Question

Let G be a group. Show that G/Z(G)≅Inn(G). (b) Let G be a group. Show that there is a non-trivial homomorphism from G to Z₂ if and only if there is H a subgroup of G such that [G:H]=2. Hint.

Answer 1

To show that G/Z(G)≅Inn(G), we need to prove that the map φ: G/Z(G) -> Inn(G), defined by φ(gZ) = ρg, where ρg(x) = gxg⁻¹, is a well-defined isomorphism. (b) To show that there is a non-trivial homomorphism from G to Z₂ if and only if there is a subgroup H of G such that [G:H] = 2.

Step-by-step explanation:

To show that G/Z(G)≅Inn(G), we need to prove that the map φ: G/Z(G) -> Inn(G), defined by φ(gZ) = ρg, where ρg(x) = gxg⁻¹, is a well-defined isomorphism.

First, we need to show that φ is well-defined. Let g₁Z = g₂Z, where g₁, g₂ ∈ G. This means that g₁⁻¹g₂ ∈ Z(G). Let x ∈ G. We have ρg₁(x) = g₁xg₁⁻¹ and ρg₂(x) = g₂xg₂⁻¹. Since g₁⁻¹g₂ ∈ Z(G), we have g₁xg₁⁻¹ = g₁(g₁⁻¹g₂)xg₁⁻¹ = g₂xg₂⁻¹, which shows that φ is well-defined.

To prove that φ is an isomorphism, we need to show that it is a homomorphism, injective, and surjective. I will provide a proof for each property:

1. Homomorphism: Let g₁Z, g₂Z ∈ G/Z(G). We have φ(g₁Zg₂Z) = φ(g₁g₂Z) = ρg₁g₂, and φ(g₁Z)φ(g₂Z) = ρg₁ρg₂ = ρg₁g₂. Therefore, φ(g₁Zg₂Z) = φ(g₁Z)φ(g₂Z), which shows that φ is a homomorphism.
2. Injective: Let g₁Z, g₂Z ∈ G/Z(G) such that φ(g₁Z) = φ(g₂Z). This means ρg₁ = ρg₂, so for any x ∈ G, we have g₁xg₁⁻¹ = g₂xg₂⁻¹. This implies that g₁⁻¹g₂ ∈ Z(G), which further implies g₁Z = g₂Z. Therefore, φ is injective.
3. Surjective: Let ρg be an arbitrary element of Inn(G). We need to find an element gZ ∈ G/Z(G) such that φ(gZ) = ρg. Let g⁻¹xg = y for some x, y ∈ G. This means ρg(x) = y. Consider the element g⁻¹Z ∈ G/Z(G). We have φ(g⁻¹Z)(x) = gxg⁻¹ = y, which shows that φ(g⁻¹Z) = ρg. Therefore, φ is surjective.

Since φ is a well-defined homomorphism that is injective and surjective, it is an isomorphism between G/Z(G) and Inn(G).

(b) To show that there is a non-trivial homomorphism from G to Z₂ if and only if there is a subgroup H of G such that [G:H] = 2, we need to prove both directions of the statement.

First, assume that there is a non-trivial homomorphism φ: G -> Z₂. Let H = φ⁻¹(1), where 1 is the identity element in Z₂. By the First Isomorphism Theorem, G/ker(φ) ≅ Im(φ), so we have [G:ker(φ)] = |Im(φ)|. Since Z₂ has only two elements, |Im(φ)| = 2, which implies [G:ker(φ)] = 2. Therefore, there exists a subgroup H = ker(φ) of G such that [G:H] = 2.

Now, assume that there is a subgroup H of G such that [G:H] = 2. Let g ∈ G, g ∉ H, and define the homomorphism φ: G -> Z₂ as φ(x) = 1 if x ∈ H and φ(x) = 0 if x ∉ H. Since H is a normal subgroup of G, we have gH = Hg. This implies that φ(g) = φ(gH) = φ(Hg) = 1. Therefore, φ is a non-trivial homomorphism from G to Z₂.