2.10 The Isomorphism Theorem
1. Define 
by 
This is a group homomorphism by direct calculation. We have 
if and only if a = 1 = c, so ker α = K.
by This is a group homomorphism by direct calculation. We have if and only if a = 1 = c, so ker α = K.
3. Since 
, H G and G/H = {H, G H} is a group with H as unity. Thus σ : G/H → {1, − 1} is an isomorphism if σ(H) = 1 and σ(G − H) = − 1. Let ϕ : G → G/H be the coset map ϕ(g) = Hg, and let α = σϕ : G → {1, − 1}. Then α is a homomorphism and: If g 
H, then α(g) = σϕ(g) = σ(H) = 1; if g ∉ H, then α(g) = σϕ(g) = σ(G − H) = − 1.
, H G and G/H = {H, G H} is a group with H as unity. Thus σ : G/H → {1, − 1} is an isomorphism if σ(H) = 1 and σ(G − H) = − 1. Let ϕ : G → G/H be the coset map ϕ(g) = Hg, and let α = σϕ : G → {1, − 1}. Then α is a homomorphism and: If g H, then α(g) = σϕ(g) = σ(H) = 1; if g ∉ H, then α(g) = σϕ(g) = σ(G − H) = − 1.
4.
a. We have 1 α−1(X) because α(1) = 1 X. If g, h α−1(X), then α(g) X and α(h) X, so α(g−1) = [α(g)]−1 X and α(gh) =
α−1(X) because α(1) = 1 X. If g, h α−1(X), then α(g) X and α(h) X, so α(g−1) = [α(g)]−1 X and α(gh) = Get Introduction to Abstract Algebra, Solutions Manual, 4th Edition now with the O’Reilly learning platform.
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