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Unformatted text preview: Math 150a: Modern Algebra Homework 8 Solutions 4.5.4 (b) Is O ( 2 ) isomorphic to the product group SO ( 2 ) ×{± I } ? Is O ( 3 ) isomorphic to SO ( 3 ) ×{± I } ? Solution: The group SO ( 2 ) is abelian [because the product of two rotations of R 2 by angles φ and ψ is a rotation by the angle ( φ + ψ ) ]. Thus SO ( 2 ) ×{± I } is abelian, but O ( 2 ) is not [since O ( 2 ) ⊃ D n ]. On the other hand, in O ( 3 ) , SO ( 3 ) is complemented by {± I } . For each X ∈ O ( 3 ) , X / ∈ SO ( 3 ) ⇒ − X ∈ SO ( 3 ) , since det ( − X ) = ( − 1 ) 3 det ( X ) and ( − X )( − X ) T = XX T = I . And clearly SO ( 3 ) ∩{± I } = { I } , since det ( − I ) = − 1. Additionally, both SO ( 3 ) and {± I } are normal in O ( 3 ) . This is because SO ( 3 ) has index 2, and because ∀ X ∈ O ( 3 ) , X ( − I ) X − 1 = − I . Hence, O ( 3 ) ∼ = SO ( 3 ) ×{± I } [see part 5 of handout on crosssections]. square 4.5.13 Prove that a rigid motion, as defined by (5.15), is bijective. Solution: Let m : R n → R n be an isometry. Then ∀ −→ x , −→ y ∈ R n  m ( −→ x ) − m ( −→ y )  =  −→ x − −→ y  . If m ( −→ x ) = m ( −→ y ) ,  −→ x − −→ y  =  m ( −→ x ) − m ( −→ y )  = ⇒ −→ x = −→ y . So, m is injective (onetoone). Proposition 4.5.20 states that m has the form m ( −→ x ) = A −→ x + −→ b for some orthogonal matrix A , and vector −→ b . Then given −→ z ∈ R n , consider −→ x = A − 1 ( −→ z − −→ b ) . We have m ( −→ x ) = A −→ x + −→ b = A ( A − 1 ( −→ z − −→ b ))+ −→ b = −→ z . So m is surjective (onto), and therefore bijective. square 5.9.3 Let O be the group of rotations of a cube. Determine the stabilizer of a (long) diagonal line. Solution: Using the correspondence between O and S 4 (described in GK4). Label the four long diagonals of the cube, and choose to determine the stabilizer of the fourth diagonal. Then we are searching for the subgroup of permutations of S 4 that fix the number 4. This subgroup is isomorphic tothat fix the number 4....
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 Spring '03
 Kuperberg
 Algebra, Angles

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