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Clearly, $G$ is a subgroup of the automorphism group of a Cayley graph $\Gamma$ of $G$, because $G$ acts on $\Gamma$. I know plenty of examples where $G = \text{Aut}(\Gamma)$, but what is an example where the automorphism group of $\Gamma$ is larger than $G$?

aras
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    Take $G=\mathbb{Z}_4$ with generator set $S={1,2,3}$. The resulting Cayley graph is the complete graph $K_4$. The automorphism group is then $S_4\neq \mathbb{Z}_4$. – Hamed Jul 10 '18 at 17:47
  • Another example. Take $G=\mathbb{Z}_2\times \mathbb{Z}_2$ with generator set $S={(1,1)}$. The resulting Cayley graph is a disjoint union of two copies of $K_2$. The resulting automorphism group has 8 elements. – Hamed Jul 10 '18 at 18:00
  • Any Cayley graph for an abelian group with exponent greater than two - the inversion map fixes 1 and has order two. – Chris Godsil Jul 10 '18 at 18:04
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    And my final comment, take $G=\mathbb{Z}_n$ with $S={1}$. This is a cyclic graph which has the dihedral group $D_n$ as its automorphism group. – Hamed Jul 10 '18 at 18:09
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    "I know plenty of examples where $G = \operatorname{Aut}(\Gamma)$...". I am skeptical here - I can only name one (the trivial group!). – user1729 Jul 10 '18 at 18:41
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    Examples where $G = \text{Aut}(\Gamma)$ seem pretty common. Some reflection groups come to mind, for instance the spherical $(2,3,5)$ group and the Euclidean $(2,3,6)$ reflection group. @user1729 – Lee Mosher Jul 11 '18 at 17:01
  • A closely related question https://math.stackexchange.com/questions/1098115/when-is-the-automorphism-group-of-the-cayley-graph-of-g-just-g – Lee Mosher Jul 11 '18 at 17:01
  • @Lee The cyclic group of order two also works! My comment started life as "it seems like a more interesting question to find examples of groups where $G=\operatorname{Aut}(\Gamma)$", but that seemed too dismissive at the time (now I feel it seems less dismissive!). I still think this is an interesting questions though, but have no time! I guess it isn't as simple as "$G=\operatorname{Aut}(\Gamma)$ if and only if $\operatorname{Out}(G)=1$"? – user1729 Jul 12 '18 at 09:30
  • @Lee So it turns out that they are very common, at least amongst finite groups. In particular, if $G$ is non-abelian with order coprime to $6$ then $G$ has some Cayley graph $\Gamma$ such that $Aut(\Gamma)=G$ (https://cms.math.ca/openaccess/cjm/v24/cjm1972v24.0993-1008.pdf MR319804). If $G$ is abelian, and not one of $7$ specified groups, then then $G$ has some Cayley graph $\Gamma$ such that the natural empbedding of $G$ in $Aut(\Gamma)$ has index at most two (Imrich, W. & Watkins, M.E. Period Math Hung (1976) 7: 243. https://doi.org/10.1007/BF02017943 ). – user1729 Jul 12 '18 at 10:03

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Assuming that $\Gamma$ is the undirected graph underlying the Cayley graph*, easy examples are found with cyclic groups:

  • The Cayley graph of $\mathbb{Z}_n=\langle 1\rangle$ is a regular $n$-gon. The automorphism group of a regular $n$-gon is the dihedral group of order $2n$, $D_{n}$.

  • The Cayley graph of $\mathbb{Z}=\langle 1\rangle$ is a straight line. The automorphism group of a straight line is the infinite dihedral group, $D_{\infty}$.

In each of these cases, $G$ has index two in $\operatorname{Aut}(\Gamma)$. The extra symmetries come from "flipping" the graph.

*Without this assumption, $G\cong\operatorname{Aut}(\Gamma)$. That is, a group $G$ is the automorphism group of any associated Cayley graph. Which is kinda the point of Cayley graphs...

user1729
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  • About the asterisk comment, this is to say that the Cayley graph as a digraph is will possess $G$ as its automorphism group. Is that correct? – Chickenmancer Jul 11 '18 at 01:11
  • @Chickenmancer No - a Cayley graph is a coloured/labelled digraph. For example, consider the Klein $4$-group $G=\mathbb{Z}_2\times\mathbb{Z}_2$, and take the the generating set $S={ (0, 1), (1, 0)}$. The corresponding digraph can be rotated, but this rotation is incompatible with the colouring. Indeed, rotation has order $4$ but $G$ contains no element of order $4$. – user1729 Jul 11 '18 at 09:39