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SOLUTION 10

# Algebra

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Math 150a: Modern Algebra Homework 10 Solutions 5.6.3 (a) Exhibit the bijective map (5.6.4) explicitly, when G is the dihedral group D 4 and S is the set of vertices of a square. Solution: Let S = { s 1 , s 2 , s 3 , s 4 } be the vertices of the square labeled counterclockwise from the top right vertex. And let D 4 be generated by r and f , where f is a reflection through a horizontal line. Then S is a D 4 -set; that is, D 4 acts on the set S . We may consider the orbit O s 1 S of the point s 1 S O s 1 = { g ( s 1 ) | g D 4 } = { s 1 , s 2 , s 3 , s 4 } ; And the stabilizer H D 4 of s 1 H = stab D 4 ( s 1 ) = ( D 4 ) s 1 = { g D 4 | g ( s 1 ) = s 1 } = { 1 , r f } . Proposition 5.6.4 states that D 4 / H and O s 1 are in bijective correspondence, where the bijection φ maps a coset to its image of s 1 , aH mapsto→ a ( s 1 ) . s 1 s 2 s 3 s 4 f r f So, to exhibit this map, verify that there are four cosets of H sending s 1 to all of the other elements of its orbit. H = r fH = { e , r f } mapsto→ e ( s 1 ) = r f ( s 1 ) = s 1 , rH = r 2 fH = { r , r 2 f } mapsto→ r ( s 1 ) = r 2 f ( s 1 ) = s 2 , r 2 H = r 3 fH = { r 2 , r 3 f } mapsto→ r 2 ( s 1 ) = r 3 f ( s 1 ) = s 3 , r 3 H = fH = { r 3 , f } mapsto→ r 3 ( s 1 ) = f ( s 1 ) = s 4 . square (b) Do the same for D n and the vertices of a regular n -gon. Solution: Let S = { s 1 , s 2 ,..., s n } be the vertices of a regular n -gon. As in part (a), we may orient the n -gon so that f sends s 1 to s n . Then O s 1 = { s 1 , s 2 ,..., s n } , and H = ( D 4 ) s 1 = { 1 , r f } . 1

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r f r f f f s 1 s 1 s 2 n 1 s 2 n D 2 n 1 D 2 n s 2 s 2 And, similarly, H = r fH = { e , r f } mapsto→ e ( s 1 ) = r f ( s 1 ) = s 1 , rH = r 2 fH = { r , r 2 f } mapsto→ r ( s 1 ) = r 2 f ( s 1 ) = s 2 , . . . r n 2 H = r n 1 fH = { r n 2 , r n 1 f } mapsto→ r n 2 ( s 1 ) = r n 1 f ( s 1 ) = s n 1 . r n 1 H = fH = { r n 1 , f } mapsto→ r n 1 ( s 1 ) = f ( s 1 ) = s n . square 5.7.3 Compute the order of the group of symmetries of a dodecahedron, when orientation reversing sym- metries such as reflections in planes, as well as rotations, are allowed. Do the same for the symmetries of a cube and of a tetrahedron. Solution: Let I , O , and T be the full symmetry groups (including orientation reversing symmetries) of the dodecahedron, cube, and tetrahedron (respectively). In each case, the symmetry group acts transitively on the faces of each polytope. And the stabilizer of each face is a dihedral group. A face of the dodecahedron is a regular pentagon, so its stabilizer is isomorphic to D 5 . Similarly, the stabilizer of a face of the cube is isomorphic to D 4 ; and the stabilizer of a face of the tetrahedron is isomorphic to D 3 . In each case, we may use the counting formula: ( order of the group ) = ( order of orbit )( order of stabilizer ) Thus, | I | = ( 12 ) | D 5 | = 12 · 10 = 120 , | O | = ( 6 ) | D 4 | = 6 · 8 = 48 , and | T | = ( 4 ) | D 3 | = 4 · 6 = 24 . square 5.7.5 Let G H K be groups. Prove the formula [ G : K ] = [ G : H ][ H : K ] without the assumption that G is finite. Solution: Since K H , each coset gK is a subset of gH —some coset of H . And each coset of H is a union of translates (in G ) of cosets of K in H : gH = uniondisplay h H g ( hK ) = uniondisplay X H / K gX .
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