, σ is a scalar matrix. Since the only scalar matrix in SO

(R) is the identity, σ is the identity. This shows that φ is injective.

To show that φ is surjective, observe that a rotation of 90 degrees around any of the x-, y- or z-axes is an element of G of order 4. Moreover, a rotation of 120 degrees around any of the diagonals has order 3. Therefore #G = #φ(G) is divisible by 12. Since #S

= 24, its subgroup φ(G) is normal and S

/φ(G) is abelian. This implies that the commutator subgroup A

is contained in φ(G). It follows that A

⊂ φ(G) ⊂ S

, so that φ(G) is either A

or S

. Since A

does not contain any elements of order 4, we must have φ(G) = S

, as required.

Lemma. Let k be a field and let σ ∈ GL

(k). Suppose that σ fixes four points in the projective plane P

(k), no three of which are on a line. Then σ is a scalar matrix.

Proof. Lift the four fixed points to vectors in k

. Since no three points are on a line, any three vectors form basis. Pick any three: e

, e

, e

∈ k

. Then the fourth vector e

is equal to a

e

+ a

e

+ a

e

with a

, a

, a

6= 0. Since σ fixes the four points in P

(k), it preserves the lines spanned by the vectors e

. This means that for each i = 1, 2, 3, 4 there is a λ

∈ k

for which σ(e

) = λ

e

. It follows that λ

a

= λ

a

for i = 1, 2, 3 and hence λ

= λ

= λ

= λ

. We conclude that σ is a scalar matrix, as required.

Exercise. Generalize the lemma to matrices A ∈ GL

(k) and all n ≥ 2.