Enumerating isomorphic subgraphs

For digraphs $$G$$ and $$H$$ if we can partition $$V(G)$$ into a family $$\{Q_t\}_{t\in V(H)}$$ indexed by $$V(H)$$ such that $$E(G)=\bigcup_{(u,v)\in E(H)}Q_u\times Q_v$$, then is every subgraph of $$G$$ isomorphic to $$H$$ induced in $$G$$ by some set of complete representatives for the partition $$P=\{Q_t:t\in V(H)\}$$?

This is not true in general. Here is a non-trivial counter example: Let $$H \sim K_{1,2}$$ and $$G\sim C_4$$, with the vertices of $$G$$ labelled $$a,b,c,d$$ around the cycle. Then $$\{a\},\{c\},\{b,d\}$$ works as a partition, but the induced subgraph $$G[a,b,d]$$ is isomorphic to $$H$$.
• And if you want to add non-trivial directedness, we can let $H$ be the graph $x \to y, z \to y$ and $G$ be $a \to b, a \to d, c\to b, c\to d$. – Puck Rombach Dec 31 '18 at 17:08
• How would you express $\{(a,b),(b,c),(c,d ),(d,a)\}$ in terms of the blocks $\{a\},\{c\},\{b,d\}$? I don't see how a cycle can be written in the form $\bigcup_{(u,v)\in E(H)}X_u\times X_v$ unless $|X_t|=1$ for all $t\in V(H)$. – Ethan Dec 31 '18 at 17:25
• I would say that $\{ ab,bc,cd,da\}=(\{a\} \times \{b,d\})\cup (\{b,d\} \times \{c\})$. Is that what you had in mind? – Puck Rombach Dec 31 '18 at 17:27
• @ Puck Rombach But $\{a\}\times \{b,d\}\cup \{b,d\}\times \{c\}=\{(a,b),(a,d),(b,c),(d,c)\}$ which isnt a cycle – Ethan Dec 31 '18 at 17:29
• I was ignoring the directions in that comment. Is that what you're worried about? In that case, we can say $\{ ab,ad,cb,cd\}= \{a\}\times \{b,d\} \cup \{c\}\times \{b,d\}$? – Puck Rombach Dec 31 '18 at 17:31