0
$\begingroup$

Let $\phi : G_1 \to G_2$ and $\psi : G_2 \to G_3$ be isomorphisms. Show that $\phi ^{-1}$ and $\psi \circ \phi$ are both isomorphisms. Using these results, show that the isomorphism of groups determines an equivalence relation on the class of all groups.

I was able to show that $\phi ^{-1}$ is an isomorphism (since $\phi$ is an isomorphism (i.e. a bijection), then the inverse exists and is also a bijection, so it's an isomorphism).

However, I'm unsure of how to show $\psi \circ \phi$ is an isomorphism. I haven't learned about homomorphisms yet, and I've seen a lot of answers using them to solve this question.

$\endgroup$
1
$\begingroup$

If $\phi $ and $\psi$ are bijections so is $\psi \circ \phi$. If these are homomorphisms then $(\psi \circ \phi) (gg') =\psi (\phi(g) \phi (g'))=\psi (\phi(g)) \psi (\phi(g'))=(\psi \circ \phi) (g)(\psi \circ \phi) (g')$. Hence $\psi \circ \phi$ is an isomorphism.

$\endgroup$
  • $\begingroup$ I haven't learned about homomorphisms in my class yet, so is there another way to show this? $\endgroup$ – Claire Apr 25 '19 at 5:09
  • $\begingroup$ What is your definition of an isomorhism? $\endgroup$ – Kavi Rama Murthy Apr 25 '19 at 5:10
  • $\begingroup$ If G and H are groups and f: H --> G is an isomorphism, then $f^{-1}$ is an isomorphism, the order of G equals the order of H, G is abelian if and only if H is abelian, G is cyclic if and only if H is cycle, and G has a subgroup of order n if and only if H has a subgroup of order n. $\endgroup$ – Claire Apr 25 '19 at 5:14
  • $\begingroup$ Also, an isomorphism exists if the function is bijective $\endgroup$ – Claire Apr 25 '19 at 5:14
  • 1
    $\begingroup$ This doesn't really define an isomorphism - it just gives consequences of $f$ being an isomorphism. $\endgroup$ – nilradical1 Apr 25 '19 at 5:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.