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Let $G$ and $H$ be isomorphic graphs. Prove that the complements of $G$ and $H$ are isomorphic.

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closed as off-topic by Zev Chonoles, Morgan Rodgers, Sally, Silvia Ghinassi, Harish Chandra Rajpoot Feb 26 at 8:18

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Zev Chonoles, Morgan Rodgers, Silvia Ghinassi, Harish Chandra Rajpoot
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What does "self-complementary both are isomorphism" mean? – Chris Eagle Oct 19 '12 at 19:33
This is rather difficult to understand. Can you restate this more clearly? – Cameron Buie Oct 19 '12 at 19:38
I edited the question to something that almost makes sense, but am not sure if that resembles the askers intention. Whatever self-complementaries are, they are probably trivially isomorphic for isomorphic inputs. – Hagen von Eitzen Oct 19 '12 at 19:53
sorry for bad english. – geni Oct 20 '12 at 7:23
"prove that the complements of G and H are isomorphic" is true. – geni Oct 20 '12 at 7:24
up vote 6 down vote accepted

I am going to answer the question that I see, which is to "Prove that the complements of $G$ and $H$ are isomorphic." I can't think of any other possible meaning to the question.

You're telling me $G$ and $H$ are isomorphic, so that means there exists a map from the vertices of $G$ to the vertices of $H$ such that $u$ is adjacent to $v$ in $G$ if and only if $f(u)$ is adjacent to $f(v)$ in $H$.

So, now you want to know if the complements of $G$ and $H$ are isomorphic?

Hint 1: If $u$ and $v$ are adjacent in $G$, what is true about $u$ and $v$ in the complement of $G$? Or, if $u$ and $v$ are not adjacent in $G$, what is true about $u$ and $v$ in the complement? Similarly, with $H$.

Hint 2: Use the same $f$ you already know exists since $G$ is isomorphic to $H$.

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protected by Zev Chonoles Feb 25 at 8:02

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