Exercise 6.4. Let G be a group and SG be the symmetric group of G. (1) Show that Autgp (G) := {σ = SG | σ(9192) = σ(91)σ (92), V91,92 € G} is a subgroup of SG, called the automorphism group of G, and every σ = Autgp (G) is called an automorphism of G. (2) For every gЄ G, recall in Exercise 3.1 that ig: GG is the conjugate (inner) automor- phism of G defined by i,(x) := g¯¹xg for every x = G. Show that I : G→ Autgp (G) defined by I(g) ig-1 for every g = G is a group homomorphism. = (3) Show that the kernel of I is the center of G, i.e. ker(I) = Z(G) = {g € G | gx = xg, Vx Є G}. (4) Show that I(G) is normal in Autgp (G). (5) Show that the symmetric group S3 is isomorphic to Autgp (S3). = (Hint: Let A {(1, 2), (1, 3), (2, 3)} C S3. Show that there exists an injective homomor- phism from Autgp (S3) into SA.)

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