Orbital diameters of the symmetric and alternating groups

Abstract

For a primitive group G acting on a finite set $\Omega$, we define the orbital diameter to be the maximum of the diameters of all orbital graphs of G. In this paper, we study the orbital diameters of symmetric and alternating groups. We give necessary numerical conditions for the orbital diameter to be bounded by some constant c and give precise descriptions of the actions for which the orbital diameter is bounded by 5. For each primitive action, we also either determine all orbital graphs of diameter 2 or give descriptions of infinite families of orbital graphs of diameter 2.

1 Introduction

Let G be a finite transitive permutation group acting on a set $\Omega$. We will denote such a group by $(G,\Omega )$. The orbitals of $(G,\Omega )$ (or just G) are the orbits of the natural action of G on $\Omega \times \Omega$. The diagonal orbital is $\Delta _{0}:=\{(\alpha ,\alpha ):\alpha \in \Omega \}$. To each orbital $\Delta$, we associate the paired orbital $\Delta ^{*}:=\{(\beta ,\alpha ):(\alpha ,\beta )\in \Delta )\}$; $\Delta$ is called self-paired if $\Delta =\Delta ^{*}$. For a given orbital $\Delta$, the orbital graph $\Gamma _{\Delta \cup \Delta ^{*}}$ is defined to be the undirected graph whose vertex set is $\Omega$ and edge set is $\Delta \cup \Delta ^{*}$. Henceforth we will assume that $\Delta \ne \Delta _{0}$ and write $\Gamma _{\Delta }$ if $\Delta$ is self-paired. Note that orbital graphs belong to a wider family of graphs known as edge-transitive graphs. By a theorem of D. G. Higman, a permutation group $(G,\Omega )$ is primitive if and only if all the orbital graphs are connected [see [5], (1.12)]. For a graph $\Gamma$, we will use $V(\Gamma )$ to denote the vertex set of $\Gamma$. The valency of $\alpha \in V(\Gamma )$ is the number of edges that are connected to $\alpha$; if all vertices of $\Gamma$ have the same valency v, then v is defined to be the valency of $\Gamma$. For a connected graph $\Gamma$, the distance between two vertices u and v is the length of a shortest path between u and v and will be denoted by $d_{\Gamma }(u,v)$; we will always assume that $u\ne v$ and use the notation d(u, v) if it is clear which graph is in question. The diameter of a connected graph is the greatest distance between any two vertices and will be denoted by $\mathrm{diam}(\Gamma )$. The orbital diameter of a primitive permutation group $(G,\Omega )$ is the maximum of the diameters of the orbital graphs of $(G,\Omega )$ and will be denoted by $\mathrm{diam_{O}}(G,\Omega )$. If ${\mathcal {C}}$ is a class of finite primitive permutation groups, then ${\mathcal {C}}$ is said to be bounded if there exists a positive integer d such that $\mathrm{diam_{O}}(G,\Omega )\le d$ for all $(G,\Omega )$ in ${\mathcal {C}}$. In this paper, we study the orbital diameters of finite symmetric and alternating groups in their primitive actions. The study is motivated by a paper [7] by Liebeck, Macpherson and Tent in which the authors describe infinite bounded classes of finite primitive permutation groups. Let ${\mathcal {C}}$ be such an infinite class. The main theorem of [7] gives tight structural information about the primitive groups in ${\mathcal {C}}$ depending on the type of the primitive group as classified in the O’Nan–Scott theorem [8]. In particular, if k is a fixed integer and ${\mathcal {C}}$ is an infinite class of finite symmetric groups $S_{n}$, or alternating groups $A_{n}$, with natural action on $I^{\{k\}}$, the set of k-subsets of $I:=\{1,\ldots ,n\}$, then ${\mathcal {C}}$ is bounded. Moreover, this is the only primitive action of $S_{n}$ and $A_{n}$ which gives rise to a bounded infinite class.

In this paper, we present versions of the results in [7] for finite symmetric and alternating groups with precise numerical bounds on the orbital diameters in question (see Theorem 1.1). As a consequence, we classify the primitive actions of the symmetric and alternating groups for which the orbital diameter is at most 5 (see Theorem 1.2). In addition, we prove results classifying orbital graphs of diameter 2 (see Theorem 1.3).

We would like to mention here that the results in this paper are part of a more general programme of classifying bounded infinite families of finite primitive permutation groups. Such results of the author on primitive actions of simple groups of Lie type will be forthcoming.

We will use the notation (G : H) to denote the set of right cosets of a subgroup H in a group G.

Let $G:=S_{n}$ or $A_{n}$ and $I:=\{1, \ldots , n\}$. Henceforth we will assume that $n\ge 5$. If H is a maximal subgroup of G then one of the following holds:

• 1. H is intransitive on I, and hence $H\cong (S_{k}\times S_{n-k})\cap G$, where $1 \le k<n/2$.

• 2. H is imprimitive on I, and hence $H\cong (S_{k} \wr S_{l})\cap G$, where $n=kl$ and $k>1, l>1$.

• 3. H is primitive on I.

Let $I=\{1,\ldots ,n\}$. If H is intransitive on I, then the action of $G:=S_{n}$ or $A_{n}$ on the set (G : H) is equivalent to the action of G on the set $I^{\{k\}}$, of k-subsets of I. The orbitals of $(G, I^{\{k\}})$ are $\Delta ^{n}_{i}:=\{(A,B):|A\cap B|=i\}$, for $i\in \{0,\ldots , k-1\}$. The associated orbital graphs are $\Gamma ^{n}_{0}, \ldots , \Gamma ^{n}_{k-1}$, where for $i\in \{0,\ldots , k-1\}$, the graph $\Gamma ^{n}_{i}$ has vertex set $I^{\{k\}}$ and edge set $\Delta ^{n}_{i}$. Note that for $k=1$, there is only one orbital graph $\Gamma ^{n}_{0}$ which has diameter 1; henceforth, we will assume that $k>1$. Lastly, if $n=2k$, then $H:=S_{k}\times S_{k}$ is not a maximal subgroup of $S_{n}$; however, the orbital graphs of the action of $S_{n}$ on the set $(S_{n}:H)$ are still $\Gamma ^{n}_{i}$, for $i\in \{0,\ldots , k-1\}$, and $\Gamma ^{n}_{i}$ is connected for all $i\ne 0$.

If H is imprimitive on I, then the action of $G:=S_{n}$ or $A_{n}$ on (G : H) is equivalent to the action of G on the set $I^{(k,l)}$, of (k, l)-partitions of I; here a (k, l)-partition of I is a partition of I into l sets $A_{1}, \ldots , A_{l}$, each of size k. Henceforth such a (k, l)-partition of I will be denoted by $A_{1}|\dots | A_{l}$. Let $A:=A_{1}|\dots |A_{l}$ and $B:=B_{1}|\dots | B_{l}$ be two (k, l)-partitions of I. We will denote by $I_{AB}$ the $l\times l$ matrix with entries ${(I_{AB})}_{ij}:=|A_{i}\cap B_{j}|$. Note that the row and columns sums of $I_{AB}$ are k, and for every $l\times l$ matrix M whose entries are non-negative integers, and whose row and column sums are k, there exists a pair (A, B) of (k, l)-partitions of I such that $I_{AB}=M$. Let $A,B,C,D \in I^{(k,l)}$. Then the pairs (A, B) and (C, D) lie in the same orbital of $S_{n}$ if and only if there exist $l\times l$ permutation matrices P and Q such that $I_{CD}=P\,I_{AB}\,Q$, i.e. $I_{AB}$ and $I_{CD}$ are equal up to permutation of rows and columns (see Proposition 4.1). Let ${\mathcal {N}}$ denote the set of all $l\times l$ matrices with non-negative integer entries, and row and column sums equal to k. Define an equivalence relation $\sim$ on ${\mathcal {N}}$, where $M\sim M^{\prime }$ if and only if $M^{\prime }$ can be obtained from M by a permutation of rows and columns. The orbitals of $(G, I^{(k,l)})$ can then be described as

$$\begin{aligned} \Delta _{[M]}:=\left\{ (A,B): A,B\in I^{(k,l)}, I_{AB}\in [M]\right\} \end{aligned}$$

where $M\in {\mathcal {N}}$, and [M] is the equivalence class of M under the equivalence relation $\sim$. Notice that for an orbital $\Delta _{[M]}$, the paired orbital $\Delta _{[M]}^{*}$ is $\Delta _{[M^{\mathrm{T}}]}$, where T in the superscript denotes taking the transpose of the matrix. The corresponding orbital graphs in this case are $\Gamma _{[M]\cup [M^{\mathrm{T}}]}$:=$\Gamma _{\Delta _{[M]}\cup \Delta _{[M]}^{*}}$. If $M^{\mathrm{T}}=M$, then we will denote the corresponding orbital graph by $\Gamma _{[M]}$.

In the case where H is primitive on I and $G:=S_{n}$, we will assume that $H\ne A_{n}$, since $\mathrm{diam_{O}}(G,(G:A_{n}))=1$, for any n.

We now state our main results.

Theorem 1.1

Let $(G,\Omega )$ be a finite primitive permutation group where $G:=S_{n}$ or $A_{n}$, and let $I=\{1,\ldots ,n\}$. Suppose there exists an integer $c\ge 1$ such that $\mathrm{diam_{O}}(G,\Omega )\le c$.

• 1. If $\Omega =I^{\{k\}}$, where $1\le k<n/2$, then $k\le c$.

• 2. If $\Omega =I^{(k,l)}$, where $n=kl$ with $l>1$ and $k>1$, then $l\le 2c$ and $k\le 2c+1$, and in particular, $n\le 2c(2c+1)$.

• 3. If $\Omega =(G:H)$, where H is primitive on I, then $n<2^{\epsilon (c+1)+3}$ with $\epsilon :=1.04$.

Notice that Theorem 1.1 agrees with the results in [7], i.e. part 1 is the only case for which n is unbounded. The proofs of Theorem 1.1(1), (2) and (3) can be found on Pages 9, 18, and 26, respectively.

For particular values of the bound c, it is possible to be more specific about the precise actions that occur in Theorem 1.1(2) and (3). Here is such a result where $c=5$.

Theorem 1.2

Let $(G,\Omega )$ be a finite primitive permutation group where $G:=S_{n}$ or $A_{n}$, and let $I=\{1,\ldots ,n\}$.

• 1. If $\Omega =I^{\{k\}}$, where $1\le k<n/2$, then $\mathrm{diam_{O}}(G,\Omega )\le 5$ if and only if $k\le 5$.

• 2. If $\Omega =I^{(k,l)}$, where $n=kl$ with $l>1$ and $k>1$, then $\mathrm{diam_{O}}(G,\Omega )\le 5$ if and only if the pair (k, l) is given by Table 1.

• 3. If $\Omega =(G:H)$, where H is primitive on I, then $\mathrm{diam_{O}}(G,\Omega )\le 5$ if and only if the pair (n, H) is given by Tables 2 and 3; where the ‘if’ direction is only known to hold for values of $n\le 12$ (see the remark below).

Table 1 Pairs (k, l) with $\mathrm {diam_{O}}(G,I^{(k,l)})\le 5$

Table 2 Pairs (n, H) with $\mathrm {diam_{O}}(S_{n},\Omega )\le 5$, where $\Omega :=(S_{n}:H)$ and H is primitive

Table 3 Pairs (n, H) with $\mathrm {diam_{O}}(A_{n},\Omega )\le 5$, where $\Omega :=(A_{n}:H)$ and H is primitive

Remark

Tables 2 and 3 include:

• 1. the rank of the action, the diameter of the orbital graphs and their valencies in each case,

• 2. four cases with $n=14$, 18 or 24, marked with question marks, where we have been unable to compute the orbital diameters.

The proofs of Theorem 1.2(1), (2) and (3) can be found on Pages 9, 25, and 29, respectively.

We also analyse orbital graphs of diameter 2. For the case where $\Omega \ne I^{(k,l)}$, we find all orbital graphs of diameter 2. For $\Omega =I^{(k,l)}$, we find an example of an infinite family of orbital graphs of diameter 2; there could be more infinite families of orbital graphs of diameter 2, but classifying all of them could take significantly more effort.

Theorem 1.3

Let $(G,\Omega )$ be a finite primitive permutation group where $G:=S_{n}$ or $A_{n}$, and let $I=\{1,\ldots ,n\}$.

• 1. If $\Omega =I^{\{k\}}$, where $1< k<n/2$, then $\mathrm{diam}(\Gamma ^{n}_{i})=2$ if and only if one of the following holds:

  • $i=0$ and $n\ge 3k-1$, or

  • $i\in \{1,\ldots , \lfloor k/2\rfloor \}$ and $n\ge 3k-2i$.

• 2. If $\Omega =I^{(k,k)}$ with $n=k^{2}$ and $k\ge 3$, let $\Gamma _{[M]}$ be the orbital graph where M is the $k\times k$ matrix with all entries equal to 1. Then $\mathrm{diam}(\Gamma _{[M]})=2$.

• 3. If $\Omega =(G:H)$, where H is primitive on I, then Tables 4 and 5 specify the precise pairs (n, H) for which $(G,\Omega )$ has an orbital graph of diameter 2 (the case n = 24 is still just a possibility; see the remark below).

Table 4 Pairs (n, H) such that $(S_{n},\Omega )$ has an orbital graph of diameter 2, where $\Omega :=(S_{n}:H)$ and H is primitive

Table 5 Pairs (n, H) such that $(A_{n},\Omega )$ has an orbital graph of diameter 2, where $\Omega :=(A_{n}:H)$ and H is primitive

Remark

• 1. Tables 4 and 5 include the rank of the action, the size of the set ${\mathcal {D}}_{2}$, of orbital graphs of diameter 2 and the valencies of the orbital graphs in each case.

• 2. Table 5 contains the case $n=24$, marked with a question mark, where we have been unable to determine or disprove the existence of an orbital graph of diameter 2.

• 3. The notation $x^{[y]}$ has been used in the ‘valencies’ column of Table 5 to denote that the valency x appears y times.

The proofs of Theorem 1.3(1), (2) and (3) can be found on Pages 9, 25, and 29, respectively.

The proofs of the main results require some analysis of the three types of primitive actions of $G:=S_{n}$ or $A_{n}$. In the first case, namely the action of G on $\Omega$ where $\Omega :=I^{\{k\}}$, some work has been done in [2] on finding the diameters of the corresponding orbital graphs. However, we believe that the main result in [2] has an error and state the correct result below.

Theorem 1.4

Let $k> 1$ and $n\ge 2k+1$ be integers. For $i\in \{0,\ldots , k-1\}$, let $\Gamma ^{n}_{i}$ be as defined previously. Then the following hold:

• 1. For $i=0, \mathrm{diam}(\Gamma ^{n}_{0})=\lceil \frac{n-k-1}{n-2k}\rceil$.

• 2. For $i\in \{1,\ldots , \left\lfloor k/2 \right\rfloor \}$,

  • (a) $\mathrm{diam}(\Gamma ^{n}_{i})=2 \Leftrightarrow n\ge 3k-2i$,

  • (b) $\mathrm{diam}(\Gamma ^{n}_{i})=3 \Leftrightarrow \frac{5(k-i)-1}{2}\le n< 3k-2i$,

  • (c) $\mathrm{diam}(\Gamma ^{n}_{i})=\lceil \frac{n-k+i-1}{n-2k+2i}\rceil >3 \Leftrightarrow n<\frac{5(k-i)-1}{2}.$

• 3. For $i> \left\lfloor k/2 \right\rfloor , \mathrm{diam}(\Gamma ^{n}_{i})=\lceil \frac{k}{k-i} \rceil$.

The error occurs in line 2 on p. 6648 of [2]. According to [2], for any $i>\left\lfloor k/2 \right\rfloor , \mathrm{diam}(\Gamma ^{n}_{i})=3$, and in Theorem 7, a construction for a path of length 3 is specified for any two vertices A and B with $m:=|A\cap B|<2i-k$. This involves constructing a set $Z=Z(n,k,i,m)$ which is, however, impossible to construct for many values of n, k, i and m; for example, let $(n,k,i,m):=(2k+1,k,k-1,0)$, where $k\ge 4$, then $|Z|:=2k-3i+m=3-k\le -1$. In fact $\mathrm{diam}(\Gamma ^{n}_{k-1})=k$.

Note that in order to prove Theorem 1.3(1), we do not require Theorem 1.4 in its entirety; however, we give a proof of the whole theorem for completeness. The proof of Theorem 1.4(1) and (2c) can be found on Page 14, and the proofs of (2a), (2b) and (3) can be found on Pages 9, 12, and 16, respectively.

To study the orbital graphs for the action of G on $\Omega$, where $\Omega :=I^{(k,l)}$ and $n=kl$, we will need Theorem 1.4(2) as well as the following results.

Theorem 1.5

Let $G:=S_{n}$ or $A_{n}$ and $\Omega :=I^{(k,l)}$, where $n=kl$ with $k>1$ and $l>1$. Then $diam_{\mathrm{O}}(G,\Omega )\ge \frac{l}{2}\left\lfloor \frac{k}{2} \right\rfloor$.

The proof of Theorem 1.5 can be found on Page 18.

Theorem 1.6

Let $G:=S_{n}$ or $A_{n}$ and $\Omega :=I^{(k,2)}$, where $n=2k$ with $k>1$. For $i\in \{1,\ldots ,\left\lfloor k/2 \right\rfloor \}$, let

$$\begin{aligned} M_{i}:=\left( \begin{array}{cc} i &{} k-i \\ k-i &{} i \end{array} \right) . \end{aligned}$$

Then the following hold:

• 1. For $k=2,3, \mathrm{diam}(\Gamma _{[M_{1}]})=1$.

• 2. For $k\ge 4$,

  • (a) $\mathrm{diam}(\Gamma _{[M_{i}]})=2$ if and only if $\frac{k-1}{4}\le i\le \frac{k}{2}$, and

  • (b) if $\mathrm{diam}(\Gamma _{[M_{i}]})\ne 2$, then $\mathrm{diam}(\Gamma _{[M_{i}]})=\left\lceil \frac{k-1}{2i} \right\rceil$.

The proof of Theorem 1.6(1) and (2a) can be found on Page 22, and the proof of (2b) can be found on Page 24.

2 Computing diameters using magma

Some of the results in this paper rely on computations which have been done using magma [1]. In this section, we describe the methods used for the computations and their implementation in magma.

2.1 Method CAM: using collapsed adjacency matrices

Let G be a transitive permutation group acting on a finite set $\Omega$. Let $\Omega _{1}:=\{\alpha \}, \Omega _{2}, \ldots ,$ $\Omega _{r}$ be the distinct orbits of $G_{\alpha }$ on $\Omega$, i.e. the suborbits of G, with representatives $\alpha _{1}:=\alpha , \alpha _{2}, \ldots , \alpha _{r}$, where r is the rank of the action. The suborbits of G are in one-to-one correspondence with the orbitals of G; a suborbit $\beta ^{G_{\alpha }}$ corresponds to the orbital $(\alpha ,\beta )^{G}:=\{(\alpha ,\beta )^{g}:g\in G\}$, where $(\alpha ,\beta )^{g}$ denotes the action of g on $(\alpha ,\beta )$. Let E be a union of orbitals of $(G,\Omega )$. Define an $r\times r$ matrix $A_{E}$ where for $k,j\in \{1,\ldots , r\}$,

$$\begin{aligned} (A_{E})_{kj}:=|E(\alpha _{k})\cap \Omega _{j}| \end{aligned}$$

and $E(\alpha _{k}):=\{\gamma \in \Omega : (\alpha _{k},\gamma )\in E\}$. Note that $A_{E}$ does not depend on the choice of the suborbit representative $\alpha _{k}$. If $E=\Delta \cup \Delta ^{*}$ where $\Delta$ is an orbital of G, then $A_{E}$ is called the collapsed adjacency matrix for the orbital graph $\Gamma _{\Delta \cup \Delta ^{*}}$ (with respect to the ordering of the suborbits). The corresponding collapsed graph is defined to be the undirected graph $\Gamma _{\Delta \cup \Delta ^{*}}^{C}$, with vertex set $\{1,\ldots , r\}$, and in which (k, j) is an edge if $(A_{\Delta \cup \Delta ^{*}})_{kj}>0$. When computing the diameter of an orbital graph, it is advantageous to work with the corresponding collapsed graph since it has significantly fewer vertices than the orbital graph but in fact the same diameter. The following lemma, which is based on Theorem 3.2 in [10], justifies the switch from $\Gamma _{\Delta \cup \Delta ^{*}}$ to $\Gamma _{\Delta \cup \Delta ^{*}}^{C}$.

Lemma 2.1

Let $(G,\Omega )$ be a finite primitive permutation group and $\Gamma$ an orbital graph of G. Then $\mathrm{diam}(\Gamma )=\mathrm{diam}(\Gamma ^{C})$.

The following result aids the computation of collapsed adjacency matrices; see Proposition 2.2 in [10].

Lemma 2.2

Let $(G,\Omega )$ be a finite primitive permutation group and $\Delta$ be an orbital of G which is not self-paired. Then $A_{\Delta \cup \Delta ^{*}}=A_{\Delta }+A_{\Delta ^{*}}$.

2.2 Implementation of CAM in magma

magma does not have an in-built function to calculate collapsed adjacency matrices. We can, however, use an in-built function to define the group G and iterate over the maximal subgroups of G to find the required subgroup $H\cong G_{\alpha }$. We can then use more in-built functions to firstly construct the permutation group $(G,\Omega )$ where $\Omega :=(G:H)$, and secondly, find the suborbits of G, with representatives $\alpha _{1}:=\alpha , \alpha _{2}, \ldots , \alpha _{r}$. Let $\Delta :=(\alpha ,\beta )^{G}$ be an orbital of G. We calculate $A_{\Delta }$ by implementing the method described in [10, p. 24]. If $\Delta$ is self-paired, then $A_{\Delta }$ is the collapsed adjacency matrix for the orbital graph $\Gamma _{\Delta \cup \Delta ^{*}}$. If $\Delta$ is not self-paired, then we can use the function $\textit{trace}:\Omega \rightarrow G$ (constructed whilst implementing the method described in [10]) where $\textit{trace}(\gamma )=g$ means that $\alpha ^{g}=\gamma$, to find $g\in G$ such that $\beta ^{g}=\alpha$. Then $\Delta ^{*}=(\alpha ,\alpha ^{g})^{G}$, and we can again use the method in [10] to compute the matrix $A_{\Delta ^{*}}$. The collapsed adjacency matrix for the orbital graph $\Gamma _{\Delta \cup \Delta ^{*}}$ is then $A_{\Delta }+A_{\Delta ^{*}}$ by Lemma 2.2. Once we have a collapsed adjacency matrix $A_{E}$ where $E:=\Delta \cup \Delta ^{*}$, the diameter of $\Gamma ^{C}_{E}$ (and hence the diameter of $\Gamma _{E}$ by Lemma 2.1) is the least number of times $A_{E}$ has to be multiplied by itself before each non-diagonal entry has taken a value greater than zero at least once.

The implemented method works well when |G : H| is small. For large values of |G : H|, the method fails at the very first step of constructing the permutation group $(G,\Omega )$.

3 Action of $S_{n}$ on k-subsets of I

In this section, we prove Theorems 1.1(1), 1.2(1), and 1.3(1). We then prove Theorem 1.4 which is used to deduce Theorem 1.6 in Sect. 4.

Proposition 3.1

Let $(G,\Omega )$ be a finite primitive permutation group where $G:=S_{n}$ or $A_{n}, I=\{1,\ldots ,n\}$, and $\Omega :=I^{\{k\}}$ with $1\le k<n/2$. Then $\mathrm{diam_{O}}(G,\Omega )\le c$ if and only if $k\le c$.

Proof

Suppose $\mathrm{diam_{O}}(G,\Omega )\le c$. Recall that in the orbital graph $\Gamma ^{n}_{k-1}$, two k-subsets are joined by an edge if they intersect in $k-1$ points. Observe that this graph has diameter k, and so $k\le c$. Conversely, suppose $k\le c$ and let $\Gamma$ be an orbital graph of $(G,\Omega )$. For a k-subset A and an integer $j\ge 1$ define

$$\begin{aligned} \Delta _{j}(A):=\{B: d_{\Gamma }(A,B)=j\}. \end{aligned}$$

If $B\in \Delta _{j}(A)$ and $g\in G_{A}$, then $B^{g}\in \Delta _{j}(A)$, hence $B^{G_{A}}\subseteq \Delta _{j}(A)$. Therefore $\Delta _{j}(A)$ is a union of non-trivial orbits of $G_{A}$ on $\Omega$. There are k non-trivial orbits of $G_{A}$ on $\Omega$, and therefore $\mathrm{diam}(\Gamma )\le k\le c$. $\square$

Proof of Theorem 1.1(1) and Theorem 1.2(1)

Theorem 1.1(1) follows from Proposition 3.1, and Theorem 1.2(1) is a special case of Theorem 1.1(1). $\square$

We now prove Theorem 1.3(1). Recall that for $i\in \{0,\ldots , k-1\}$, the graph $\Gamma ^{n}_{i}$ has vertex set $I^{\{k\}}$ where $I=\{1,\ldots , n\}$, and edge set $\{(A,B): |A\cap B|=i\}$. Henceforth, for two sets A and B, the notation $A{\setminus }B$ will be used to refer to the complement of B in A. The following lemma follows from Lemmas 1 and 2 in [3] and the fact that $\Gamma ^{n}_{i}$ is a subgraph of the graph K(n, k, i) from [3].

Lemma 3.2

Let $n\ge 2k, i\in \{0,\ldots , k-1 \}$ ($i\ne 0$, if $n=2k$), and $r\ge 1$. Let $A, B\in V(\Gamma ^{n}_{i})$.

• 1. If A and B are connected by a path of length 2r, then

  $$\begin{aligned}|A \cap B| \ge (2r+1)k-rn-2ri.\end{aligned}$$

• 2. If A and B are connected by a path of length $2r+1$, then

  $$\begin{aligned}|A \cap B| \le rn-2rk+(2r+1)i.\end{aligned}$$

Lemma 3.3

Let $n\ge 2k, i\in \{0,\ldots , \left\lfloor k/2\right\rfloor \}$ ($i\ne 0$, if $n=2k$), and $A, B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$.

• 1. If $n<3k-2i$, then

  $$\begin{aligned} d(A,B)=2 \Leftrightarrow m\ge 3k-n-2i. \end{aligned}$$

• 2. If $n\ge 3k-2i$, then $d(A,B)=2$.

Proof

Let $I=\{1, \ldots , n\}$. Suppose $n<3k-2i$. If $d(A,B)=2$, then by Lemma 3.2(1), we have $m\ge 3k-n-2i$. Conversely, assume $m\ge 3k-n-2i$. If $i\le k-m$, then choose a k-subset C consisting of i elements from $A\setminus B, i$ elements from $B\setminus A$, and $k-2i$ elements from $I\setminus (A\cup B)$ (note that $|I\setminus (A\cup B)|\ge k-2i$, since $n\ge 3k-m-2i$). If $i>k-m$, then choose a k-subset C consisting of $i-k+m$ elements from $A\cap B, k-m$ elements from $A\setminus B, k-m$ elements from $B\setminus A$, and $m-i$ elements from $I\setminus (A\cup B)$ (note that $|I\setminus (A\cup B)|\ge m-i$ since $n\ge 2k-i$). Then $A-C-B$ is a path of length 2, and hence $d(A,B)=2$. This proves the first part of the lemma. Now suppose $n\ge 3k-2i$. Then $m\ge 3k-n-2i$, for all $m\ge 0$. Now as in the proof of the converse of the first part, we can choose a k-subset C such that $A-C-B$ is a path of length 2, and hence $d(A,B)=2$. $\square$

Corollary 3.4

Let $n\ge 2k$ and $i\in \{0,\ldots , \left\lfloor k/2\right\rfloor \}$ ($i\ne 0$, if $n=2k$). Suppose that $\mathrm{diam}(\Gamma ^{n}_{i})\ge 3$. Then $n<3k-2i$ and for $A, B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$,

$$\begin{aligned} d(A,B)>2 \Leftrightarrow m<3k-n-2i. \end{aligned}$$

Proof of Theorem 1.3(1) and Theorem 1.4(2a)

Note that $\mathrm{diam}(\Gamma ^{n}_{i})=2$ if and only if for all $A\ne B\in V(\Gamma ^{n}_{i})$ such that $|A\cap B|\ne i$, it holds that $d(A,B)=2$. Hence using Lemma 3.3, $\mathrm{diam}(\Gamma ^{n}_{0})=2$ if and only if either $n\ge 3k$ or $3k-n=1$, and if $i\in \{1,\ldots , \lfloor k/2 \rfloor \}$, then $\mathrm{diam}(\Gamma ^{n}_{i})=2$ if and only if $n\ge 3k-2i$. Now suppose $i>\lfloor k/2 \rfloor$ and let $A,B\in V(\Gamma ^{n}_{i})$ with $|A\cap B|=0$. If C is a common neighbour of A and B, then $|C|\ge 2i>k$ which contradicts the fact that $|C|=k$. Hence $d(A,B)>2$ and $\mathrm{diam}(\Gamma ^{n}_{i})\ge 3$. $\square$

We now continue proving the rest of Theorem 1.4. We will need to prove the following lemma.

Lemma 3.5

Let $n\ge 2k, i\in \{0,\ldots , \left\lfloor k/2\right\rfloor \}$ ($i\ne 0$, if $n=2k$), and $d:=\mathrm{diam}(\Gamma ^{n}_{i})$. Assume $d\ge 3$. Let $A, B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$. Then for $v\in \{3,\ldots , d\}$,

$$\begin{aligned} P(v): \quad d(A,B) = v \Leftrightarrow f(n,k,i,v)\le m \le g(n,k,i,v), \end{aligned}$$

where f and g are as in Tables 6 and 7, for v even and v odd, respectively.

Table 6 $v=2r, r\ge 2$

Table 7 $v=2r+1, r\ge 1$

In order to prove Lemma 3.5, we now prove the following sequence of results.

Lemma 3.6

Let $n\ge 2k, i\in \{0,\ldots , \left\lfloor k/2\right\rfloor \}$ ($i\ne 0$, if $n=2k$), and $t\ge 3$. Suppose the following conditions hold:

• 1. $\mathrm{diam}(\Gamma ^{n}_{i})\ge t+1$,

• 2. $n<\frac{(2t-1)(k-i)}{t-1}$,

• 3. P(v) holds for all $v\in \{3,\ldots , t\}$.

Let $A, B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$. Then for t even ($t=2r$),

$$\begin{aligned} d(A,B)>2r \Leftrightarrow (r-1)n-(2r-2)k+(2r-1)i<m<(2r+1)k-rn-2ri, \end{aligned}$$

and for t odd ($t=2r+1$),

$$\begin{aligned} d(A,B)>2r+1 \Leftrightarrow rn-2rk+(2r+1)i<m<(2r+1)k-rn-2ri. \end{aligned}$$

Remark

By Lemma 3.7, if (1) and (3) of Lemma 3.6 hold, then so does (2); hence, after Lemma 3.7 has been proved, we may omit (2) from the statement of Lemma 3.6. Similarly, after Lemma 3.5 has been proved, we may also omit (3).

Proof of Lemma 3.6

We proceed by induction on t. For the base case, suppose the conditions of the lemma hold for $t=3$. Then by Corollary 3.4 and P(3),

$$\begin{aligned} d(A,B)= & {} 3 \Leftrightarrow 0\le m\le n-2k+3i,\\ d(A,B)= & {} 2 \Leftrightarrow m\ge 3k-n-2i. \end{aligned}$$

Therefore the result for $t=3$ follows. Now let $t=2r>3$ and suppose the lemma holds for $t-1$. Suppose also that the conditions of the lemma hold for t. Then the conditions of the lemma also hold for $t-1$, and so by the inductive assumption,

$$\begin{aligned} d(A,B)> & {} 2r-1 \Leftrightarrow (r-1)n-(2r-2)k\nonumber + (2r-1)i<m\nonumber \\< & {} (2r-1)k-(r-1)n-(2r-2)i. \end{aligned}$$

(1)

Now by P(2r), we have that

$$\begin{aligned} d(A,B)= & {} 2r \Leftrightarrow (2r+1)k-rn-2ri\le m \nonumber \\&\le (2r-1)k-(r-1)n-(2r-2)i-1, \end{aligned}$$

(2)

and therefore we obtain the result for $t=2r$ by combining (1) and (2). To complete the induction, it remains to show the result for $t+1=2r+1$ which can be done using similar arguments to the above and is left to the reader. $\square$

Lemma 3.7

Let $n\ge 2k, i\in \{0,\ldots , \left\lfloor k/2\right\rfloor \}$ ($i\ne 0$, if $n=2k$), and $t\ge 3$. Suppose P(v) holds for all $v\in \{3,\ldots t\}$ and $\mathrm{diam}(\Gamma ^{n}_{i})\ge t+1$. Then

$$\begin{aligned} n<\frac{(2t-1)(k-i)}{t-1}. \end{aligned}$$

Proof

We proceed by induction on t. For the base case, assume that P(3) holds and $\mathrm{diam}(\Gamma ^{n}_{i})\ge 4$. Suppose $n\ge \frac{5(k-i)}{2}$. Then using Corollary 3.4 and P(3) we see that $d(A,B)\le 3$ for all $A,B\in V(\Gamma ^{n}_{i})$, which contradicts the fact that $\mathrm{diam}(\Gamma ^{n}_{i})\ge 4$. Hence $n< \frac{5(k-i)}{2}$. This completes the base case. Now let $t=2r\ge 4$ and assume that the lemma holds for $t-1$. Suppose P(v) holds for all $v\in \{3,\ldots t\}$ and $\mathrm{diam}(\Gamma ^{n}_{i})\ge t+1$. Then P(v) holds for all $v\in \{3\ldots , t-1\}$ and $\mathrm{diam}(\Gamma ^{n}_{i})\ge t$; hence, by the inductive assumption, we have $n<\frac{(2t-3)(k-i)}{t-2}$. Now by Lemma 3.6, for $A, B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$, we see that (1) holds. Suppose $n\ge \frac{(2t-1)(k-i)}{t-1}$. Then using P(2r), we see that $d(A,B)\le 2r$, for all $A,B\in V(\Gamma ^{n}_{i})$, which contradicts the fact that $\mathrm{diam}(\Gamma ^{n}_{i})\ge 2r+1$. Hence $n<\frac{(2t-1)(k-i)}{t-1}$. To complete the induction, it remains to show the result for $t+1=2r+1$ which can be done using similar arguments to the above and is left to the reader. $\square$

Lemma 3.8

Suppose $n\ge 2k, i\in \{0,\ldots , \left\lfloor k/2\right\rfloor \}$ ($i\ne 0$, if $n=2k$), and $\mathrm{diam}(\Gamma ^{n}_{i})\ge ~3$. Let $A, B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$. If $n<\frac{5(k-i)}{2}$, then

$$\begin{aligned} d(A,B) = 3 \Leftrightarrow 0\le m \le n-2k+3i, \end{aligned}$$

and if $n\ge \frac{5(k-i)}{2}$, then

$$\begin{aligned} d(A,B) = 3 \Leftrightarrow 0\le m \le 3k-n-2i-1. \end{aligned}$$

Proof

Suppose $n<\frac{5(k-i)}{2}$. If $d(A,B)=3$, then by Lemma 3.2(2), we have $m\le n-2k+3i$. For the converse, assume $m\le n-2k+3i$. Since $n<\frac{5(k-i)}{2}$, we have that $m<3k-n-2i$, and so by Corollary 3.4, we have $d(A,B)>2$. Choose a neighbour $A_{1}$ of A such that $|A_{1}\cap B|$ is maximal. If $i<m$, then $|A_{1}\cap B|=i+k-m$, and if $i>m$, then $|A_{1}\cap B|=m+k-i$. In both cases (for $i<m$ use $m\le n-2k+3i$, and for $i>m$ use $m\ge 0$), we have $|A_{1}\cap B|\ge 3k-2i-n$. Hence by Lemma 3.3(1), we have $d(A_{1},B)=2$, and so $d(A,B)=3$. Now suppose $n\ge \frac{5(k-i)}{2}$. If $d(A,B)=3$, then by Corollary 3.4, we have $m<3k-n-2i$. For the converse, assume $m<3k-n-2i$. Then by Corollary 3.4, we have $d(A,B)>2$, and using the fact that $n\ge \frac{5(k-i)}{2}$, we see that $m\le n-2k+3i$. Now as above, we show that $d(A,B)=3$ by constructing a path of length 3 between A and B. $\square$

Corollary 3.9

Let $n>2k$. For $i=0$,

$$\begin{aligned} \mathrm{diam}(\Gamma ^{n}_{0})=3 \Leftrightarrow \frac{5k-1}{2}\le n<3k-1, \end{aligned}$$

and for $i\in \{1,\ldots , \left\lfloor k/2 \right\rfloor \}$,

$$\begin{aligned} \mathrm{diam}(\Gamma ^{n}_{i})=3 \Leftrightarrow \frac{5(k-i)-1}{2}\le n<3k-2i. \end{aligned}$$

Proof

Suppose $\mathrm{diam}(\Gamma ^{n}_{0})=3$. By Theorem 1.3(1), we have $n<3k-1$ and by Corollary 3.4, for $A, B\in V(\Gamma ^{n}_{0})$ with $m:=|A\cap B|\ne 0$,

$$\begin{aligned} d(A,B)>2 \Leftrightarrow m<3k-n. \end{aligned}$$

(3)

Now $\mathrm{diam}(\Gamma ^{n}_{0})=3$ implies that for all $A, B\in V(\Gamma ^{n}_{0})$ with $d(A,B)>2$, we must have $d(A,B)=3$. Thus using Lemma 3.8, we deduce that either $n \ge \frac{5k}{2}$ or $3k-n-1=n-2k$. Conversely, suppose $\frac{5k-1}{2}\le n<3k-1$. Then by Theorem 1.3(1), we have $\mathrm{diam}(\Gamma ^{n}_{0})\ge 3$. Hence by Corollary 3.4, for $A, B\in V(\Gamma ^{n}_{0})$ with $m:=|A\cap B|\ne 0$, we see that (3) holds. Now using Lemma 3.8 and the fact that $n\ge \frac{5k-1}{2}$, we see that $d(A,B)\le 3$ for all $A, B\in V(\Gamma ^{n}_{0})$. Hence $\mathrm{diam}(\Gamma ^{n}_{0})=3$. This proves the first statement of the corollary. The statement for $i>0$ can be proved using similar arguments to the above and is left to the reader. $\square$

Proof of Theorem 1.4(2b)

See Corollary 3.9. $\square$

We need one final lemma before we can prove Lemma 3.5.

Lemma 3.10

Let $n\ge 2k, i\in \{0,\ldots , \left\lfloor k/2\right\rfloor \}$ ($i\ne 0$, if $n=2k$), and $r\ge 2$. Let $A, B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$.

• 1. Suppose $d(A,B)\ge 2r$, P(v) holds for $v\in \{3,\ldots , 2r-1\}$, and

  $$\begin{aligned} (2r+1)k-rn-2ri\le m < (2r-1)k-(r-1)n-(2r-2)i. \end{aligned}$$

  (4)

  Then $d(A,B)=2r$.

• 2. Suppose $d(A,B)\ge 2r+1$, P(v) holds for $v\in \{3,\ldots , 2r\}$, and

  $$\begin{aligned} (r-1)n-(2r-2)k+(2r-1)i< m \le rn-2rk+(2r+1)i. \end{aligned}$$

  (5)

  Then $d(A,B)= 2r+1$.

Proof

Suppose the conditions in the first part of the lemma hold. Choose a neighbour $A_{1}$ of A such that $|A_{1}\cap B|$ is minimal. If $k-m<i$, then using Lemma 3.3, we have $d(A,B)=2$; therefore $k-m\ge i$, in which case $|A_{1}\cap B|=3k-i-n-m$. Now rearranging (4) gives

$$\begin{aligned} (2r-3)i+(r-2)n-(2r-4)k< & {} |A_{1}\cap B| \\\le & {} (2r-1)i+(r-1)n-(2r-2)k. \end{aligned}$$

Therefore using Lemma 3.7 (with $t=2r-1$) and $P(2r-1)$, we have $d(A_{1},B)=2r-1$, and hence $d(A,B)=2r$. This proves the first part of the lemma. Now suppose the conditions in the second part of the lemma hold. Choose a neighbour $A_{1}$ of A such that $|A_{1}\cap B|$ is maximal. Then (5) implies $m>i$, and so $|A_{1}\cap B|=i+k-m$. Now rearranging (5) implies

$$\begin{aligned} (2r+1)k-rn-2ri\le |A_{1}\cap B| < (2r-1)k-(2r-2)i-(r-1)n . \end{aligned}$$

Therefore using Lemma 3.7 (with $t=2r$) and P(2r), we have $d(A_{1},B)=2r$, and hence $d(A,B)=2r+1$. $\square$

Proof of Lemma 3.5

For $v=3$, the result is proved in Lemma 3.8. For $v\ge 4$, we proceed by induction on v. Here $d=\mathrm{diam}(\Gamma ^{n}_{i})\ge v$. Firstly, we prove the base case $v=4$. Suppose $n<\frac{7(k-i)}{3}$. If $d(A,B)=4$, then by Lemma 3.2(1), we have $m\ge 5k-2n-4i$. Since $d(A,B)>2$, by Corollary 3.4, we also have $m<3k-n-2i$. For the converse, assume $5k-2n-4i\le m<3k-n-2i$. Note that since P(3) already holds, by the remark following Lemma 3.6, in order to use Lemma 3.6 (with $t=3$), we only need to ensure that the first condition of the lemma is satisfied, which is by the assumption of Lemma 3.5. Now using Lemma 3.6 and the fact that $n<\frac{7(k-i)}{3}$, we see that $d(A,B)>3$, and so by Lemma 3.10, we have $d(A,B)=4$. Now suppose $n\ge \frac{7(k-i)}{3}$. If $d(A,B)=4$, then using Lemma 3.6 (with $t=3$), we have $m>n-2k+3i$. Also $d(A,B)>2$, so by Corollary 3.4, we have $m<3k-n-2i$. For the converse, assume $n-2k+3i< m<3k-n-2i$. Then using Lemma 3.6, we see that $d(A,B)>3$. Now using the fact that $n\ge \frac{7(k-i)}{3}$, we have $5k-2n-4i\le m<3k-n-2i$, and so by Lemma 3.10, we have $d(A,B)=4$. This completes the base case.

Now let $r\ge 2$ and assume that P(v) holds for all $v\in \{3,\ldots , 2r\}$. We will show that $P(2r+1)$ holds. Suppose $n<\frac{(4r+1)(k-i)}{2r}$. If $d(A,B)=2r+1$, then by Lemma 3.2(2), we have $m\le rn-2rk+(2r+1)i$. Since $d(A,B)>2r-1$, by Lemma 3.6 we also have $m>(r-1)n-(2r-2)k+(2r-1)i$. For the converse, assume that

$$\begin{aligned} (r-1)n-(2r-2)k+(2r-1)i<m\le rn-2rk+(2r+1)i. \end{aligned}$$

(6)

Then using Lemma 3.6 and the fact that $n<\frac{(4r+1)(k-i)}{2r}$, we have $d(A,B)>2r$, and so by Lemma 3.10, we have $d(A,B)=2r+1$. Now suppose that $n\ge \frac{(4r+1)(k-i)}{2r-1}$. If $d(A,B)=2r+1$, then by Lemma 3.6, we have $m<(2r+1)k-rn-2ri$. Since $d(A,B)>2r-1$, by Lemma 3.6 we also have $m>(r-1)n-(2r-2)k+(2r-1)i$. For the converse, assume that

$$\begin{aligned}(r-1)n-(2r-2)k+(2r-1)i<m<(2r+1)k-rn-2ri.\end{aligned}$$

Then using Lemma 3.6, we have $d(A,B)>2r$. Now using the fact that $n\ge \frac{(4r+1)(k-i)}{2r-1}$, we see that (6) holds, and so by Lemma 3.10, we have $d(A,B)=2r+1$. This shows $P(2r+1)$. To complete the induction, it remains to show $P(2r+2)$ which can be done using similar arguments to the above and is left to the reader. $\square$

Proof of Theorem 1.4(1) and (2c)

By Theorem 1.3(1) and Corollary 3.9, we see that Theorem 1.4(1) holds when $\mathrm{diam}(\Gamma ^{n}_{0})$ is 2 or 3. For $t\ge 4$, an equivalent restatement of Theorem 1.4(1) and (2c) is that for $i\in \{0,\ldots , \left\lfloor k/2 \right\rfloor \}$,

$$\begin{aligned} R(t): \quad \mathrm{diam}(\Gamma ^{n}_{i})=t \Leftrightarrow \frac{(2t-1)(k-i)-1}{t-1}\le n< \frac{(2t-3)(k-i)-1}{t-2}. \end{aligned}$$

(7)

We prove (7) by induction on t. The base case will be to show that

$$\begin{aligned} \mathrm{diam}(\Gamma ^{n}_{i})=4 \Leftrightarrow \frac{7(k-i)-1}{3}\le n < \frac{5(k-i)-1}{2}. \end{aligned}$$

(8)

Suppose $\mathrm{diam}(\Gamma ^{n}_{i})=4$. Then by Theorem 1.3(1) and Corollary 3.9, $n < \frac{5(k-i)-1}{2}$. Note that by the remark following Lemma 3.6, in order to use Lemma 3.6, we only need to ensure that the first condition of the lemma is satisfied. Now by Lemma 3.6, for $A, B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$, we have

$$\begin{aligned} d(A,B)>3 \Leftrightarrow n-2k+3i<m<3k-n-2i. \end{aligned}$$

(9)

Now since $\mathrm{diam}(\Gamma ^{n}_{i})=4, P(4)$ holds by Lemma 3.5 and using P(4), we obtain that either $n\ge \frac{7(k-i)}{3}$ or $n-2k+3i+1=5k-2n-4i$. Conversely, suppose that the right-hand side of (8) holds. Since $n < \frac{5(k-i)-1}{2}$, by Theorem 1.3(1) and Corollary 3.9, we have $\mathrm{diam}(\Gamma ^{n}_{i})\ge 4$, and by Lemma 3.6, we see that (9) holds. Now using (9), P(4) and the fact that $n\ge \frac{7(k-i)-1}{3}$, we obtain that for all $A,B\in V(\Gamma ^{n}_{i})$ with $d(A,B)>3$, we have $d(A,B)=4$, and thus $\mathrm{diam}(\Gamma ^{n}_{i})=4$. This completes the base case.

Now let $r\ge 3$ and suppose R(t) holds for all $t\in \{4,\ldots , 2r-2\}$. We will show that $R(2r-1)$ holds. Suppose $\mathrm{diam}(\Gamma ^{n}_{i})=2r-1$. Then by Theorem 1.3(1), Corollary 3.9, and the inductive assumption, we have $n< \frac{(4r-5)(k-i)-1}{2r-3}$. Now by Lemma 3.6, for $A, B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$, we have

$$\begin{aligned} d(A,B)> & {} 2r-2 \Leftrightarrow (r-2)n-(2r-4)k+(2r-3)i\nonumber \\<&m<(2r-1)k-(r-1)n-(2r-2)i. \end{aligned}$$

(10)

Now $\mathrm{diam}(\Gamma ^{n}_{i})=2r-1$ implies that for all $A,B\in V(\Gamma ^{n}_{i})$ with $d(A,B)>2r-2$, we must have $d(A,B)=2r-1$. Hence using $P(2r-1)$, we obtain that either $(2r-2)n\ge (4r-3)(k-i)$, or

$$\begin{aligned} (2r-1)k-(r-1)n-(2r-2)i-1=(r-1)n-(2r-2)k+(2r-1)i. \end{aligned}$$

Conversely, suppose that the right-hand side of (7) holds with $t=2r-1$. Since $n< \frac{(4r-5)(k-i)-1}{2r-3}$, using Theorem 1.3(1), Corollary 3.9 and the inductive assumption, we have $\mathrm{diam}(\Gamma ^{n}_{i})\ge 2r-1$, and by Lemma 3.6, we see that (10) holds. Now using (10), $P(2r-1)$ and the fact that $n\ge \frac{(4r-3)(k-i)-1}{2r-2}$, we obtain that for all $A,B\in V(\Gamma ^{n}_{i})$ with $d(A,B)>2r-2$, we have $d(A,B)=2r-1$, and thus $\mathrm{diam}(\Gamma ^{n}_{i})=2r-1$. This shows $R(2r-1)$. To complete the induction, it remains to show R(2r) which can be proved similarly and is left to the reader. $\square$

We now prove some lemmas required to prove Theorem 1.4(3).

Lemma 3.11

Let $n\ge 2k, i\in \{\left\lfloor k/2\right\rfloor +1,\ldots , k-1\}$, and suppose $A, B\in V(\Gamma ^{n}_{i})$ with $|A\cap B|=:m>i$. Then $d(A,B)=2$.

Proof

Since $m>i$, we can choose a k-subset C containing i elements from $A\cap B$, and since $i>k/2$ implies $|I\setminus (A\cup B)|=n-2k+m> k-i$, we can also choose C to contain $k-i$ elements from $I\setminus (A\cup B)$. Then $|C|=k$ and $A-C-B$ is a path of length 2, and hence $d(A,B)=2$. $\square$

Now let $n\ge 2k, i\in \{\left\lfloor k/2\right\rfloor +1,\ldots , k-1\}$, and suppose that $A, B\in V(\Gamma ^{n}_{i})$ with $|A\cap B|=:m<i$. Let $j_{m}$ be the least positive integer such that $j_{m}(k-i)+m\ge i$, i.e. $j_{m}=\lceil \frac{i-m}{k-i} \rceil$. Construct a path

$$\begin{aligned} A-A_{1}-\dots -A_{j_{m}}, \end{aligned}$$

(11)

where we set $A_{0}:=A$, and for $l\in \{1,\ldots , j_{m}\}, A_{l}$ is chosen so that $|A_{l-1}\cap A_{l}|=i$ and $|A_{l}\cap B|$ is maximal. Observe that $|A_{l}\cap B|=|A_{l-1}\cap B|+(k-i)=m+l(k-i)$, for any $l\in \{1,\ldots , j_{m}\}$.

Lemma 3.12

Let $n\ge 2k, i\in \{\left\lfloor k/2\right\rfloor +1,\ldots , k-1\}$, and suppose $A, B\in V(\Gamma ^{n}_{i})$ with $|A\cap B|=: m<i$. Then $d(A,B)\le j_{m}+1$.

Proof

Consider the path in (11). If $j_{m}(k-i)+m= i$, then $|A_{j_{m}}\cap B|=i$. Hence $A-A_{1}-\dots - A_{j_{m}}-B$ is a path of length $j_{m}+1$. If $|A_{j_{m}}\cap B|=j_{m}(k-i)+m> i$, then let $\alpha :=j_{m}(k-i)+m-i>0$. Since $i>k/2$, this implies

$$\begin{aligned} |A_{j_{m}-1}\cup B|+\alpha =3k-2i\le n, \end{aligned}$$

and hence we can choose a vertex $A^{\prime }_{j_{m}}$ which is the vertex $A_{j_{m}}$ but with $\alpha$ elements from $(A_{j_{m}}\cap B)\setminus (A_{j_{m}}\cap A_{j_{m}-1})$ replaced with $\alpha$ elements which are not in $A_{j_{m}-1}\cup B$. Then $A-A_{1}-\dots - A_{j_{m}-1} - A^{\prime }_{j_{m}}-B$ is a path of length $j_{m}+1$, and hence $d(A,B)\le j_{m}+1$.

$\square$

By Lemma 3.12 we know that if $A, B\in V(\Gamma ^{n}_{i})$ with $|A\cap B|:=m<i$, then $d(A,B)\le j_{m} +1 = \lceil \frac{k-m}{k-i} \rceil$. We will now show that this inequality is in fact an equality.

Lemma 3.13

Let $n\ge 2k, i\in \{\left\lfloor k/2\right\rfloor +1,\ldots , k-1\}$ and suppose $A, B\in V(\Gamma ^{n}_{i})$ with $|A\cap B|:=m$. Then $d(A,B)\ge \lceil \frac{k-m}{k-i} \rceil .$

Proof

Suppose $d(A,B)=l+1$ and $A-C_{1}-\dots - C_{l}-B$ is a shortest path. It can be seen using a short inductive argument that $C_{l}$ can differ from A by at most $l(k-i)$ points. Using this and the fact that $|C_{l}\cap B|\le |A\cap B|+|C_{l}\setminus A|$, we get $|C_{l}\cap B|\le m+l(k-i)$. On the other hand, $|C_{l}\cap B|=i$ so $i\le m+l(k-i)$. This implies

$$\begin{aligned} l+1\ge \frac{k-m}{k-i}, \end{aligned}$$

from which the claim follows. $\square$

Remark

• 1. If $m> i$, then $0<\frac{k-m}{k-i}<1$, so Lemma 3.13 just states that $d(A,B)\ge 1$. However if $m<i$, then $\frac{k-m}{k-i}>1$ and so Lemma 3.13 provides useful information.

• 2. Lemma 3.13 also holds for $i\le \left\lfloor k/2 \right\rfloor$. For $m>i$, this is obvious by the first part of the remark. For $m<i$, this can be seen as follows. Since $m<i$, we have $d(A,B)\ge 2$. Also, $k-i\ge k/2$ which implies that

  $$\begin{aligned} \frac{k-m}{k-i}\le \frac{2(k-m)}{k}=2\left( 1-\frac{m}{k}\right) \le 2. \end{aligned}$$

  Hence $d(A,B)\ge 2\ge \frac{k-m}{k-i}$, from which the claim follows.

The following corollary is a consequence of Lemmas 3.12 and 3.13.

Corollary 3.14

Let $n\ge 2k, i\in \{\left\lfloor k/2\right\rfloor +1,\ldots , k-1\}$, and suppose $A, B\in V(\Gamma ^{n}_{i})$ with $|A\cap B|:=m<i$. Then $d(A,B)=\lceil \frac{k-m}{k-i} \rceil$.

Proof of Theorem 1.4(3)

Let $A, B\in V(\Gamma ^{n}_{i})$ with $|A\cap B|=m\ge 0$. If $m<i$, then by Corollary 3.14, $d(A,B)=\lceil \frac{k-m}{k-i} \rceil \ge 2$, and if $m>i$, then by Lemma 3.11, $d(A,B)=2$. As m increases, the value of the function $\lceil \frac{k-m}{k-i} \rceil$ decreases. Hence the diameter of $\Gamma ^{n}_{i}$ is obtained when $m=0$, which gives $\lceil \frac{k}{k-i} \rceil$. $\square$

4 Action of $S_{n}$ on (k, l)-partitions of I

We begin this section by finding the orbitals of the action of $S_{n}$ on $I^{(k,l)}$. We then prove a result from which Theorem 1.5 is deduced and then use Theorem 1.5 to deduce Theorem 1.1(2). We continue by proving a sequence of results which lead up to the proof of Theorem 1.2(2), and we conclude the section by proving Theorem 1.3(2).

Proposition 4.1

Suppose that $n=kl$ where $k>1,l>1$. Let $A,B,C,D \in I^{(k,l)}$. Then the pairs (A, B) and (C, D) lie in the same orbital of $S_{n}$ if and only if there exist $l\times l$ permutation matrices P and Q such that $I_{CD}=P\,I_{AB}\,Q$.

Proof

Write $A=A_{1}|\dots |A_{l}, B=B_{1}|\dots |B_{l}, C=C_{1}|\dots |C_{l}$, and $D=D_{1}|\dots |D_{l}$. Suppose (A, B) and (C, D) lie in the same orbital of $S_{n}$. Then there exists $g\in S_{n}$ such that $A^{g}=C$ and $B^g=D$. Let $C^{\prime }:=C_{1}^{\prime }|\dots | C_{l}^{\prime }$ be the partition C but with the components $C_{i}$ permuted in such a way that $A_{i}^{g}=C_{i}^{\prime }$. Similarly, let $D^{\prime }$ be the partition $D:=D_{1}^{\prime }|\dots | D_{l}^{\prime }$ but with the components $D_{i}$ permuted in such a way that $B_{i}^{g}=D_{i}^{\prime }$. Then there exists permutation matrices $P^{-1}$ and $Q^{-1}$ such that $I_{C^{\prime }D^{\prime }}=P^{-1}I_{CD}Q^{-1}$. Now for any $i,j\in \{1,\ldots , l\}$, we have that $|C_{i}^{\prime }\cap D_{j}^{\prime }|=|A_{i}\cap B_{j}|$, and hence $I_{CD}=PI_{C^{\prime }D^{\prime }}Q=PI_{AB}Q$. For the converse, suppose $I_{CD}=PI_{AB}Q$ for some $l\times l$ permutation matrices P and Q. Then after permuting the $A_{i}$s and $B_{j}$s suitably, we may suppose $I_{AB}=I_{CD}$. Now since $S_{n}$ is n-transitive, we can pick $g\in S_{n}$ such that $(A_{i}\cap B_{j})^{g}=C_{i}\cap D_{j}$ for all $i,j\in \{1,\ldots ,l\}$ (note that $|\cup _{i,j\in \{1,\ldots ,l\}}{(A_{i}\cap B_{j})}|=n$). Then $(A,B)^{g}=(C,D)$, and so (A, B) and (C, D) lie in the same orbital. $\square$

Definition 4.2

Let $n=kl$, where $k,l>1$. We define the $l\times l$ matrix

$$\begin{aligned} M_{kl}:=\left( \begin{array}{c|cc} kI_{l-2} &{} 0 &{} 0\\ \hline 0 &{} 1 &{} k-1\\ 0 &{} k-1 &{} 1\\ \end{array} \right) . \end{aligned}$$

Remark

If $A, B\in V(\Gamma _{[M_{kl}]})$ are adjacent, then there exists a transposition in $S_{n}$ sending A to B. Hence every path between A and B can be represented by a sequence of transpositions and the number of transpositions is the length of the path. For a path P between A and B, we will use $\sigma _{P}$ to denote the product of the corresponding sequence of transpositions.

Proposition 4.3

Let $n=kl$, where $k,l>1$. Then diam($\Gamma _{[M_{kl}]})\ge \frac{l}{2}\left\lfloor \frac{k}{2} \right\rfloor$.

Proof

Let A and B be two (k, l)-partitions such that

$$\begin{aligned} I_{AB}\sim \left( \begin{array}{cccccc} \left\lfloor \frac{k}{2} \right\rfloor &{} \left\lceil \frac{k}{2} \right\rceil &{} 0 &{} \ldots &{} 0 &{} 0 \\ 0 &{}\left\lfloor \frac{k}{2} \right\rfloor &{} \left\lceil \frac{k}{2} \right\rceil &{} 0 &{} \ldots &{} 0 \\ \vdots &{} \ddots &{} \ddots &{} \ddots &{} &{} \vdots \\ \vdots &{} &{}\ddots &{} \ddots &{} \ddots &{} 0\\ 0 &{}\ldots &{}\ldots &{} 0 &{}\left\lfloor \frac{k}{2} \right\rfloor &{} \left\lceil \frac{k}{2} \right\rceil \\ \left\lceil \frac{k}{2} \right\rceil &{} 0 &{} \ldots &{} \ldots &{} 0 &{} \left\lfloor \frac{k}{2} \right\rfloor \end{array} \right) . \end{aligned}$$

(12)

Write $A=A_{1}|\dots |A_{l}$ and $B=B_{1}|\dots |B_{l}$. If $d(A,B)=r$ and P is a path of length r between A and B, then $\sigma _{P}$ sends A to B and for each $i\in \{1,\ldots ,l\}$ there exists $j\in \{1, \ldots , l\}$ such that $\sigma _{P}$ sends $A_{i}$ to $B_{j}$. Now $|A_{i}\cap B_{j}|\le \left\lceil \frac{k}{2} \right\rceil$ and hence $\sigma _{P}$ must move at least $k-\left\lceil \frac{k}{2} \right\rceil = \left\lfloor \frac{k}{2} \right\rfloor$ elements from each $A_{i}$. Therefore $\sigma _{P}$ must move at least $l \left\lfloor \frac{k}{2} \right\rfloor$ elements in total. Any permutation which moves x elements cannot be expressed as a product of fewer than x / 2 transpositions. Hence we must have $r\ge \frac{l}{2} \left\lfloor \frac{k}{2} \right\rfloor$, from which the result follows. $\square$

Proof of Theorem 1.5

This follows as a corollary of Proposition 4.3. $\square$

Proof of Theorem 1.1(2)

Suppose $\mathrm{diam_{O}}(G,\Omega )\le c$. Then by Theorem 1.5, we must have

$$\begin{aligned} \frac{l}{2} \left\lfloor \frac{k}{2} \right\rfloor \le c. \end{aligned}$$

(13)

Now since $\left\lfloor \frac{k}{2} \right\rfloor \ge 1$, (13) implies $l\le 2c$, and since $l\ge 2$ it also implies $k\le 2c+1$, and hence $n=kl\le 2c(2c+1)$. $\square$

Next we prove Theorem 1.2(2). Suppose $G:=S_{n}$ or $A_{n}$ and the orbital diameter of $(G, I^{(k,l)})$ is bounded by 5. By Theorem 1.1(2), we must have $1<l\le 10$ and $1<k\le 11$, and we can make these values more precise using Proposition 4.3. In order to prove Theorem 1.2(2), for each possible pair (k, l), we must either determine that the diameters of all orbital graphs are bounded by 5 or find an orbital graph of diameter greater than 5. In order to reduce the possibilities for the pair (k, l) we prove the following results. In what follows, for a permutation $\sigma$, the support of $\sigma$ is $\mathrm{supp}(\sigma ):=\{x: x^{\sigma }\ne x\}$. For $l=3$, we prove

Proposition 4.4

$\mathrm{diam}(\Gamma _{[M_{k3}]})\ge 2 \left\lfloor \frac{k}{2} \right\rfloor$.

Proof

Let $A:=A_{1}|A_{2}|A_{3}$ and $B:=B_{1}|B_{2}|B_{3}$ be two (k, 3)-partitions with

$$\begin{aligned} I_{AB}= \left( \begin{array}{ccc} \left\lfloor \frac{k}{2} \right\rfloor &{} 0 &{}\left\lceil \frac{k}{2} \right\rceil \\ \left\lceil \frac{k}{2} \right\rceil &{}\left\lfloor \frac{k}{2} \right\rfloor &{}0\\ 0 &{} \left\lceil \frac{k}{2} \right\rceil &{} \left\lfloor \frac{k}{2} \right\rfloor \end{array} \right) . \end{aligned}$$

Let $d(A,B)=r$ and suppose that P is a path of length r between A and B. As in the proof of Proposition 4.3, $\sigma _{P}$ must move at least $\left\lfloor \frac{k}{2} \right\rfloor$ elements from each $A_{i}$. If there exists i such that $\sigma _{P}$ moves all k elements from $A_{i}$, then $\sigma _{P}$ moves at least $k+2\left\lfloor \frac{k}{2} \right\rfloor$ elements in total, and hence $r\ge \left\lceil \frac{k}{2} + \left\lfloor \frac{k}{2} \right\rfloor \right\rceil =k \ge 2\left\lfloor \frac{k}{2} \right\rfloor$. Now assume that $\sigma _{P}$ fixes at least one element from each $A_{i}$. Set $X_{1}:=A_{1}\cap B_{1}, Y_{1}:=A_{2}\cap B_{2}$ and $Z_{1}:=A_{3}\cap B_{3}$, and let $X_{2}, Y_{2}$ and $Z_{2}$ be their complements in $A_{1}, A_{2}$ and $A_{3}$, respectively. Then the following information can be read off $I_{AB}$ ($\sqcup$ denotes disjoint union):

$$\begin{aligned} |X_{1}|= & {} |Y_{1}|=|Z_{1}|=\left\lfloor \frac{k}{2} \right\rfloor , |X_{2}|=|Y_{2}|=|Z_{2}|=\left\lceil \frac{k}{2} \right\rceil ,\\ A_{1}= & {} X_{1}\sqcup X_{2}, A_{2}=Y_{1}\sqcup Y_{2}, A_{3}=Z_{1}\sqcup Z_{2},\\ B_{1}= & {} X_{1}\sqcup Y_{2}, B_{2}=Y_{1}\sqcup Z_{2}, B_{3}=Z_{1}\sqcup X_{2}. \end{aligned}$$

Now either ${A_{1}}^{\sigma _{P}}=B_{1}$ or ${A_{1}}^{\sigma _{P}}=B_{3}$. Suppose ${A_{1}}^{\sigma _{P}}=B_{1}$. Then ${A_{i}}^{\sigma _{P}}=B_{i}$ for all i. Now if $\alpha \in X_{2}$ and $\beta :=\alpha ^{\sigma _{P}}$, then $\beta \in X_{1}\sqcup Y_{2}$, so $\beta \ne \alpha$. Also, if $\beta \in Y_{2}$ then $\beta ^{\sigma _{P}}\in Y_{1}\sqcup Z_{2}$, so $\beta ^{\sigma _{P}}\ne \alpha$, and if $\beta \in X_{1}$, then $\beta ^{\sigma _{P}}\in X_{1}\sqcup Y_{2}$, so again $\beta ^{\sigma _{P}}\ne \alpha$. Hence $\alpha$ is contained in a cycle of $\sigma _{P}$, of length at least 3. Repeating similar arguments, we see that if $\alpha \in Y_{2}$ or $\alpha \in Z_{2}$, then $\alpha$ is contained in a cycle of $\sigma _{P}$, of length at least 3. By the remark following Definition 4.2, we have that r is greater then or equal to the minimal length of a factorisation of $\sigma _{P}$ by transpositions. Using the fact that the minimal length of a factorisation of a permutation $\sigma$ by transpositions is $|\mathrm{supp}(\sigma )|-c_{\sigma }$, where $c_{\sigma }$ is the number of non-trivial cycles in the disjoint cycle decomposition of $\sigma$, we get $r\ge |\mathrm{supp}(\sigma _{P})|-c_{\sigma _{P}}$. Suppose that there are s disjoint cycles of length at least 3 in the disjoint cycle decomposition of $\sigma _{P}$. Let $I_{3}\subseteq \{1,\ldots , n\}$ be the union of the supports of these cycles. Then $|\mathrm{supp}(\sigma _{P})|-\sigma _{P}\ge |I_{3}|-s$. Furthermore, $s\le \frac{1}{3} |I_{3}|$ and hence we get $r\ge \frac{2}{3} |I_{3}|$. On the other hand, the preceding discussion implies $|I_{3}|\ge |X_{2}\cup Y_{2}\cup Z_{2}|=3\left\lceil \frac{k}{2} \right\rceil$. Combining the two, we obtain $r\ge 2\left\lceil \frac{k}{2} \right\rceil \ge 2\left\lfloor \frac{k}{2} \right\rfloor$. Now suppose ${A_{1}}^{\sigma _{P}}=B_{3}$. Then ${A_{2}}^{\sigma _{P}}=B_{1}$ and ${A_{3}}^{\sigma _{P}}=B_{2}$. Repeating similar arguments to the above, we can see that every element of $X_{1}\cup Y_{1}\cup Z_{1}$ must be contained in a cycle of length at least 3, and so as above, we have that

$$\begin{aligned} r\ge \frac{2}{3}\,3\left\lfloor \frac{k}{2} \right\rfloor =2\left\lfloor \frac{k}{2} \right\rfloor . \end{aligned}$$

Therefore $d(A,B)\ge 2\left\lfloor \frac{k}{2} \right\rfloor$, from which the result follows. $\square$

For $k=3$, we have the following result.

Proposition 4.5

For $l\ge 4, \mathrm{diam}(\Gamma _{[M_{3l}]})\ge \left\lceil \frac{l}{2}\right\rceil +1$.

Proof

Let A and B be two (3, l)-partitions of I with

$$\begin{aligned} I_{AB}= \left( \begin{array}{cccccc} 1 &{} 2 &{} 0 &{} \ldots &{} &{} 0 \\ 0 &{} 1 &{} 2 &{} 0 &{} \ldots &{} 0 \\ \vdots &{} \ddots &{} \ddots &{} \ddots &{} &{} \vdots \\ \vdots &{} &{} \ddots &{} \ddots &{} \ddots &{} 0 \\ 0 &{} \ldots &{} \dots &{} 0 &{} 1 &{} 2 \\ 2 &{} 0 &{} \ldots &{} \ldots &{} 0 &{} 1 \end{array} \right) \end{aligned}$$

as in (12). By Proposition 4.3, $d(A,B)\ge \lceil \frac{l}{2}\rceil$. We will prove the claim by showing that $d(A,B)> \lceil \frac{l}{2}\rceil$. Assume by contradiction that P is a path of length $\lceil \frac{l}{2}\rceil$ between A and B. Write $\sigma _{P}=a_{1}\ldots a_{\left\lceil \frac{l}{2}\right\rceil }$, where each $a_{i}$ is a transposition, and write $A=A_{1}|\dots |A_{l}$. By the proof of Proposition 4.3, $\sigma _{P}$ moves at least one element from each $A_{i}$. It follows that if l is even, then $|\mathrm{supp}(\sigma _{P})|=l$ and if l is odd, then $|\mathrm{supp}(\sigma _{P})|\in \{l,l+1\}$. In all cases, $\mathrm{supp}(\sigma _{P})=\cup _{i=1}^{\left\lceil \frac{l}{2} \right\rceil }{\mathrm{supp}(a_{i})}$, and now a careful analysis, whose details are left to the reader, shows that in all cases, there exists $1\le i \le \left\lceil \frac{l}{2} \right\rceil$ and $1\le r\ne s \le l$ such that $a_{i}=(x\, y)$ for some $x\in A_{r}$ and $y\in A_{s}$; moreover, the supports of $a_{i}$ and $a_{j}$ are disjoint for all $1\le j\le \left\lceil \frac{l}{2} \right\rceil$ such that $j\ne i$, and $|\mathrm{supp}(\sigma )\cap A_{r}|=|\mathrm{supp}(\sigma )\cap A_{s}|=1$. In particular, $a_{i}$ commutes with every other transposition and we can write $\sigma _{P}=a_{i}\sigma ^{\prime }_{P}$ where $\sigma ^{\prime }_{P}:=a_{1}\cdots a_{i-1}a_{i+1}\cdots a_{\left\lceil \frac{l}{2} \right\rceil }$. Thus $\sigma _{P}$ sends $A_{r}$ to $(A_{r}\backslash \{x\}) \cup \{y\}$ and $A_{s}$ to $(A_{s}\backslash \{y\})\cup \{x\}$. Therefore $I_{AB}\sim N$, where N is the $l\times l$ matrix $\left( \begin{array}{cc} M &{} 0 \\ 0 &{} X \end{array} \right) ,$ with $M:=\left( \begin{array}{cc} 2 &{} 1 \\ 1 &{} 2 \end{array}\right)$, and where X is an $(l-2)\times (l-2)$ matrix. Then there exist permutation matrices P and Q such that $P^{-1}I_{AB}=NQ$. Recall that multiplication by $P^{-1}$ on the left permutes the rows of $I_{AB}$, and multiplication by Q on the right permutes the columns of N. Note that the entries of $P^{-1}I_{AB}=NQ$ take values from the set $\{0,1,2\}$. Firstly, suppose $(NQ)_{11}=2$. Then the first and second rows of $P^{-1}I_{AB}$ must be $(2,0,\ldots , 0,1)$ and $(1,2,0,\ldots , 0)$, respectively. This implies that $(NQ)_{12}=0$ and $(NQ)_{22}=2$, which is not possible since N does not contain a column of the form $(0,2,\ldots )^{\mathrm{T}}$. Now suppose that $(NQ)_{11}=1$. Then the first and second rows of $P^{-1}I_{AB}$ must be $(1,2,0,\ldots , 0)$ and $(2,0,\ldots , 0,1)$, respectively. This implies that $(NQ)_{12}=2$ and $(NQ)_{22}=0$, which is again not possible. Finally suppose that $(NQ)_{11}=0$. Then $(NQ)_{21}=0$, and there exist $i\in \{2,\ldots , l\}$ such that $(NQ)_{1i}=1$ and $(NQ)_{2i}=2$. However this is a contradiction since the matrix $I_{AB}$ has the property that if $(I_{AB})_{r1}=(I_{AB})_{s1}=0$ and $(I_{AB})_{rt}=1$, then $(I_{AB})_{st}=0$, where $t\in \{2,\ldots , l\}$. Therefore we must have $d(A,B)\ge \left\lceil \frac{l}{2}\right\rceil +1$. $\square$

We also have a small improvement in the lower bound in Proposition 4.3 which is obtained as follows.

Proposition 4.6

Let $A, B\in V(\Gamma _{[M_{kl}]})$ with $I_{AB}$ as in (12). If P is a path between A and B and $\sigma _{P}\in S_{kl}$ moves exactly $l \left\lfloor \frac{k}{2} \right\rfloor$ points from A, then $d(A,B)\ge \left\lfloor \frac{k}{2} \right\rfloor (l-1)$, and hence $\mathrm{diam}(\Gamma _{[M_{kl}]})\ge \left\lfloor \frac{k}{2} \right\rfloor (l-1)$.

Proof

Let $A:=A_{1}|\dots |A_{l}$ and $B:=B_{1}|\dots |B_{l}$. Since $I_{AB}$ is given by (12), we can write $A_{i}=X_{i}\sqcup Y_{i}$ for all $1\le i \le l$, where $|X_{i}|=\left\lfloor \frac{k}{2} \right\rfloor$. Then $B_{i}=X_{i+1}\sqcup Y_{i}$ for all $1\le i \le l-1$, and $B_{l}=X_{1}\sqcup Y_{l}$. By the proof of Proposition 4.3, $\sigma _{P}$ moves at least $\left\lfloor \frac{k}{2} \right\rfloor$ elements from each $A_{i}$. Hence by the assumption, exactly $\left\lfloor \frac{k}{2} \right\rfloor$ elements are moved from each $A_{i}$. Therefore ${A_{i}}^{\sigma _{P}}=B_{i}$, for all i. In particular, $\sigma _{P}$ sends $X_{i}$ to $X_{i+1}$, for $i\in \{1,\ldots , l-1\}$, and $X_{l}$ to $X_{1}$. Hence if we write $\sigma _{P}=C_{1}\cdots C_{r}$, where the $C_{i}$ are non-trivial disjoint cycles, then we can conclude that each $C_{i}$ has length which is a multiple of l, say $t_{i}l$, and $r\le \left\lfloor \frac{k}{2} \right\rfloor$. Then $\sum _{i=1}^{r}{t_{i}l}= \left\lfloor \frac{k}{2} \right\rfloor l$ and $\sigma _{P}$ cannot be expressed as a product of fewer than

$$\begin{aligned} \sum _{i=1}^{r}{(t_{i}l-1)} = \left\lfloor \frac{k}{2} \right\rfloor l -r \end{aligned}$$

transpositions. The expression on the right-hand side is minimal when $r=\left\lfloor \frac{k}{2} \right\rfloor$ and hence $d(A,B)\ge \left\lfloor \frac{k}{2} \right\rfloor (l-1)$. $\square$

Corollary 4.7

${ diam}(\Gamma _{[M_{kl}]})\ge { min}\left\{ \left\lceil \frac{l}{2} \left\lfloor \frac{k}{2} \right\rfloor + \frac{1}{2}\right\rceil , \left\lfloor \frac{k}{2} \right\rfloor (l-1) \right\}$.

Proof

If $A, B\in V(\Gamma _{[M_{kl}]})$ and P is a path between A and B, then $\sigma _{P}$ moves at least $l \left\lfloor \frac{k}{2} \right\rfloor$ points from A. Now if $\sigma _{P}$ moves exactly $l \left\lfloor \frac{k}{2} \right\rfloor$ points, then by Proposition 4.6, $\mathrm{diam}(\Gamma _{[M_{kl}]})\ge \left\lfloor \frac{k}{2} \right\rfloor (l-1)$. Otherwise, $\sigma _{P}$ moves at least $l \left\lfloor \frac{k}{2} \right\rfloor +1$ points and hence cannot be expressed as a product of fewer than $\frac{l \left\lfloor \frac{k}{2} \right\rfloor +1}{2}$ transpositions. Therefore $\mathrm{diam}(\Gamma _{[M_{kl}]})\ge \left\lceil \frac{l}{2} \left\lfloor \frac{k}{2} \right\rfloor + \frac{1}{2}\right\rceil$. $\square$

We conjecture that Proposition 4.4 can be generalised; more precisely that $\mathrm{diam}(\Gamma _{[M_{kl}]})\ge \left\lfloor \frac{k}{2} \right\rfloor (l-1)$ for arbitrary values of k and l.

Using Propositions 4.4, 4.5, and Corollary 4.7, we obtain the following result.

Proposition 4.8

Let $G:=S_{n}$ or $A_{n}$ and $\Omega :=I^{(k,l)}$, where $n=kl, k>1, l>1$, and suppose that (k, l) is not given by one of the values in Table 8. Then $\mathrm{diam_{O}}(G,\Omega )\ge 6$.

Table 8 Possible pairs (k, l) with $\mathrm {diam_{O}}(G,I^{(k,l)})\le 5$

For the case $l=2$, we find the diameters of all orbital graphs of $(S_{k2}, I^{(k,2)})$ as stated in Theorem 1.6. Recall that for $l=2$, the orbital graphs are $\Gamma _{[M_{i}]}$, where $i\in \{1,\ldots , \left\lfloor k/2\right\rfloor \}$ and

$$\begin{aligned} M_{i}=\left( \begin{array}{cc} i &{} k-i \\ k-i &{} i \end{array} \right) . \end{aligned}$$

Note that the matrix $M_{i}$ is determined by $(M_{i})_{11}$ and if $A, B\in V(\Gamma _{[M_{i}]})$, then the matrix $I_{AB}$ is determined by $(I_{AB})_{11}$. In order to prove Theorem 1.6 we will need to prove the following criterion which relates distances in the graphs $\Gamma _{[M_{i}]}$ to distances in the graphs $\Gamma ^{2k}_{i}$ from Sect. 3. In what follows, for a k-subset X of $I:=\{1,\dots ,2k\}$, we will use the notation $X^{c}$ to refer to the complement in I of the set X.

Lemma 4.9

Let $i\in \{1,\dots , \left\lfloor k/2 \right\rfloor \}$ and $A,B\in V(\Gamma _{[M_{i}]})$, where $A:=A_{1}|A_{2}$ and $B:=B_{1}|B_{2}$. Suppose $d_{\Gamma _{[M_{i}]}}(A,B)\ge r\ge 2$. Then $d_{\Gamma _{[M_{i}]}}(A,B)=r$ if and only if there exist $l,m\in \{1,2\}$ such that $d_{\Gamma ^{2k}_{i}}(A_{l},B_{m})=r$. In particular, $diam(\Gamma _{[M_{i}]})\ge r$ if and only if $diam(\Gamma ^{2k}_{i})\ge r$.

Proof

Suppose $d_{\Gamma _{[M_{i}]}}(A,B)=r$ and let $A-C_{1}-\dots - C_{r-1}-B$ be a shortest path, where $C_{i}:=X_{i}|Y_{i}$. Then there exist $\{l,m\}\in \{1,2\}$, and $Z_{i}\in \{X_{i},Y_{i}\}$ for $i\in \{1,\ldots , r-1\}$, such that $|A_{l}\cap Z_{1}|=i, |Z_{i}\cap Z_{i+1}|=i$ for $i\in \{1,\ldots , r-2\}$, and $|Z_{r-1}\cap B_{m}|=i$. Hence $A_{l}-Z_{1}-\dots - Z_{r-1}-B_{m}$ is a path of length r in $\Gamma ^{2k}_{i}$. If $d_{\Gamma ^{2k}_{i}}(A_{l},B_{m})=:t<r$ and $A_{l}-D_{1}-\dots D_{t-1}-B_{m}$ is a shortest path, then

$$\begin{aligned} A=A_{l}|A_{l}^{c}-D_{1}|D_{1}^{c}-\dots - D_{t-1}|D_{t-1}^{c}-B_{m}|B_{m}^{c}=B \end{aligned}$$

is a path of length t in $\Gamma _{[M_{i}]}$, which contradicts the fact that $d_{\Gamma _{[M_{i}]}}(A,B)<r$. Hence $d_{\Gamma ^{2k}_{i}}(A_{l},B_{m})=r$. Conversely, suppose there exist $l,m\in \{1,2\}$ such that $d_{\Gamma ^{2k}_{i}}(A_{l},B_{m})=r$. Let $A_{l}-X_{1}-\dots - X_{r-1}-B_{m}$ be a shortest path in $\Gamma ^{2k}_{i}$. Then

$$\begin{aligned} A=A_{l}|A_{l}^{c}-X_{1}|X_{1}^{c}-\dots -X_{r-1}|X_{r-1}^{c}-B_{m}|B_{m}^{c}=B \end{aligned}$$

is a path of length r in $\Gamma _{[M_{i}]}$, and since $d_{\Gamma _{[M_{i}]}}(A,B)\ge r$, we have $d_{\Gamma _{[M_{i}]}}(A,B)=r$.

$\square$

The following lemma follows directly from Lemmas 3.3 and 3.5.

Lemma 4.10

Suppose $n=2k$. Let $i\in \{1,\ldots , \left\lfloor k/2\right\rfloor \}$ and $A,B\in V(\Gamma ^{n}_{i})$ with $m:=|A\cap B|\ne i$. Suppose $d:=\mathrm{diam}(\Gamma ^{n}_{i})\ge 2$. If $2i<k$, then

$$\begin{aligned} d(A,B)=2 \Leftrightarrow m\ge k-2i, \end{aligned}$$

and if $2i\ge k$, then $d(A,B)=2$. Furthermore for $v\in \{3.\ldots , d\}$, if $(2v-1)i<k$, then

$$\begin{aligned} d(A,B)&=v \Leftrightarrow k-vi\le m<k-(v-2)i,&v \text { even}\\ d(A,B)&=v \Leftrightarrow (v-2)i<m\le vi,&v \text { odd}, \end{aligned}$$

and if $(2v-1)i\ge k$, then

$$\begin{aligned} d(A,B)= & {} v \Leftrightarrow (v-1)i<m<k-(v-2)i, \quad v \text { even}\\ d(A,B)= & {} v \Leftrightarrow (v-2)i<m< k-(v-1)i, \quad v \text { odd}.\end{aligned}$$

Proof of Theorem 1.6(1) and (2a)

Firstly, note that for $k<4$, all orbital graphs have diameter 1. Suppose $k\ge 4$ and let $A,B\in V(\Gamma _{[M_{i}]})$ with $(I_{AB})_{11}=: j$, where $j\in \{1,\ldots , \left\lfloor k/2\right\rfloor \}$ and $j\ne i, k-i$. Then $d_{\Gamma _{[M_{i}]}}(A,B)\ge 2$ and by Lemma 4.9, $d_{\Gamma _{[M_{i}]}}(A,B)=2$ if and only if there exist $l,m\in \{1,2\}$ such that $d_{\Gamma ^{2k}_{i}}(A_{l},B_{m})=2$. Suppose $2i<k$. Then by Lemma 4.10,

$$\begin{aligned} d_{\Gamma _{[M_{i}]}}(A,B)=2 \Leftrightarrow \text { either } j\ge k-2i \text { or } j\le 2i. \end{aligned}$$

(14)

Therefore $\mathrm{diam}(\Gamma _{[M_{i}]})=2$ if and only if $2i\ge k-2i-1$. Now if $2i\ge k$, then by Lemma 4.10, we have $\mathrm{diam}(\Gamma _{[M_{i}]})=2$. We now obtain the result by gathering together the values of i for which the diameter is 2. $\square$

Lemma 4.11

Let $i\in \{1,\ldots , \left\lfloor k/2\right\rfloor \}$ and $A,B\in V(\Gamma _{[M_{i}]})$ with $(I_{AB})_{11}=: j$, where $j\in \{1,\ldots , \left\lfloor k/2\right\rfloor \}$ and $j\ne i,k-i$. Suppose $d_{\Gamma _{[M_{i}]}}(A,B)\ge 3$. If $5i<k$, then

$$\begin{aligned} d_{\Gamma _{[M_{i}]}}(A,B)=3 \Leftrightarrow \text { either } 2i< j\le 3i \text { or } k-3i\le j < k-2i, \end{aligned}$$

(15)

and if $5i\ge k$, then

$$\begin{aligned} d_{\Gamma _{[M_{i}]}}(A,B)=3 \Leftrightarrow 2i< j< k-2i. \end{aligned}$$

(16)

Proof

Note that we must have $2i<k$ (else the diameter is 2), and hence (14) holds. Then the proof follows from Lemmas 4.9, 4.10, and (14). $\square$

In order to prove Theorem 1.6(2b), we need the following.

Lemma 4.12

Let $t\ge 2$ and suppose $\mathrm{diam}(\Gamma _{[M_{i}]})\ge t+1$. If $t=2$, then $2i<k$ and if $t\ge 3$, then $(2t-1)i<k$.

Proof

If $2i\ge k$, then by Lemma 4.10, we have $\mathrm{diam}(\Gamma _{[M_{i}]})=2$, hence the result for $t=2$. If $t\ge 3$, then by Lemma 4.9, $\mathrm{diam}(\Gamma ^{2k}_{i})\ge t+1$ and the result follows from Lemma 3.7. $\square$

Proposition 4.13

Let $i\in \{1,\ldots , \left\lfloor k/2\right\rfloor \}$. Then

$$\begin{aligned} \mathrm{diam}(\Gamma _{[M_{i}]})=3 \Leftrightarrow \frac{k-1}{6}\le i< \frac{k-1}{4}. \end{aligned}$$

Proof

Suppose $\mathrm{diam}(\Gamma _{[M_{i}]})=3$. Then $\mathrm{diam}(\Gamma _{[M_{i}]})>2$, so by Theorem 1.6(2a), we must have $i<\frac{k-1}{4}$. Now by Lemma 4.12, we see that (14) holds. Using (14) and Lemma 4.11, we see that $3i\ge k-3i-1$. Conversely if $\frac{k-1}{6}\le i< \frac{k-1}{4}$, then by Theorem 1.6(2a), $\mathrm{diam}(\Gamma _{[M_{i}]})\ge 3$. Now by Lemma 4.12, we see that (14) holds. Using (14) and Lemma 4.11, and the fact that $i\ge \frac{k-1}{6}$, we see that for all $A,B\in V(\Gamma ^{n}_{i})$ with $d(A,B)>2$, we must have $d(A,B)=3$, and thus $\mathrm{diam}(\Gamma _{[M_{i}]})=3$. $\square$

Lemma 4.14

Let $r\ge 3$ and $A,B\in V(\Gamma _{[M_{i}]})$ with $(I_{AB})_{11}=: j$, where $j\in \{1,\ldots ,k/2\}$ and $j \ne i, k-i$. Suppose $d_{\Gamma _{[M_{i}]}}(A,B)\ge r$.

• 1. If $(2r-1)i<k$, then $d_{\Gamma _{[M_{i}]}}(A,B)=r$ if and only if either $k-ri\le j<k-(r-1)i$ or $(r-1)i<j\le ri$.

• 2. If $(2r-1)i\ge k$, then $d_{\Gamma _{[M_{i}]}}(A,B)=r$ if and only if $(r-1)i<j<k-(r-1)i$.

Proof

We proceed by induction or r. The base case $r=3$ has been done in Lemma 4.11. Now suppose $d_{\Gamma _{[M_{i}]}}(A,B)\ge r\ge 4$ and that the lemma holds for all $s<r$. Then using Lemma 4.12 and the inductive hypothesis to determine the pairs of vertices for which the distance is strictly less than r, we obtain that

$$\begin{aligned} d_{\Gamma _{[M_{i}]}}(A,B)\ge r\Leftrightarrow (r-1)i<j<k-(r-1)i. \end{aligned}$$

(17)

For j as in (17), by Lemmas 4.9 and 4.10, if $(2r-1)i<k$, then

$$\begin{aligned} d_{\Gamma _{[M_{i}]}}(A,B)=r \Leftrightarrow \text { either } k-ri\le j<k-(r-2)i \text { or } (r-2)i<j\le ri, \end{aligned}$$

and if $(2r-1)i\ge k$, then

$$\begin{aligned} d_{\Gamma _{[M_{i}]}}(A,B)= & {} r \Leftrightarrow \text { either } (r-1)i< j<k-(r-2)i \text { or } \\&(r-2)i<j<k-(r-1)i. \end{aligned}$$

We obtain the result by combing the inequalities above with (17). $\square$

The next corollary is a direct consequence of the proof of Lemma 4.14.

Corollary 4.15

Let $r\ge 3$ and $A,B\in V(\Gamma _{[M_{i}]})$ with $(I_{AB})_{11}=: j$, where $j\in \{1,\ldots ,k/2\}$ and $j \ne i, k-i$. If $\mathrm{diam}(\Gamma _{[M_{i}]})\ge r$, then

$$\begin{aligned} d_{\Gamma _{[M_{i}]}}(A,B)\ge r\Leftrightarrow (r-1)i<j<k-(r-1)i. \end{aligned}$$

We now prove Theorem 1.6(2b).

Proof of Theorem 1.6(2b)

Suppose $\mathrm{diam}(\Gamma _{[M_{i}]})\ne 2$. It suffices to show that for $r\ge ~3$,

$$\begin{aligned} \mathrm{diam}(\Gamma _{[M_{i}]})=r \Leftrightarrow \frac{k-1}{2r}\le i < \frac{k-1}{2r-2}. \end{aligned}$$

We proceed by induction on r. The base case $r=3$ is done in Proposition 4.13. Now let $r\ge 4$ and suppose for the inductive hypothesis that

$$\begin{aligned} \mathrm{diam}(\Gamma _{[M_{i}]})=s \Leftrightarrow \frac{k-1}{2s}\le i < \frac{k-1}{2s-2}, \end{aligned}$$

(18)

for all $s\in \{3,\ldots , r-1\}$. Suppose $\mathrm{diam}(\Gamma _{[M_{i}]})=r$. Then $\mathrm{diam}(\Gamma _{[M_{i}]})>r-1$, and so by Theorem 1.6(2a), Proposition 4.13 and (18), we must have $i < \frac{k-1}{2r-2}$. By Corollary 4.15, we have that for all $A,B\in V(\Gamma _{[M_{i}]})$ with $(I_{AB})_{11}=: j$, where $j\in \{1,\ldots , \left\lfloor k/2\right\rfloor \}$ and $j\ne i,k-i$,

$$\begin{aligned} d_{\Gamma _{[M_{i}]}}(A,B)\ge r\Leftrightarrow (r-1)i<j<k-(r-1)i. \end{aligned}$$

Now $\mathrm{diam}(\Gamma _{[M_{i}]})=r$ implies that for all $A,B\in V(\Gamma _{[M_{i}]})$ with $d(A,B)\ge r$, we must have $d(A,B)=r$. Using this and Lemma 4.14, we see that either $(2r-1)i\ge k$, or $(2r-1)i<k$ and $ri\ge k-ri-1$. In either case, $i\ge \frac{k-1}{2r}$. Conversely if $\frac{k-1}{2r}\le i < \frac{k-1}{2r-2}$, then again by Theorem 1.6(2a), Proposition 4.13 and (18), we have $\mathrm{diam}(\Gamma _{[M_{i}]})\ge r$. Then using Corollary 4.15, Lemma 4.14, and the fact that $i\ge \frac{k-1}{2r}$, we see that for all $A,B\in V(\Gamma _{[M_{i}]})$ with $d(A,B)\ge r$, we have $d(A,B)=r$, and thus $\mathrm{diam}(\Gamma _{[M_{i}]})=r$. $\square$

To complete the proof of Theorem 1.2(2), we need to establish which of the values in Table 8 in Proposition 4.8 actually yield $\mathrm{diam_{O}}(G,I^{(k,l)})\le 5$. To do this, we have used magma. We have devised two codes; the first takes as inputs the values k and l and returns the diameters of all orbital graphs of $(G, I^{(k,l)})$. In order to be able to deal with as high values of $n=kl$ as possible, the code is designed to calculate the collapsed adjacency matrix of each orbital graph as outlined in Sect. 2. As mentioned in Sect. 2, this code has limitations when the index gets large. In this case, we have devised a second code; fix two (k, l)-partitions A and B (usually with $I_{AB}$ is as in (12)), and let $\sigma \in G:=S_{kl}$ be a permutation sending A to B. Then any permutation sending A to B is of the form $w\sigma$, where $w\in G_{A}\cong S_{k}\wr S_{l}$. Let ${\mathcal {R}}$ be the set of permutations sending A to B. The code takes as inputs the values of k, l and $\sigma$ and calculates for each $\tau \in {\mathcal {R}}$ the minimal number of transpositions $T(\tau )$ that $\tau$ can be a product of. The code returns the value $min_{\tau \in {\mathcal {R}}}{T(\tau )}$, which is (by the remark following Definition 4.2) the value of d(A, B) in the orbital graph $\Gamma _{[M_{kl}]}$, and hence a lower bound for $\mathrm{diam}(\Gamma _{[M_{kl}]})$. Note that one can obtain the exact value of $\mathrm{diam}(\Gamma _{[M_{kl}]})$ by calculating d(A, B) for all pairs (A, B), but for the results in this paper, only a lower bound is required.

Proof of Theorem 1.2(2)

Table 8 lists the values of k and l which remain to be considered. For the case where $l=2$, using Theorem 1.6 it can be seen that $\mathrm{diam_{O}}(G,\Omega )\le 5$ if and only if $2\le k\le 11$. For $l\ge 3$, magma has been used to verify the result as outlined in the paragraph after the proof of Theorem 1.6(2b). The data obtained can be found in Appendix 1. $\square$

Although as $n=kl$ tends to infinity the diameters of the corresponding orbital graphs are unbounded, there do exist infinite families of orbital graphs with small diameters. We will now present one infinite family of diameter 2 graphs and prove Theorem 1.3(2). In order to do this, we will need the following preliminary definitions and lemma.

Definition 4.16

Let $A\in M_{n}({\mathbb {R}}_{\ge 0})$ be an $n\times n$ matrix with non-negative real entries, and let $k\ge 1$ be an integer. We say that A is a k-doubly stochastic matrix if each row sum and each column sum is k. We say that A is doubly stochastic if A is 1-doubly stochastic.

Definition 4.17

Let $A\in M_{n}({\mathbb {R}}_{\ge 0})$. A diagonal $D:=(a_{1},\ldots , a_{n})$ of A is a sequence of n entries of A such that no two entries lie in the same row or column of A. A diagonal D is called positive if $a_{i}>0$ for all i.

The following lemma can be found as Lemma 7.4.4 in [6].

Lemma 4.18

Every doubly stochastic matrix has a positive diagonal.

Proof of Theorem 1.3(2)

Let $A, B\in V(\Gamma _{[M]})$ and write $A=A_{1}|\dots | A_{k}, B=B_{1}|\dots | B_{k}$. We will show that there exists a vertex C such that $I_{AC}\sim M\sim I_{CB}$. We will do this by describing an explicit construction for C. Define $I^{\prime }_{AB}:=\frac{1}{k}I_{AB}$. Then $I^{\prime }_{AB}$ is a doubly stochastic matrix. By Lemma 4.18, $I^{\prime }_{AB}$, and therefore $kI^{\prime }_{AB}$, has a positive diagonal. Moreover, the latter diagonal has positive integer entries, in particular at least 1. Denote this diagonal by $D=(a_{i_{1}j_{1}}, \ldots , a_{i_{k}j_{k}} )$, where for $r\in \{1,\ldots , k\}, a_{i_{r}j_{r}}: =|S_{i_{r}j_{r}}|>0$ and $S_{i_{r}j_{r}}:=A_{i_{r}}\cap B_{j_{r}}$. Let $C_{1}$ be a k-subset which consists of exactly one element from each intersection $S_{i_{r}j_{r}}$, for $r\in \{1,\ldots ,k \}$. Now $|C_{1}|=k$ and $|A_{i}\cap C_{1}|=1=|C_{1}\cap B_{i}|$, for all $i\in \{1,\ldots ,k \}$. Note that when we subtract 1 from each of the entries $a_{i_{r}j_{r}}$ in $I_{AB}$ then we are left with a $(k-1)$-doubly stochastic matrix $I_{AB}^{(1)}$ where ${(I_{AB}^{(1)})}_{i_{r}j_{r}}=|(A_{i_{r}}\cap B_{j_{r}})\setminus C_{1}|$, for $r\in \{1,\ldots ,k \}$, and ${(I_{AB}^{(1)})}_{xy}=(I_{AB})_{xy}$, otherwise. As before we can use Lemma 4.18 to deduce that $I_{AB}^{(1)}$ has a positive integer diagonal, say $D^{\prime }=(b_{t_{1}u_{1}}, \ldots , b_{t_{k}u_{k}} )$, where for $r\in \{1,\ldots ,k\}$, either $b_{t_{r}u_{r}}=|(A_{t_{r}}\cap B_{u_{r}}){\setminus }C_{1}|$ or $b_{t_{r}u_{r}}=|A_{t_{r}}\cap B_{u_{r}}|$. In the former case, define $T_{t_{r}u_{r}}:=(A_{t_{r}}\cap B_{u_{r}}){\setminus }C_{1}$, and in the latter case, define $T_{t_{r}u_{r}}:=A_{t_{r}}\cap B_{u_{r}}$. Note that in either case $T_{t_{r}u_{r}}\cap C_{1}=\emptyset$. We let $C_{2}$ be the set which consists of exactly one element from each of the sets $T_{t_{r}u_{r}}$, for $r\in \{1,\ldots , k\}$. Then $|C_{2}|=k, |A_{i}\cap C_{2}|=1=|C_{2}\cap B_{i}|$ for all $i\in \{1,\ldots ,k \}$, and the matrix $I_{AB}^{(2)}$ is $(k-2)$-doubly stochastic. We may repeat the above process to construct $C_{j}$ at which stage we are left with a $(k-j)$-doubly stochastic matrix. Then we can construct $C_{j+1}$ as long as $k-j\ge 1$, i.e. $j+1\le k$. So we may construct sets $C_{1}, \ldots , C_{k}$ such that $|C_{j}|=k$ and $|A_{i}\cap C_{j}|=1=|C_{j}\cap B_{i}|$ for all $i,j\in \{1,\ldots ,k \}$. The sets $C_{j}$ are disjoint by construction. Now define $C:=C_{1}|\dots | C_{k}$. Then $I_{AC}\sim M \sim I_{CB}$, and hence $A-C-B$ is a path of length 2. $\square$

5 Action of G on (G : H), H primitive on I

In this section, we prove Theorems 1.1(3), 1.2(3) and 1.3(3). The analysis of the actions of $G:=S_{n}$ or $A_{n}$ on $\Omega :=(G:H)$, where H is primitive on $I=\{1,\ldots ,n\}$, rely on the following bound.

Lemma 5.1

Let $(G,\Omega )$ be a finite primitive permutation group and $H:=G_{\alpha }$. Let $\Delta :=(\alpha ,\beta )^{G}$ and $d:=diam(\Gamma _{\Delta \cup \Delta ^{*}})$. Then

$$\begin{aligned} |\Omega |\le 1+2|Y|+2|Y|(2|Y|-1)+\dots +(2|Y|)(2|Y|-1)^{d-1}, \end{aligned}$$

where $Y:=\beta ^{H}$. In particular (since $|Y|\le |H|$), we have that

$$\begin{aligned} |\Omega |\le 1+2|H|+2|H|(2|H|-1)+\dots +(2|H|)(2|H|-1)^{d-1}. \end{aligned}$$

Proof

Let v denote the valency of $\Gamma _{\Delta \cup \Delta ^{*}}$. If $\Delta$ is self-paired, then $v=|\beta ^{H}|$, and if $\Delta$ is not self-paired, then $v=2|\beta ^{H}|$. Now the result follows from the following inequality:

$$\begin{aligned} |\Omega |\le 1+v+v(v-1)+v(v-1)^{2}+\dots +v(v-1)^{d-1}. \end{aligned}$$

$\square$

We will also need the following result (Corollary 1.2 in [9]) concerning the orders of primitive subgroups of $S_{n}$ and $A_{n}$.

Lemma 5.2

If H is a primitive subgroup of $S_{n}$ not containing $A_{n}$, then $|H|< 3^{n}$. Moreover, if $n>24$, then $|H|<2^{n}$.

Proof of Theorem 1.1(3)

By Lemma 5.1

$$\begin{aligned} |G:H|&\le 1+2|H|+2|H|(2|H|-1)+\dots +2|H|(2|H|-1)^{c-1} < (2|H|)^{c+1}. \end{aligned}$$

(19)

Then using Lemma 5.2 and (19), either $n\le 24$ or $n!/2 < 2^{(c+1)(n+1)+n}$. Now by (8.8) in [4, p.161], we have that $n!/2>(n/e)^{n}$. If $n>24$, then using this and the fact that $(n+1)/n\le \epsilon :=1.04$, we obtain

$$\begin{aligned} n<e\,2^{\epsilon (c+1)+1}<2^{\epsilon (c+1) + 3}. \end{aligned}$$

(20)

Note that (20) also holds for $n\le 24$ (using $c\ge 1$) hence holds for all n. $\square$

Next we prove Theorem 1.2(3). By Lemma 5.1, if $(G,\Omega )$ has an orbital graph of diameter d, then

$$\begin{aligned} |G|\le |H|\left( 1+ 2|H|\frac{(2|H|-1)^{d}-1}{2|H|-2} \right) . \end{aligned}$$

(21)

So either $n\le 24$ or, using Lemma 5.2,

$$\begin{aligned} |G|\le 2^{n}\left( 1+ 2^{n+1}\frac{(2^{n+1}-1)^{d}-1}{2^{n+1}-2} \right) . \end{aligned}$$

(22)

Using (22) with $d=5$, we see that $n\le 173$ for $G=S_{n}$ and $n\le 174$ for $G=A_{n}$. We can then use in-built functions in magma to run through values of $n\le 174$ and for each such n, iterate through the maximal subgroups of G and output only the primitive subgroups H which satisfy (21) with $d=5$. This way we limit the number of possible pairs (n, H) for which (G, (G : H)) has an orbital graph of diameter $d\le 5$; to 22 and 30 for $G=S_{n}$ and $G=A_{n}$, respectively. Now if $\mathrm{diam_{O}}(G,\Omega )\le 5$, then there must exist an orbital graph of diameter d where $d\le 5$, and hence we also obtain all possibilities of (G, (G : H)) for which the orbital diameter is bounded by 5. In order to rule out some of these possibilities, we prove the following result.

Proposition 5.3

Let $(G,\Omega )$ be a finite primitive permutation group where $G:=S_{n}$ or $A_{n}$ and $\Omega :=(G:H)$, where H is primitive on $I=\{1,\ldots ,n\}$. Suppose there exists a subgroup $X\le H$ such that $N_{G}(X)>N_{H}(X)$. If $N_{G}(X)\setminus N_{H}(X)$ contains an element of order 2, then set $a:=1$, otherwise set $a:=2$. There exists an orbital graph of diameter d, where

$$\begin{aligned} d>\frac{log(|G:H|)}{log(a|H|)-log(|X|)}-1. \end{aligned}$$

Proof

Pick $g\in N_{G}(X)\setminus N_{H}(X)$. Let $\alpha :=H, \beta :=Hg$, and $k:=|\beta ^{G_{\alpha }}|$. Consider the orbital $\Delta :=(\alpha ,\beta )^{G}$ and let $d:=diam(\Gamma _{\Delta \cup \Delta ^{*}})$. If g has order 2, then g swaps $\alpha$ and $\beta$, and hence $\Delta$ is self-paired. By the proof of Lemma 5.1, we have that

$$\begin{aligned} |G:H|\le 1+ ak+ak(ak-1)+\dots + ak(ak-1)^{d-1}<(ak)^{d+1}, \end{aligned}$$

where a is as in the statement of the lemma. Rearranging for d, this implies

$$\begin{aligned} d>\frac{log(|G:H|)}{log(ak)}-1. \end{aligned}$$

(23)

Now $k=|G_{\alpha }:G_{\alpha }\cap g^{-1}G_{\alpha }g|$ and $|G_{\alpha }\cap g^{-1}G_{\alpha }g|\ge |X|$. Hence $k\le \frac{|H|}{|X|}$. Using this in (23), we obtain the result. $\square$

Table 9 Possible pairs (n, H) with $\mathrm {diam_{O}}(S_{n},\Omega )\le 5$, where $\Omega :=(S_{n}:H)$ and H is primitive

Table 10 Possible pairs (n, H) with $\mathrm {diam_{O}}(A_{n},\Omega )\le 5$, where $\Omega :=(A_{n}:H)$ and H is primitive

In light of Proposition 5.3, picking a large subgroup $X\le H$ such that $N_{G}(X)>N_{H}(X)$ can determine the existence of an orbital graph of large diameter. We have used magma to devise a code which, given G and H, finds the largest such X and calculates the lower bounds in Proposition 5.3. This code has been used with the possibilities of (G, (G : H)) obtained using (21) with $d\le 5$, to obtain the following result.

Proposition 5.4

Let $(G,\Omega )$ be a finite primitive permutation group where $G:=S_{n}$ or $A_{n}$ and $\Omega :=(G:H)$, where H is primitive on $I=\{1,\ldots ,n\}$ and $H\ne A_{n}$. If n and H are in not in Tables 9 or 10, then $\mathrm{diam_{O}}(G,\Omega )\ge 6$.

To deal with the remaining cases in Tables 9 and 10, we have again used magma to devise a code which takes as input a value of n and outputs each possible primitive group H from Tables 9 and 10, the diameters of the corresponding orbital graphs, and their valencies. Again the method uses the collapsed adjacency matrix of each orbital graph to calculate the diameter, as outlined in Sect. 2.

Table 11 Possible pairs (n, H) such that $(S_{n},\Omega )$ has an orbital graph of diameter 2, where $\Omega :=(S_{n}:H)$ and H is primitive

Proof of Theorem 1.2(3)

By Proposition 5.4, the only possible pairs (n, H) for which $\mathrm{diam_{O}}(G,\Omega )\le 5$ are those in Tables 9 and 10. Finally, magma has been used to verify (by implementing method CAM as in Sect. 2) that out of the possible cases, the only cases for which $\mathrm{diam_{O}}(G,\Omega )\le 5$ are the ones listed in the theorem. The data obtained is given in Appendix 1. $\square$

Table 12 Possible pairs (n, H) such that $(A_{n},\Omega )$ has an orbital graph of diameter 2, where $\Omega :=(A_{n}:H)$ and H is primitive

Proof of Theorem 1.3(3)

Using (22) with $d=2$, we obtain $n\le 24$ for both $G=S_{n}$ and $G=A_{n}$. We then use in-built functions in magma to run through values of $n\le 24$ and for each such n, iterate through the maximal subgroups of G to obtain only the primitive ones H which satisfy (21) with $d=2$. We conclude that if there exists an orbital graph of diameter 2, then n and H must be in Tables 11 or 12. Finally, magma has been used to verify (by implementing method CAM as in Sect. 2) that the only pairs (n, H) for which diameter 2 graphs occur are the ones stated in the theorem. $\square$

Appendix 1: Bounds on orbital diameters

Table 13 contains results (obtained using magma) which are used to verify that if $G:=S_{kl}$ or $A_{kl}$, then $\mathrm{diam}(G,I^{(k,l)})\le 5$ only for values of k and l listed in Theorem 1.2(2). The table specifies the values of k and l, the rank of the action and the value of $\mathrm{diam_{O}}(G, I^{(k,l)})$.

Table 13 Orbital diameters for $(G,I^{(k,l)})$

Tables 14 and 15 contain results (obtained using magma) which are used to verify that if $G:=S_{n}$ or $A_{n}$ and $\Omega :=(G:H)$, where H is primitive on $I=\{1,\ldots , n\}$ ($H\ne A_{n}$), then $\mathrm{diam}(G,\Omega )\le 5$ only for the pairs (n, H) listed in Theorem 1.2(3). The tables specify the pairs (n, H), the rank of the action, and the value of $\mathrm{diam_{O}}(G,\Omega )$.

Table 14 Orbital diameters for $(S_{n},\Omega )$, where $\Omega :=(S_{n}:H)$ and H is primitive

Table 15 Orbital diameters for $(A_{n},\Omega )$, where $\Omega :=(A_{n}:H)$ and H is primitive