Algebra Homework 5

Hence, we see the existence of π ∈ Sym(I) with α(g) = πg for all g ∈ G. (⇐) Assume π ∈ Sym(I) exists such that α(g)(i) = (π ◦ g ◦ π−1)(i).† By definition of the symmetric group, π : I → I, i → π(i) is a bijection. Let g ∈ G and i ∈ I. Observe π(g ¦ρ i) = π(g(i)) = (π ◦ g) ◦ (π−1 ◦ π)(i) = (π ◦ g ◦ π−1) ◦ π(i) = α(g) ◦ π(i)† = g ¦ρ◦α π(i) Hence, by definition 2.10.13(a), π is G-equivariant. We already noted that π is bijective, so by 2.10.13(b), π is a G-isomorphism. This means ¦ρ is isomorphic to ¦ρ◦α by 2.10.13(c). ¤