0
$\begingroup$

Say we have a (simple) graph $\Gamma$, and G=Aut($\Gamma$) .

Is it true (in general) that 2 induced subgraphs of $\Gamma$, say $\Gamma_1$ and $\Gamma_2$, are isomorphic iff they are in the same orbit of the action of G?

I suspect that the answer is 'no'.

First, I think one side is true and is trivial: If they are in the same orbit then there is an automorphism that, restricted to the vertices of $\Gamma_1$ and $\Gamma_2$ is an isomorphism.

However, I don't know, that given an isomorphism between $\Gamma_1$ and $\Gamma_2$ if we can extend it to an automorphism on $\Gamma$.

Am I correct so far?

Also, given a graph, how do I go about to show that for this specific graph this argument is true (while not being true in general)? I suspect that it has some connection to the cycle index of the action of G on V.

I know that $Z(G,1+x) = 1+x+2x^2+4x^3+5x^4+5x^5+4x^6+...+x^9$

(the graph in question is $L_2(3)$)

Thanks in advance!

Shay

$\endgroup$
1
  • 1
    $\begingroup$ A vertex or an edge gives you a subgraph. Isn't it obvious that the automorphism group need not act transitively on vertices or edges? It would have to if your statement were true. Maybe I misunderstand the meaning of "induced subgraph", though. $\endgroup$
    – t.b.
    Jul 16, 2011 at 12:02

1 Answer 1

1
$\begingroup$

No. Take two copies $H_1, H_2$ of the same graph $H$ and join them by two vertices $v_1, v_2$, then attach an extremely long path to $v_1$.

$\endgroup$

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.