# Let $G,G'$ be two digraphs, show that $\phi^{-1}$ is an isomorphism

Problem So let $$\phi : G \rightarrow G'$$ be an isomorphism between two directed graphs. Prove that $$\phi^{-1}$$ is an isomorphism. Also prove if $$H \leq Aut(G')$$ then $$\phi^{-1}H\phi\leq Aut(G)$$.

My solution So two digraphs $$G,G'$$ are isomorphic, means that $$\phi:V(G)\rightarrow V(G')$$ is a bijection between vertex sets of $$G,G'$$ and $$(u,v)$$ is an arc in $$G$$ if and only if $$(\phi(u),\phi(v))$$ is an arc in $$G'$$. Hence to show that $$\phi^{-1} : G' \rightarrow G$$ is a digraph isomorphism we need to show that $$(u,v)$$ is an arc in $$G'$$ if and only if $$(\phi^{-1}(u),\phi^{-1}(v))$$is an arc in G. Since obviously inverse of a bijection between vertex sets is a bijection as well. So $$(u,v)$$ is an arc in $$G'$$ $$\Leftrightarrow$$ $$(\phi\phi^{-1}u,\phi\phi^{-1}v)$$ is an arc in $$G'$$ $$\Leftrightarrow$$ $$(\phi(\phi^{-1}u),\phi(\phi^{-1}v))$$ is an arc in $$G'$$ $$\Leftrightarrow$$ $$(\phi^{-1}u,\phi^{-1}v)$$ is an arc in $$G$$. Hence we showed what we wanted to show.

Now I am stuck in the second part of the problem. Should I simply show all the subgroup properties or the is some other way?