Verified answer

Let's assume that you mean it is a group of permutations.

S[A] is the permutations of A, but what is a permutation?

It is a one-to-one, onto function from A to A.

Same for S[B].

Then what can we say about ψ? Is it a group homomorphism?

First you need to prove that. To do so, show that for any σ and τ in S[A], the following equation is true:

      ψ( σ τ ) = ψ( σ ) ψ (τ)

But that is not hard to show since:

      ψ( σ ) ψ (τ) = (f σ f⁻¹ ) (f τ f⁻¹ ) = f σ f⁻¹ f τ f⁻¹ = f σ τ f⁻¹ = ψ( σ τ )

Now is it one-to-one and onto? That would make it an isomorphism.

You can either prove this by showing that it is one-to-one and onto, or you could simply construct a well-defined inverse function.

POSSIBILITY ONE:

To show it is one-to-one, assume ψ(h) = ψ(k).

Then f h f⁻¹ = f k f⁻¹

Compose on both sides to get f⁻¹ f h f⁻¹ f = f⁻¹ f k f⁻¹ f

Then cancel to get h=k.

To show it is onto, notice that for any g in S[B], ψ(f⁻¹ g f ) = g.

POSSIBILITY 2:

The inverse function is ψ⁻¹(g) = f⁻¹ g f.

(f o g) = 3((5x-8)^2) - 2(5x-8) + 7 = 25x^2 - 80x + sixty 4 (h o f o g) = (25x^2 - 80x + sixty 4)^2 - 3(25x^2 - 80x + sixty 4) = 625x^4 - 4000x^3 + 6400x^2 - 10240x + 4096 What you do is plug g(x) into the x's in the f(x) equation to get (f o g). then you definitely plug (f o g) into the x's in the h(x) equation to get (h o f o g). i wish that replaced into not too puzzling :)