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Let $G$ and $H$ be two self complimentary graphs with with disjoint vertex sets, where H has even order n. Let F be the graph obtained from $G \cup H$ by joining each vertex in G to every vertex of degree less than $\frac{n}{2}$ in $H$. Show that $G \cup H$ is selfcomplimentary.

My Attemp at proof so far: We consider $\overline{G \cup H}$, to do so we consider $G$ and $H$ seperately. Since $G \cong \overline{G}$ and $H \cong \overline{H}$ we know that each connected component is self complementary and so we must now consider the connections between the 2 components. We know that a self complimentary graph has exactly $\frac{1}{2}\binom{n}{2}$ edges, with exactly $\frac{n}{2}$ vertices above and below degree $\frac{n}{2}$. This implies that there is exactly the same number of connecctions between $G$ and $H$ as there are between $\overline{G}$ and $\overline{H}$.
From here i'm not completely certain where to go, any help is appreciated!

2 Answers 2

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Let $\varphi$ be the isomorfism $\varphi: G\rightarrow \overline G$ and let $\theta$ be the isomorphism $\theta:H\rightarrow \overline H$.

Consider the function $\sigma: F\rightarrow \overline F: f(v)=\left\{ \begin{array}{ll} \varphi(v) & v\in G \\ \theta(v) & v\in H \end{array}\right .$

We have to show that $\sigma(uv)$ is not an edge of $F$ if and only if $uv$ is not an edge of $F$ ( this would prove $\sigma$ is an isomorphism).

We clearly only need to check the case $u\in G$ and $v\in H$.

Notice that $d(\theta(v))$=n-d(v)-1$ for all $v\in H$. So we have:

$uv\in F \iff d(v)

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Your proof looks like it should work out! Let me add a bit of notation to help clarify things for myself.

Definition 1: For two graphs $G_1 = (V(G_1),E(G_1))$ and $G_2 = (V(G_2),E(G_2))$ , a graph isomorphism is a bijective function $\phi:V(G_1)\to V(G_2)$ such that $\{v,w\}\in E(G_1)$ if and only if $\{\phi(v), \phi(w)\}\in E(G_2)$.

So you are using Definition 1 with $G_1 = G\cup H$ and $G_2 = \overline{G\cup H}$.

Like you said, there exists a graph isomorphism $\phi_G:V(G)\to V(\overline{G})$ between $G$ and $\overline{G}$ and an isomorphism $\phi_H:V(H)\to V(\overline{H})$ between $H$ and $\overline{H}$. Define $\phi_{G\cup H}:V(G\cup H)\to V(\overline{G\cup H})$ to be $$\phi_{G\cup H}(v) = \begin{cases} \phi_G(v) &~\text{if}~v\in V(G)\\ \phi_H(v)&~\text{if}~v\in V(H).\end{cases}$$

From here, you would like to show

  1. $\phi_{G\cup H}$ is bijective (this follows since $\phi_G$ and $\phi_H$ are bijective).
  2. $\{v,w\}\in E(G\cup H)$ if and only if $\{\phi_{G\cup H}(v), \phi_{G\cup H}(w)\}\in E(\overline{G\cup H})$.

You have done this so far (in not as many words), and we have to show 2. is true. For this, we just have to show that 2. holds for the following cases

  • Case 1: $u,v\in V(G)$
  • Case 2: $u,v\in V(H)$
  • Case 3: $u\in V(G)$ and $v\in V(H)$.

Only the last case isn't trivial. Note $\{v,w\}\in E(G\cup H)$ with $v\in V(G)$ and $w\in V(H)$ if and only if deg$(w) < n/2$ in $H$ iff and only if deg$(\phi_{H}(w)) < n/2$ in $\overline{H}$ iff deg$(\phi_{H}(w)) \geq n/2$ in $H$. Note deg$(\phi_{H}(w)) \geq n/2$ in $H$ if and only if $\phi_{G\cup H}(v)$ is not connected to $\phi_{H}(w)$ in $G\cup H$. This gives the result.