TITLE: Show that $\sigma\circ\phi : G_1 \to G_3$ is an isomorphism
QUESTION [5 upvotes]: Let $G_1, G_2$ and $G_3$ be groups.  Let $\phi: G_1 \to G_2$ and $\sigma: G_2 \to G_3$ be isomorphisms of groups.  Show that $$\sigma\circ\phi: G_1 \to G_3$$ is an isomorphism.
I understand to prove the composition is a homomorphism (operation preserving) and a bijection.  I need help with notation and how to show the operation preserving, onto and one-to-one.
Operation preserving:  $\sigma\circ\phi(ab)=\sigma(\phi(ab))=\sigma(\phi(a)\phi(b))=(\sigma\circ\phi(a))(\sigma\circ\phi(b))$
Am I on the right track?

REPLY [3 votes]: Yes, now you just need that the composition of bijections is a bijection. 
Perhaps by showing that $\phi^{-1}\circ \sigma^{-1}: G_3 \to G_1$ is the inverse of $\sigma \circ \phi$ (which holds for any bijections $\sigma$ and $\phi$, not necessarily between algebraic structures). Just compose this with $\sigma \circ \phi$ (on the left and on the right) to verify that you get identity maps in both cases.

REPLY [2 votes]: That is correct. All that remains is to show that the composition is a bijection.
Personally, it suffices to say "the composition of two bijections is a bijection." It is not difficult to show directly, however, either by showing subjectivity and infectivity or by showing that the inverse exists.