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\title{\bf Embedding factorizations for 3-uniform hypergraphs II:  $r$-factorizations into $s$-factorizations}

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\author{Amin Bahmanian\\%\thanks{Supported by NASA grant ABC123.}\\
\small Department of Mathematics\\[-0.8ex]
\small Illinois State University\\[-0.8ex] 
\small Normal, Illinois, U.S.A.\\
\small 61790-4520\\
\and 
Mike Newman \thanks{Research supported by NSERC}\\
\small Department of Mathematics and Statistics\\[-0.8ex]
\small University of  Ottawa\\[-0.8ex]
\small Ottawa, Ontario, Canada\\
\small\tt mnewman@uottawa.ca
}


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\date{\dateline{November 17, 2015}{May 17, 2016}\\
\small Mathematics Subject Classifications: 05C70, 05C65, 05C15, 05B40, 05B05}

\begin{document}

\maketitle

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\begin{abstract}
 Motivated by a 40-year-old problem due to Peter Cameron on extending partial parallelisms,  we provide necessary and sufficient    conditions under which one can  extend  an $r$-factorization of a complete $3$-uniform hypergraph on $m$ vertices, $K_m^3$,  to an $s$-factorization of $K_n^3$. This generalizes an existing result of Baranyai and Brouwer--where they proved it for the case $r=s=1$. 



  \bigskip\noindent \textbf{Keywords:} factorizations; embedding; detachments; amalgamations; edge-colorings; hypergraphs
\end{abstract}

\section{Introduction}


 Let $V$ be a given finite set of cardinality $n$; the elements of $V$ will be called points. We denote the set of all $h$-subsets of $V$ by $\binom{V}{h}$. A {\it parallelism} of $\binom{V}{h}$ is a partition of $\binom{V}{h}$ whose classes are themselves partitions of $V$; the classes are called {\it parallel classes}. Note that a parallelism satisfies the usual Euclidean axiom for parallels: for every point $v \in V$ and for each $h$-subset $U$ of $V$, there is exactly one $h$-subset $U'$ which is parallel to $U$ (that is, contained in the same parallel class as $U$) and contains $V$. Obviously, a parallelism can exist only if $h$ is a divisor of $n$. It was  conjectured by Sylvester that this condition is sufficient as well, and Baranyai proved this conjecture \cite{MR0416986}. The direction of research in similar subjects such as Steiner triple systems and Latin squares for which general existence theorems have been proved suggests the following problem. 
\begin{question} \textup{(Cameron \cite[Question 1.2]{MR0419245})} \label{cameronbookprob}
 Under what conditions can partial parallelisms be extended to parallelisms?
\end{question}
There are different possible interpretations based on the precise notion of ``partial'' and ``extend''.  To formulate this problem  precisely, let us first introduce some basic terminology. 

Let $K_n^h$ denote the {\it complete $h$-uniform} hypergraph on $n$ vertices which is a hypergraph on $n$ vertices whose edges are all $h$-subset of the vertex set. An {\it $r$-factorization} of a hypergraph is a partition (coloring) of the edges into $r$-regular spanning sub-hypergraphs. The  following %more general and more precise 
formulation of Problem \ref{cameronbookprob} was investigated by the first author and Rodger in \cite{MR3056885}:
\begin{question} \label{arbitToRfac} %\textup{(Bahmanian, Rodger \cite[Question xxx]{MR3056885})}
 Under what conditions can arbitrary edge-colorings of $K_m^h$ be extended to $r$-factorizations of $K_n^h$?
\end{question}
In this direction, it has been proven that 
\begin{theorem}\textup{(Bahmanian, Rodger \cite[Theorem 3.1]{MR3056885})}\label{facembedgen1}
Suppose that  $n\geq (2+\sqrt{2})m$. Then a $q$-edge-coloring of $\mathcal F=K_m^3$ can be extended to an $r$-factorization of $K_{n}^3$ if and only if 
\begin{itemize}
\item [\textup {(i)}] $3\divides rn,$
\item [\textup {(ii)}]  $r \divides \binom{n-1}{2},$ 
\item [\textup {(iii)}] $q\leq\binom{n-1}{2}/r,$ and 
\item [\textup {(iv)}] $d_{\mathcal F(j)}(v)\leq r$ for each $v\in V(\mathcal F)$ and $1\leq j \leq q.$ 
\end{itemize}
\end{theorem}
Here $d_{\mathcal F(j)}(v)$ is the degree of vertex $v$ in the sub-hypergraph of $\mathcal F$ induced by color $j$. 

In this paper, we investigate the following formulation of Cameron's problem which is a special case of Problem \ref{arbitToRfac}:  We are given $K_n^h$ which has $K_m^h$ as a sub-hypergraph, and the edges of $K_m^h$ have been colored so that the degree of each vertex within each color class is $r$ (so that we have an $r$-factorization of $K_m^h$). Can we color the remaining edges of $K_n^h$ so as to achieve an $s$-factorization of $K_n^h$? 
\begin{question} \label{r-fac2s-fac} %\textup{(Bahmanian, Rodger \cite[Question xxx]{MR3056885})}
 Under what conditions can an $r$-factorization of $K_m^h$ be extended to an $s$-factorization of $K_n^h$?
\end{question}
Baranyai and Brouwer \cite{BaranBrouwer77} conjectured that a 1-factorization of $K_m^h$ can be extended to a 1-factorization of $K_n^h$ if and only if $h$ divides $m,n$, and $n\geq 2m$. They proved this for $h=2,3$, and for arbitrary $h$ when $n$ is sufficiently large. 
This conjecture of Baranyai and Brouwer was beautifully settled by H{\"a}ggkvist and Hellgren \cite{MR1249714}.
\begin{theorem}\textup{(H{\"a}ggkvist, Hellgren \cite[Theorem 2]{MR1249714})}\label{HaggHell}
Let $n=qt$ and $m=pt$, where $p\leq q/2$. Suppose that we are given a coloring of a subgraph $K_m^t$, using $\binom{m-1}{t-1}$ colors. Then this coloring can be extended to a coloring of $K_n^t$ using $\binom{n-1}{t-1}$ colors.
\end{theorem}

In an attempt to generalize this result and extend Theorem \ref{facembedgen1} for larger values of $h$, in an earlier paper we showed that  %\cite{BahNewm1} 
\begin{theorem}\textup{(Bahmanian, Newman \cite[Theorem 1.7]{BahNewm1})}\label{thm:main-r}
If  $\gcd(m,n,h) = \gcd(n,h)$, then an $r$-factorization of $K_m^h$ can be extended to an $r$-factorization of $K_n^h$ if and only if
\item [\textup {(G1)}] $h\divides rm$,  $h\divides rn;$  
\item [\textup {(G2)}]  $r\divides\binom{m-1}{h-1}$, $r\divides\binom{n-1}{h-1};$
\item [\textup {(G3)}] $n\geq 2m$.
\end{theorem}
In this paper, we completely  solve Problem \ref{r-fac2s-fac} for   $h=3$, which can be seen as an improvement of Theorem \ref{facembedgen1} for the case when the arbitrary edge-coloring of $K_m^h$ is replaced by a regular edge-coloring (see Theorem  \ref{embedfac1}). Studying embedding factorization of graphs dates back to over 40 years ago, see for example the classical paper by Cruse \cite{MR0329925}, and its extensions by Andersen and Hilton \cite{MR584118}. For results concerning embedding connected factorization of graphs we refer the reader to \cite{MR1983358, MR2325799, MR1315436}. 

This paper is organized as follows. In Section \ref{neccsec}, we discuss the necessary conditions. In Section \ref{detachsec}, we give the prerequisites, and in Section \ref{emb3graphsec}, we prove our main result.
 
\section{General Necessary Conditions} \label{neccsec}
Throughout this paper we assume that $m,n,r,s,h\in \mathbb N$. Moreover, in order to avoid trivial cases we assume that 
\begin{equation}\label{basicassump}
h\geq 2, \mbox{  and  } n> m>h.
\end{equation}
\begin{lemma} \label{embedfac1nec}
If an $r$-factorization of $K_m^h$ can be embedded into an $s$-factorization of $K_{n}^h$, then
\begin{itemize}
\item [\textup {(N1)}] $h\divides rm$,  $h\divides sn;$  
\item [\textup {(N2)}]  $r\divides\binom{m-1}{h-1}$, $s\divides\binom{n-1}{h-1};$
\item [\textup {(N3)}] $1\leq  s/r\leq \binom{n-1}{h-1}/\binom{m-1}{h-1};$
\item [\textup {(N4)}] $n\geq \frac{h}{h-1}m $ if  $1<s/r< \binom{n-1}{h-1}/\binom{m-1}{h-1};$
\item [\textup {(N5)}] $n\geq 2m$ if $s=r.$
\end{itemize}
\end{lemma}
\begin{proof}
Suppose that an $r$-factorization of $K_m^h$ can be embedded into an $s$-factorization of $K_{n}^h$. The degree sum of  each $r$-factor in an $r$-factorization of $K_m^h$ is $rm$, which must be divisible by the size of each edge, $h$. On the other hand the degree of each vertex in $K_m^h$ is $\binom{m-1}{h-1}$ which must be divisible by $r$. A similar argument shows that $h\divides sn$, and $s\divides\binom{n-1}{h-1}$. This proves (N1) and (N2). 

Let $q=\binom{m-1}{h-1}/r, k=\binom{n-1}{h-1}/s$. One can think of an $r$-factorization of $K_m^h$ as a $q$-edge-coloring in which each color class induces an $r$-factor. So we are extending a $q$-edge-coloring of $K_m^h$ to a $k$-edge-coloring of $K_n^h$ by extending each $r$-factor in $K_m^h$ to an $s$-factor in $K_n^h$, thus $s\geq r$ and $k\geq q$. In other words, $1\leq  s/r\leq \binom{n-1}{h-1}/\binom{m-1}{h-1}$. This proves (N3).   

For convenience, let us refer to the vertices in $V(K_n^h)\backslash V(K_m^h)$ as  the new vertices, the edges in $E(K_n^h)\backslash E(K_m^h)$ as  the new edges, and the colors in $\{q+1,\dots, k\}$ as new colors if $k>q$. 

Let $e_j$ be the number of edges of color $j$ in $K_m^h$ for $1\leq j\leq k$.  In an $s$-factorization of $K_n^h$, each of the $n-m$ new vertices   is adjacent with exactly $s$ edges of each color class, therefore all the $n-m$ new vertices are adjacent with at most $s(n-m)$ edges of each color class. Since in an $s$-factorization of $K_n^h$ the number of hyperedges of each color class is $sn/h$, for $1\leq j\leq k$ we have $$s(n-m)+e_j\geq sn/h.$$
 
If $1<s/r< \binom{n-1}{h-1}/\binom{m-1}{h-1}$ (or $s>r$ and $k>q$), then since $e_j=0$ for $q+1\leq j\leq k$,  we have $s(n-m)\geq sn/h$ which  proves (N4).


If $s/r=1$ (or $s=r$), fix a color $j\in\{1,\dots,q\}$. Since $r=s$,  there is no edge colored $j$ between $V(K_m^h)$ and the new vertices. Therefore, in order to to form an $s$-factor in $K_n^h$, there must be $r(n-m)/h$ edges colored $j$ in $K_{n-m}^h$ (the subgraph induced by the new vertices). But the total number of edges in $K_{n-m}^h$ is $\binom{n-m}{h}$. Therefore
$$\binom{n-m}{h}\geq \frac{\binom{m-1}{h-1}}{r}\frac{r(n-m)}{h}.$$
Thus $\frac{h}{n-m}\binom{n-m}{h}\geq \binom{m-1}{h-1}$ which implies $\binom{n-m-1}{h-1}\geq \binom{m-1}{h-1}$, and so $n-m-1\geq m-1$, and so $n\geq 2m$. This proves (N5) and the proof is complete. 
\end{proof}
\begin{remark}\textup{ Note that if $1=  s/r= \binom{n-1}{h-1}/\binom{m-1}{h-1}$, then $n=m$ which is a trivial case. 
}\end{remark}

\section{Fair Detachments of Hypergraphs} \label{detachsec}
If $x,y\in \mathbb R$, by $x \approx y$ we mean that $\lfloor y \rfloor \leq x \leq  \lceil y \rceil$. For the purpose of this paper, a {\it hypergraph} $\mathcal G$ is a pair $(V(\mathcal G),E(\mathcal G))$ where $V(\mathcal G)$ is a finite set called the {\it vertex} set, $E(\mathcal G)$ is the {\it edge} multiset, where every edge is itself a multi-subset of $V(\mathcal G)$. This means that not only can an edge  occur multiple times in $E(\mathcal G)$, but also each vertex can have multiple occurrences within an edge.  By an edge of the form $\{u_1^{m_1},u_2^{m_2},\dots,u_r^{m_r}\}$, we mean an edge in which vertex $u_i$ occurs $m_i$ times for $1\leq i\leq r$. 
The total number of occurrences of a  vertex $v$ among all edges of $E(\mathcal G)$ is called the {\it degree}, $d_{\mathcal G}(v)$ of $v$ in $\mathcal G$. The {\it multiplicity} of an edge $e$ in $\mathcal G$, written $m_{\mathcal G} (e)$, is the number of repetitions of $e$ in $E(\mathcal G)$ (note that $E(\mathcal G)$ is a multiset, so an edge may appear multiple times). If $\{u_1^{m_1},u_2^{m_2},\dots,u_r^{m_r}\}$ is an edge in $\mathcal G$, then we abbreviate $m_{\mathcal G}(\{u_1^{m_1},u_2^{m_2},\dots,u_r^{m_r}\})$ to $m_{\mathcal G}(u_1^{m_1},u_2^{m_2},\dots,u_r^{m_r})$. 
If $U_1,\dots,U_r$ are multi-subsets of $V(\mathcal G)$, then $m_{\mathcal G}(U_1,\dots,U_r)$ means $m_{\mathcal G}(\bigcup_{i=1}^r U_i)$, where the union of $U_i$s is the usual union of multisets. Whenever it is not ambiguous, we drop the subscripts; for example we write $d(v)$ and $m(e)$ instead of $d_{\mathcal G}(v)$ and $m_{\mathcal G} (e)$, respectively. 

For a positive integer $h$, $\mathcal G$ is said to be $h$-{\it uniform} if $|e|=h$ for each $e\in E$.   For  a positive integer $r$, an $r$-factor in a hypergraph $\mathcal G$ is a spanning  $r$-regular sub-hypergraph, and an {\it $r$-factorization}  is a partition of the edge set of $\mathcal G$ into $r$-factors. 
The hypergraph $K_n^h:=(V,\binom{V}{h})$ with $|V|=n$ (by $\binom{V}{h}$ we mean the collection of all $h$-subsets of $V$) is called a  {\it complete} $h$-uniform hypergraph. 
A {\it $k$-edge-coloring} of $\mathcal G$ is a mapping $f:V(\mathcal G)\rightarrow C$ (often the set of colors $C$ is $\{1,\dots,k\}$) and color class $j$ of $\mathcal G$, written $\mathcal G(j)$, is the sub-hypergraph of $\mathcal G$ induced by the edges of color $j$. 


Let $\mathcal G$ be a hypergraph, let $U$ be
some finite set, and let $\Psi : V(\mathcal G) \to U$ be a surjective mapping.  The map $\Psi$ extends
naturally to $E(\mathcal G)$.  For $A \in E(\mathcal G)$ we define $\Psi(A) = \{\Psi(x) : x \in A\}$.
Note that $\Psi$ need not be injective, and $A$ may be a multiset.  Then we define the
hypergraph $\mathcal F$ by taking $V(\mathcal F)=U$ and $E(\mathcal F)=\{ \Psi(A) : A \in E(\mathcal G) \}$.  We say
that $\mathcal F$ is an {\it amalgamation} of $\mathcal G$, and that $\mathcal G$ is a {\it detachment} of
$\mathcal F$.  Associated with $\Psi$ is a function $g$ defined by $g(u)=|\Psi^{-1}(u)|$; to be
more specific we will say that $\mathcal G$ is a $g$-detachment of $\mathcal F$.  Then $\mathcal G$ has $\sum_{u\in
  V(\mathcal F)} g(u)$ vertices.  Note that $\Psi$ induces a bijection between the edges of $\mathcal F$ and
the edges of $\mathcal G$, and that this bijection preserves the size of an edge.  We adopt the
convention that it preserves the color also, so that if we amalgamate or detach an
edge-colored hypergraph the amalgamation or detachment preserves the same coloring on the
edges.  We make explicit a straightforward observation: Given $\mathcal G$, $V(\mathcal F)$ and $\Psi$ the
amalgamation is uniquely determined, but given $\mathcal F$, $V(\mathcal G)$ and $\Psi$ the detachment is in
general far from uniquely determined.

We need the following special case of a general result in \cite{MR2942724}.
\begin{theorem}\textup{(Bahmanian \cite[Theorem 4.1]{MR2942724})}\label{mainthhypgen1}
Let $\mathcal F$ be a $k$-edge-colored hypergraph and let $g:V(\mathcal F)\rightarrow {\mathbb N}$. Then there exists a  $g$-detachment $\mathcal G$ (possibly with multiple edges) of $\mathcal F$ whose edges are all sets, with amalgamation function $\Psi:V(\mathcal G)\rightarrow V(\mathcal F)$,  $g$  being the number function associated with $\Psi$, such that:
\begin{enumerate}
\item [\textup{(F1)}] $d_{\mathcal G(j)}(v) \approx d_{\mathcal F(j)}(u)/g(u)$ for each $u\in V(\mathcal F)$, each $v\in \Psi^{-1}(u)$ and $1\leq j\leq k;$
\item [\textup{(F2)}] $m_\mathcal G(U_1,\dots,U_r) \approx m_\mathcal F(u_1^{m_1},\dots,u_r^{m_r})/\Pi_{i=1}^r\binom {g(u_i)}{m_i} $ for  distinct $u_1,\dots,u_r\in V(\mathcal F)$ and  $U_i\subset \Psi^{-1}(u_i)$ with $|U_i|=m_i\leq g(u_i)$ for $1\leq i\leq r$. 
\end{enumerate}
\end{theorem}
An immediate consequence of Theorem \ref{mainthhypgen1} is the following that will be  most useful throughout this paper. 
\begin{corollary} \label{maindetachcor}
Let $\mathcal F$ be a hypergraph with vertex set $\{u,v\}$ such that $m(u^{i},v^{h-i})=\binom{m}{i}\binom{n-m}{h-i}$ for $0\leq i\leq h-1$. Then an $r$-factorization of $K_m^h$ can be embedded into an $s$-factorization of $K_n^h$ if and only if we can color the edges of $\mathcal F$ with $k$ colors so that 
\begin{equation}  \label{maindetachcoreq}
d_j(u) = \left \{ \begin{array}{ll}
m(s-r) & \mbox { for } 1\leq j\leq q,  \\
sm & \mbox { for } q+1\leq j\leq k, \mbox { if } k>q, \end{array} \right.\\
\end{equation}
\begin{equation}  \label{maindetachcoreq2}
 d_j(v)=s(n-m) \mbox{\quad   for }  1\leq j\leq k. 
\end{equation}
where $q=\binom{m-1}{h-1}/r, k=\binom{n-1}{h-1}/s$, and $q,k\in \mathbb N$.
\end{corollary}
\begin{proof} First, suppose that an $r$-factorization of $K_m^h$ can be embedded into an $s$-factorization of $K_n^h$.  By Lemma \ref{embedfac1nec}, $q,k$ both are integers. By removing the edges of $K_m^h$ from $K_n^h$, amalgamating those $m$ vertices in $K_n^h$ that belong to  $K_m^h$ into a single vertex $u$, and the remaining $n-m$ vertices of $K_n^h$ into a vertex $v$,   we obtain the hypergraph $\mathcal F$. The $k$-edge-coloring of $K_n^h$ (in which each color class is an $s$-factor) induces a $k$-edge-coloring in $\mathcal F$ that satisfies  (\ref{maindetachcoreq}) and (\ref{maindetachcoreq2}).

Conversely, suppose that an $r$-factorization of $K_m^h$ is given, and the edges of  $\mathcal F$ are colored with $k$ colors so that (\ref{maindetachcoreq}) and (\ref{maindetachcoreq2}) are satisfied. We show that we can embed the given $r$-factorization of $K_m^h$ into as $s$-factorization of $K_n^h$. 
Let $g:V(\mathcal F)\rightarrow {\mathbb N}$ with $g(u)=m$, $g(v)=n-m$. By Theorem \ref{mainthhypgen1}, there exists a $g$-detachment $\mathcal G$ of $\mathcal F$ such that:
\begin{enumerate} [(a)]
\item By (F1), for each  $w\in \Psi^{-1}(u)$
\begin{equation*} 
d_{\mathcal G(j)}(w) \approx d_{j}(u)/g(u)= \left \{ \begin{array}{ll}
m(s-r)/m=s-r & \mbox { for } 1\leq j\leq q,  \\
sm/m=s & \mbox { for } q+1\leq j\leq k, \mbox { if } k>q, \end{array} \right.\\
\end{equation*}
and for each $w\in \Psi^{-1}(v)$, 
\begin{equation*} 
d_{\mathcal G(j)}(w) \approx d_{j}(v)/g(v)=s(n-m)/(n-m)=s  \mbox { for }  1\leq j\leq k.
\end{equation*}
\item By (F2),  $m_\mathcal G(U,V) \approx \frac{m(u^{i},v^{h-i})} { {\binom {g(u)}{i}}{\binom {g(v)}{h-i}}}=\frac{\binom{m}{i}\binom{n-m}{h-i}}{\binom{m}{i}\binom{n-m}{h-i}}=1$ for   $U\subset \Psi^{-1}(u),V\subset \Psi^{-1}(v)$ with $|U|=i, |V|=h-i$, for $0\leq i\leq h-1$. 
\end{enumerate}
Let us assume that $V(K_m^h)=\Psi^{-1}(u)$, and think of the given $r$-factorization of $K_m^h$ as a $q$-edge-coloring of $K_m^h$ so that each color class induces an $r$-factor. Let $\mathcal H$ be a hypergraph whose vertex set is $V(\mathcal G)$, whose edges are $E(K_m^h)\cup E(\mathcal G)$, and its edges are colored according to the colors of edges of $K_m^h$ and $\mathcal G$.  Obviously, $\mathcal H$ contains an $r$-factorization of $K_m^h$. Moreover, the definition of $\mathcal H$ together with $(a)$ and (b) respectively implies that $d_{\mathcal H(j)}(x)=s$ for $1\leq j\leq k$, and $\mathcal H\cong K_n^h$. This completes the proof.
\end{proof}
\section{The Main Result} \label{emb3graphsec}
In order to prove our main result, let us first review the obvious necessary conditions. 
\begin{lemma} \label{embedfaclemma1}
If an $r$-factorization of $K_m^3$ can be embedded into an $s$-factorization of $K_{n}^3$, then
\begin{itemize}
\item [\textup {(C1)}] $3\divides rm$,  $3\divides sn;$  
\item [\textup {(C2)}]  $r\divides\binom{m-1}{2}$, $s\divides\binom{n-1}{2};$
\item [\textup {(C3)}] $1\leq  s/r\leq \binom{n-1}{2}/\binom{m-1}{2};$
\item [\textup {(C4)}] $n\geq 3m/2 $ if  $1<s/r< \binom{n-1}{2}/\binom{m-1}{2};$
\item [\textup {(C5)}] $n\geq 2m$ if $s=r;$
\item [\textup {(C6)}] $sm\binom{n-m}{2}\geq \binom{n-1}{2}$ if $m(s-r)$ is odd and $s/r= \binom{n-1}{2}/\binom{m-1}{2}$.
\end{itemize}
\end{lemma}
\begin{proof}
Taking $h=3$ in Lemma \ref{embedfac1nec} proves (C1)--(C5). To prove (C6), suppose $m(s-r)$ is odd and $s/r= \binom{n-1}{2}/\binom{m-1}{2}$. If by contrary, $m\binom{n-m}{2}< \binom{n-1}{2}/s$, and if $\mathcal F$ is the hypergraph described in  Corollary \ref{maindetachcor}, then there exists a color $j$ for which $m_j(u,v^2)=0$. Therefore, $m(s-r)=d_j(u)=2m_j(u^2,v)$, contradicting the fact that $m(s-r)$ is odd.
\end{proof}
For the rest of this section, we assume that (C1)--(C6) are satisfied, and that $$q:=\binom{m-1}{2}/r, k:=\binom{n-1}{2}/s$$.
\begin{remark}\textup{ A similar argument shows that it is necessary that
\begin{equation*}
m\binom{n-m}{2} \geq \left \{ \begin{array}{ll}
k & \mbox { if } m,s \mbox { are odd and } r \mbox { is even},  \\
q &  \mbox { if } m,r \mbox { are odd and } s \mbox { is even},  \\
k-q & \mbox { if } m,r,s \mbox { are odd}. \end{array} \right.
\end{equation*}
However, in Lemma \ref{elem3} we will show that in most cases, (C4) implies this general necessary condition. 
}\end{remark}

In order to prove that (C1)--(C6) are also sufficient for an $r$-factorization of $K_m^3$ to be embedded into an $s$-factorization of $K_{n}^3$, we need to prove a few elementary results.
\begin{lemma} \label{elem0} \leavevmode
\begin{itemize}
\item [\textup{(a)}] $m[\binom{n-1}{2}-\binom{m-1}{2}]=2(n-m)\binom{m}{2}+m\binom{n-m}{2}$
\item [\textup{(b)}] $(n-m)\binom{m}{2}=m\binom{n-1}{2}-\binom{n}{3}-2\binom{m}{3}+\binom{n-m}{3}$
\end{itemize}
\end{lemma}
\begin{proof} Let $\mathcal F$ be a hypergraph with vertex set $\{u,v\}$ such that  $m(u^{i},v^{3-i})=\binom{m}{i}\binom{n-m}{3-i}$ for $0\leq i\leq 2$.
Counting the degree of $u$ in two different ways proves (a). Using part (a), we have the following that proves (b).
{\small\begin{eqnarray*}
m\binom{n-1}{2}-\binom{n}{3}-2\binom{m}{3}+\binom{n-m}{3} & = & m\binom{m-1}{2}+2(n-m)\binom{m}{2}+m\binom{n-m}{2}-2\binom{m}{3}\\
&-&[\binom{m}{3}+\binom{n-m}{3}+(n-m)\binom{m}{2}+m\binom{n-m}{2}]\\
&-&2\binom{m}{3}+\binom{n-m}{3}\\
&=&(n-m)\binom{m}{2}.
\end{eqnarray*}}
\end{proof}
\begin{lemma} \label{elem1} 
\begin{equation}\label{maineq2rf2} 
(n-m)\binom{m}{2} \geq \left \{ \begin{array}{ll}
q(sm-sn/3-2rm/3)+(k-q)(sm-sn/3)\\% & \mbox { if } n\geq 3m/2,  \\
(k-q)(rm-rn/3) & \mbox { if } r=s. \end{array} \right.
\end{equation}
\end{lemma}
\begin{proof}
To prove the first inequality, we have
\begin{eqnarray*}\label{longeq0}
q(sm-sn/3-2rm/3)+(k-q)(sm-sn/3) & = & k(sm-sn/3)-q(2rm/3)\nonumber\\
%&=&ks(m-n/3)-qr(2m/3)\\
&=&(m-n/3)\binom{n-1}{2}-(2m/3)\binom{m-1}{2}\nonumber\\
&=&m\binom{n-1}{2}- \binom{n}{3}-2\binom{m}{3}\nonumber\\
&<&(n-m)\binom{m}{2},
\end{eqnarray*}
where the last inequality is true by Lemma \ref{elem0}(b). 

If $r=s$, then by (C5) $n\geq 2m$, and the following  proves the second inequality.
\small{\begin{eqnarray*}
(n-m)\binom{m}{2}&\geq& (k-q)(rm-rn/3)\\
&=&(m-n/3)[\binom{n-1}{2}-\binom{m-1}{2}]\iff \\
3(n-m)m(m-1)&\geq &(3m-n)[(n-1)(n-2)-(m-1)(m-2)]\\
&=& (3m-n)(n-m)(n+m-3)\iff \\
(n-m)[3m(m-1)-(3m-n)(n+m-3)]&\geq& 0 \iff\\
 (n-3)(n-m)(n-2m)&\geq& 0.
\end{eqnarray*}}
\end{proof}
\begin{lemma}\label{elem3} 
\begin{equation} \label{elem3eq}
(n-m)\binom{m}{2} \leq q\lfloor m(s-r)/2\rfloor+(k-q)\lfloor ms/2\rfloor 
\end{equation}
\end{lemma}
\begin{proof}
Let $\alpha=m(s-r)/2-\lfloor m(s-r)/2\rfloor, \beta=sm/2-\lfloor sm/2 \rfloor$. Note that $\alpha, \beta\in \{0,1/2\}$, and 
\begin{eqnarray*}
2q\lfloor m(s-r)/2\rfloor+2(k-q)\lfloor ms/2\rfloor & = & 2q[ m(s-r)/2-\alpha]+2(k-q)( ms/2-\beta)\\
&=&kms-qmr/-2\alpha q-2\beta(k-q)\\
&=&m\binom{n-1}{2}-m\binom{m-1}{2}-2\alpha q-2\beta(k-q)\\
&\mathop= \limits^{\scriptsize{\mbox {lem. }} \ref{elem0}(a)}&2(n-m)\binom{m}{2}+m\binom{n-m}{2}-2\alpha q-2\beta(k-q).   
\end{eqnarray*}
Therefore, (\ref{elem3eq}) is equivalent to 
\begin{equation}\label{maineq2rf1}
m\binom{n-m}{2}\geq 2\alpha q+2\beta(k-q).
\end{equation}
If $k>q$, there are four cases to consider. 
\begin{enumerate}[(a)]
\item $\alpha=\beta=0$: In this case $m$ is even or $r,s$ are even, and so  (\ref{maineq2rf1}) is equivalent to $m\binom{n-m}{2}\geq 0$ which is trivial. 
\item $\alpha=0,\beta=1/2$: In this case $m,s,r$ are odd, and so  (\ref{maineq2rf1}) is equivalent to $m\binom{n-m}{2}\geq k-q$, and we have
\small{ \begin{eqnarray}
m\binom{n-m}{2}&\geq&  \frac{\binom{n-1}{2}}{s}-\frac{\binom{m-1}{2}}{r} \iff \nonumber \\ 
 rsm\binom{n-m}{2} &\geq& r \binom{n-1}{2}-s\binom{m-1}{2} \nonumber \\
  &=&r[ \binom{m-1}{2} +\binom{n-m}{2} +(n-m)(m-1)  ] -s\binom{m-1}{2} \iff \nonumber \\
 \quad\quad\quad r(n-m)(m-1)&\leq & r\binom{n-m}{2}( sm-1 ) +(s-r)\binom{m-1}{2} \label{lastmsrodd}.   
\end{eqnarray}}
By (\ref{basicassump}) $m\geq 4$, but $m$ is odd, and so $m\geq 5$, which implies that $n-m\geq 3$. Therefore $\binom{n-m}{2} \geq n-m$, which proves (\ref{lastmsrodd}).  
\item  $\alpha=\beta=1/2$: In this case $m,s$ are odd and $r$ is even, and so (\ref{maineq2rf1}) is equivalent to $m\binom{n-m}{2}\geq k$, and we have
 \begin{eqnarray*}
 sm\binom{n-m}{2}\geq \binom{n-1}{2}=\binom{m-1}{2}+\binom{n-m}{2}+(n-m)(m-1) &\iff& \\
 (sm-1)\binom{n-m}{2} \geq  \binom{m-1}{2}+(n-m)(m-1)&\iff& \\
  (sm-1)(n-m)(n-m-1) \geq (m-1)(m-2)+2(n-m)(m-1)&=&\\
  =(m-1)(2n-m-2). 
\end{eqnarray*}
Since $m$ is odd, by (\ref{basicassump}) $m\geq 5$, and we have $m^2-4m+1\geq 0$ or $(m+1)(m-3)\geq 2(m-2)$. But $m$ is odd and so by (C4) $n-m\geq \frac{m+1}{2}$  which implies $2(n-m)(n-m-2)\geq \frac{m+1}{2}(\frac{m+1}{2}-2) \geq m-2$ and so $2(n-m)^2-4(n-m)\geq m-2$. Thus, $$2(n-m)(n-m-1)=2(n-m)^2-2(n-m)\geq 2n-m-2.$$ Since $r$ is even, and $s$ is odd, we have  $s>r\geq 2$. Therefore 
$$(sm-1)(n-m)(n-m-1)> 2(m-1)(n-m)(n-m-1)\geq  (m-1)(2n-m-2).$$

\item  $\alpha=1/2,\beta=0$: In this case $m,r$ are odd and $s$ is even , and thus  (\ref{maineq2rf1}) is equivalent to $m\binom{n-m}{2}\geq q$. So we need to show that 
$rm\binom{n-m}{2}\geq \binom{m-1}{2}$ or equivalently,  $rm(n-m)(n-m-1)\geq (m-1)(m-2)$. Since $m^2-4m+7 \geq 0$, we have $(m+1)(m-1) \geq 4m-8$, so $\frac{m+1}{2}(\frac{m+1}{2}-1) \geq m-2$, and since $r\geq 1$ and for $m$ odd by (C4), $n\geq m+\frac{m+1}{2}$, we have 
\begin{eqnarray*}
rm(n-m)(n-m-1)&>& (m-1)(n-m)(n-m-1)\\
& \geq& (m-1)\frac{m+1}{2}(\frac{m+1}{2}-1)\geq (m-1)(m-2).
\end{eqnarray*}
\end{enumerate}
If $k=q$, there are two cases to consider. 
\begin{enumerate}[(a)]
\item If $m(s-r)$ is even, then   (\ref{maineq2rf1}) is equivalent to $m\binom{n-m}{2}\geq 0$ which is trivial. 
\item If $m(s-r)$ is odd, then (\ref{maineq2rf1}) is equivalent to $m\binom{n-m}{2}\geq q$ which is true by (C6). 
\end{enumerate}
 \end{proof} 
Case $r=s$ of the following result is proved using a different method by the authors in \cite{BahNewm1}. 
%%%%%%%%%%%% MAIN RESULT %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem} \label{embedfac1}
An $r$-factorization of $K_m^3$ can be embedded into an $s$-factorization of $K_{n}^3$ if and only if
\begin{itemize}
\item [\textup {(C1)}] $3\divides rm$,  $3\divides sn;$  
\item [\textup {(C2)}]  $r\divides\binom{m-1}{2}$, $s\divides\binom{n-1}{2};$
\item [\textup {(C3)}] $1\leq  s/r\leq \binom{n-1}{2}/\binom{m-1}{2};$
\item [\textup {(C4)}] $n\geq 3m/2 $ if  $1<s/r< \binom{n-1}{2}/\binom{m-1}{2};$
\item [\textup {(C5)}] $n\geq 2m$ if $s=r;$
\item [\textup {(C6)}] $sm\binom{n-m}{2}\geq \binom{n-1}{2}$ if $m(s-r)$ is odd and $s/r= \binom{n-1}{2}/\binom{m-1}{2}$.
\end{itemize}
\end{theorem}
\begin{proof}
The necessity is obvious by Lemma \ref{embedfaclemma1}. To prove the sufficiency,  let $\mathcal F$ be a hypergraph with vertex set $\{u,v\}$ such that 
 $m(u^{i},v^{3-i})=\binom{m}{i}\binom{n-m}{3-i}$ for $i=0,1,2$.  By Corollary \ref{maindetachcor}, it is enough to find a $k$-edge-coloring of $\mathcal F$ such that (\ref{maindetachcoreq}) and (\ref{maindetachcoreq2}) are satisfied. In what follows, we find such a coloring. Observe that in any $k$-edge-coloring of $\mathcal F$,  for $1\leq j\leq k$ we have
\begin{eqnarray} \label{degreeid}
d_j(u)&=&2m_j(u^2,v)+m_j(u,v^2), \mbox { and } \nonumber \\ 
d_j(v)&=&2m_j(u,v^2)+m_j(u^2,v)+3m_j(v^3).
\end{eqnarray} 
There are two cases to consider. 
 
 \noindent {\bf Case 1. $\boldsymbol{s>r}$}: We color the edges of the form $\{u^2,v\}$ so that 
\begin{eqnarray} \label{f-edges1}
sm-sn/3-2rm/3 & \leq  m_j(u^2,v)  \leq &  m(s-r)/2  \quad\mbox {for } 1\leq j\leq q, \nonumber\\
sm-sn/3           & \leq   m_j(u^2,v)  \leq &  ms/2       \quad\quad \quad \mbox {for } q+1\leq j\leq k,  \mbox { if } k>q.
\end{eqnarray}
In order to show that such a coloring is possible, first note that $ms/2\geq sm-sn/3$ is equivalent to $n\geq 3m/2$, which is true  if $k>q$ (by (C4)). Moreover, 
$m(s-r)/2\geq sm-sn/3-2rm/3$ is equivalent to $n\geq \frac{m}{2}(3-r/s)$ which is true by (\ref{basicassump}), and the following sequence of equivalences.
\begin{eqnarray*}
n\geq \frac{m}{2}(3-r/s) =\frac{m}{2}[3-\binom{m-1}{2}/\binom{n-1}{2}]&\iff& \\
2n\geq 3m-\frac{m(m-1)(m-2)}{(n-1)(n-2)}&\iff& \\
2n(n-1)(n-2)\geq 3m(n-1)(n-2)-m(m-1)(m-2)&\iff&\\
2(n-m)(n-m-1)(2n+m-4)\geq 0.
\end{eqnarray*}

 Therefore, it is enough to show that 
\begin{equation*}
q(sm-sn/3-2rm/3)+(k-q)(sm-sn/3)\leq m(u^2,v) \leq  q\lfloor m(s-r)/2\rfloor+(k-q)\lfloor ms/2\rfloor,
\end{equation*}
which is true by Lemmas \ref{elem1} and \ref{elem3}. 
 
Then, we color the edges of the form $\{u,v^2\}$ so that 
\begin{equation*} 
m_j(u,v^2) =\left \{ \begin{array}{ll}
m(s-r)-2m_j(u^2,v) & \mbox { for } 1\leq j\leq q,  \\
sm-2m_j(u^2,v)& \mbox { for } q+1\leq j\leq k,  \mbox { if } k>q.  \end{array} \right.
\end{equation*}
This is possible, because by (\ref{f-edges1})  $m_j(u,v^2)\geq 0$ for $1\leq j\leq k$ , and 
\begin{eqnarray*}
\sum_{j=1}^k m_j(u,v^2) & = & qm(s-r)+sm(k-q)-2\sum_{j=1}^k m_j(u^2,v)\\
&=&ksm-qrm-2(n-m)\binom{m}{2}\\
&=&m\binom{n-1}{2}-m\binom{m-1}{2}-2(n-m)\binom{m}{2}\\
&\mathop= \limits^{\scriptsize{\mbox {lem. }} \ref{elem0}(a)}&m\binom{n-m}{2}=m(u,v^2).
\end{eqnarray*}
Finally, we color the edges of the form $\{v^3\}$ so that  
\begin{equation*} 
m_j(v^3) =\left \{ \begin{array}{ll}
sn/3-sm+m_j(u^2,v)+2rm/3 & \mbox { for } 1\leq j\leq q,  \\
sn/3-sm+m_j(u^2,v)& \mbox { for } q+1\leq j\leq k,  \mbox { if } k>q.  \end{array} \right.
\end{equation*}
This coloring is possible, because by (C1)  $m_j(v^3)$ is an integer for $1\leq j\leq k$, by (\ref{f-edges1}) $m_j(v^3)\geq 0$ for $1\leq j\leq k$, and
\begin{eqnarray*}
\sum_{j=1}^k m_j(v^3) & = & q (sn/3-sm+2rm/3)+(k-q)(sn/3-sm)+\sum_{j=1}^k m_j(u^2,v)\\
&=&(n-m)\binom{m}{2}+skn/3-skm+2qrm/3\\
&=&(n-m)\binom{m}{2}+n\binom{n-1}{2}/3-m\binom{n-1}{2}+2m\binom{m-1}{2}/3\\
&=&(n-m)\binom{m}{2}+\binom{n}{3}-m\binom{n-1}{2}+2\binom{m}{3}\\
&\mathop= \limits^{\scriptsize{\mbox {lem. }} \ref{elem0}(b)}&\binom{n-m}{3}=m(v^3).
\end{eqnarray*}
Using (\ref{degreeid}), we verify that  the described edge-coloring satisfies (\ref{maindetachcoreq}) and (\ref{maindetachcoreq2}). 
\begin{equation*} 
d_j(u) =\left \{ \begin{array}{ll}
2m_j(u^2,v)+m(s-r)-2m_j(u^2,v)=m(s-r) & \mbox { for } 1\leq j\leq q,  \\
2m_j(u^2,v)+sm-2m_j(u^2,v)=sm& \mbox { for } q+1\leq j\leq k,  \mbox { if } k>q.  \end{array} \right.
\end{equation*}
For $1\leq j\leq q$,
  \begin{equation*} 
d_j(v) =3(sn/3-sm+m_j(u^2,v)+2rm/3)+2(sm-rm-2m_j(u^2,v))+m_j(u^2,v)=s(n-m),
\end{equation*}
and for $q+1\leq j\leq k$, if $k>q$
\begin{equation*} 
d_j(v) =3(sn/3-sm+m_j(u^2,v))+2(sm-2m_j(u^2,v))+m_j(u^2,v)=s(n-m).
\end{equation*}

\noindent {\bf Case 2.   $\boldsymbol{r=s}$:} We color the edges of the form $\{u^2,v\}$ so that 
\begin{eqnarray} \label{f-edges2}
&m_j(u^2,v)  = &  0  \quad\quad\quad\mbox {for } 1\leq j\leq q, \nonumber\\
rm-rn/3            \leq &  m_j(u^2,v)  \leq &  rm/2        \quad \mbox {for } q+1\leq j\leq k.
\end{eqnarray}
In order to show that such a coloring is possible, first note that $rm/2\geq rm-rn/3$ is equivalent to $n\geq 3m/2$,  which is true by (C5). Therefore, it is enough to show that 
$$
(k-q)(rm-rn/3)\leq m(u^2,v) \leq  (k-q)\lfloor rm/2\rfloor,
$$
which is true by Lemmas \ref{elem1} and \ref{elem3}. 
 
Then, we color the edges of the form $\{u,v^2\}$ so that 
\begin{equation*} 
m_j(u,v^2) =\left \{ \begin{array}{ll}
0 & \mbox { for } 1\leq j\leq q,  \\
rm-2m_j(u^2,v)& \mbox { for } q+1\leq j\leq k.  \end{array} \right.
\end{equation*}
This is possible, because by (\ref{f-edges2})  $m_j(u,v^2)\geq 0$ for $1\leq j\leq k$ , and 
\begin{eqnarray*}
\sum_{j=1}^k m_j(u,v^2) & = & rm(k-q)-2\sum_{j=q+1}^k m_j(u^2,v)\\
&=&m\binom{n-1}{2}-m\binom{m-1}{2}-2(n-m)\binom{m}{2}\\
&\mathop= \limits^{\scriptsize{\mbox {lem. }} \ref{elem0}(a)}&m\binom{n-m}{2}=m(u,v^2).
\end{eqnarray*}
Finally, we color the edges of the form $\{v^3\}$ so that  
\begin{equation*} 
m_j(v^3) =\left \{ \begin{array}{ll}
r(n-m)/3 & \mbox { for } 1\leq j\leq q,  \\
rn/3-rm+m_j(u^2,v)& \mbox { for } q+1\leq j\leq k.  \end{array} \right.
\end{equation*}
This coloring is possible, because by (C1)  $m_j(v^3)$ is an integer for $1\leq j\leq k$, by (\ref{f-edges2}) $m_j(v^3)\geq 0$ for $1\leq j\leq k$, and
\begin{eqnarray*}
\sum_{j=1}^k m_j(v^3) & = & \sum_{j=1}^q m_j(v^3)+\sum_{j=q+1}^k m_j(v^3)\\
&=&qr(n-m)/3+(k-q)(rn/3-rm)+m(u^2,v)\\
&=&(n-m)\binom{m}{2}+2qrm/3+krn/3-krm\\
&=&(n-m)\binom{m}{2}+2\binom{m}{3}+\binom{n}{3}-m\binom{n-1}{2}\\
&\mathop= \limits^{\scriptsize{\mbox {lem. }} \ref{elem0}(b)}&\binom{n-m}{3}=m(v^3).
\end{eqnarray*}
Using (\ref{degreeid}), we verify that  the described edge-coloring satisfies (\ref{maindetachcoreq}) and (\ref{maindetachcoreq2}). 
\begin{equation*} 
d_j(u) =\left \{ \begin{array}{ll}
0 & \mbox { for } 1\leq j\leq q,  \\
2m_j(u^2,v)+rm-2m_j(u^2,v)=rm& \mbox { for } q+1\leq j\leq k.  \end{array} \right.
\end{equation*}
For $1\leq j\leq q$,
  \begin{equation*} 
d_j(v) =3r(n-m)/3=r(n-m),
\end{equation*}
and for $q+1\leq j\leq k$,
\begin{equation*} 
d_j(v) =3(rn/3-rm+m_j(u^2,v))+2(rm-2m_j(u^2,v))+m_j(u^2,v)=r(n-m).
\end{equation*}
Applying Corollary \ref{maindetachcor} to $\mathcal F$, completes the proof.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\subsection*{Acknowledgements}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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