= Solution
The inclusion of a <full subgraph> $\Gamma'\subseteq\Gamma$ sends each generator of $A_{\Gamma'}$ to the equally named generator of $A_\Gamma$. Define
$$
r:A_\Gamma\longrightarrow A_{\Gamma'}
$$
by fixing the generators in $\Gamma'$ 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 $A_{\Gamma'}\to A_\Gamma$ is the identity. The natural map has a left inverse and is therefore injective, proving that $A_{\Gamma'}$ is isomorphic to a subgroup of $A_\Gamma$.
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