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Let $G=(V,E)$ be a simple graph which has the following property. If $(u,v)\in E$ and $(v,s)\in E$ then $(u,s)\in E$. What kind of graph is this? Does it have any name and special property?

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  • $\begingroup$ Per @bof, you may want to change the hypotheses so as to stop all such graphs being merely collections of disconnected points. $\endgroup$ – Patrick Stevens Feb 9 '18 at 9:09
  • $\begingroup$ FYI, strictly speaking, to write '$(u,v)\in E$' is wrong, at least according to what is currently the usual formalisation of 'graph' via set theory (see e.g. Diestel: Graph Theory); in that formalism, a simple graph has two-sets as edges, and $(u,v)$ in set-theory is not equal to the two-set containing only $u$ and $v$. (Of course, $(u,v)$ in the usual formalisation is a two-set, but another one: $(u,v)=\{u,\{u,v\}\}$. $\endgroup$ – Peter Heinig Feb 9 '18 at 18:32
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Such a graph is a disjoint union of complete graphs. Indeed, it's easy to prove by induction on the number of vertices of any connected subgraph that any connected subgraph is complete.

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  • $\begingroup$ A simple group with the stated property is an empty graph, a graph with no edges. Because, if $(u,v)\in E,$ then $(v,u)\in E$ and $(u,u)\in E,$ so the graph is not simple. $\endgroup$ – bof Feb 9 '18 at 8:24
  • $\begingroup$ @bof True. My statement is still correct though ;) $\endgroup$ – Patrick Stevens Feb 9 '18 at 9:08

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