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Let a Cayley graph $G$ of a group $H$ with respect to the generating set $\{s_i\}$ have a clique of order $> 2$. In addition assume the graph $G$ is non-complete. If the clique size is less than half the order of $G$, then is it possible for some group $H$ that $G$ has a unique "disjoint maximal clique". By "disjoint maximal clique", I mean a clique equal to the clique size of the graph, and such that any other clique of same order would not be vertex disjoint with the prior clique.

I don't think so. For, if $(e),(s_1),(s_1\cdot s_2),(s_1\cdot s_2\cdot s_3),\ldots,(s_1\cdot s_2\cdots s_n)$ be the sequence of vertices in a maximal clique, then I think even $(s_1^2),(s_1^3),(s_1^2\cdot s_2),\ldots,(s_1^2\cdot s_2\cdots s_n)$ would also be a sequence of vertices in a maximal clique, where $e$ denotes the identity element. But, what if $s_1$ is an order $2$ or $3$ element. How do we ensure that there always exist a disjoint clique apart from the clique $(e),(s_1),(s_1\cdot s_2),(s_1\cdot s_2\cdot s_3),\ldots,(s_1\cdot s_2\cdots s_n)$? Will this be true at least for the case when $H$ is an abelian/cyclic group? Any hints? Thanks beforehand.

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  • $\begingroup$ Your question is not clear. "Disjoint" is a property of two things, not a property of one thing. $\endgroup$ – Brendan McKay May 23 at 9:04
  • $\begingroup$ @BrendanMcKay edited the post. $\endgroup$ – vidyarthi May 23 at 11:17
  • $\begingroup$ Your terminology looks confusing. "Unique maximal clique" usually means that it is a maximal clique and there is no other maximal clique. $\endgroup$ – Fedor Petrov May 23 at 11:23
  • $\begingroup$ @FedorPetrov no, my meaning is that there is a maximal clique and no other "vertex disjoint" maximal clique $\endgroup$ – vidyarthi May 23 at 11:41
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    $\begingroup$ Rephrased: "Do all maximal cliques in $G$ pairwise intersect"? (Also I don't see why you need to assume $G$ is non-complete: then your condition is trivially satisfied.) $\endgroup$ – Sam Hopkins May 23 at 12:44
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Let $G$ be the linegraph of the complete graph $K_n$ for $n\geq 5$. For some but not all $n$, $G$ is a Cayley graph, see Chris Godsil's answer to another question.

$G$ has $\binom n2$ vertices and degree $2n-4$. The maximum cliques of $G$ correspond to the edges incident with one vertex and so they have size $n-1$. Moreover, the cliques corresponding to two different vertices of $K_n$ have one point in common, namely the edge between those two vertices.

Therefore, $G$ is an example of a Cayley graph for which any two maximum cliques intersect, even though the maximum cliques only have size about the square root of the number of vertices.

I wonder if this example is optimal in some sense.


ADDED: Here is an exposition of Ilya's argument from the comments.

Theorem. If a vertex-transitive graph with $N$ vertices has cliques of size $k$ such that $k^2<N$, then there are two such cliques which are disjoint.

Proof. Take a fixed $k$-clique $C$ and apply a random automorphism $\gamma$. The expected number of elements of $C$ that map to an element of $C$ is $k^2/N$, so $k^2<N$ implies that $C$ must sometimes map to a clique disjoint from itself.

In the case of a Cayley graph of a group $\varGamma$, we can use a random non-identity element of $\varGamma$ to improve the inequality to $k(k-1)<N-1$.

There is a clique size gap of about $\sqrt 2$ between these bounds and the linegraph of a complete graph. So the problem is still missing a complete solution.

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    $\begingroup$ It is optimal in some sense: if a Cayley graph contains a clique of size $<\sqrt n$, then it contains its isomorphic copy, from probabilistic reasons $\endgroup$ – Ilya Bogdanov May 23 at 16:01
  • $\begingroup$ @IlyaBogdanov you mean if a cayley graph has clique size $< \sqrt n$, then it has disjoint cliques of maximal size? Any reference for this? $\endgroup$ – vidyarthi May 23 at 16:51
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    $\begingroup$ If a clique size is $k$, there are $n$ copies of one clique (starting from any vertex), and only $k(k-1)+1$ of them intersects the initial one (any vertex of a copy may coincide with any vertex of the origin). $\endgroup$ – Ilya Bogdanov May 23 at 16:54
  • $\begingroup$ will the property hold for cayley graphs on cyclic or abelian groups at least? $\endgroup$ – vidyarthi May 23 at 16:56
  • $\begingroup$ @IlyaBogdanov in saying that there are $n$ cliques of size $k$, you are using the vertex transitivity of the graph, isnt it? So this should hold for vertex transitive graphs also, right? $\endgroup$ – vidyarthi May 23 at 17:02

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