Isomorphism Classes for Order 4 Size 3 Simple Graphs

Theorem
There are $3$ equivalence classes for simple graphs of order $4$ and size $3$ under isomorphism:


 * Isomorphism-Classes-Order4-Size3-Simple.png

Proof
The fact that the $3$ graphs given are not isomorphic follows from Vertex Condition for Isomorphic Graphs.

The vertices have degrees as follows:
 * Graph $1$: $2, 2, 1, 1$
 * Graph $2$: $3, 1, 1, 1$
 * Graph $3$: $2, 2, 2, 0$

The fact that there are no more isomorphism classes of such graphs could be proved constructively.

Let the $4$ vertices be named $A, B, C$ and $D$.

Lemma: There must be intersections among the edges.

Proof: If there were no intersection at all, it requires at least $3 \times 2 = 6$ vertices.

Hence,, $2$ of the edges are $AB$ and $AC$.

To place the last edge, there are $\binom{4}{2}=6$ potential choices:


 * $AB$: this makes the graph not simple;
 * $AC$: this makes the graph not simple;
 * $AD$: this is isomorphic to graph $2$;
 * $BC$: this is isomorphic to graph $3$;
 * $BD$: this is isomorphic to graph $1$;
 * $CD$: this is isomorphic to graph $1$.

Hence, by Proof by Cases these $3$ are the only isomorphism classes.