MATH 401, ASSIGNMENT 3

DANIEL R. GRAYSON

1.7: 21. Let SO(3) = {g ∈ O(3) | det g = 1}. Let V = {(±1, ±1, ±1)} be the set of vertices of the cube, and regard the group of rigid motions of the cube as G = {g ∈ SO(3) | gV = V }, and let D be the set of lines L = ℝv through a point of v ∈ V and the origin. The group G also acts on D, since gL = gℝv = ℝgv ∈ D, and that action amounts to a homomorphism ϕ : G → S_{D} which we have to show is injective. (The map ϕ is defined by defining ϕ(g) ∈ S_{D} to be the function L ↦ gL. We see that ϕ is a homomorphism by appealing to the associativity g′(gL) = (g′g)L.) So consider g ∈ ker ϕ. For each v ∈ V we have ℝgv = gℝv = ℝv, so v and gv are vectors of the same length lying in the same line; thus either gv = v or gv = −v. If v and v′ are adjacent vertices of the cube separated by a distance 2, then so are gv and gv′; but v and −v′ are not adjacent, and −v and v′ are not adjacent. Thus either gv = v and gv′ = v′, or gv = −v and gv′ = −v′. Since v and the vertices adjacent to v span ℝ^{3}, any linear transformation is determined

by its action on them; hence g = (\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}) or g = (\begin{matrix}−1 & 0 & 0 \\ 0 & −1 & 0 \\ 0 & 0 & −1 \end{matrix}). But the

latter matrix has determinant −1, so is not in G, and hence we must have g = 1. Now that we know ϕ is injective, it follows that ϕ is an isomorphism, because both groups have order 24. (The number of rotational symmetries of the cube is twice the number of edges.)

University of Illinois at Urbana-Champaign

   E-mail address: dan@math.uiuc.edu
   URL: http://www.math.uiuc.edu/~dan

Date: September 23, 1999.