Answer to Question #162936 in Abstract Algebra for K

Question

Prove that if phi : S ---> T is an isomorphism of <S,*> with <T, #> and psi: T ---> U is an isomorphism of <T, #> with <U, diamond> then the composite function psi composed with phi is an isomorphism of <S, *> with <U, diamond>

Answer 1

To show that $\psi \circ \phi$ is an isomorphism, we show that $\psi \circ \phi$ is a homomorphism and then we show that it is bijective.

To show homomorphism: Let $x,y \in S$ consider,

$\psi \circ \phi (xy) = \psi (\phi (xy))$ . Since $\phi$ is a homomorphism, we have that

$\psi (\phi (xy)) = \psi (\phi(x) \phi(y))$

And since $\psi$ is a homomorphism, we have that

$\psi (\phi(x) \phi(y)) = \psi(\phi(x))\psi(\phi(y)) = \psi \circ \phi(x)\psi \circ \phi(y)$

as desired. Hence $\psi \circ \phi$ is a homomorphism.

To show that it is bijective, we show that $\psi \circ \phi$ is both injective and surjective.

Let $\psi \circ \phi (x) = \psi \circ \phi (y)$ for $x,y \in S$

$\implies \psi (\phi(x)) = \psi (\phi(y))$

And since $\phi$ is injective, we have that

$\psi(x) = \psi(y)$

And since $\psi$ is injective, we have that

$x=y$ .

Let $z \in U$ , then $\exist p \in T \ni \psi(p) = z$

(Since $\psi$ is surjective).

Since $p \in T$ , by the surjectivity of $\phi$, we have that $\exist q \in S \ni \phi(q) =p$

So we have that $z = \psi(p) = \psi(\phi(q)) = \psi \circ \phi(q)$

And this implies that $\psi \circ \phi$ is surjective. Hence we can safely conclude that $\psi \circ \phi$ is an isomorphism.