Hw3 Problem 7

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Problem 7

Prove that $\phi: S \rightarrow S'$ is an isomorphism of $\langle S,*\rangle$ with $\langle S',*'\rangle$ and $\psi: S' \rightarrow S''$ is an isomorphism of $\langle S',*'\rangle$ with $\langle S'',*''\rangle$, then the composite function $\psi \circ \phi$ is an isomorphism of $\langle S,*\rangle$ with $\langle S'',*''\rangle$.

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Solution:

Suppose $\phi: S \rightarrow S'$ is an isomorphism of $\langle S,*\rangle$ with $\langle S',*'\rangle$ and $\psi: S' \rightarrow S''$ is an isomorphism of $\langle S',*'\rangle$ with $\langle S'',*''\rangle$. Now, for the composite function $\psi \circ \phi$ to be an isomorphism, three properties must be satisfied.

1-1: Let $a,b \in S$. If$(\psi \circ \phi)(a) = (\psi \circ \phi)(b)$, that is $\psi(\phi(a)) = \psi(\phi(b))$, we get $\phi(a) = \phi(b)$ (since $\psi: S' \rightarrow S''$ is an isomorphism of $\langle S',*'\rangle$ with $\langle S'',*''\rangle$,$\psi$ is 1-1 ). Because $\phi: S \rightarrow S'$ is an isomorphism of $\langle S,*\rangle$ with $\langle S',*'\rangle$,we know $\phi$ is 1-1. And $\phi(a) = \phi(b)$. So $a=b$. Hence, $\psi \circ \phi$ is 1-1.

Onto: Let $a'' \in S''$. Now, we know $\psi: S' \rightarrow S''$ so there must exist an element $a' \in S'$ such that $\psi(a')=a''$. We also know $\phi; S \rightarrow S'$, so there also exists $a \in S$ such that $\phi(a)=a'$. Now, $(\psi \circ \phi)(a) = \psi(\phi(a)) = \psi(a') = a''$. Hence, $\psi \circ \phi$ is onto.

Homomorphism: Let $a,b \in S$. We know $\phi$ and $\psi$ are isomorphisms, so

(1)

\begin{align} (\psi \circ \phi)(a * b) = \psi(\phi(a*b) = \psi(\phi(a) *' \phi(b)) = \psi(\phi(a)) *'' (\phi(b)) = (\psi \circ \phi)(a) *'' (\psi \circ \phi)(b) \end{align}

Hence, $\psi \circ \phi$ is isomorphic.

Therefore, the composite function $\psi \circ \phi$ is an isomorphism of $\langle S,*\rangle$ with $\langle S'',*''\rangle$.