Homepage Solution manuals David S. Dummit Abstract Algebra Exercise 1.7.21 (The group of rigid motions of a cube is isomorphic to $S_4$)

Exercise 1.7.21 (The group of rigid motions of a cube is isomorphic to $S_4$)

Show that the group of rigid motions of a cube is isomorphic to S 4 . [This group acts on the set of four pairs of opposite vertices.]

Answers

(See the figure, and complements, in the solution of Exercise 7.5.10 of “Cox, Galois Theory” on this site)

Proof. Consider the cube C = { A 1 , A 2 , , A 8 } with vertices

A 1 = ( 1 , 1 , 1 ) , A 2 = ( 1 , 1 , 1 ) , A 3 = ( 1 , 1 , 1 ) , A 4 = ( 1 , 1 , 1 ) , A 5 = ( 1 , 1 , 1 ) , A 6 = ( 1 , 1 , 1 ) , A 7 = ( 1 , 1 , 1 ) , A 8 = ( 1 , 1 , 1 ) .

Let G be the group of rigid motions of a cube. Every rigid motion g G gives a permutation of the 8 vertices of the cube, thus fixes their gravity center O = ( 0 , 0 , 0 ) , so g is an axial rotation, or g = e .

We give an arbitrary numbering of the four long diagonals (pairs of opposite vertices):

D 1 = { A 2 , A 7 } , D 2 = { A 3 , A 8 } , D 3 = { A 4 , A 5 } , D 4 = { A 1 , A 6 } .

Since g is an isometry, | | g ( A 2 ) g ( A 7 ) | | = | | A 2 A 7 | | = 2 3 , hence the pair { g ( A 2 ) , g ( A 7 ) } is a long diagonal, and it is the same for the other long diagonals. Therefore G acts on the set

A = { { A 2 , A 7 } , { A 3 , A 8 } , { A 4 , A 5 } , { A 1 , A 6 } }

of the four long diagonals by

g { A i , A j } = { g ( A i ) , g ( A j ) } , g G , ( i , j ) { ( 2 , 7 ) , ( 3 , 8 ) , ( 4 , 5 ) , ( 1 , 6 ) } .

The associate permutation representation is

φ { G S A g φ g , where φ g { A A D = { A i , A j } φ g ( D ) = g D = { g ( A i ) , g ( A j ) } ,

We search the kernel of φ . g ker ( φ ) if and only if

{ g ( A 2 ) , g ( A 7 ) } = { A 2 , A 7 } , { g ( A 3 ) , g ( A 8 ) } = { A 3 , A 8 } , { g ( A 4 ) , g ( A 5 ) } = { A 4 , A 5 } , { g ( A 1 ) , g ( A 6 ) } = { A 1 , A 6 } .

so g maps the four lines containing the 4 long diagonals on themselves.

Assume, for the sake of contradiction, that g e . Then g is an axial rotation. If every vertex A i maps to the opposite vertice, then g = s O is the symmetry through O ( g and s O are affine transformations which send at least 4 non coplanar points on the same images). But s 0 is not a rigid motion. Therefore there is some long diagonal { A i , A j } such that g ( A i ) = A i , g ( A j ) = A j . Then the axis of the rotation is A i A j , a long diagonal. The only lines globally invariant by such a rotation are the axis, and perhaps the lines in the orthogonal plane (if g is a rotation of π radians). Therefore the 3 others long diagonals, which are fixed by g , would be orthogonal to the axis, but the four diagonals are not mutually orthogonal. This contradiction shows that g = e , so

ker ( φ ) = { e } .

Therefore φ is injective (the action is faithful) and G φ ( G ) S G .

Since | A | = 4 , S A S 4 . We prove in Exercise 1.2.10 that | G | = 24 = | S 4 | = | S G | . Therefore φ : G S A is bijective: it is an isomorphism, so

G S 4 .

(another proof is given in Exercise 7.5.10 of “Cox, Galois Theory”) □

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2025-10-06 10:06
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