Let g ∈ G, where G is a group. Prove that the function f : G → G given by f (h) = ghg−1 isan isomorphism from G to itself (i.e. an automorphism of G).

To prove that the function f: G → G given by f(h) = ghg−1 is an isomorphism from G to itself, we need to show that it is both a homomorphism and a bijection.

1. Homomorphism: A function f: G → G is a homomorphism if for all a, b in G, f(ab) = f(a)f(b). Let's check this for our function:

   f(ab) = gabg−1 = gahbg−1 = ghg−1 * gbg−1 = f(a)f(b)

   So, f is a homomorphism.

2. Bijection: A function f: G → G is a bijection if it is both injective (one-to-one) and surjective (onto).

   • Injective: Assume f(a) = f(b) for some a, b in G. Then, g a g−1 = g b g−1. Multiplying both sides on the right by g gives g a = g b, and then multiplying both sides on the left by g−1 gives a = b. So, f is injective.

   • Surjective: For any element a in G, we can find an element b in G such that f(b) = a. Let b = g−1 a g. Then, f(b) = g g−1 a g g−1 = a. So, f is surjective.

Therefore, the function f: G → G given by f(h) = ghg−1 is an isomorphism from G to itself.