2.1 Proof of Theorem 1.6

Let G be a graph on n vertices with girth at least $2k+1$ , and let G contain t 2-factors. We aim to show that G contains an $S_k$ -set of size at least $t/2^{n-1}$ . Orient each edge of G uniformly and independently to obtain a random directed graph $G'.$ Say that $G'$ respects a 2-factor F if no two edges in F have the same initial vertex or the same terminal vertex in $G'$ . Note that, for each component C of F (which is a cycles), the probability that any two edges in C satisfy this condition is $2\cdot 2^{-|V(C)|}=2^{1-|V(C)|}.$ Thus,

$$ \begin{align*}\mathbb P[G'\text{ respects }F]=\sum_C 2^{1-|V(C)|}=2^{c(F)}\cdot 2^{-n}\ge 2^{1-n},\end{align*} $$

where $c(F)$ is the number of components of F:

$$ \begin{align*}\mathbb E[|\{F:G'\text{ respects }F\}|]\ge t/2^{n-1}.\end{align*} $$

Taking some orientation which respects at least as many 2-factors as the expectation, we obtain a family $\mathcal F$ of at least $t/2^{n-1}$ directed 2-factors. To each of these, we associate the permutation $\pi _{F}\in \mathrm {Sym}(n)$ such that $\pi _{F}(i)=j$ if $(i,j)\in E(F).$ These permutations are all distinct, so if $A=\{\pi _{F}:F\in \mathcal F\},$ then $|A|\ge t/2^{n-1}$ .

Now suppose $\alpha _1,\ldots ,\alpha _k,\beta _1,\ldots ,\beta _k\in A$ satisfy

$$ \begin{align*}\alpha_k\dots\alpha_1=\beta_k\dots\beta_1.\end{align*} $$

Let $i\in [n]$ . For $\ell \in [0,k]$ , let $x_\ell =(\alpha _\ell \dots \alpha _1)(i)$ and $y_\ell =(\beta _\ell \dots \beta _1)(i)$ , so that $x_0=y_0=i$ and $x_k=y_k=(\alpha _k\dots \alpha _1)(i).$ Since $G'$ has no opposite edges and the $\alpha _\ell ,\beta _\ell $ have no cycles of length 1 or 2 in their decompositions, there is no pausing or backtracking in the walks $x_0\dots x_k$ and $y_0\dots y_k$ :

$$ \begin{align*}x_\ell\not\in\{x_{\ell-1},x_{\ell-2}\},\ y_\ell\not\in\{ y_{\ell-1},y_{\ell-2}\},\ \ell=2,\ldots,k.\end{align*} $$

This implies that, if for some $\ell <\ell '$ , we have $x_\ell =x_{\ell '}$ , then $G[\{x_\ell ,\ldots ,x_{\ell '}\}]$ contains a cycle, which contradicts that the girth of G is at least $2k+1$ . Thus, $x_0,\ldots ,x_k$ are all distinct and similarly so are $y_0,\ldots ,y_k$ . Moreover, $y_1=x_1$ , for otherwise $x_k=y_k$ implies that $G[\{x_0,\ldots ,x_k,y_0,\ldots ,y_k\}]$ contains a cycle, contradicting that the girth of G is at least $2k+1$ . Thus, $\alpha _1(i)=\beta _1(i)$ , and this holds for all i so that $\alpha _1=\beta _1$ . We obtain

$$ \begin{align*}\alpha_k\dots\alpha_2=\beta_k\dots\beta_2\end{align*} $$

and repeating the argument k times proves that for all $\ell $ , $\alpha _\ell =\beta _\ell $ .

2.2 Proof of Theorem 1.3

In part (a), we claim that the maximum size of an $S_k$ -set in $\mathrm {Sym}(n)$ is $(n!)^{1/k+O(1/\log n)}.$ The upper bound is immediate from the universal upper bound of $|\Gamma |^{1/k}$ , valid for any group $\Gamma $ . To prove the lower bound, we begin by describing a blow-up construction to obtain from an arbitrary group $\Gamma $ an $S_k$ -set in the symmetric group on the elements of $\Gamma $ .

Suppose $|\Gamma |=n$ and $A\subseteq \Gamma $ is an $S_k$ -set of size a. For $g\in \Gamma $ , we write $gA:=\{ga:a\in A\}$ . For a set X, we will denote by $\mathrm {Sym}(X)$ the group of permutations of X. Let

$$ \begin{align*}A'=\{\pi\in \mathrm{Sym}(\Gamma):\text{for all } x\in\Gamma,\ \pi(x)\in xA\}.\end{align*} $$

Let M be the $\Gamma \times \Gamma $ matrix where $M_{xy}=1$ if $x^{-1}y\in A$ and $M_{xy}=0$ otherwise. Then $A'$ is the set of permutations $\pi $ satisfying $M_{x\pi (x)}=1$ for all $x\in \Gamma $ , and so $|A'|=\mathrm {per}(M)$ . The matrix $M/a$ is doubly stochastic, so we can estimate $\mathrm {per}(M/a)$ using the Egorychev–Falikman theorem.

Theorem 2.1 (Egorychev–Falikman [Reference Egorychev17, Reference Falikman20])

If M is an $n\times n$ doubly stochastic matrix, then

$$ \begin{align*}\mathrm{per}(M)\ge\frac{n!}{n^n}\end{align*} $$

with equality if and only if M is the constant matrix $n^{-1}J$ .

We obtain

$$ \begin{align*}\mathrm{per}(M)= a^n\text{per}(M/a)\ge a^n\frac{n!}{n^n}\ge a^{n-O(n/\log n)},\end{align*} $$

where the last inequality holds as long as $a\geq n^\epsilon $ , for some fixed positive constant $\varepsilon $ and sufficiently large n (which will hold as we will obtain A using Theorem 1.2). Now we claim that $A'$ is an $S_k$ -set in $\mathrm {Sym}(n)$ . If $\alpha _1\dots \alpha _k=\beta _1\dots \beta _k$ with $\alpha _i,\beta _i\in A'$ , then for all $x\in \Gamma $ ,

$$ \begin{align*}\alpha_1\dots\alpha_k(x)=\beta_1\dots\beta_k(x).\end{align*} $$

By the definition of $A'$ , there exist $a_1,\ldots ,a_k,b_1,\ldots ,b_k\in A$ such that $\alpha _i(x)=xa_i$ and $\beta _i(x)=xb_i$ for all $i\in \{1,\ldots ,k\}$ . Thus,

$$ \begin{align*}xa_k\dots a_1=xb_k\dots b_1\end{align*} $$

implying

$$ \begin{align*}a_k\dots a_1=b_k\dots b_1\end{align*} $$

and so

$$ \begin{align*}(a_k,\ldots,a_1)=(b_k,\ldots,b_1).\end{align*} $$

In particular, $a_k=b_k$ implies $\alpha _k(x)=\beta _k(x)$ . This holds for all $x\in \Gamma $ , so $\alpha _k=\beta _k$ and thus $\alpha _1\dots \alpha _{k-1}=\beta _1\dots \beta _{k-1}$ . Repeating this argument k times, using the fact that the $S_k$ -set A is also an $S_{\ell }$ -set for $\ell <k$ , we find that $(\alpha _1,\ldots \alpha _k)=(\beta _1,\ldots ,\beta _k)$ .

We have shown that whenever such $\Gamma ,A$ exist for given n, we have

$$ \begin{align*}M_k(\mathrm{Sym}(n))\ge a^{n-O(n/\log n)}.\end{align*} $$

To obtain good $\Gamma ,A$ , we apply Theorem 1.2. If $n=(p^k-1)k$ for some prime p with $k|(p-1)$ , we may take $a\ge cn^{1/k}$ (where c depends only on k). Thus, for such n,

$$ \begin{align*}M_k(\mathrm{Sym}(n))\ge (cn^{1/k})^{n-O(n/\log n)}=n^{n/k-O(n/\log n)}=(n!)^{1/k-O(1/\log n)}.\end{align*} $$

Now let $n\in \mathbb N$ be arbitrary. We refer to the following density-of-primes result which will also be useful later.

Theorem 2.2 (Baker–Harman–Pintz [Reference Baker, Harman and Pintz4])

Let $\pi (x;q,a)$ denote the number primes $p\le x$ with $p\equiv a\pmod q$ . If $(a,q)=1$ , $x^{0.55+\varepsilon }\le M\le x/\log x$ , $q\le \log ^Ax$ (for constants $A,\varepsilon>0$ ) and x is large enough,

$$ \begin{align*}\frac{0.99 M}{\log x}<\pi(x;q,a)-\pi(x-M;q,a)<\frac{1.01M}{\log x}.\end{align*} $$

Proof Applying Theorem 2.2 with $a=1$ , $q=k$ , $x=(n/k+1)^{1/k}$ , and $M=x^{0.56}$ gives that for large $n,$ there is a prime

$$ \begin{align*}p\in\left((n/k+1)^{1/k}-(n/k+1)^{0.56/k},(n/k+1)^{1/k}\right].\end{align*} $$

Then

$$ \begin{align*}p^k\in \left(n/k+1-(n/k+1)^{1-0.43/k},(n/k+1)\right]\end{align*} $$

and so

$$ \begin{align*}(p^k-1)k\in\left(n-n^{1-0.42/k},n\right].\end{align*} $$

It is clear that $M_k(\mathrm {Sym}(n))$ is increasing in n since $\mathrm {Sym}(n)\subseteq \mathrm {Sym}(n+1)$ . By Claim 1, there exists $m\in (n-n^{1-0.42/k},n]$ of the form $m=(p^k-1)k$ for prime p and $k|(p-1)$ . Then

$$ \begin{align*}\begin{aligned} M_k(\mathrm{Sym}(n))&\ge (n-n^{1-0.42/k})!^{1/k-O(1/\log(n-n^{1-0.42/k}))}\\&\ge n^{-n^{1-0.42/k}/k-o(1)}(n!)^{1/k-O(1/\log n)}\\ &\ge(n!)^{1/k-O(1/\log n)}. \end{aligned}\end{align*} $$

In part (b), we claim that the maximum size of an $S_k$ -set in the alternating group $\mathrm {Alt}(n)$ is $(n!)^{1/k+O(1/\log n)}$ . The upper bound is immediate from $M_k(\mathrm {Alt}(n))\le |\mathrm {Alt}(n)|^{1/k}$ . For the lower bound, note by Theorem 1.3(a) that there exists an $S_k$ -set $A\subseteq \mathrm {Sym}({n-2})$ with $|A|=(n!)^{1/k+O(1/\log n)}.$ Consider A as a subset of $\mathrm {Sym}(n)$ in the natural embedding $\mathrm {Sym}({n-2})\subseteq \mathrm {Sym}(n)$ , and let $\tau $ be the transposition $((n-1)\ n)$ . If at least half the permutations in A are even, then $|A\cap \mathrm {Alt}(n)|\ge n/2$ . Otherwise, at least half the permutations in A are odd, implying that $|A\tau \cap \mathrm {Alt}(n)|\ge |A|/2$ . Clearly, $A\cap \mathrm {Alt}(n)$ is an $S_k$ -set. Since every $\alpha \in A$ fixes both $n-1$ and n, $\alpha $ commutes with $\tau $ . Thus, if $\alpha _1,\ldots ,\alpha _k,\beta _1,\ldots ,\beta _k\in A$ satisfy $\alpha _1\tau \dots \alpha _k\tau =\beta _1\tau \dots \beta _k\tau ,$ we have

$$ \begin{align*}\begin{aligned} \tau^k\alpha_1\dots\alpha_k=\alpha_1\tau\dots\alpha_k\tau=\beta_1\tau\dots\beta_k\tau=\tau^k\beta_1\dots\beta_k \end{aligned}\end{align*} $$

so $\alpha _1\dots \alpha _k=\beta _1\dots \beta _k$ , implying $A\tau \cap \mathrm {Alt}(n)$ is an $S_k$ -set. In either case, we have

$$ \begin{align*}M_k(\mathrm{Alt}(n))\ge |A|/2\ge((n-2)!)^{1/k-O(1/\log(n-2))}/2=(n!)^{1/k-O(1/\log n)}.\end{align*} $$

6.1 Proof of Theorem 1.7

We first show that $(1-o(1))n^{1/k}\le m^0(n,\mathcal F_k)$ . For the time being suppose that $n=m^k$ for some integer m. We consider the following n-vertex digraph, which is a well-known generalization of a construction of de Bruijn [Reference De Bruijn14]: let $G=(V,E)$ , where $V=[m]^k$ , the set of all k-tuples with coordinates in $[m]$ , and let $((x_1,\ldots ,x_k),(y_1,\ldots ,y_k))$ be an edge if and only if $(x_2,\ldots ,x_k)=(y_1,\ldots ,y_{k-1})$ . The digraph G is $\mathcal F_k$ -free, implying that $m^0(n,\mathcal F_k)\ge m=n^{1/k}$ . Now let $n\in \mathbb N$ be arbitrary. Let m be the minimum integer such that $m^k\ge n$ , and observe that $m^k=(1+o(1))n$ . Let G be the construction described above on $m^k$ vertices. By Lemma 3, there is a directed subgraph $G'\subseteq G$ on n vertices with $\delta ^0(G')\ge (1-o(1))\delta ^0(G)=(1-o(1))m\ge (1-o(1))n^{1/k}$ . Thus, $m^0(n,\mathcal F_k)\ge (1-o(1))n^{1/k}.$

We now show that $m^0(n,\mathcal F_k)\le m^+(n,\mathcal C_{k,k})$ . If a graph with minimum outdegree $\delta ^+\ge 1$ contains some $C_{\ell ,\ell }$ with $2\le \ell \le k$ , say formed by the directed paths $x_0,\ldots , x_{\ell }$ and $y_0,\ldots , y_\ell $ (where $x_0=y_0$ and $x_\ell =y_\ell $ ), then there exists some directed walk $z_\ell =x_\ell ,z_{\ell +1},\ldots ,z_k$ . Then $x_0,\ldots ,x_{\ell },z_{\ell +1},\ldots ,z_{k}$ and $y_0,\ldots ,y_\ell ,z_{\ell +1},\ldots ,z_k$ form a graph in $\mathcal F_k$ (we will use this fact of “extending the walks” frequently below). Thus, $m^+(n,\mathcal F_k)\le m^+(n,C_{\ell ,\ell })$ .

We now show that $m^+(n,\mathcal C_{k,k})\le (2k+o(1))m^+(n,\mathcal F_k)$ . Let G be an n-vertex $\mathcal C_{k,k}$ -free directed graph with minimum degree $\delta ^+$ . Our first step is to modify the graph G so that only certain orientations of short cycles are allowed. Given a closed walk $W=v_0e_0v_1e_1\dots v_{k-1}e_{k-1}v_0$ in the underlying graph of a directed graph, its type $t(W)$ is the absolute value of

$$ \begin{align*}|\{i:e_i=(v_i,v_{i+1})\}|-|\{i:e_i=(v_{i+1},v_i)\}|\end{align*} $$

with the sum $i+1$ taken modulo k; in other words, it is the “net number of forward steps” in the walk.

Proof Randomly partition the vertices of G as $V(G)=V_0\sqcup \cdots \sqcup V_{2h-1}$ so that each vertex v is assigned to one part $P(v)$ , uniformly and independently, and let $G'$ be the graph obtained by keeping only the edges from $V_i$ to $V_{i+1}\pmod {2h}$ . For each $v\in V(G),$ we have that $d_{G'}^+(v)=\sum _{w:(v,w)\in E(G)}\textbf {1}_{P(w)=P(v)+1}$ , where the $\textbf {1}_{P(w)=P(v)+1}$ are $d_G^+(v)$ independent Bernoulli random variables with parameter $1/(2h).$ The Chernoff bound [Reference Chernoff11] gives

$$ \begin{align*}\mathbb P\left[d_{G'}^+(v)<\frac{1-\varepsilon}{2h}d_G^+(v)\right]\le e^{-\frac{\varepsilon^2d_G^+(v)}{4h}}\le e^{-\frac{\varepsilon^2\delta^+}{4h}}.\end{align*} $$

Therefore,

$$ \begin{align*}\mathbb P\left[\exists v\ d_{G'}^+(v)<\frac{1-\varepsilon}{2h}d_G^+(v)\right]\le ne^{\frac{-\varepsilon^2\delta^+}{4h}}<1.\end{align*} $$

Thus, with positive probability $d_{G'}^+(v)\ge \frac {1-2\varepsilon }{2h}\delta ^+$ for every v. Now the definition of $G'$ guarantees that every cycle in $G'$ of length at most $2h-1$ has type $0$ .

Consider the graph $G'$ obtained from G by Lemma 4, with $h=k$ . We claim that $G'$ is $\mathcal F_k$ -free. For if $x_0,\ldots ,x_k$ and $y_0,\ldots ,y_k$ are two walks with the same initial and same terminal vertices, there exists a minimum i such that $x_i\ne y_i$ . If $x_i=y_{i'}$ for some $i'\ne i$ , then $x_0,\ldots ,x_i=y_{i'},y_{i'-1},\ldots ,y_0=x_0$ is a closed walk of type $\ne 0$ and of length at most $2k-1$ , contradicting the definition of $G'$ . So, $x_i\not \in \{y_0,\ldots ,y_k\}$ . There exists a minimum $j>i$ such that $x_j\in \{y_0,\ldots ,y_k\}$ . Let $x_j=y_{j'}$ . If $j=j'$ , then $x_i,\ldots ,x_j,y_{j-1},\ldots ,y_{i-1}$ is a $C_{j-i+1,j-i+1}$ , where $2\le j-i+1\le k$ , a contradiction. If $j\ne j'$ , then $j<k$ or $j'<k$ and so $x_0,\ldots ,x_j,y_{j'-1},\ldots ,y_0$ is a closed walk of type $\ne 0$ and length at most $2k-1$ , a contradiction. Thus,

$$ \begin{align*}m^+(\mathcal F_k)\ge\frac{1-\varepsilon}{2k}\delta^+(G)\end{align*} $$

and the inequality follows.

Finally, we show that $m^+(n,\mathcal F_k)\le n^{1/k}$ . Let G be an $\mathcal F_k$ -free graph with minimum outdegree $\delta ^+\ge 1$ . Let $v\in V$ . Let $L_i$ be the set of vertices x for which there exists a directed walk of length i from v to x. If $i<k$ and there exist distinct directed walks of length i in G with the same initial and same terminal vertices, then the condition $\delta ^+\ge 1$ allows us to extend the walks to length k while still having the same terminal vertices. Thus, G is also $\mathcal F_i$ -free for $i\le k$ . Hence, for $x,y\in L_i$ with $i\le k-1,$ we have $N^+(x)\cap N^+(y)=\emptyset $ , and so $|L_{i+1}|\ge \delta ^+|L_i|$ . It follows that

$$ \begin{align*}n\ge|L_k|\ge(\delta^+)^k.\end{align*} $$

6.2 Proof of Theorem 1.8

We first prove the lower bound

$$\begin{align*}\left(\frac{1}{(2\ell-2)^{1/2}}-o(1)\right)n^{1/2}\le m^0(n,C_{\ell,\ell}).\end{align*}$$

We begin by defining the digraphs that will prove this inequality.

Definition 6.1 Let $\ell ,m\in \mathbb N$ . We define a digraph $G=G_{\ell ,m}$ on a vertex set $V=V_0\sqcup \cdots \sqcup V_{\ell -2}\sqcup W_0\sqcup \cdots \sqcup W_{\ell -2}$ , where for each $i,$ we have $V_i=\{v_{ijk}:j,k\in [m]\}$ and $W_i=\{w_{ijk}:j,k\in [m]\}$ . The edges of G are defined as follows: for $0\le i\le \ell -3$ and $j,k\in [m],$ let

(2) $$ \begin{align}\begin{aligned}\text{for all } j'&\ (v_{ijk},v_{(i+1)j'k})\in E(G),\\ &\ (v_{(\ell-2)jk}, w_{0j'k}) \in E(G);\\ \text{for all } k'&\ (w_{ijk}, w_{(i+1)jk'})\in E(G),\\ &\ (w_{(\ell-2)jk}, v_{0jk'}) \in E(G) .\end{aligned}\end{align} $$

Figure 1:

The digraph $G_{\ell ,m}$ when $\ell =4$ and $m=2$ .

Figure 1 Long description

The diagram consists of five vertical columns of nodes, with each column containing four nodes arranged in two pairs (top pair and bottom pair). Directed arrows connect nodes from left to right.

* Column 1 (labeled W sub 2): Contains four nodes. Horizontal and diagonal arrows connect these to Column 2.

* Column 2 (labeled V sub 0): Contains four nodes. Each node has a horizontal arrow to the next column and a diagonal arrow that crosses to the opposite pair (top to bottom or bottom to top).

* Column 3 (labeled V sub 1): Contains four nodes. Similar to the previous stage, arrows connect horizontally and diagonally to Column 4.

* Column 4 (labeled V sub 2): Contains four nodes. Arrows connect horizontally and diagonally to Column 5.

* Column 5 (labeled W sub 0): Contains four nodes. Arrows connect horizontally and diagonally to a final unlabeled column on the far right.

* Column 6 (labeled W sub 1): The final visible set of four nodes receiving the edges from the previous stage.

The connections create a symmetrical butterfly-like pattern where paths converge and diverge across the horizontal axis.

Assume for a contradiction that G contains a $C_{\ell ,\ell }$ composed of the two directed paths $x_0,\ldots ,x_\ell $ and $y_0,\ldots ,y_{\ell }$ where $x_0=y_0$ and $x_\ell =y_\ell $ . By the symmetry of the sets $V_i$ and $W_i$ , we may assume that $x_0\in V_i$ for some $0\le i\le \ell -2$ . Then $V(C_{\ell ,\ell })\cap W_0=\{x_{\ell -1-i},y_{\ell -1-i}\}.$ Since $x_0=y_0$ and $x_{\ell -1-i}\ne y_{\ell -1-i}$ , the structure of $G[V_0\cup \dots V_{\ell -2}\cup W_0]$ guarantees that $x_{\ell -1-i}=w_{0jk}$ and $y_{\ell -1-i}=w_{0j'k}$ for some $j\ne j'$ . Now $x_\ell \in W_{i+1}$ (with the convention $W_{\ell -1}=V_0$ ). Then the structure of $G[W_0,\ldots ,W_{\ell -1}]$ implies that $x_\ell =w_{(i+1)jk'}$ and $y_\ell =w_{(i+1)j'k"}$ for some $k,k"$ ; but this contradicts $x_\ell =y_\ell $ (Figure 1).

Thus, $G_{\ell ,m}$ is a $C_{\ell ,\ell }$ -free digraph on $(2\ell -2)m^2$ vertices with minimum indegree and minimum outdegree m. This proves that for every m, $m^0((2\ell -2)m^2,C_{\ell ,\ell })\ge m$ . Given any $n\in \mathbb N$ , there is some $n'$ of the form $n'=(2\ell -2)m^2$ with $n'\in (n,n +o(n))$ . By applying Lemma 3, we obtain

$$ \begin{align*}m^0(n,C_{\ell,\ell})\ge(1-o(1))\left\lfloor \left(\frac{n'}{2\ell-2}\right)^{1/2}\right\rfloor\ge\left(\frac{1}{(2\ell-2)^{1/2}}-o(1)\right)n^{1/2}.\end{align*} $$

The second claim in Theorem 1.8, that $m^0(n,C_{\ell ,\ell })\le m^+(n,C_{\ell ,\ell })$ , is immediate.

Finally, we show that $m^+(n,C_{\ell ,\ell })\le (2\ell +o(1))n^{1/2}$ . Let G be an n-vertex $C_{\ell ,\ell }$ -free digraph with minimum outdegree $\delta ^+$ . Using Lemma 4, we pass to the directed graph $G'$ with $d:=\delta ^+(G')\ge \frac {1-\varepsilon }{2\ell }\delta ^+$ in which every closed walk of length at most $2\ell -1$ has type 0. Let $v\in V(G')$ , and note there is a set $L_1$ of d vertices in $N_{G'}^+(v)$ . Assume we have constructed a set $L_i$ ( $i\le \ell -2)$ of d vertices such that for any $x,y\in L_i$ , there are paths $P_1,P_2$ on i edges oriented from v to $x,y,$ respectively, so that

(3) $$ \begin{align} V(P_1),V(P_2)\subseteq\{v\}\cup L_1\cup\cdots\cup L_{i}\text{ and }V(P_1)\cap V(P_2)=\{v\}. \end{align} $$

For $x\in L_i$ , we have $|N^+(x)|\ge d$ so we can greedily choose distinct vertices $L_{i+1}=\{f(x):x\in L_i\}$ such that $(x,f(x))\in E(G')$ for all $x\in L_i$ . Moreover, we have $f(x)\not \in \{v\}\cup L_1\cup \cdots \cup L_i$ or else $G'$ would contain a cycle C of length at most $2i+1$ with $t(C)\ne 0$ , a contradiction. Thus, for $x,y\in L_i$ , we can extend the paths $P_1$ and $P_2$ by the edges $(x,f(x)),(y,f(y))$ to satisfy Equation (3). We arrive by induction at the set $L_{\ell -1}$ . If there exist $x,y\in L_{\ell -1}$ and $z\in V(G')$ such that $(x,z),(y,z)\in E(G')$ , then similarly to the above, we have $z\not \in \{v\}\cup L_1\cup \cdots \cup L_{\ell -1}$ . Thus, applying Equation (3) to the vertices $x,y$ and extending the paths by $xz,yz$ gives a copy of $C_{\ell ,\ell }$ , a contradiction. Hence, $N^+_{G'}(x)\cap N_{G'}^+(y)=\emptyset $ so

(4) $$ \begin{align} n\ge\left|\bigcup_{x\in L_{\ell-1}}N_{G'}^+(x)\right|\ge d\cdot d, \end{align} $$

which gives $\left (\frac {1-\varepsilon }{2\ell }\delta ^+\right )^2\le n$ and the result follows.

We observe that, in the digraph $G_{\ell ,m}$ , with $L_i$ defined as in the previous paragraph, we have $\left |\bigcup _{x\in L_{\ell -1}}N^+(x)\right |=n/(2\ell -2)$ , matching the bound in Equation (4) up to a constant factor.

6.3 Proof of Corollary 1

We say that Hamilton paths $H_1,H_2$ in $K_n$ are said to create a two-part $C_\ell $ if there exist nonempty paths $P_1\subseteq H_1$ , $P_2\subseteq H_2$ such that the edge union of $P_2$ and $P_2$ is isomorphic to $C_\ell $ . We recall that $\widehat M(n,\ell )$ denotes the maximum size of a family of Hamilton paths in $K_n$ which pairwise create a two-part $C_\ell $ . Our goal is to show that if $\ell \ge 4$ is even, then $\widehat M(n,\ell )\le (n!)^{1/2+O(1/\log n)}.$

Let $\ell =2r$ . Assume for the time being that $(2r-2)(2r+1)|n$ . First, we count Hamilton cycles in the digraph $G=G_{r,m}$ from Definition 6.1 with $m^2=n/(2r-2)$ . Note that $G[V_0,\ldots ,V_{\ell -1}]$ and $G[W_0,\ldots ,W_{\ell -1}]$ are each m disjoint copies of a blowup of a directed $P_{\ell -1}$ . Let $X_1,\ldots ,X_m$ be the components of $G[V_0,\ldots ,V_{\ell -1}]$ and let $Y_1,\ldots ,Y_m$ be the components of $G[W_0,\ldots ,W_{\ell -1}]$ . A transition vector is a word

$$ \begin{align*}t=X_{f(1)}Y_{g(1)}X_{f(2)}Y_{g(2)}\ldots X_{f(m^2)}Y_{g(m^2)}\end{align*} $$

with properties

• • $f,g:[m^2]\to [m];$

• • for each $i,j\in [m]$ , the contiguous subwords $X_iY_j$ and $Y_jX_i$ each occur exactly once (we consider the vector cyclically, so that $Y_{f(m^2)}X_{f(1)}$ is a contiguous subword);

• • $f(1)=g(1)=1$ .

(The importance of the second property comes from the fact that, for any $X_i$ and $Y_j$ , there are exactly two vertices in $X_i\cap Y_j$ , one in $V_0$ and one in $W_0$ . As we will see below, the second property is used to guarantee that certain walks associated with the transition vector visit each vertex in $V_0\cup W_0$ exactly once.) A transition vector is equivalent to an Eulerian circuit in the bidirected $K_{m,m}$ . To enumerate Eulerian circuits, we refer to the famous result of de Bruijn, van Aardenne-Ehrenfest, Smith, and Tutte. If G is a digraph and $v\in V(G)$ , then an oriented spanning subtree of G with root v is a spanning subdigraph T of G with the following properties: (1) the underlying graph of T is a tree and (2) for any vertex $u\in V(G)$ , T contains a directed walk from u to v.

Theorem 6.1 (de Bruijn, van Aardenne-Ehrenfest, Smith, and Tutte [Reference de Bruijn and van Aardenne-Ehrenfest15, Reference Tutte and Smith40])

Let H be a strongly connected digraph in which every vertex v has $d^+(v)=d^-(v)$ . Let $t_v(H)$ denote the number of oriented spanning subtrees with root v. Then for any $v\in V(H),$ the number of Eulerian circuits of H is

$$ \begin{align*}\mathrm{ec}(H)=t_v(H)\prod_{v\in V(H)}(d^+(v)-1)!.\end{align*} $$

There are $m^{2(m-1)}$ spanning trees of $K_{m,m}$ [Reference Abu-Sbeih1]. For any fixed vertex $v\in V(K_{m,m})$ , each such spanning tree has a unique orientation with root v, obtained by orienting every edge so that the head is closer to v than the tail is. By Theorem 6.1, we conclude that there are $m^{2(m-1)}(m-1)!^{2m}$ transition vectors. We say that a directed Hamilton cycle in G

$$ \begin{align*}H=v_{0j_1k_1}\dots w_{0j_2k_1}\dots v_{0j_2k_2}\dots w_{0j_3k_2}\dots\dots v_{0j_1k_1}\end{align*} $$

(making no assumptions on the hidden portions of the vertex sequence) follows the transition vector t if $v_{0j_sk_s}\in X_{f(s)}$ and $w_{0j_{s+1}k_s}\in Y_{g(s)}$ for every $s\in [m^2]$ , that is, $k_s=f(s)$ and $j_{s+1}=g(s)$ (see Figure 2).

Claim 2 For any transition vector $t,$ there are exactly $(m!)^{2m(r-2)}$ directed Hamilton cycles in G which follow t.

Figure 2:

The solid lines describe a directed Hamilton cycle in $G_{4,2}$ which follows the transition vector $X_1Y_1X_2Y_2X_1Y_2X_2Y_1$ .

Figure 2 Long description

The diagram consists of nodes represented by black dots connected by solid and dashed directed arrows. The nodes are arranged in four horizontal rows and six vertical columns. The columns are labeled at the bottom from left to right as W sub 2, V sub 0, V sub 1, V sub 2, W sub 0, and W sub 1.

At column V sub 0, the top two nodes are labeled X sub 1 and X sub 2, and the bottom two nodes are also labeled X sub 1 and X sub 2. At column W sub 0, the top two nodes are labeled Y sub 1 and Y sub 1, while the bottom two nodes are labeled Y sub 2 and Y sub 2.

The solid lines indicate the directed Hamilton cycle. Starting from the left at W sub 2, the path enters the top two rows horizontally. Between W sub 2 and V sub 0, the path crosses in the bottom two rows. From V sub 0 to V sub 1, all four rows move horizontally. Between V sub 1 and V sub 2, the path remains horizontal in the top and bottom rows but crosses diagonally between the middle two rows. From V sub 2 to W sub 0, the path moves horizontally. Between W sub 0 and W sub 1, the path crosses diagonally in the top two rows and moves horizontally in the bottom two rows. Dashed lines represent other possible directed edges in the graph that are not part of this specific cycle.

Note that every Hamilton cycle follows exactly one transition vector. Therefore, the number of Hamilton cycles in G is

$$ \begin{align*}\begin{aligned}m^{2(m-1)}(m-1)!^{2m}m!^{2m(r-2)}=m^{(2r-2)m^2+O(m^2/\log m)}=n^{n/2+O(n/\log n)}.\end{aligned}\end{align*} $$

For each Hamilton cycle G, we delete a single edge and take the resulting directed path; thus, there is a family $\mathcal P$ of $n^{n/2+O(n/\log n)}$ directed Hamilton paths in G. Suppose $P,Q\in \mathcal P$ and their underlying paths create a 2-part $\ell $ -cycle C. Since $G_{r,m}$ contains no $C_{r,r}$ , and $C_{r,r}$ is the unique 2-part cycle of type 0, we have $t(C)\ne 0$ . We will filter out these remaining $\ell $ -cycles. Partition $[n]$ into equal parts $[n]=N_0\sqcup \cdots \sqcup N_{2r}$ . Let $\Sigma $ be the set of all directed Hamilton paths in $K_n$ starting in $N_0$ and whose $i^{th}$ vertex belongs to $N_{i\pmod {2r+1}}.$ Clearly, no two paths in $\Sigma $ create an $\ell $ -cycle of type $\ne 0$ , and

$$ \begin{align*}|\Sigma|=\left(\frac{n}{2r+1}\right)^{2r+1}\left(\frac{n}{2r+1}-1\right)^{2r+1}\dots(1)^{2r+1}=(n/(2r+1))!^{2r+1}=n^{n+O(n/\log n)}.\end{align*} $$

Let $\pi $ be a random relabeling of $[n]$ , then taking an outcome in which $|\pi \mathcal P\cap \Sigma |$ is at least average, we obtain a family $\mathcal P':=\pi \mathcal P\cap \Sigma $ of directed Hamilton paths in $K_n$ no two of which create any two-part cycle, with $|\mathcal P'|=n^{n/2+O(n/\log n)}$ . To convert this to an upper bound on $\widehat M(n,\ell ),$ we refer to a folklore lemma about vertex-transitive graphs (see, e.g., [Reference Godsil and Royle23], Lemma 7.2.2).

Lemma 5 If a graph G is vertex-transitive, then

$$ \begin{align*}\alpha(G)\omega(G)\le|V(G)|.\end{align*} $$

Consider the auxiliary graph whose vertices are all Hamilton paths in $K_n$ where two paths are adjacent if they create a two-part $\ell $ -cycle. The set of underlying graphs of the elements of $\mathcal {P}'$ corresponds to an independent set (and, by construction, every element of $\mathcal P'$ has a different underlying graph). Applying Lemma 5 to this auxiliary graph, we obtain

$$ \begin{align*}\widehat M(n,\ell)\le\frac{n!}{n^{n/2+O(n/\log n)}}=(n!)^{1/2+O(1/\log n)}.\end{align*} $$

Now consider general $n\in \mathbb N$ . Note that $\widehat M(n,\ell )$ is increasing. Thus, taking the smallest $n'>n$ satisfying $(2r-2)(2r+1)|n'$ gives

$$ \begin{align*}\widehat M(n,\ell)\le (n+O(1))!^{1/2+O(1/\log n)}=(n!)^{1/2+O(1/\log n)}.\end{align*} $$