Unformatted text preview: 148 (ix) The alternating group A 5 is a solvable group. Solution. False. 5.16 Let ϕ : A → H be a group homomorphism. If B C A and B ≤ ker ϕ , prove that the induced map ϕ ∗ : A / B → H , given by aB 7→ ϕ( a ) , is a well-de fi ned homomorphism with im ϕ ∗ = im ϕ . Solution. If aB = cB , then c − 1 a ∈ B ≤ ker ϕ , and so ϕ( c − 1 a ) = 1. Therefore, ϕ( c ) = ϕ( a ) , and ϕ ∗ is well-de fi ned. 5.17 If z ∈ C is a constructible number, prove that Q ( i , z )/ Q is a radical exten- sion. Solution. Absent. 5.18 Let k be a fi eld and let f ( x ) ∈ k [ x ] . Prove that if E and E are splitting fi elds of f ( x ) over k , then Gal ( E / k ) ∼ = Gal ( E / k ) . Solution. Absent. 5.19 Prove that F 3 [ x ] /( x 3 − x 2 − 1 ) ∼ = F 3 [ x ] /( x 3 − x 2 + x − 1 ) . Solution. We saw, in Chapter 3, that both cubics are irreducible in F 3 [ x ] , and so the quotient rings are fi elds of order 3 3 = 27. But any two fi nite fi elds of the same order are isomorphic.elds of the same order are isomorphic....
View Full Document
- Fall '11
- Group Theory, Normal subgroup, Solution., fQ, Endomorphism, ker φ