## For two graphs, $H$ and $G$ let $\bar{H}$ and $\bar{G}$ be their respective compliments. Then if $H$ is a subgraph of $G$, then $\bar{H}$ is a subgraph of $\bar{G}$. [closed]

Hello,

Thank you for taking a look at this post. I am trying to prove the statement as given in the title. I have managed to disprove it but I am yet not sure if it is correct. I would appreciate it if somebody could comment if the above statement is true or false. Thank you for your time.

-

## closed as off topic by Andreas Blass, Benjamin Steinberg, Steven Landsburg, Brendan McKay, Dan PetersenApr 20 2012 at 14:15

This statement is not true. For example, let $G=K_4$ and $H=C_4$, where $G$ and $H$ are complete graph and cycle graph with four vertices, respectively. It is easy to check that, $H$ is a subgraph of $G$, but $\overline{H}$ is not a subgraph of $\overline{G}$.
 Thank you for your reply. This statement is indeed not true. I see I wrote prove in my question, it should be disprove. I will correct it now. – Samrat Roy Apr 20 2012 at 13:07 Okay, I have corrected my statement. But please consider $G = K_3$ and $H = K_4$. Then, $\bar{G}$ is indeed subgraph of $\bar{H}$. – Samrat Roy Apr 20 2012 at 13:18 Actually, I have not been able to show that if the subgraph, $G$ in question is complete, then somehow this statement will also fail. I think that I would need to make the complete graph as a subgraph a exception – Samrat Roy Apr 20 2012 at 13:24