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 $P=\{Q_t:t\in V(H)\}$?
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This is not true in general. Here is a nontrivial 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 $H$.

$\begingroup$ And if you want to add nontrivial directedness, we can let $H$ be the graph $x \to y, z \to y$ and $G$ be $a \to b, a \to d, c\to b, c\to d$. $\endgroup$ – Puck Rombach Dec 31 '18 at 17:08

$\begingroup$ 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 $t\in V(H)$. $\endgroup$ – Ethan Dec 31 '18 at 17:25

$\begingroup$ I would say that $\{ ab,bc,cd,da\}=(\{a\} \times \{b,d\})\cup (\{b,d\} \times \{c\})$. Is that what you had in mind? $\endgroup$ – Puck Rombach Dec 31 '18 at 17:27

$\begingroup$ @ Puck Rombach But $\{a\}\times \{b,d\}\cup \{b,d\}\times \{c\}=\{(a,b),(a,d),(b,c),(d,c)\}$ which isnt a cycle $\endgroup$ – Ethan Dec 31 '18 at 17:29

$\begingroup$ I was ignoring the directions in that comment. Is that what you're worried about? In that case, we can say $\{ ab,ad,cb,cd\}= \{a\}\times \{b,d\} \cup \{c\}\times \{b,d\}$? $\endgroup$ – Puck Rombach Dec 31 '18 at 17:31