 Solution

The inclusion of a full subgraph ${\Gamma }^{\prime }\subseteq \Gamma$ sends each generator of $A_{{\Gamma }^{\prime }}$ to the equally named generator of $A_{\Gamma }$. Defineby fixing the generators in ${\Gamma }^{\prime }$ and sending every other generator to the identity. Every commutator relation of $A_{\Gamma }$ maps to a valid relation, so $r$ is a group homomorphism. Its composite with the natural map $A_{{\Gamma }^{\prime }}\to A_{\Gamma }$ is the identity. The natural map has a left inverse and is therefore injective, proving that $A_{{\Gamma }^{\prime }}$ is isomorphic to a subgroup of $A_{\Gamma }$.