\section{Introduction}
Throughout the paper, we let $n\geq 2$ denote a positive integer and let $N=\{1, \dots, n\}$.\footnote{We exclude case $n=1$ as the only group action on the set \{1\} is the trivial action.}  By a \textit{permutation group on n letters} we mean a subgroup $G$ of $S_n$ endowed with the natural action $(g,x) \mapsto gx \coloneqq g(x) : G \times N \to N$. We call $n$ the \emph{degree} of the permutation group $G$. The action of $G$ on $N$ induces an action of $G$ on $\mathscr{P}(N)$, given by $(g,X)\mapsto gX =\{gx: x\in X\} : G \times \mathscr{P}(N) \to \mathscr{P}(N)$. In this case, the elements being acted on are subsets of $N$. Accordingly, we shall call the orbits under this action \textit{set-orbits}. Note that for all $g \in G$ and $X \subseteq N$, $|gX|=|X|$. Since there are $n+1$ distinct sizes of subsets of $N$, it follows that there are at least $n+1$ distinct set-orbits. Additionally, this shows that all sets in the same set-orbit will have the same cardinality. A set-orbit containing sets of size $t$ will be called a $t$-\textit{set-orbit}. For a given permutation group $G$ on $n$ letters, the number of distinct $t$-set-orbits under the action of $G$ on $\mathscr{P}(N)$ will be denoted $s_t(G)$ and the total number of set-orbits will be denoted $s(G)$. Clearly $s(G)=\sum_{s=0}^ns_t(G)$.

\begin{define}[Beaumont and Peterson, \cite{BP55}]
Given an integer $0\leq t\leq n$, a permutation group $G$ on $n$ letters is called \emph{t-set-transitive} if for all $t$-element subsets $S,T \subseteq N$, there exists $g\in G$ such that $g S=T$.
\end{define}

In terms of the action of $G$ on $\mathscr{P}(N)$, we see that $G$ is $t$-set-transitive if and only if $s_t(G)=1$. In other words, a $t$-set-transitive group is a permutation group with exactly one $t$-set-orbit. Clearly, all permutation groups are $0$-set-transitive and $n$-set-transitive. A $1$-set-transitive permutation group is simply transitive on $N$.

\begin{define}[Beaumont and Peterson, \cite{BP55}]\label{def2.2}
A permutation group $G$ on $n$ letters is called \emph{set-transitive} if $G$ is $t$-set-transitive for all integers $0\leq t\leq n$.
\end{define}

Set-transitive groups were studied as early as 1944 by Neumann and Morgenstern \cite{MORNEU}. In \cite{BP55}, Beaumont and Peterson proved that a set-transitive permutation group on $n$ letters, with $n \notin \{5,6,9\}$, always contains the alternating group $A_n$. There has also been significant research devoted to bounding the number of set orbits $s(G)$ of a degree $n$ permutation group $G$. A trivial lower bound is $s(G) \geq 2^n/|G|$. In \cite{CAMERON}, Cameron proved that if $G$ has order $\exp(o(n^{1/2}))$, then $s(G)= (2^{n}/|G|)(1 + o(1))$. In \cite{BAPYBER}, Babai and Pyber showed that if $G$ does not contain $A_l$ ($l>t\geq 4$) as a composition factor, then $\frac {\log_2 s(G)} n \geq \frac {c} t$ for some positive constant $c$. In the same paper, they raised the following question: what is $\inf(\frac {\log_2 s(G)} n)$ over all solvable degree $n$ permutation groups $G$? This question was answered in a recent paper by the sixth author \cite{YY11}.\\
\indent On the other hand, there has been relatively little work done on the problem of classifying groups in terms of their number of set-orbits. Beaumont and Peterson successfully classified all set-transitive permutation groups in \cite{BP55}, and Kantor classified $2,3,4$-set transitive groups which are not $2,3,4$-transitive in \cite{KAN}. These classifications lend itself to a natural generalization in the following sense. Viewed in terms on the action of $G$ on $\mathscr{P}(N)$, we see that $G$ is set-transitive if and only if $s(G)=n+1$. Our paper seeks to classify completely the permutation groups $G$ on $n$ letters satisfying $s(G)=n+r$, for small positive integers $r$. The paper is laid out in the following manner. In Sections 2 and 3, we give necessary definitions and useful facts. In Section 4, we develop a general method mimicking the strategy in \cite{BP55}. In Section 5, we exemplify the method by classifying groups with $n+2$, $n+3$, $n+4$ and $n+5$ set-orbits. Then we use GAP to calculate all such groups for $r \leq 11$. Before continuing, we state the following useful facts from \cite{BP55} and \cite{LW65}. Given a permutation group $G$ on $n$ letters:
\begin{enumerate}
    \item If $G$ contains a $t$-set-transitive subgroup $H$, then $G$ is $t$-set-transitive.
    \item If $G$ is $t$-transitive, then $G$ is $s$-set-transitive for all positive integers $s\leq t$.
    \item The symmetric group $S_n$ is set-transitive.
    \item The alternating group $A_n$ is set-transitive for all $n\geq 3$.
    \item If $s_t(G) = 1$ for some $2 \leq t \leq \left \lfloor \frac{n}{2} \right \rfloor$, then $G$ is primitive.\footnote{Throughout this paper, we let $\lfloor x \rfloor$ be the floor function.}
\end{enumerate}

\section{Main Theorems}

Theorem 3 in \cite{BP55} states that if $G$ is $t$-set-transitive, then $G$ is $(n-t)$-set-transitive for any integer $1\leq t\leq n-1$. We generalize this result in the following theorem.

\begin{lem}
    Let $G$ be a permutation group on $n$ letters and let $0\leq t\leq n$ be an integer. Then $s_t(G)=s_{n-t}(G)$.
\end{lem}

\begin{proof}
  This is an easy observation.
\end{proof}
  


Note that there are $n+1$ distinct sizes of sets in $\mathscr{P}(N)$. Hence, if $n$ is odd, there are an even number of distinct set sizes, and so Lemma 3.1 implies that a permutation group on $n$ letters must have an even number of set-orbits. Therefore, if $n$ is odd and a permutation group $G$ on $n$ letters has $s(G)=n+r$, then $r$ will have to be odd as well. Now consider the situation where $n$ is even and a permutation group $G$ on $n$ letters has an odd number of set-orbits. In this case, since $s_t(G)=s_{n-t}(G)$ for all $0\leq t \leq n$, it follows that $s_{n/2}(G)$ must be odd. We summarize these results in the following lemma.

\begin{lem}\label{cor3.2}
Let $G$ be a permutation group on $n$ letters and suppose $s(G)=n+r$.
\begin{enumerate}
    \item If $r$ is even, then $n$ is even.
    \item If $r$ is odd and $n$ is even, then $s_{n/2}(G)$ is odd.
\end{enumerate}
\end{lem}

Observe the following theorem from \cite{LW65}.

\begin{thm}[Livingstone and Wagner, \cite{LW65}]
    Given a permutation group $G$ on $n$ letters and an integer $1\leq t\leq \frac{n}{2}$, we have $s_{t-1}(G) \leq s_t(G)$.
\end{thm}

\begin{remark}
This theorem implies that if $1\leq t <\left\lfloor \frac{n}{2} \right\rfloor - 1$ then $s_t(G) > 1$ only if $s_{t-1}(G) > 1$. We will make frequent use of this fact in our classification.
\end{remark}

\begin{lem}\label{lem3.4}
    Let $G$ be $\left (\left \lfloor\frac{n}{2} \right \rfloor+k \right)$-set-transitive for a positive integer $k$. If there exists a prime $p$ such that $\left \lfloor\frac{n}{2} \right \rfloor+k < p \leq n$, then $G$ is $(n-p+1)$-transitive.
\end{lem}

\begin{proof}
    By Theorem 7, Corollary 1 in \cite{BP55}, it suffices to show that $p> \max \left( \left \lfloor\frac{n}{2} \right \rfloor+k, n- \left \lfloor\frac{n}{2} \right \rfloor-k \right) = \left \lfloor\frac{n}{2} \right \rfloor+k$. This is true by assumption, so we are done.
\end{proof}

\begin{lem} \label{lem3.5}
    Let $k$ be a positive integer and let $G$ be a permutation group on $n$ letters that does not contain $A_n$. If there exists a prime $p$ such that $\left \lfloor\frac{n}{2} \right \rfloor+k < p < \frac{2n}{3}$, then $G$ is not $\left (\left \lfloor\frac{n}{2} \right \rfloor+k \right)$-set-transitive.
\end{lem}

\begin{proof}
    Assume for contradiction that $G$ is $\left (\left \lfloor\frac{n}{2} \right \rfloor+k \right)$-set-transitive, then by Lemma \ref{lem3.4} such a group $G$ is $\left(n-p+1 \right)$-transitive. Since $G$ does not contain $A_n$, $G$ is at most $(\frac{n}{3}+1)$-transitive (see \cite[page 152]{Burnside}). Now notice that $n-p+1 > n-\frac{2n}{3} + 1$ since $p<\frac{2n}{3}$ and since  $n-\frac{2n}{3} + 1= \frac{n}{3}+1$, we have reached a contradiction and thus are done.
\end{proof}

Note that both Lemmas \ref{lem3.4} and \ref{lem3.5} hold for $k=0$ when $n$ is even. If we find a maximum $k_0$ for which there exists a prime $p$ such that $\left\lfloor\frac{n}2 \right\rfloor + k_0 < p < \frac23 n$, then $G$ cannot be $\left\lfloor\frac{n}2 \right\rfloor + k$ set-transitive for any $k \leq k_0$. If $n$ is odd and such a $k_0$ exists, then $G$ is not $\left\lfloor\frac{n}2 \right\rfloor + 1$ set-transitive. Thus it is not $\left\lfloor\frac{n}2 \right\rfloor$ set-transitive either. These results are summarized in the following corollary.

\begin{cor}\label{cor3.6}
    Let $G$ be a permutation group on $n$ letters not containing $A_n$, where $n$ is even (odd). Let $k_0$ be the greatest non-negative (positive) integer such that there exists a prime $p$ with $\left\lfloor \frac{n}{2} \right\rfloor + k_0 < p < \frac{2n}{3}$. Then for all $0 \leq k \leq k_0$, $G$ is not $\left\lfloor\frac{n}2 \right\rfloor + k$ set-transitive.
\end{cor}
    \begin{proof}
        If $G$ is even, then the existence of such a non-negative $k_0$ shows that there exists a prime $p$ such that $\frac{1}{2}+k \leq \frac{n}{2} + k_0 < p < \frac{2n}{3}$. By Lemma \ref{lem3.5}, $G$ is not $\frac{n}{2} + k$ set transitive. For the case of $G$ being odd, a positive $k_0$ shows that there exists a prime $p$ such that $\left\lfloor\frac{n}{2} \right\rfloor+k \leq \left\lfloor  \frac{n}{2} \right\rfloor + k_0 < \frac{2n}{3} $. Since this holds at least for $k = 1$, it holds for $k = 0$ since $n - \left( \left\lfloor\frac{n}{2} \right\rfloor+1\right) = \left\lfloor\frac{n}{2} \right\rfloor$.
    \end{proof}


\begin{thm}\label{thm3.7}
    A permutation group $G$ on $n$ letters not containing $A_n$ is not $\left \lfloor \frac{n}{2} \right \rfloor + k$ set-transitive for any positive integer values of $k \leq k_0$ where $k_0$ is the largest integer such that $48 - \frac{n+1}{2} \leq k_0 \leq \frac{5}{54}n - \frac12$.
\end{thm}
    \begin{proof}
     First note that $48 \leq \frac{n+1}2 + k_0$. Due to a result by Breusch \cite{Breusch}, which states that there exists a prime between $x$ and $\frac{9x}8$ for $x \geq 48$, there exists a prime $p$ between $\frac{n+1}2 + k_0$ and $\frac9{16}n + \frac{9k_0}8 + \frac9{16}$. Note that $p >  \left \lfloor \frac{n}2 \right \rfloor + k$ for all $k \leq k_0$, and that $p < \frac9{16}n + \frac{9}8 k_0 + \frac9{16} \leq \frac{9}{16}n + \frac{9}{8}\left( \frac5{54}n - \frac12 \right) + \frac9{16} = \frac23 n.$ Thus by Lemma \ref{lem3.5}, $G$ cannot be $\left \lfloor \frac{n}{2} \right \rfloor + k$ set-transitive for any $k \leq k_0$.
    \end{proof}

\begin{remark}
It is worth pointing out that for $k = 0$ the theorem still holds true for even $n$. For odd $n$, $G$ is not $\left \lfloor\frac{n}{2}\right\rfloor$ set-transitive when a positive $k_0$ exists because $G$ will not be $\left\lfloor \frac{n}{2}\right\rfloor+1$ set-transitive.
This theorem is powerful as it shows that often sets of the same size lie in different orbits. The larger $n$ is, the more set-orbits $G$ will have. The theorem can be applied to make an upper bound on the amount of letters $G$ can permute and have exactly $n+r$ set-orbits. For example, the first value of $n$ for which we get an applicable $k_0$ is $n=81$, which gives $k_0 = 7$. This gives that a permutation group $G$ on $n$ letters not containing $A_n$ has at least $14$ additional set-orbits, which leads to a corollary.
\end{remark}

\begin{cor}
If a permutation group $G$ on $n$ letters that does not contain $A_n$ has less than $n+16$ set-orbits, then $n \leq 81$.
\end{cor}

\section{Outline of Methods}
    In this section we will outline a step-by-step method on how we fully classify groups with $n+r$ set-orbits for $2\leq r \leq 5$ which can also be applied to classify groups with $n+r$ set-orbits for even greater $r$. To outline our method we first reduce the amount of letters $n$ on which $G$ could act, then once we have a reasonable sized list we can test a number of specific permutation groups for the remaining $n$ values. Throughout the whole method we assume that any permutation group we consider does not contain $A_n$ because if it did then $s(G) = n+1$. We will exemplify how to do some of the steps in the calculation sections of this paper.

    \subsection*{Step 0}
        Choose the $r$ value for which you want to classify all groups with $s(G) = n+r$. Let $k_0 =\left\lfloor \frac{r-1}{2} \right\rfloor$ and find the smallest $n$ such that $k_0$ fits in the bounds specified by Theorem \ref{thm3.7}. The smallest $n$ for which any $k_0$ appears is $n=81$ with $k_0 = 7$. This is ``Step $0$" because for any reasonable $r$, i.e. $r<16$, we know that $n \leq 81$.
    \subsection*{Step 1}
        Now that we have an upper bound on $n$ we can start eliminating some of the possible $n$. Right away, if $r$ is even then we can eliminate all the odd $n$ by Lemma \ref{cor3.2}. Since we need $n+r$ set-orbits, we know that $G$ cannot be $s$ set-transitive for at most $r-1$ different set sizes $s$. This is where we can use Corollary \ref{cor3.6}. If $n$ is odd then we need a $k_0$ value such that there is a prime $\frac{n-1}{2}+ k_0 < p< \frac{2n}{3}$ and $2k_0> r-1$. If $n$ is even then we need a $k_0$ value such that there is a prime $\frac{n}{2} + k_0 < p < \frac{2n}{3}$ and $2k_0 + 1 > r-1$. This is because then we would know that $s_t(G) > 1$ for $2k_0$ different $s$ values in the odd case, and $2k_0+1$ different $s$ values in the even case (due to the fact that $s_\frac{n}{2}(G) = s_{n - \frac{n}{2}}(G)$). If we look at a table of primes and find any such $p$ values for the necessary $k_0$ for a given $n$, then we can remove that $n$ from the list of candidates.
    \subsection*{Step 2: Miller's Method}
        Now we look at our remaining $n$ values and apply a theorem of Miller \cite[vol. \RNum{3}, p. 439]{Miller}  which states that if $n=mp_0+r$, $p_0$ is prime, $m \in\mathbb{N}$, $p_0>m$, $r>m$, then a group $G$ on $n$ symbols, not containing $A_n$, cannot be more than $r$-transitive. We decompose $n$ so we have values of $m$, $p_0$, and $r$ that fit the conditions and we try to find a sufficiently small $r$. Using Lemmas \ref{lem3.4} and \ref{lem3.5}, we see if we can find a small enough $p$ such that $\left\lfloor  \frac{n}{2}\right\rfloor + k_1 < p \leq n $, then $n - p + 1 > r$ will contradict that $G$ is $\left\lfloor  \frac{n}{2}\right\rfloor + k_1$ set-transitive. For this, $k_1$ is the same as $k_0$ in the last step but it is not necessary that $\left\lfloor\frac{n}{2} \right\rfloor + k_1 < \frac{2n}{3}$. If we reach a contradiction, then we can remove that $n$ from the list.
    \subsection*{Step 3}
        After applying Miller's Method we have reduced the number of possible groups on $n$-letters that can have $n+r$ set orbits. We will now reduce the number of groups even further using the following argument. For a given $n$ and $r$ where $r<n-4$, if $s_2(G)>1$ then it follows that $s(G) > n+r$, in which case $G$ does not have set orbit $n+r$. Thus, we can assume $s_2(G) = 1$ which implies that $G$ is primitive by Theorem 6 in \cite{BP55}. Similarly, if $r<n-2$, $s_1(G)>1$ implies that $s(G) > n+r$, in which case $G$ does not have set orbits $n+r$. By assuming $s_1(G)=1$ in this case, we know that $G$ is transitive by Theorem 5 in \cite{BP55}. In these two cases, we will use \cite{TGps} to find the structure of the transitive groups. To reduce even further, we will introduce a fact from \cite{BP55} which follows simply from the orbit stabilizer theorem.
            \begin{fact}
                If a permutation group $G$ on $n$ letters is $s$ set-transitive, then $n \choose s$ divides $|G|$.
            \end{fact}

        \noindent Since we know $s_{k_1}(G) = 1$, then $n \choose {k_1}$ must divide $|G|$.  Under the above assumption, $G$ is primitive or transitive, in which case one only needs to check a few groups. If the above restriction on $r$ and $n$ does not hold, then we cannot use the fact that $s_t(G) = 1$ for any $t$. Under this situation, one must check all non-trivial subgroups of $S_n$. We can use GAP to compute these cases.

    \subsection*{Step 4: Computation} At this point, we have a list of possible $n$ for the degree of $G$, and a list of possible groups for each $n$. Now we can simply run a GAP program to calculate $s(G)$ for each candidate, and list out the ones that have $s(G) = n+r$ as desired.

    \begin{remark}
        We will do this process by hand for $r = 2,3,4,5$ to exemplify the process and then use a GAP program to go up to $r = 11$. Since we are going in a linear order for $r = 2,3,4,5$, we will often run into the same group twice. An example of this is when we consider all primitive groups on $8$ letters such that ${8 \choose 3} = 56$ divides $|G|$, and in a later section we consider all primitive groups such that ${8 \choose 2} = 28$ divides $|G|$. We will not consider the same groups twice if we have already calculated $s(G)$, but rather just list the groups we know have $s(G) = n+r$ from previous sections and then only consider the groups whose order is divisible by $28$ but not $56$. There will be several different instances where we can use previous knowledge to reduce the possible number of groups with $n+r$ set-orbits.
    \end{remark}

\section{Groups With Few Set-Orbits}

\subsection{Groups with $n+2$ set-orbits}

Assume a group $G$ on the set $N = \lbrace 1, \ldots, n\rbrace$ has $n+2$ set-orbits.

Looking at only even $n$ and applying Step 1 with $k_0=1$, we are left with the possibilities: $$n = 2,4,6,8,10,12, 14,16, 24.$$
To exemplify Step 2, we will show one of the calculations done. Since $24 = 1 \times 19 + 5$, then $G$ is at most $5$-transitive. Using $k_1 = 1$  we need to find the smallest prime $p$ such that $13 < p \leq  24$, so $p = 17$. Thus, if $G$ was $13$ set-transitive, then it would be $n-p+1 = 8$-transitive, which would contradict that it is at most $5$-transitive. Thus, we know a permutation group on $24$ letters cannot have $s(G) = n+2$.

We spare the reader from having to see any more of these calculations. At the end of this method, we are left with $$n = 2, 4, 6, 8, 12.$$

Before continuing to Step 3, we take care of the trivial case of $n=2$. The only permutation groups on $2$ letters are the trivial group and $S_2$. It happens that the trivial group has $s(G) = 4$, so we must include it. We will no longer consider $n=2$ for any $r$.

Now we move on to Step 3. We consider transitive groups on $4$ letters and primitive groups on $6,8,12$ letters such that $15,56,792$ divides the group orders, respectively.

Now we move on to Step 4, computation. We show all of the groups for which $s(G) = n+2$ in the following table. For the GAP ID we let nTr be TransitiveGroup(n,r), nPr be PrimitiveGroup(n,r). We let nSr be ConjugacyClassesSubgroups($S_n$)$[\mathrm{r}]$.

\begin{center}
\begin{tabular}{|c|c|c|c|}
   \hline
    $n$ & $G$ & $|G|$ & GAP ID \\ \hline
     $2$& $\{e\}$ & $1$ & 2S1 \\ \hline
     $4$ & $C_4$ &$4$ & 4T1 \\ \hline
     $4$ & $D_8$ & $8$ & 4T3 \\ \hline
     6 & $PSL(2,5)$ & $60$ & 6P1 \\ \hline
     $8$ & $AGL(1,8)$ & $56$ & 8P1 \\ \hline
     $8$ & $A\Gamma L(1,8)$ & $168$ & 8P2 \\ \hline
     $8$ & $PGL(2,7)$ & $336$ & 8P5 \\ \hline
     $8$ & $ASL(3,2)$ & $1344$ & 8P3 \\ \hline
   $12$ & $M_{12}$ & $95040$ & 12P2 \\ \hline
\end{tabular}
\end{center}

\subsection{Groups with $n+3$ set-orbits}

From the previous computations we know that $C_2 \times C_2$ on $4$ letters and $PSL(2,7)$ on 8 letters have $s(G) = n+3$.

For Step 1, we use a $k_0$ value of 2 for odd $n$ and 1 for even $n$. We find primes in the necessary range and are left with $$n = 3,4 \dots, 16, 19, 23, 24, 25, 43.$$

Applying Step 2, we are left with $$n = 3,4,5,6,7,8,9,11,12.$$

For $n=3$, we check all subgroups of $S_3$. For $n=4,5$ we check the transitive groups. For $n=6,7,8,9,11,12$ we check primitive groups with order divisible by $15,21,56, 84,330, 792$, respectively. Note that for the even $n$ we are already done from the previous section. We show the results for all groups with $n+3$ set-orbits in a table, using the same GAP identification key as in the previous section.

\begin{center}

\begin{tabular}{|c|c|c|c|}
   \hline
    $n$ & $G$ & $|G|$ & GAP ID \\ \hline
     $3$& $C_2$ & $2$ & 3S2 \\ \hline
    $4$ & $C_2 \times C_2$ & $4$ & 4T2 \\ \hline
     $5$ & $C_5$ & $5$ & 5T1 \\ \hline
     5 & $D_{10}$ & $10$ & 5T2 \\ \hline
     $7$ & $AGL(1,7)$ & $42$ & 7P4 \\ \hline
     $7$ & $L(3,2)$ & $168$ & 7P5 \\ \hline
     $8$ & $PSL(2,7)$ & 168 & 8P4 \\ \hline
      $11$& $M_{11}$ & 7920 & 11P6 \\ \hline
\end{tabular}
\end{center}


\subsection{Groups with $n+4$ set-orbits}

We know that the same numbers we were unable to remove in the earlier sections will reappear. Since for even numbers we need $k_0 = 2$ the following $n$'s can no longer be removed using Step 1: $n=18,22,34,42$. So we must continue with $$n = 4, 6,8,10,12,14,16,18,22,24,34,42.$$

After Step 2 we have only $$n = 4,6,8,10,12.$$

So we must check all subgroups of $S_4$, the transitive groups on $6$ letters, and the primitive groups on $8,10,12$ letters that whose order is divisible by $28, 120, 495$. We do not recheck the primitive groups on $8$ letters divisible by 56.

The table of all groups with $s(G) = n+4$ is listed below.

\begin{center}

\begin{tabular}{|c|c|c|c|}
   \hline
    $n$ & $G$ & $|G|$ & GAP ID \\ \hline
     4& $C_3$ & $3$ & 4S4 \\ \hline
     4 & $S_3$ & $6$ & 4T8 \\ \hline
     $6$ & $C_3 \times S_3$ & 18 & 6T5 \\ \hline
     $6$ & $S_4$ & $24$ & 6T8 \\ \hline
     $6$ & $S_3 \times S_3$ & 36 & 6T9 \\ \hline
     $6$& $C_3^2 \rtimes C_4$ & 36 & 6T10 \\ \hline
     $6$ & $C_2 \times S_4$ & 48 & 6T11 \\ \hline
     6 & $C_3^2 \rtimes D_8$ & 72 & 6T13 \\ \hline
     $10$ & $PGL(2,9)$ & $720$ & 10P4 \\ \hline
     $10$ & $P\Gamma L(2,9)$ & $1440$ & 10P7 \\ \hline
\end{tabular}
\end{center}


\subsection{Groups with $n+5$ set-orbits}

From previous sections we have found five groups with $n+5$ set-orbits. For even $n$ we continue to use $k_0=2$, and for odd $n$ we take $k_0=3$. After Step 1 $\&$ 2 we have the same even numbers as in the above section, so $n=4,6,8,10,12$. The odd numbers after Step 1 will be the same as in the $n+3$ case, but now we must add $n=17,21,33,41$. We must apply Step 2 to the odd numbers $$n=5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 33, 41, 43.$$

After applying Step 2 we have $$n= 4,5,6,7,8,9,10,11,12.$$

Note that once again, we handled all the necessary computations for the even $n$ in the previous section. So we must check all subgroups of $S_5$, the transitive subgroups of $S_7$, and the primitive groups on $9,11$ letters whose order is divisible by $36,165$ but not $84, 330$, respectively. We list all groups with $n+5$ set-orbits in the table below.

\begin{center}

\begin{tabular}{|c|c|c|c|}
   \hline

 $n$ & $G$ & $|G|$ & GAP ID \\ \hline
     3& $e$ & $1$ & 3S1 \\ \hline
     4 & $C_2 \times C_2$ & 4 & 4S6 \\ \hline
     5 & $A_4$ & $12$ & 5S14 \\ \hline
     5 & $S_4$ & $24$ & 5S17 \\ \hline
     $6$ & $C_2 \times A_4$ & 24 & 6T6 \\ \hline
     $6$ & $S_4$ & $24$ & 6T7 \\ \hline
     $7$ & $C_7 \times C_3$ & 21 & 7P3 \\ \hline
     $9$& $ASL(2,3)$ & 216 & 9P6 \\ \hline
     $9$ & $AGL(2,3)$ & 432 & 9P7 \\ \hline
     $10$ & $M_{10}$ & 720 & 10P6 \\ \hline
\end{tabular}
\end{center}

\subsection*{Remaining Computations}
We indeed classify all the cases till $r=11$ using the same method. We will just list the result here.

Groups with $n+6$ set-orbits:

\begin{center}
\begin{longtable}{|c|c|c|c|}
   \hline
   $n$ & $G$ & $|G|$ & GAP ID \\ \hline
    4 &   $C_{2} $ & 2 & 4S2 \\ \hline
6 &   $A_{4} $ & 12 & 6S31 \\ \hline
6 &   $C_{5} \rtimes C_{4} $ & 20 & 6S38 \\ \hline
6 &   $A_{5} $ & 60 & 6S50 \\ \hline
6 &   $S_{5} $ & 120 & 6S53 \\ \hline
\end{longtable}
\end{center}

Groups with $n+7$ set-orbits:

\begin{center}
\begin{longtable}{|c|c|c|c|}
   \hline
   $n$ & $G$ & $|G|$ & GAP ID \\ \hline
    5 &   $C_{4} $ & 4 & 5S6 \\ \hline
5 &   $S_{3} $ & 6 & 5S10 \\ \hline
5 &   $C_{6} $ & 6 & 5S11 \\ \hline
5 &   $D_{8} $ & 8 & 5S12 \\ \hline
5 &   $D_{12} $ & 12 & 5S15 \\ \hline
6 &   $D_{12} $ & 12 & 6S33 \\ \hline
7 &   $S_{5} $ & 120 & 7S89 \\ \hline
7 &   $A_{6} $ & 360 & 7S93 \\ \hline
7 &   $S_{6} $ & 720 & 7S94 \\ \hline
8 &   $\left( \left( \left( C_{2} \times C_{2} \times C_{2} \times C_{2} \right) \rtimes C_{2} \right) \rtimes C_{2} \right) \rtimes C_{3} $ & 192 & 8T38 \\ \hline
8 &   $\left( \left( \left( C_{2} \times C_{2} \times C_{2} \right) \rtimes \left( C_{2} \times C_{2} \right) \right) \rtimes C_{3} \right) \rtimes C_{2} $ & 192 & 8T40 \\ \hline
8 &   $\left( \left( \left( C_{2} \times C_{2} \times C_{2} \times C_{2} \right) \rtimes C_{3} \right) \rtimes C_{2} \right) \rtimes C_{3} $ & 288 & 8T42 \\ \hline
8 &   $\left( \left( \left( \left( C_{2} \times C_{2} \times C_{2} \times C_{2} \right) \rtimes C_{2} \right) \rtimes C_{2} \right) \rtimes C_{3} \right) \rtimes C_{2} $ & 384 & 8T44 \\ \hline
8 &   $\left( \left( A_{4} \times A_{4} \right) \rtimes C_{2} \right) \rtimes C_{2} $ & 576 & 8T45 \\ \hline
8 &   $\left( A_{4} \times A_{4} \right) \rtimes C_{4} $ & 576 & 8T46 \\ \hline
8 &   $\left( S_{4} \times S_{4} \right) \rtimes C_{2} $ & 1152 & 8T47 \\ \hline
9 &   $\left( C_{3} \times C_{3} \right) \rtimes C_{8} $ & 72 & 9T15 \\ \hline
9 &   $\left( C_{3} \times C_{3} \right) \rtimes QD_{16} $ & 144 & 9T19 \\ \hline
12 &   $M_{11} $ & 7920 & 12P1 \\ \hline
\end{longtable}
\end{center}

Groups with $n+8$ set-orbits:

\begin{center}
\begin{longtable}{|c|c|c|c|}
   \hline
   $n$ & $G$ & $|G|$ & GAP ID \\ \hline
   4 &   $C_{2} $ & 2 & 4S3 \\ \hline
6 &   $C_{6} $ & 6 & 6S17 \\ \hline
8 &   $\left( \left( C_{2} \times C_{2} \times C_{2} \right) \rtimes \left( C_{2} \times C_{2} \right) \right) \rtimes C_{3} $ & 96 & 8S242 \\ \hline
8 &   $\left( \left( C_{2} \times C_{2} \times C_{2} \times C_{2} \right) \rtimes C_{2} \right) \rtimes C_{3} $ & 96 & 8S247 \\ \hline
8 &   $\left( \left( \left( C_{2} \times C_{2} \times C_{2} \right) \rtimes \left( C_{2} \times C_{2} \right) \right) \rtimes C_{3} \right) \rtimes C_{2} $ & 192 & 8S268 \\ \hline
8 &   $\left( \left( \left( C_{2} \times C_{2} \times C_{2} \times C_{2} \right) \rtimes C_{2} \right) \rtimes C_{3} \right) \rtimes C_{2} $ & 192 & 8S270 \\ \hline
8 &   $A_{7} $ & 2520 & 8S293 \\ \hline
8 &   $S_{7} $ & 5040 & 8S294 \\ \hline
12 &   $PSL(2,11) \rtimes C_{2} $ & 1320 & 12T218 \\ \hline
\end{longtable}
\end{center}

Groups with $n+9$ set-orbits:

\begin{center}
\begin{longtable}{|c|c|c|c|}
   \hline
   $n$ & $G$ & $|G|$ & GAP ID \\ \hline
   5 &   $C_{2} \times C_{2} $ & 4 & 5S5 \\ \hline
6 &   $C_{2} \times A_{4} $ & 24 & 6S40 \\ \hline
6 &   $S_{4} $ & 24 & 6S41 \\ \hline
6 &   $C_{2} \times S_{4} $ & 48 & 6S49 \\ \hline
7 &   $A_{5} $ & 60 & 7S81 \\ \hline
8 &   $\left( \left( C_{2} \times C_{2} \times C_{2} \times C_{2} \right) \rtimes C_{3} \right) \rtimes C_{2} $ & 96 & 8S240 \\ \hline
9 &   $\left( C_{3} \times C_{3} \right) \rtimes Q_{8} $ & 72 & 9S370 \\ \hline
9 &   $A_{8} $ & 20160 & 9S551 \\ \hline
9 &   $S_{8} $ & 40320 & 9S552 \\ \hline
10 &   $S_{6} $ & 720 & 10T32 \\ \hline

\end{longtable}
\end{center}

Groups with $n+10$ set-orbits:

\begin{center}
\begin{longtable}{|c|c|c|c|}
   \hline
   $n$ & $G$ & $|G|$ & GAP ID \\ \hline
6 &   $C_{5} $ & 5 & 6S14 \\ \hline
6 &   $S_{3} $ & 6 & 6S16 \\ \hline
6 &   $C_{3} \times C_{3} $ & 9 & 6S28 \\ \hline
6 &   $D_{10} $ & 10 & 6S29 \\ \hline
6 &   $\left( C_{3} \times C_{3} \right) \rtimes C_{2} $ & 18 & 6S35 \\ \hline
6 &   $C_{3} \times S_{3} $ & 18 & 6S37 \\ \hline
6 &   $S_{3} \times S_{3} $ & 36 & 6S45 \\ \hline
8 &   $GL(2,3) $ & 48 & 8S216 \\ \hline
10 &   $A_{6} $ & 360 & 10S1396 \\ \hline
10 &   $PSL\left( 2,8\right) $ & 504 & 10S1448 \\ \hline
10 &   $PSL\left(2,8 \right) \rtimes C_{3} $ & 1512 & 10S1539 \\ \hline
10 &   $A_{9} $ & 181440 & 10S1590 \\ \hline
10 &   $S_{9} $ & 362880 & 10S1591 \\ \hline
12 &   $PSL(2,11) $ & 660 & 12T179 \\ \hline
\end{longtable}
\end{center}

Groups with $n+11$ set-orbits:

\begin{center}
\begin{tabular}{|c|c|c|c|}
   \hline
   $n$ & $G$ & $|G|$ & GAP ID \\ \hline
   5 &   $C_{3} $ & 3 & 5S4 \\ \hline
5 &   $S_{3} $ & 6 & 5S9 \\ \hline
7 &   $D_{14} $ & 14 & 7S48 \\ \hline
7 &   $C_{2} \times \left( C_{5} \rtimes C_{4} \right) $ & 40 & 7S75 \\ \hline
7 &   $S_{5} $ & 120 & 7S87 \\ \hline
7 &   $C_{2} \times A_{5} $ & 120 & 7S88 \\ \hline
7 &   $C_{2} \times S_{5} $ & 240 & 7S92 \\ \hline
8 &   $SL(2,3) $ & 24 & 8S154 \\ \hline
9 &   $\left( C_{2} \times C_{2} \times C_{2} \right) \rtimes C_{7} $ & 56 & 9S355 \\ \hline
9 &   $\left( \left( C_{3} \times C_{3} \times C_{3} \right) \rtimes C_{3} \right) \rtimes C_{2} $ & 162 & 9S457 \\ \hline
9 &   $\left( \left( C_{3} \times C_{3} \times C_{3} \right) \rtimes C_{3} \right) \rtimes C_{2} $ & 162 & 9S458 \\ \hline
9 &   $\left( C_{2} \times C_{2} \times C_{2} \right) \rtimes \left( C_{7} \rtimes C_{3} \right) $ & 168 & 9S462 \\ \hline
9 &   $\left( \left( C_{3} \times C_{3} \times C_{3} \right) \rtimes C_{3} \right) \rtimes \left( C_{2} \times C_{2} \right) $ & 324 & 9S497 \\ \hline

9 & $PSL(3,2) \rtimes C_2$ & 336 & 9S499 \\ \hline

9 &   $\left( \left( \left( C_{3} \times C_{3} \times C_{3} \right) \rtimes \left( C_{2} \times C_{2} \right) \right) \rtimes C_{3} \right) \rtimes C_{2} $ & 648 & 9S522 \\ \hline
9 &   $\left( \left( \left( C_{3} \times C_{3} \times C_{3} \right) \rtimes \left( C_{2} \times C_{2} \right) \right) \rtimes C_{3} \right) \rtimes C_{2} $ & 648 & 9S524 \\ \hline
9 &   $\left( \left( \left( \left( C_{3} \times C_{3} \times C_{3} \right) \rtimes \left( C_{2} \times C_{2} \right) \right) \rtimes C_{3} \right) \rtimes C_{2} \right) \rtimes C_{2} $ & 1296 & 9S534 \\ \hline
9 &   $\left( C_{2} \times C_{2} \times C_{2} \right) \rtimes PSL \left( 3,2 \right) $ & 1344 & 9S535 \\ \hline
10 &   $\left( C_{5} \times C_{5} \right) \rtimes \left( \left( C_{4} \times C_{4} \right) \rtimes C_{2} \right) $ & 800 & 10S1496 \\ \hline
10 &   $\left( C_{2} \times C_{2} \times C_{2} \times C_{2} \right) \rtimes S_{5} $ & 1920 & 10S1542 \\ \hline
10 &   $C_{2} \times \left( \left( C_{2} \times C_{2} \times C_{2} \times C_{2} \right) \rtimes A_{5} \right) $ & 1920 & 10S1543 \\ \hline
10 &   $C_{2} \times \left( \left( C_{2} \times C_{2} \times C_{2} \times C_{2} \right) \rtimes S_{5} \right) $ & 3840 & 10S1561 \\ \hline
10 &   $\left( A_{5} \times A_{5} \right) \rtimes C_{2} $ & 7200 & 10S1569 \\ \hline
10 &   $\left( A_{5} \times A_{5} \right) \rtimes \left( C_{2} \times C_{2} \right) $ & 14400 & 10S1576 \\ \hline
10 &   $\left( A_{5} \times A_{5} \right) \rtimes C_{4} $ & 14400 & 10S1577 \\ \hline
10 &   $\left( A_{5} \times A_{5} \right) \rtimes D_{8} $ & 28800 & 10S1584 \\ \hline
11 &   $A_{10} $ & 1814400 & 11S3091 \\ \hline
11 &   $S_{10} $ & 3628800 & 11S3092 \\ \hline


\end{tabular}
\end{center}


 \section{Closing Remarks}

Now that a general method is developed for calculating all groups with $n+r$ orbits where $r$ is not too large. The GAP code used for the calculation is available at ~\cite{GAPcode}. We have successfully classified all the cases for $r \leq 11$, but it appears our personal computers cannot handle the case when $r=12$. We remark that by far the most computationally taxing step is finding all the subgroups of $S_n$. If one could come up with a method to circumvent this, the classification could go much farther. Another computationally taxing step is calculating how many set orbits a large group has. If both of these steps can be improved upon, the classification could go much further.

\section*{Acknowledgement}

This research was conducted under NSF-REU grant DMS-1757233 by the first, second, third, and fifth authors during
the Summer of 2019 under the supervision of the fourth and sixth authors. The authors gratefully acknowledge the financial support of NSF and also thank Texas State University for providing a great working environment and support. The fourth and sixth authors were also partially supported by grants from the Simons Foundation (\#280770 to Thomas M. Keller, \#499532 to Yong Yang).


