More odd graph theory from another point of view

TL;DR

This paper proves that certain Kneser graphs, especially odd graphs of even order, are vertex-transitive non-Cayley graphs using elementary number theory and group theory, expanding understanding of their symmetry properties.

Contribution

It provides a new, elementary proof that specific Kneser graphs are vertex-transitive non-Cayley graphs, avoiding deep group-theoretic methods used in prior work.

Findings

Almost all odd graphs are of even order

Such graphs are vertex-transitive non-Cayley graphs

Line graphs of certain odd graphs are also vertex-transitive non-Cayley graphs

Abstract

The Kneser graph $K(n, k)$ has as vertices all $k$-element subsets of $[n]=\{1,2,...,n \}$ and an edge between any two vertices that are disjoint. If $n=2k+1$, then $K(n, k)$ is called an odd graph. Let $ n >4$ and $1< k < \frac{n}{2} $. In the present paper, we show that if the Kneser graph $K(n,k)$ is of even order where $n$ is an odd integer or both of the integers $n,k $ are even, then $K(n,k)$ is a vertex-transitive non Cayley graph. Although, these are special cases of Godsil [8], unlike his proof that uses some very deep group-theoretical facts, ours uses no heavy group-theoretic facts. We obtain our results by using some rather elementary facts of number theory and group theory. We show that "almost all" odd graphs are of even order, and consequently are vertex-transitive non Cayley graphs. Finally, we show that if $ k > 4 $ is an even integer such that $k$ is not of the form $ k = 2^t $ for some $ t> 2 $, then the line graph of the odd graph $O_{k+1}$ is a vertex-transitive non Cayley graph.

Equations

$$\theta (v)=\{\theta (x_1),\theta (y_1),...,\theta (x_a),\theta (y_a),\theta (i_1),...,\theta (i_t)\}=$$

$$\{y_1,x_1,...,y_a,x_a,i_1,i_2,...,i_t\}=v$$

$$\theta (g)\{u,v\}=\{g(u),g(v)\},g\in Aut(\Gamma ),\{u,v\}\in E(\Gamma )$$

Full text

Keywords : Kneser graph, Odd graph, Permutation group, Cayley graph, Line graph

AMS subject classifications. 05C25, 05C69, 94C15

1. Introduction and Preliminaries

In this paper, a graph $\Gamma=(V,E)$ is considered as a finite, undirected, connected graph, without loops or multiple edges, where $V=V(\Gamma)$ is the vertex-set and $E=E(\Gamma)$ is the edge-set. For every terminology and notation not defined here, we follow [2, 6, 7, 11].

The study of vertex-transitive graphs has a long and rich history in discrete mathematics. Prominent examples of vertex-transitive graphs are Cayley graphs which are important in both theory as well as applications. Vertex-transitive graphs that are not Cayley graphs, for which we use the abbreviation VTNCG, have been an object of a systematic study since 1980 [3,8]. In trying to recognize whether or not a vertex-transitive graph is a Cayley graph, we are left with the problem of determining whether the automorphism group contains a regular subgroup [2]. The reference [1] is an excellent source for studying graphs that are VTNCG.

Let $n>4$ be an integer and $1<k<\frac{n}{2}$. The Kneser graph $K(n,k)$ is the graph with the $k$-element subsets of $[n]=\{1,2,...,n\}$ as vertices, where two such vertices are adjacent if and only if they are disjoint. If $n=2k+1$, then the graph $K(2k+1,k)$ is called an odd graph and is denoted by $O_{k+1}$. There are several good reasons for studying these graphs. One is that the questions which arise are related to problems in other areas of combinatorics, such as combinatorial set theory, coding theory, and design theory. A second reason is that the study of odd graphs tends to highlight the strengths and weaknesses of the techniques currently available in graph theory, and that many interesting problems and conjectures are encountered [3,4].

Amongst the various interesting properties of the Kneser graph $K(n,k)$, we interested in the automorphism group of it and we want to see how it acts on its vertex set. If $\theta\in Sym([n])$,  then

$f_{\theta}:V(K(n,k))\longrightarrow V(K(n,k))$, $f_{\theta}(\{x_{1},...,x_{k}\})=\{\theta(x_{1}),...,\theta(x_{k})\}$

is an automorphism of $K(n,k)$ and the mapping $\psi:Sym([n])\longrightarrow Aut(K(n,k))$, defined by the rule $\psi(\theta)=f_{\theta}$ is an injection. In fact, $Aut(K(n,k))=\{f_{\theta}|\theta\in Sym([n])\}\cong Sym([n])$ [7], and for this reason we identify $f_{\theta}$ with $\theta$ when $f_{\theta}$ is an automorphism of $K(n,k)$, and in such a situation we write $\theta$ instead of $f_{\theta}$. It is an easy task to show that the Kneser graph $K(n,k)$ is a vertex-transitive graph [7].

In 1979 Biggs [3] asked whether there are many values of $k$ for which the odd graphs $O_{k+1}$ are Cayley graphs. In 1980 Godsil [8] proved (by using some very deep group-theoretical facts of group theory [9,10]) that for ’almost all’ values of $k$, the Kneser graph $K(n,k)$ is a $VTNCG$. In the present paper, we show that if the Kneser graph $K(n,k)$ is of even order where $n$ is an odd integer or both of the integers $n,k$ are even, then $K(n,k)$ is a $VTNCG$. We call the odd graph $O_{k+1}$ an even-odd graph when its order is an even integer. We show that ‘almost all’ odd graphs are even-odd graphs, and consequently ’almost all’ odd graphs are $VTNCG$. We obtain our results, by using some rather elementary facts of number theory and group theory.

Finally, we show that if $k>4$ is an even integer and $k$ is not of the form $k=2^{t}$ for some $t>2$, then the line graph of the odd graph $O_{k+1}$ is a $VTNCG$.

2. Main results

Theorem 2.1**.**

Let $n,k$ are integers, $n>4$, $2\leq k<\frac{n}{2}$ and ${n}\choose{k}$ is an even integer. Then the Kneser graph $K(n,k)$ is a vertex transitive non Cayley graph If one of the following conditions holds:

$(I)$*      $n$ is an odd integer;*

$(II)$*    $n$ and $k$ are even integers.*

Proof.

We know that the Kneser graph $K(n,k)$ is a vertex-transitive graph (for every positive integer $n$) [7], hence it is sufficient to show that it is a non Cayley graph.

On the contrary, we assume that the Kneser graph $K(n,k)$ is a Cayley graph. Then the group $Aut(K(n,k))=Sym([n])$, $[n]=\{1,2,...,n\}$ has a subgroup $R$, such that $R$ acts regularly on the vertex-set of $K(n,k)$. In particular, the order of $R$ is ${n\choose k}$, and since ( by assumption ) this number is an even integer, then $2$ divides $|R|$. Therefore, by the Cauchy’s theorem the group $R$ has an element $\theta$ of order $2$. We know that each element of $Sym([n])$ has a unique factorization into disjoint cycles of $Sym([n])$, hence we can write $\theta=\rho_{1}\rho_{2}...\rho_{h}$, where each $\rho_{i}$ is a cycle of $Sym([n])$ and $\rho_{i}\cap\rho_{j}={\varnothing}$ when $i\neq j$. We also know that if $\theta=\rho_{1}\rho_{2}...\rho_{h}$, where each $\rho_{i}$ is a cycle of $Sym([n])$ and $\rho_{i}\cap\rho_{j}={\varnothing}$, then the order of the permutation $\theta$ is the least common multiple of the integers, $|\rho_{1}|,\ |\rho_{2}|,...,|\rho_{h|}$. Since $\theta$ is of order $2$, then the order of each $\rho_{i}$ is $2$ or $1$, say, $|\rho_{i}|\in\{1,2\}$. In other words, each $\rho_{i}$ is a transposition or a cycle of length $1$.  Let $\theta=\tau_{1}\tau_{2}...\tau_{a}(i_{1})(i_{2})...(i_{b})$, where each $\tau_{r}$ is a transposition and each $i_{s}\in[n]$. We now argue the cases (I) and (II).

(I)  Let $n=2m+1,\ m>1$. Therefore, we have $2a+b=n=2m+1$, where $b$ is an odd integer, and hence it is non-zero. Since $b$ is a positive odd integer, then $b-1$ is an even integer. we let $d=\frac{b-1}{2}$, so that $d$ is a non-negative integer, $d<b$ and $m=a+d$. Let $\tau_{r}=(x_{r}y_{r})$, $1\leq r\leq a$, where $x_{r},y_{r}\in[n]$. Now, there are two cases:

(i)     $2a\leq k$,    (ii)   $2a>k$.

(i)   Suppose $2a\leq k$. Then there is some integer $t$ such that $2a+t=k$, and since $2a+b=2m+1$, then $t\leq b$. Thus, for transpositions $\tau_{1},\tau_{2},...,\tau_{a}$ and cycles $(i_{1}),...,(i_{t})$ of the cycle factorization of $\theta$, the set $v=\{x_{1},y_{1},...,x_{a},y_{a},i_{1},i_{2},...,i_{t}\}$ is a $k$ subset of the set $[n]$, and thus it is a vertex of the Kneser graph $K(n,k)$. Therefore, we have;

[TABLE]

[TABLE]

(ii)    Suppose $2a>k$. Then there is some integer $c$ such that $2c\leq k$ and $2(c+1)>k$. If $2c=k$, then we take the vertex $v=\{x_{1},y_{1},...,x_{c},y_{c}\}$, and hence we have;

$\theta(v)=\{\theta(x_{1}),\theta(y_{1}),...,\theta(x_{c}),\theta(y_{c})\}=\{y_{1},x_{1},...,y_{c},x_{c}\}=v$

We now assume $2c<k$, then $2c+1=k$. Since $b\geq 1$, then for transpositions $\tau_{1},\tau_{2},...,\tau_{c}$ and cycle $(i_{1})$ of the cycle factorization of $\theta$, the set $v=\{x_{1},y_{1},...,x_{c},y_{c},i_{1}\}$ is a $k$-subset of the set $[n]$, and therefore it is a vertex of the Kneser graph $K(n,k)$. Thus, we have;

$\theta(v)=\{\theta{(x_{1})},\theta(y_{1}),...,\theta(x_{c}),\theta(y_{c}),\theta(i_{1})\}=\{y_{1},x_{1},...,y_{c},x_{c},i_{1}\}=v$

(II) Let $n=2m,\ m>2$ and $k=2e,\ 0<2e<m$. Since $\theta=\tau_{1}\tau_{2}...\tau_{a}(i_{1})(i_{2})...(i_{b})$,  $\tau_{r}=(x_{r}y_{r})$, $1\leq r\leq a$, where $x_{r},y_{r}\in[n]$, then we have $2a+b=n=2m$, where $b$ is an even integer. We now consider the following cases.

If $a<e$, then $2a<2e$, and hence there is some integer $t$ such that $2a+t=2e=k$. Since $2a+b=n>2k=4e$, then $t<b$. Therefore, $v=\{x_{1},y_{1},...,x_{a},y_{a},i_{1},...,i_{t}\}$ is a vertex of $K(n,k)$.

If $a\geq e$, then $v=\{x_{1},y_{1},...,x_{e},y_{e}\}$ is a vertex of $K(n,k)$.

On the other hand, we can see that in every case we have $\theta(v)=v$.

From the above argument, it follows that $\theta$ fixes a vertex of the Kneser graph $K(n,k)$, which is a contradiction, because $R$ acts regularly on the vertex-set of $K(n,k)$ and $\theta\in R$ is of order 2. This contradiction shows that the assertion of our theorem is true.

∎

Remark 2.2*.*

Since $1<k<\frac{n}{2}$, hence if in the case (I) of the above theorem, we add the condition,

’and assume that $k$ is an even integer’

then we can construct two vertices $v,w$ of $K(n,k)$ such that $\theta(v)=v$, $\theta(w)=w$ and $v\cap w=\varnothing$. In fact, if $k=2l,\ l>0$, and $2a\leq k=2l$, we have no problem for constructing the vertices $v$ and $w$. If $2a>k=2l$, then we can construct, by transpositions, $\tau_{1},...,\tau_{l}$, the vertex $v$, and hence we now have enough transpositions and fixed points of $\theta$, namely $\tau_{l+1},...,\tau_{a}$ and $i_{1},...,i_{b}$, for constructing the vertex $w$.

In particular, if the integers $n,k$ are even, then there are vertices $v,w$ of $K(n,k)$ such that $\theta(v)=v$, $\theta(w)=w$ and $v\cap w=\varnothing$.

Since ${5}\choose{2}$=10 is an even integer and 5 is an odd integer, then we can conclude from the above theorem that $K(5,2)$, namely, the Petersen graph is a vertex transitive non-Cayley graph.

Let $k$ be a positive integer. We know from the definition of the odd graph $O_{k+1}$ that its order is ${2k+1}\choose{k}$.

Definition 2.3**.**

We define the odd graph $O_{k+1}$ as an even-odd graph, when it has an even order, namely, if the number ${2k+1}\choose{k}$ is an even integer.

For example, we know that the Petersen graph has ${5}\choose{2}$ = 10 vertices, so by our definition it is an even-odd graph. In the first step, we want to show that ‘almost all’ odd graphs are even-odd graphs. In fact, we will show that if $k>1$ and $k$ is not of the form $k=2^{t}-1$ for some $t\geq 2$, then the number ${2k+1}\choose{k}$ is an even integer. For our purpose, we need the following fact which was discovered by Lucas [5, chapter 3]. We assume here the usual conventions for binomial coefficients, in particular, ${a}\choose{b}$=0 if $a<b$.

Theorem 2.4**.**

Let $p$ be a prime number and let $m=a_{0}+a_{1}p+...+a_{k}p^{k}$, $n=b_{0}+b_{1}p+...+b_{k}p^{k}$, where $0\leq a_{i},b_{i}<p$ for $i=0,...,k$. Then we have;

${m}\choose{n}$ $\equiv\prod_{i=0}^{k}$ ${a_{i}}\choose{b_{i}}$   ( mod   $p$ )

We now are ready to prove the following assertion.

Lemma 2.5**.**

If $k>1$ and $k$ is not of the form $k=2^{t}-1$ for some $t\geq 2$, then the number ${2k+1}\choose{k}$ is an even integer.

Proof.

Let $k=a_{0}+a_{1}2+...+a_{n}2^{n}$, where $0\leq a_{i}<2$ for $i=0,...,n$ with $a_{n}\neq 0$. If for every $i$, $a_{i}=1$, then we have $k=1+2+2^{2}+...+2^{n}=2^{n+1}-1$, which is impossible. Therefore, we must have $a_{r}=0$, for some $r=0,...,n-1$. If $j$ is the largest index such that $a_{j}=0$, then for all $i$ such that $i>j$ we have $a_{i}=1$. It is obvious that $j\neq n$. If we let $2k+1=b_{0}+b_{1}2+...+b_{n}2^{n}+b_{n+1}2^{n+1}$, then we must have $b_{i+1}=a_{i}$ for $i=0,...,n$. Therefore, ${b_{j+1}}\choose{a_{j+1}}$ = ${a_{j}}\choose{a_{j+1}}$ = ${0}\choose{1}$ =0. By Theorem 2.4. we have

${2k+1}\choose{k}$ \equiv\prod_{i=0}^{n+1}$${b_{i}}\choose{a_{i}}   ( mod   $2\ )$

where $a_{n+1}=0$. Now since ${b_{j+1}}\choose{a_{j+1}}$ = 0, then

${2k+1}\choose{k}$ \equiv\prod_{r=0}^{n+1}$${b_{i}}\choose{a_{i}} =0 ( mod   $2\ )$

Thus, for such a $k$ the integer ${2k+1}\choose{k}$ is an even integer.

∎

Remark 2.6*.*

It follows from the proof of the above lemma that, if $k=2^{t}-1$, for some positive integer $t$, then the integer ${2k+1}\choose{k}$ is an odd integer.

The following result follows from Lemma 2.5.

Corollary 2.7**.**

‘Almost all’ odd graphs are even-odd graphs, in other words, if $k>1$ and $k$ is not of the form $k=2^{t}-1$ for some $t\geq 2$, then the odd graph $O_{k+1}$ is an even-odd graph.

Now, it follows from Theorem 2.1. the following result.

Theorem 2.8**.**

Let $k$ be a positive integer. If $O_{k+1}$ is an even-odd graph, then it is a vertex-transitive non Cayley graph, namely, $O_{k+1}$ is a VTNCG.

Let $\Gamma$ be a graph, then the line graph $L(\Gamma)$ of the graph $\Gamma$ is constructed by taking the edges of $\Gamma$ as vertices of $L(\Gamma)$, and joining two vertices in $L(\Gamma)$ whenever the corresponding edges in $\Gamma$ have a common vertex. There is an important relation between $Aut(\Gamma)$ and $Aut(L(\Gamma))$. In fact, we have the following result [3, chapter 15].

Theorem 2.9**.**

The mapping $\theta:Aut(\Gamma)\rightarrow Aut(L(\Gamma))$ defined by;

[TABLE]

is a group homomorphism and in fact we have;

$(i)$*      $\theta$ is a monomorphism provided $\Gamma\neq K_{2}$;*

$(ii)$*    $\theta$ is an epimorphism provided $\Gamma$ is not $K_{4}$, $K_{4}$ with one edge deleted, or $K_{4}$ with two adjacent edges deleted.*

In the sequel, for an application of our method in proving the previous results, we show that the line graphs of some classes of odd graphs are vertex-transitive non Cayley graphs.

The graph $\Gamma$ is called a distance-transitive graph if for all vertices $u,v,x,y,$ of $\Gamma$ such that $d(u,v)=d(x,y)$, there exists some $g\in Aut(\Gamma)$ satisfying $g(u)=x$ and $g(v)=y$. It is well known that every odd graph is a distance-transitive graph [2]. The action of $Aut(\Gamma)$ on $V(\Gamma)$ induces an action on $E(\Gamma)$, by the rule $\beta\{x,y\}=\{\beta(x),\beta(y)\}$, $\beta\in Aut(\Gamma)$, and $\Gamma$ is called edge-transitive if this action is transitive. It is clear that a distance-transitive graph is an edge-transitive graph, and hence an odd graph is an edge-transitive graph. If a graph $\Gamma$ is edge-transitive, then its line graph is vertex transitive. Thus the line graph of an odd graph is a vertex-transitive graph.

Theorem 2.10**.**

Let $k>4$ be an even integer. If the integer ${2k+1}\choose{k}$ is a multiple of $4$, then $L(O_{k+1})$, the line graph of the odd graph $O_{k+1}$ is a vertex-transitive non Cayley graph.

Proof.

We let $\Gamma=L(O_{k+1})$. Then, each vertex of $\Gamma$ is of the form $\{v,w\}$, where $v,w$ are $k$-subsets of $[n]=\{1,2,...,n\}$ and $v\cap w=\varnothing$. Since each vertex of $O_{k+1}$ is of degree $k+1$, then the order of $\Gamma$ is \frac{k+1}{2}$${2k+1}\choose{k} and hence $\Gamma$ is of even order, because by assumption, ${2k+1}\choose{k}$ is a multiple of 4. Note that by Theorem 2.9. each automorphism of the graph $\Gamma$ is of the form $f_{\theta}$, $\theta\in Sym([n])$, where $f_{\theta}(\{v,w\})=\{\theta(v),\theta(w)\}$ for every vertex $\{v,w\}$ of $\Gamma$.

Now, On the contrary, assume that $\Gamma$ is a Cayley graph. Then the automorphism group of the graph $\Gamma$ has a subgroup $R$ such that $R$ acts regularly on the vertex set of $\Gamma$ and hence $R$ is of order \frac{k+1}{2}$${2k+1}\choose{k}. Therefore, $R$ has an element $f_{\theta}$ of order 2, where $\theta\in Sym([n])$, and hence $\theta$ is of order 2. Thus, by Remark 2.2. there are $k$-subsets $v,w$ of $[n]$ such that $v\cap w=\varnothing$ and $\theta(v)=v,\ \theta(w)=w$. Consequently, for the vertex $\{v,w\}$ of the graph $\Gamma$, we have,

$f_{\theta}(\{v,w\})=\{\theta(v),\theta(w)\}=\{v,w\}$

which is a contradiction. ∎

Remark 2.11*.*

Let the group $R$ be as in the above theorem. We do not know whether $R$ acts transitively on the family of $k$-subsets of $[n]$, hence it seems that we cannot use the methods of Godsil [8] for proving the above theorem.

Lemma 2.12**.**

Let $k>4$ be an even integer and $k$ is not of the form $k=2^{t}$ for some $t\geq 2$, then the number ${2k+1}\choose{k}$ is a multiple of $4$.

Proof.

We know that;

${2k+1}\choose{k}$ = $\frac{(2k+1)!}{(k!)(k+1)!}$ = $\frac{(k+2)...(2k)(2k+1)}{2.3....k}=2(2k+1)\frac{(k+2)(k+3)...(2k-1)}{2.3...(k-1)}$

If we let $t=\frac{(k+2)(k+3)...(2k-1)}{2.3...(k-1)}$, then we have

$(k+1)t=\frac{(k+1)(k+2)...(2k-1)}{2.3...(k-1)}$ = ${2(k-1)+1}\choose{k-1}$ = $\frac{(2k-1)!}{(k!)(k-1)!}$

If we show that $t$ is a multiple of 2, then we conclude that $2(2k+1)t$= ${2k+1}\choose{k}$ is a multiple of 4. Note that by our assumption $k\neq 2^{i}$ for each positive integer $i$, hence $(k-1)$ is not of the form $2^{i}-1$ for some positive integer $i$, thus by Lemma 2.5. the integer ${2(k-1)+1}\choose{k-1}$ is an even integer, and therefore 2 divides the integer $(k+1)t$. On the other hand, $k$ is an even integer, so $k+1$ is an odd integer, and hence 2 and $(k+1)$ are relatively prime, thus 2 divides the integer $t$. ∎

Now, by Theorem 2.10. and Lemma 2.12. the following theorem follows.

Theorem 2.13**.**

Let $k>4$ be an even integer such that $k$ is not of the form $k=2^{t}$ for some integer $t>2$. Then the line graph of the odd graph $O_{k+1}$ is a vertex-transitive non Cayley graph, namely, for such an integer $k$, the graph $L(O_{k+1})$ is a VTNCG.

**Acknowledgements **

The author is thankful to the anonymous reviewers for valuable comments and suggestions.