This comes from a book called Introduction to Graph Theory (Dover Books on Mathematics) at the end of the first chapter we are asked to draw all 17 subgraphs of k3 which is pretty easy to do. however the next problem 22 asks The number of non-isomorphic subgraphs of K3 is only 7. Draw them. I do not quite understand what I am comparing the subgraphs to. I could for instance say compare K3 with say another graph proving the 2 graphs are not isomorphic by comparing their subgraphs alternatively I could compare the k3 subgraphs to each other in which case I end up with 7 distinct types of subgraphs:

  1. 3 vertices 3 edges
  2. 3 vertices 2 edges
  3. 3 vertices 0 edges
  4. 2 vertices 1 edges
  5. 2 vertices 0 edges
  6. 1 vertices 0 edges
  7. 3 vertices 1 edges

is the list above the correct answer?


You are correct that you are comparing the subgraphs to one another. The goal is to decide which of the 17 labelled (i.e., the vertices have labels) subgraphs of $K_3$ are isomorphic to one another. The isomorphism relation is an equivalence relation, and the really important aspect of that for us is transitivity. That means that if $G_1$ is isomorphic to $G_2$ and $G_2$ is isomorphic to $G_3$, then $G_1$ is isomorphic to $G_3$. This drastically cuts down on the number of pairs of graphs you have to investigate.

Here are three subgraphs of $K_3$ (the numbers are labeled vertices and the hyphens are edges):




If you remove the vertex labels, however, they all look like:

x-x-x (the path on three vertices, $P_3$)

Since the graphs are identical when labels are removed, they are all isomorphic to one another. You can say they belong to the same isomorphism class.

Another subgraph of $K_3$ is $K_3$ itself. Even when you remove the vertex labels, you will be able to tell that it is a different graph than $P_3$. It is not isomorphic to $P_3$, and therefore belongs in a different isomorphism class.

Continue making these sorts of comparisons until all 17 labelled graphs live in exactly one isomorphism class.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.