0
$\begingroup$

Is there any research work about automorphism group of non-commutating graph?

$\endgroup$
1
  • $\begingroup$ do you mean the non-commuting graph of a group? If not, could you define what you mean? $\endgroup$
    – Nick Gill
    Nov 9, 2012 at 14:14

1 Answer 1

4
$\begingroup$

Given that you're based in Esfahan, and given that there are a number of Iranian mathematicians working on the non-commuting graphs of groups, I'll assume this is what you're asking about.

So the non-commuting graph of a group is the graph with vertices equal to elements of $G$, and two elements connected if they do not commute. Let's throw away the central elements of $G$ as these are just isolated vertices that will only confuse things.

Clearly the automorphism group of the non-commuting graph of a group $G$ will be the same as the automorphism group of the commuting graph of $G$ since the latter is the complement of the former.

To my knowledge the only completely general statement that one can make about this automorphism group is that it contains $G/Z(G)$ as a subgroup, since $G$ acts naturally on the vertices of the graph by conjugation.

Generally research on commuting graphs has not focused on the automorphism group. Instead people have been interested in recognition questions (when does the commuting graph of a group characterise the group), or to graph-theoretic questions (e.g. computing the diameter, chromatic number, clique number etc). I asked a MO-question on this recently that may be helpful.

The only specific reference I have been able to find to automorphism groups of commuting graphs is from this conference in which Mirzargar gave a talk showing that the automorphism group "is a non-abelian group such that its order is not prime power and square-free number." (I don't quite know how to interpret this statement, as the use of English is a little confusing.) So far as I'm aware this result has not appeared in the literature.

$\endgroup$
2
  • $\begingroup$ soroosh, my pleasure. if you like the answer click the tick' to the side of it. this will accept' the answer and take if off the unanswered stack. $\endgroup$
    – Nick Gill
    Nov 14, 2012 at 12:12
  • $\begingroup$ @ Nick! OK! so thanks for your attetion! $\endgroup$
    – sebastian
    Jan 13, 2013 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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