# Show that $\psi \circ \phi$ is an isomorphism.

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.

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.
• 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. – Claire Apr 25 '19 at 5:14
• This doesn't really define an isomorphism - it just gives consequences of $f$ being an isomorphism. – nilradical1 Apr 25 '19 at 5:16