0
$\begingroup$

If $\phi: S \rightarrow S' $ is an isomorphism of $\langle S,* \rangle$ with $\langle S',*' \rangle$, and $\psi: S' \rightarrow S''$ is of $\langle S',*' \rangle$ with $\langle S'', *''\rangle$

Prove that $\psi \circ \phi$ is and isomorphism of $\langle S, * \rangle$ with $\langle S'', *'' \rangle $

I get one-to-one here: $\psi(a)=\psi(b)$ then $a=b$, since, as the problem mentioned, $\psi:S' \rightarrow S''$ is an isomorphism. But since $\phi:S \rightarrow S'$ is also an isomorphism, we can conclude that $\phi(\psi(a))=\phi(\psi(b))$ then $a = b$

I'm running into a bit of a problem showing that $\psi \circ \phi$ is onto $S''$ and that it is actually a homomorphism, as well.

Much appreciated.

  • 1
    Pick a point $x''$ in $S''$ and show that it has a $\psi$-pre-image in $S'$. Denote that pre-image by $x'$. Now find a $\phi$-pre-image of $x'$ in $S$. – 2017-02-04
  • 0
    To see that $\phi\circ\psi$ is an homomorphism: $\phi(\psi(ab))=\phi(\psi(a)\psi(b))=\phi(\psi(a))\phi(\psi(b))$ using that both $\phi$ and $\psi$ are homomorphisms. – 2017-02-04
  • 0
    Is $\psi $ an isomorphism? – 2017-02-04

0 Answers 0