Let me denote the graph in the picture by $\Gamma$ and I will refer to the points as numbers $1-9$. I need to find Aut($\Gamma$).

enter image description here

Looking at this graph, it seems that there can be a permutation on the outer most edges 17 and 96, then both, then a symmetric mirror of this graph. So we have id,(17),(96),(17)(96),(25)(16)(97),(25)(19)(67),(25)(16)(97)(34),(25)(19)(67)(34). Is this correct? are there any other automorphisms I'm missing? thanks.

  • 2
    $\begingroup$ Why are you using a different labelling for vertices than the one on the picture? What is the correspondence? $\endgroup$ – verret Apr 16 at 9:45
  • 5
    $\begingroup$ The automorphism group of this graph does indeed have order $8$. $\endgroup$ – Derek Holt Apr 16 at 9:49
  • $\begingroup$ @verret A-G as 1-9. It's easier like this. $\endgroup$ – mandella Apr 16 at 9:54

Your Answer

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

Browse other questions tagged or ask your own question.