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I am working on the statement " A graph $G$ is a $nKn$ (collection of $n$ number of complete graph each of order $n$) graph if and only if $\bar{G}$ is $α(G) = α(\bar{G}) = n$ and $(p-n)$-regular , where $p$ is the order of $G$ and $α(G)$ is the independence number of $G$ with $n\geqslant 1$."

The converse part seem to be a confusion. Kindly help me with it.

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Perhaps you could edit the meaning of $a(G)$ into your question. –  Gerry Myerson May 9 '12 at 12:58
    
α(G) is the independence number of G –  James Alex May 9 '12 at 14:07
    
I'm not sure I'm understanding the question properly. Can you clarify why the graph $G$ consisting of a 3-cycle and a 4-cycle would not be a counterexample? It would seem to satisfy $n=3$, $p=7$, but it is obviously not a $3K_3$. –  Will Orrick May 10 '12 at 16:48
    
@WillOrrick The graph you mentioned is 2 regular, not (7-3)=4 regular. –  Graphth May 10 '12 at 20:20
    
@Graphth : Hmmm. I had understood that it was $\overline{G}$ that was required to be $(p-n)$-regular, rather than $G$. Perhaps this is the root of my confusion. I'm now not sure that I understand how $3K_3$ is defined. Wouldn't $p$ equal 9 in that case, and $G$ have to be 6-regular? –  Will Orrick May 10 '12 at 21:41
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1 Answer

(Extended compilation of comments.)

While it is true that the disjoint union of $n$ copies of $K_n$ has the properties

  1. $\alpha(G)=\alpha(\overline G)=n$ (where $\alpha$ is the independence number)
  2. $\overline{G}$ is $p-n$ regular (where $p$ is the number of vertices of $G$)

the converse is not true. not every graph with properties 1 and 2. Will Orrick gave the following counterexample in comments: let $G$ be the disjoint union of two cycles $C_3 $ and $C_4$; then $\alpha(G)=\alpha(\overline G)=3$ and $\overline{G}$ is $4$-regular where $4=7-3$.

Brian M. Scott then observed that if $\alpha(\overline G)=n$ is strengthened to

(*) every maximal independent set in $\overline G$ has cardinality $n$

then the conclusion $G=nK_n$ follows. Indeed, (*) says that every maximal clique of $G$ has cardinality $n$. Therefore, every vertex is contained in a copy of $K_n$. Since $G$ is $n$-regular (by virtue of 2), it is a disjoint union of some copies of $K_n$. The condition $\alpha(G)=n$ places the number of the copies of $K_n$ at $n$.

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