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I am thankful for feedback given to me, through comments on one question on MSE that I have deleted and also through responses to my question on Meta. With this feedback, I am rewriting another question that I have deleted, adding some context, and hopefully this will be a better question on MSE.

Is there a non-complete ternary Cayley graph that is a core with $3^3 = 27$ vertices?

Definition. By a ternary Cayley graph, I mean a (simple, undirected) graph whose vertex set is an elementary finite abelian $3$-group, thus its vertex set is necessarily isomorphic to $\mathbb{Z}_3^n := \bigoplus_{i = 1}^n \mathbb{Z}/3\mathbb{Z}$ for some nonnegative integer $n$, and whose edges are determined by a so-called connection set $C \subseteq \mathbb{Z}_3^n \setminus \{0\}$ that is closed under taking negation, i.e. $-C = C$. Namely, two vertices $x, y \in \mathbb{Z}_3^n$ are adjacent if and only if $x - y \in C$.

Note that the condition $-C = C$ ensures that the adjacency relation is symmetric, thus the Cayley graph is undirected. That $C$ does not contain the zero vector ensures that there are no loops. The Cayley graph on $\mathbb{Z}_3^n$ with connection set $C$ is denoted by $\mathrm{Cay}(\mathbb{Z}_3^n, C)$.

Definition. A graph is a core if all its endomorphisms are automorphisms.

By an endomorphism of a graph $X = (V, E)$ is meant an endomap $f : V \to V$ of its vertex set such that edges are sent to edges, i.e. if $x$ and $y$ are adjacent in $X$, then $f(x)$ and $f(y)$ are adjacent in $X$. An automorphism is simply an invertible endomorphism, as usual.

The context of this question is that,

it can be shown that every ternary Cayley graph core with at most $3^2= 9$ vertices is necessarily complete. I won't give a proof here, but if there is sufficient interest in a proof, I can add one in a later edit.

I conjecturely propose that the following Cayley graph with $3^3$ vertices is a noncomplete core: $$\mathrm{Cay}(\mathbb{Z}_3^3, \{i, 2i, j, 2j, k, 2k, i +j, 2i + 2j, i + k, 2i + 2k, i + j + k, 2i + 2j + 2k\}).$$ where $i = (1, 0, 0)$, $j = (0, 1, 0)$, $k = (0, 0, 1)$. However, I am happy to get any other construction on $3^3$ vertices.

Colin Tan
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  • What means "complete"? – Alex Ravsky Nov 12 '24 at 08:00
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    A graph is complete if any two distinct vertices are adjacent. – Colin Tan Nov 12 '24 at 08:48
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    For $d < 3$, the only ternary Cayley graph with $3^d$ vertices that is a core is the complete graph $K_{3^d}$. – Colin Tan Nov 12 '24 at 08:50
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    When $d=3$, the complete graph $K_{3^d} = K_{27}$ is a ternary Cayley graph core. My question asks whether there is any other ternary Cayley graph core with 27 vertices. – Colin Tan Nov 12 '24 at 12:13
  • I don't understand how the complete graph $K_{27}$ is a core. According to the definition, in this case any map $f:V\to V$ sends edges to edges (since every pair is an edge). On the other hand, the empty graph on $27$ vertices is a core – Ewan Delanoy Nov 12 '24 at 14:19
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    A complete (simple) graph is a core since two distinct vertices cannot be mapped onto the same vertex as there are no loops. – Colin Tan Nov 12 '24 at 15:44
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    The empty graph with more than one vertex is not a core since the map that sends all the vertices to a fixed vertex is a non-surjective endomorphism. – Colin Tan Nov 12 '24 at 15:45
  • @ColinTan You could try contacting Gordon Royle, at the University of Western Australia. He works on this kind of stuff and has lots of computational tools. He could tell you which Cayley graphs on 27 vertices are cores in a matter of seconds (or maybe minutes). – verret Apr 03 '25 at 05:53
  • Thanks verret, for your advice. Professor Royle has indeed done so at my reposting of this question on MO (https://mathoverflow.net/questions/482524/is-there-a-ternary-cayley-graph-on-27-vertices-that-is-a-non-complete-core) – Colin Tan May 06 '25 at 03:51

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