Math 661 Fall 2013 Homework 1 Solutions
نویسنده
چکیده
Proof. First suppose that a ∼H b, so that a = bk for some k ∈ H, and let ah (with h ∈ H) be an artibrary element of aH. Then we have ah = bkh = b(kh) ∈ bH, hence aH ⊆ bH. The proof of bH ⊆ aH is similar. Conversely, suppose that aH = bH. Since a ∈ aH = bH we have a = bk for some k ∈ H. Then a−1b = k−1 ∈ H, hence a ∼H b. (d) Prove that the map g 7→ ag is a bijection from H to aH. Proof. Consider the map G→ G defined by g 7→ a−1g. Since an arbitrary element of aH looks like ah for some h ∈ H we see that the map sends aH → H. Since this maps also inverts the map g 7→ ag we conclude that both maps are bijective.
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Math 661 Fall 2013 Homework 5 Drew Armstrong
(a) Prove that [G,G]CG. (b) Prove that the quotient Gab := G/[G,G] (called the abelianization of G) is abelian. (c) If N CG is any normal subgroup such that G/N is abelian, prove that [G,G] ≤ N . (d) Put everything together to prove the universal property of abelianization: Given a homomorphism φ : G→ A to an abelian group A, there exists a unique homomorphism φ̄ := Gab → A such that φ = φ̄ ◦ π, ...
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