Frucht's theorem states that every group can be realized as an automorphism group of a graph. I am interested in which subgroups of $S_n$ can be realized as an automorphism group of a graph on $n$ vertices. Are there any things we can say about these groups or their actions which don't tautologically follow from the definition of automorphism groups of graphs?
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1Do you consider graph as oriented? Or you view them as a (reflexive?) incidence relation? Do you allow selfloops/multi-edges? In addition, you probably require that the action on the $n$ vertices is the given original permutation action. The answer to the question "Which subgroups of $S_n$ are equal to the automorphism group of a graph structure on ${1,\dots,n}$ is highly sensitive to these details. – YCor Jun 24 '18 at 06:53
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Now as regards the question, does it mean anything else than "what can we say that is non-trivial"? I think you could include some context, and possibly some trivial examples. In this kind of question, there can be somewhat trivial observations, and when thinking about such a question I like to have other preoccupations than "is this thing considered as trivial by the OP". In any case, if you think a little about your own problem, you can end up with some more precise and well-thought question. – YCor Jun 24 '18 at 06:56
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A graphical regular representation (GRR) of a group $G$ is a Cayley graph $X$ such that $\mathrm{Aut}(X)\cong G$. We know that if $|G| > 32$ and $G$ is neither abelian with exponent greater then two, nor generalized dicyclic, then $G$ has a GRR. A generalized dicyclic group is a non-abelian group that admits an automorphism that maps each element to itself or its inverse; they necessarily have an abelian subgroup of index two. (There is an explicit presentation.)
If $G$ is cyclic with order a prime power $q$ and $G$ acts as a group of automorphisms of a graph $X$, then $G$ must have an orbit of length $q$. f $|V(X)|=q$, then $X$ is a Cayley graph for $G$ and $|\mathrm{Aut}(X)| \ge 2q$. So cyclic groups of prime power order $q$ cannot be realized as automorphism groups of graphs with at most $q$ vertices.
Babai (On the minimum order of graphs with given group https://cms.math.ca/10.4153/CMB-1974-082-9) shows that if $G$ is not cyclic of order 3, 4 or 5, then it is the full automorphism group of a graph with at most $2|G|$ vertices. We can use this to realize more groups as automorphism groups of small graphs. For example if $G$ is cyclic of order 63, take the disjoint union of graphs on 14 and 18 vertices (with automorphism groups cyclic of order 7 and 9 respectively) along with an asymmetric graph on 41 vertices to get a graph on 63 vertices with automorphism group isomorphic to $G$.
Chris Godsil
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