Order 1 Simple Graph is Unique up to Isomorphism

Theorem

Let $G_1 = \struct {\map V {G_1}, \map E {G_1} }$ and $G_2 = \struct {\map V {G_2}, \map E {G_2} }$ be simple graphs of order $1$.

Then $G_1$ and $G_2$ are isomorphic.

Proof

There is only one bijection from $\map V {G_1}$ to $\map V {G_2}$.

There are no vertices adjacent to the sole vertex in $\map V {G_1}$

There are no vertices adjacent to the sole vertex in $\map V {G_2}$.

Hence the result, vacuously.

$\blacksquare$

Sources

• 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.2$: Isomorphic Graphs