Homepage Solution manuals David S. Dummit Abstract Algebra Exercise 1.2.10 (Group of rigid motions of a cube)

Exercise 1.2.10 (Group of rigid motions of a cube)

Let G be the group of rigid motions in 3 of a cube. Show that | G | = 24 .

Answers

Outline of a proof.

Proof. Let 𝒞 = { A , B , C , D , E , F , G , H } a cube (for instance the points of coordinates ( ± 1 , ± 1 , ± 1 ) ), and G be the group of rigid motions of 𝒞 .

G acts on the set 𝒞 by g M = g ( M ) ( g G , M 𝒞 ) . Using rotations of angles ± π 2 and axe 0 I , where I is a centre of a face, we can send A on any vertex. So the action is transitive, there is a unique orbit O A , and | O A | = | 𝒞 | = 8 .

The stabilizer G A of A has three elements: the identity e and the two rotations of angle ± 2 π 3 and axe OA . By the fundamental theorem of group actions, ( G : G A ) = | O A | , so

| G | = | G A | | O A | = 3 8 = 24 .

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2025-09-09 08:03
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