existence of disconnected, self-complementary graph Does there exist a disconnected self-complementary graph?
I think no, because if graph is disconnected(connected), then the complementary will be connected(disconnected). Am I proceeding in the right direction. Thanks for your help.
 A: You are proceeding in the right direction.
However, if the graph is connected, then its complemented may be connected too; think of $K_5$, split in the pentagon and the five-pointed star.

To show that indeed the complement of a disconnected graph is connected, let $u, v$ be vertices of $G$.
If $u,v$ are in different connected components (CCs), then there is an edge between them in $G^c$. If they are in the same CC, then there is a vertex $w$ in a different CC since $G$ is not connected. Both $u$ and $v$ have an edge to $w$ in $G^c$, hence there is a path from $u$ to $v$ in $G^c$.
A: Disconnecton of base graph not implies that complementary graph is connected and vice versa.
Let $G=K_4$ You can decompose it into 2 connected graphs which are self-component(to each other) and also isomorphic.
A: Does this work?
o----------o
o----------o
^above is your G
thus, G bar =
o                        o                 
|                        |
|                        |
o                        o
both graphs are disconnected and both are equal.
