Question

Definition. If f:G→H is a homomorphism of groups, then the kernel of f is K={a∈G∣f(a)=Eh }.

Let G be a group and let f:G→Inn(G) be the map defined by f(a) = Ca . Prove that f is a surjective homomorphism with kernel Z(G).

Answer 1

To prove that the map f: G → Inn(G), defined by f(a) = Ca, is a surjective homomorphism with kernel Z(G), we need to show two things: first, that f is a homomorphism, and second, that f is surjective and its kernel is Z(G).

Step-by-step explanation:

To prove that the map f: G → Inn(G), defined by f(a) = Ca, is a surjective homomorphism with kernel Z(G), we need to show two things: first, that f is a homomorphism, and second, that f is surjective and its kernel is Z(G).

Proving that f is a homomorphism:

To show that f is a homomorphism, we need to demonstrate that it preserves the group structure. Let's take two elements a, b ∈ G. Then f(ab) = Cab = CaCb. Using the property of conjugation, we can rewrite this as f(a)f(b), showing that f is indeed a homomorphism.

Proving that f is surjective with kernel Z(G):

To prove that f is surjective, we need to show that for every element h ∈ Inn(G), there exists an element g ∈ G such that f(g) = h. By definition, for any element g ∈ G, f(g) = Cg. Since every inner automorphism corresponds to a conjugation, it follows that f is surjective.

Now, let's prove that the kernel of f is Z(G). The kernel of f consists of all elements a ∈ G such that f(a) = E, the identity element of Inn(G). Using the definition of f, we have f(a) = Ca = E. This implies that a ∈ Z(G), the center of G. Therefore, the kernel of f is Z(G).