Let $n = 4k$ or $n = 4k+1$ for non-negative $k$. Prove that there exist a self-complementary graph $G = (V, E)$, where $|V| =n$

Question

Letting $\bar{G}$ be the complement graph of $G$.

I know that if $G$ is self-complementary then $n=4k$ or $n=4k+1$ is true. However, I am having trouble proving the other way around. I have tried to use a proof by induction but at some point in the proof it became too complicated to complete.

We know that self-complementary means that graph $G$ and $\bar{G}$ are both isomorphic and complement of each others.

It is easy to prove that they are complement of each other, but I am finding difficulty proving that they are isomorphic.

Answer 1

This idea maybe work: Consider it in four groups with same vertecis, and try to make the $\bar{G}$ in this way.