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\title{\bf The cycle polynomial of a permutation group}

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% the street address 

\author{Peter J. Cameron \\
\small School of Mathematics and Statistics\\[-0.8ex]
\small University of St Andrews\\[-0.8ex]
\small North Haugh \\[-0.8ex]
\small St Andrews, Fife, U.K.\\
\small\tt pjc20@st-andrews.ac.uk\\
\and
Jason Semeraro \\
\small Department of Mathematics\\[-0.8ex]
\small University of Leicester\\[-0.8ex]
\small Leicester, U.K.\\
\small\tt jpgs1@leicester.ac.uk
}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{Sep 8, 2017}{Dec 28, 2017}{Jan 25, 2018}\\
\small Mathematics Subject Classifications: 20B05, 05C31}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
% succinct statement of background followed by a listing of the
% principal new results that are to be found in the paper. The abstract
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\begin{abstract}
 The cycle polynomial of a finite permutation group $G$ is the generating
function for the number of elements of $G$ with a given number of cycles:
\[F_G(x) = \sum_{g\in G}x^{c(g)},\]
where $c(g)$ is the number of cycles of $g$ on $\Omega$. In the first part
of the paper, we develop basic properties of this polynomial, and give a
number of examples.

In the 1970s, Richard Stanley introduced the notion of reciprocity for
pairs of combinatorial polynomials. We show that, in a considerable number
of cases, there is a polynomial in the reciprocal relation to the cycle 
polynomial of $G$; this is the \emph{orbital chromatic polynomial} of 
$\Gamma$ and $G$, where $\Gamma$ is a $G$-invariant graph, introduced by the
first author, Jackson and Rudd. We pose the general problem of finding all
such reciprocal pairs, and give a number of examples and characterisations:
the latter include the cases where $\Gamma$ is a complete or null graph or
a tree.

The paper concludes with some comments on other polynomials associated with
a permutation group.
\end{abstract}

\section{The cycle polynomial and its properties}\label{sec:cyclebasic}

 The \emph{cycle index} of a permutation group $G$ acting on a set
$\Omega$ of size $n$ is a polynomial
in $n$ variables which keeps track of all the cycle lengths of elements. If the variables are $s_1,\ldots,s_n$, then
the cycle index is given by
\[Z_G(s_1,\ldots,s_n) = \sum_{g\in G}\prod_{i=1}^n s_i^{c_i(g)},\]
where $c_i(g)$ is the number of cycles of length $i$ in the cycle decomposition
of $G$. (It is customary to divide this polynomial by $|G|$ but we prefer not to do so here.)

Define the \emph{cycle polynomial} of a permutation group $G$ to be $F_G(x):=Z_G(x,x,\ldots,x)$; that is
\[F_G(x) = \sum_{g\in G}x^{c(g)},\]
where $c(g)$ is the number of cycles of $g$ on $\Omega$ (including fixed
points). Clearly the cycle polynomial is a monic polynomial of degree $n$.

\begin{proposition}\label{p:polya}
If $a$ is an integer, then $F_G(a)$ is a multiple of $|G|$.
\end{proposition}

\begin{proof}
Consider the set of colourings of $\Omega$ with $a$ colours (that is, functions
from $\Omega$ to $\{1,\ldots,a\}$. There is a natural action of $G$ on this
set. A colouring is fixed by an element $g\in G$ if and only if it is constant
on the cycles of $g$; so there are $a^{c(g)}$ colourings fixed by $g$. Now the
orbit-counting Lemma shows that the number of orbits of $G$ on colourings is
\[\frac{1}{|G|}\sum_{g\in G}a^{c(g)};\]
and this number is clearly a positive integer. The fact that $f(a)$ is an integer for all $a$  follows from \cite[Proposition 1.4.2]{rps}.
\end{proof}

Note that the combinatorial interpretation of $F_G(a)/|G|$ given in the proof of Proposition \ref{p:polya} is the most common application of P\'{o}lya's theorem. 

\begin{proposition}
$F_G(0)=0$; $F_G(1)=|G|$; and $F_G(2)\ge(n+1)|G|$, with equality if and only
if $G$ is transitive on sets of size $i$ for $0\le i\le n$.
\end{proposition}

\begin{proof}
The first assertion is clear.

There is only one colouring with a single colour. 

If there are two colours, say red and blue, then the number of orbits on
colourings is equal to the number of orbits on (red) subsets of $\Omega$.
There are $n+1$ possible cardinalities of subsets, and so at least $n+1$
orbits, with equality if and only if $G$ is is transitive on sets of size $i$ for $0\le i\le n$. (Groups with this property are called set-transitive and were determined by Beaumont and
Petersen \cite{bp}; there are only the symmetric and alternating groups and
four others with $n=5,6,9,9$.)
\end{proof}

Now we consider values of $F_G$ on negative integers. Note that the \emph{sign} $\sgn(g)$ of the permutation $g$ is $(-1)^{n-c(g)}$; a permutation is even or odd 
according as its sign is $+1$ or $-1$. If $G$ contains odd permutations, then
the even permutations in $G$ form a subgroup of index~$2$.

\begin{proposition}\label{p:minus}
If $G$ contains no odd permutations, then $F_G$ is an even or odd function 
according as $n$ is even or odd; in other words,
\[F_G(-x)=(-1)^nF_G(x).\]
\label{p:parity}
\end{proposition}

\begin{proof}
The degrees of all terms in $F_G$ are congruent to $n$ mod~$2$.
\end{proof}

In particular, we see that if $G$ contains no odd permutations, then $F_G(x)$
vanishes only at $x=0$. However, for permutation groups containing odd
permutations, there may be negative roots of $F_G$. 

\begin{theorem}
Suppose that $G$ contains odd permutations, and let $N$ be the subgroup of
even permutations in $G$. Then, for any positive integer $a$,
we have $0\le(-1)^nF_G(-a)<F_G(a)$, with equality if and only if $G$ and $N$
have the same number of orbits on colourings of $\Omega$ with $a$ colours.
\end{theorem}

\begin{proof}
Let $\Delta$ denote the set of $a$-colourings of $\Omega$. Say that an orbit $\O$ of $G$ on $\Delta$ is split if the $G$- and $N$-orbits of $\omega$ do not coincide for some (and hence any) $\omega \in \O$. The number of split orbits is the number of $N$-orbits on $\Delta$ less the number of $G$-orbits on $\Delta$, namely $$\frac{1}{|G|}\sum_{g \in N} 2\cdot|\Fix_\Delta(g)|-\frac{1}{|G|}\sum_{g \in G} |\Fix_\Delta(g)|,$$ where $\Fix_\Delta(g)$ denotes the number of fixed points of $g$ on $\Delta$. This, in turn, is equivalent to $$\frac{1}{|G|} \sum_{g \in G} \sgn(g)a^{c(g)}=\frac{1}{|G|}(-1)^nF_G(-a),$$ and the result follows from this. 
%We have
%\begin{eqnarray*}
%\sum_{g\in N}a^{c(g)} + \sum_{g\in G\setminus N}a^{c(g)} &=& F_G(a),\\
%\sum_{g\in N}a^{c(g)} - \sum_{g\in G\setminus N}a^{c(g)} &=& (-1)^nF_G(-a).
%\end{eqnarray*}
%Moreover, $N$ has at least as many orbits on colourings as does $G$, and so
%we have
%\[\sum_{g\in N}a^{c(g)}/|N| \ge \sum_{g\in G}a^{c(g)}/|G|.\]
%So $F_N(a)$ is at least half of $F_G(a)$, showing that the left-hand side
%of the second equation above is at least zero. This proves the inequality;
%we see that equality holds if and only if $G$ and $N$ have equally many orbits
%on colourings, as required.
\end{proof}


\begin{proposition}
If $G$ is a permutation group containing odd permutations, then the set of
negative integer roots of $F_G$ consists of all integers $\{-1,-2,\ldots,-a\}$
for some $a\ge1$.
\end{proposition}

\begin{proof}
$F_G(-1)=0$, since $G$ and $N$ have equally many orbits (namely $1$) on
colourings with a single colour.

Now suppose that $F_G(-a)=0$, so that $G$ and $N$ have equally many orbits on
colourings with $a$ colours; thus every $G$-orbit is an $N$-orbit. Now every
colouring with $a-1$ colours is a colouring with $a$ colours, in which the
last colour is not used; so every $G$-orbit on colourings with $a-1$ colours
is an $N$-orbit, and so $F_G(-a+1)=0$. The result follows.
\end{proof}

The property of having a root $-a$ is preserved by overgroups:

\begin{proposition}\label{p:overgroups}
Suppose that $G_1$ and $G_2$ are permutation groups on the same set, with
$G_1\le G_2$. Suppose that $F_{G_1}(-a)=0$, for some positive integer $a$.
Then also $F_{G_2}(-a)=0$.
\end{proposition}

\begin{proof}
It follows from the assumption that $G_1$ (and hence also $G_2$) contains odd
permutations. Let $N_1$ and $N_2$ be the subgroups of even permutations in
$G_1$ and $G_2$ respectively. Then $N_2\cap G_1=N_1$, and so $N_2G_1=G_2$.
By assumption, $G_1$ and $N_1$ have the same orbits on $a$-colourings. Let
$K$ be an $a$-colouring, and $g\in G_2$; write $g=hg'$, with $h\in N_2$ and
$g'\in G_1$. Now $Kh$ and $Khg'$ are in the same $G_1$-orbit, and hence in the
same $N_1$-orbit; so there exists $h\in N_1$ with $Kg=Khg'=Khh'$. Since
$hh'\in N_2$, we see that the $G_2$-orbits and $N_2$-orbits on $a$-colourings
are the same. Hence $F_{G_2}(-a)=0$.
\end{proof}

The cycle polynomial has nice behaviour under direct product, which shows that
the property of having negative integer roots is preserved by direct product.

\begin{proposition}\label{p:prod}
Let $G_1$ and $G_2$ be permutation groups on disjoint sets $\Omega_1$ and
$\Omega_2$. Let $G=G_1\times G_2$ acting on $\Omega_1\cup\Omega_2$. Then
\[F_G(x)=F_{G_1}(x)\cdot F_{G_2}(x).\]
In particular, the set of roots of $F_G$ is the union of the sets of roots of
$F_{G_1}$ and $F_{G_2}$.
\end{proposition}

\begin{proof}
This can be done by a calculation, but here is a more conceptual proof. It
suffices to prove the result when a positive integer $a$ is substituted for
$x$. Now a $G$-orbit on $a$-colourings is obtained by combining a $G_1$-orbit
on colourings of $\Omega_1$ with a $G_2$-orbit of colourings of $\Omega_2$; so
the number of orbits is the product of the numbers for $G_1$ and $G_2$.
\end{proof}

The result for the wreath product, in its imprimitive action, is obtained in a
similar way.

\begin{proposition}\label{p:wreath}
\[F_{G\wr H}(x) = |G|^mF_H(F_G(x)/|G|).\]
\end{proposition}

\begin{proof} Again it suffices to prove that, for any positive integer $a$, the
equation is valid with $a$ substituted for $x$.

Let $\Delta$ be the domain of $H$, with $|\Delta|=m$. An orbit of the base
group $G^m$ on $a$-colourings is an $m$-tuple of $G$-orbits on $a$-colourings,
which we can regard as a colouring of $\Delta$, from a set of colours whose
cardinality is the number $F_G(a)/|G|$ of $G$-orbits on $a$-colourings. Then an
orbit of the wreath product on $a$-colourings is given by an orbit of $H$ on
these $F_G(a)/|G|$-colourings, and so the number of orbits is
$(1/|H|)F_H(F_G(a)/|G|)$. Multiplying by $|G\wr H|=|G|^m|H|$ gives the
result.
\end{proof}

\begin{corollary}
If $n$ is odd, $m > 1$, and $G=S_n\wr S_m$, then $F_G(x)$ has roots $-1,\ldots,-n$.
\end{corollary}

\begin{proof}
In Proposition \ref{p:sym} to come, we show that $F_{S_n}(x)=x(x+1)\cdots(x+n-1)$. Therefore $F_{S_n}(x)$  divides $F_G(x)$, so we have roots
$-1,\ldots,-n+1$. Also, there is a factor
\[|S_n|(F_{S_n}(x)/|S_n|+1) = x(x+1)\cdots(x-n+1)+n!.\]
Substituting $x=-n$ and recalling that $n$ is odd, this is $-n!+n!=0$.
\end{proof}

The next corollary shows that there are imprimitive groups with arbitrarily
large negative roots.
 
\begin{corollary}
Let $n$ be odd and let $G$ be a permutation group of degree $n$ which contains
no odd permutations. Suppose that $F_G(a)/|G|=k$. Then, for $m>k$, the
polynomial $F_{G\wr S_m}(x)$ has a root $-a$.
\end{corollary}

\begin{proof}
By Proposition~\ref{p:parity}, $F_G(-a)/|G|=-k$. Now for $m>k$, the
polynomial $F_{S_m}(x)$ has a factor $x+k$; the expression for $F_{G\wr S_m}$
shows that this polynomial vanishes when $x=-a$.
\end{proof}

The main question which has not been investigated here is:
\begin{quote}
What about non-integer roots?
\end{quote}

It is clear that $F_G(x)$ has no positive real roots; so if $G$ contains no
odd permutations, then $F_G(x)$ has no real roots at all, by
Proposition~\ref{p:parity}. When $G=A_n$ we have the following result which was communicated to us by Valentin F\'{e}ray \cite{F}.




\begin{theorem}
If $F_{A_n}(a)=0$ for some complex number $a$ then $\Re(a)=0$.
\end{theorem}

\begin{proof}
We show that this result is a particular case of \cite[Theorem 3.2]{stanley2}. In the notation of that theorem, set $d:=n-1$ and $g(t):=t^d+1$ so that the hypotheses are clearly satisfied and $m=0$. Now $$P(q)=(E^d+1) \prod_{i=0}^{n-1}(q+i) = \prod_{i=0}^{n-1}(q-i) + \prod_{i=0}^{n-1}(q+i)=2F_{A_n}(q),$$ where $E$ is the backward shift operator on polynomials in $q$ given by  $Ef(q)=f(q-1)$. \cite[Theorem 3.2(a)]{stanley2} now delivers the result.
\end{proof}



\section{Some examples}

\begin{proposition}\label{p:sym}
For each $n$ we have, $$F_{S_n}(x)=\prod_{i=0}^{n-1} (x+i).$$
\end{proposition}

\begin{proof}
By induction and Proposition \ref{p:overgroups}, $F_{S_n}(-a)=0$ for each $0 \le a \le n-2$. Since $F_{S_n}(x)$ is a polynomial of degree $n$ and the coefficient of $x$ in $F_{S_n}(x)$ is $(n-1)!$ we must have that $n-1$ is the remaining root.

Several other proofs of this result are possible.
We can observe that the number of orbits of $S_n$ on
$a$-colourings of $\{1,\ldots,n\}$ (equivalently, $n$-tuples chosen from the
set of colours with order unimportant and repetitions allowed) is
$n+a-1\choose n$. Or we can use the fact that the number of permutations of
$\{1,\ldots,n\}$ with $k$ cycles is the unsigned \emph{Stirling number of the
first kind} $u(n,k)$, whose generating function is well known to be
\[\sum_{k=1}^nu(n,k)x^k=x(x+1)\cdots(x+n-1).\qedhere\]
\end{proof}

\begin{proposition}
For each $n$ we have $$F_{C_n}(x)=\sum_{d \mid n} \phi(d)x^{n/d},$$ where $\phi$ is  Euler's totient function.
\end{proposition}

\begin{proof}
This is a consequence of the well-known formula for the cycle index of a cyclic group.
\end{proof}


\begin{proposition}\label{prop:fg}
Let $p$ be an odd prime and $G$ be the group $\PGL_2(p)$ acting on a set of size $p+1$. Then $F_G(x)$ is given by
$$\frac{p(p+1)}{2}  x^2F_{C_{p-1}}(x)+\frac{p(p-1)}{2} F_{C_{p+1}}(x) + (p^2-1)(x^2-x^{p+1})$$
\end{proposition}

\begin{proof}
Each semisimple element of $\GL_2(p)$ either has eigenvalues in $\mathbb{F}_p$ and lies in a torus isomorphic to $\mathbb{F}_p^\times \times \mathbb{F}_p^\times$ or else it has eigenvalues in $\mathbb{F}_{p^2} \backslash \mathbb{F}_p$ and lies in a torus isomorphic to $\mathbb{F}_{p^2}^\times$. If $T$ is either kind of torus then $N_{\GL_2(p)}(T)$ is generated by $T$ and an automorphism which inverts each element, so there are $|\GL_2(p)|/2|T|$ conjugates of $T$. Distinct tori intersect in the subgroup of scalar matrices $Z(\GL_2(p))$. Hence, ignoring the identity, the images of semisimple elements of $\GL_2(p)$ in $G$ can be partitioned into $|\GL_2(p)|/2(p-1)^2=p(p+1)/2$ tori of the first type, each generated by an element of cycle type $(p-1,1^2)$ and $|\GL_2(p)|/2(p^2-1)=p(p-1)/2$ tori of the second type, each generated by an element of cycle type $(p+1)$. This leaves the elements of order $p$, all self-centralizing and of cycle type $(p,1)$. Therefore $$F_G(x)=x^2 \frac{p(p+1)}{2}F_{C_{p-1}}(x)+\frac{p(p-1)}{2}F_{C_{p+1}}(x)+(p-1)(p+1)x^2-ax^{p+1},$$ where $a=p^2-1$ corrects for over-counting the identity. 
%We count the elements in $G$ of order $k$. When $k=2$, there are two conjugacy classes of involutions of lengths $\frac{p(p-1)}{2}$ and $\frac{p(p+1)}{2}$ with the first class consisting of elements which act fixed-point freely and the second consisting of elements which fix exactly two points. For $\epsilon = \pm 1$, and each non-trivial odd divisor $k$ of $p+\epsilon$, there are $$\phi(k)\frac{p(p-\epsilon)}{2}$$ elements of order $k$ with 
%$$
%c(g)=
%\begin{cases}
% \frac{p+1}{k},& \mbox{ if $\epsilon = 1$ }%\\
% \frac{p-1}{k} + 2,& \mbox{ if $\epsilon = -1$ }.\\
%\end{cases}
%$$
%Finally, there is a single conjugacy class of elements of order $p$ of length $p^2-1$. Putting this altogether, we obtain
%\begin{eqnarray*}
%F_G(x) &=& x^{p+1}+\frac{p(p-1)}{2}x^{\frac{p+1}{2}}+\frac{p(p+1)}{2}x^{\frac{p+3}{2}}+ \frac{p(p-1)}{2} \sum_{1,2 \neq k \mid p+1} \phi(k)x^{\frac{p+1}{k}}\\
%&&\quad+\frac{p(p+1)}{2} \sum_{1,2 \neq k \mid p-1} \phi(k)x^{\frac{p-1}{k}+2}+(p^2-1)x^2.
%\end{eqnarray*}
%Simplifying gives the result.
\end{proof}

\section{Reciprocal pairs}

Richard Stanley, in a 1974 paper \cite{stanley}, explained (polynomial)
combinatorial reciprocity thus:
\begin{quote}
A \emph{polynomial reciprocity theorem} takes the following form. Two
combinatorially defined sequences $S_1$, $S_2$, \ldots\ and $\bar S_1$,
$\bar S_2$, \ldots\ of finite sets are given, so that the functions
$f(n)=|S_n|$ and $\bar f(n)=|\bar S_n|$ are polynomials in $n$ for all
integers $n\ge1$. One then concludes that $\bar f(n)=(-1)^df(-n)$, where
$d=\deg f$.
\end{quote}

We will see that, in a number of cases, the cycle polynomial satisfies a
reciprocity theorem.

\subsection{The orbital chromatic polynomial}

First, we define the polynomial which will serve as the reciprocal polynomial
in these cases. A \emph{(proper) colouring} of a graph $\Gamma$ with $q$
colours is a map from the vertices of $\Gamma$ to the set of colours having
the property that adjacent vertices receive different colours. Note that,
if $\Gamma$ contains a loop (an edge joining a vertex to itself), then it has
no proper colourings. Birkhoff observed that, if there are no loops, then
the number of colourings with $q$ colours is the evaluation at $q$ of a
monic polynomial $P_\Gamma(x)$ of degree equal to the number of vertices, the
\emph{chromatic polynomial} of the graph.

Now suppose that $G$ is a group of automorphisms of $\Gamma$. For $g\in G$,
let $\Gamma/g$ denote the graph obtained by ``contracting'' each cycle of $g$
to a single vertex; two vertices are joined by an edge if there is an edge
of $\Gamma$ joining vertices in the corresponding cycles. The chromatic
polynomial $P_{\Gamma/g}(q)$ counts proper $q$-colourings of $\Gamma$ fixed
by $g$. If any cycle of $g$ contains an edge, then $\Gamma/g$ has a loop, and
$P_{\Gamma/g}=0$. Now (with a small modification of the definition
in~\cite{cjr}) we define the \emph{orbital chromatic polynomial} of the pair
$(\Gamma,G)$ to be

\begin{equation}\label{e:orbchrom}
P_{\Gamma,G}(x)=\sum_{g\in G}P_{\Gamma/g}(x).
\end{equation}

The orbit-counting Lemma immediately shows that $P_{\Gamma,G}(q)/|G|$ is equal to
the number of $G$-orbits on proper $q$-colourings of $\Gamma$.

Now, motivated by Stanley's definition, we say that the pair $(\Gamma,G)$,
where $\Gamma$ is a graph and $G$ a group of automorphisms of $\Gamma$, is
a \emph{reciprocal pair} if
\[P_{\Gamma,G}(x)=(-1)^nF_G(-x),\]
where $n$ is the number of vertices of $\Gamma$.

\begin{remark}
An alternative definition of reciprocality is obtained using the polynomial $$\bar{P}_{\Gamma,G}(x)=\sum_{g \in G} \sgn(g)\bar{P}_{\Gamma/g}(x)$$ where $\bar{P}_\Delta(x)=(-1)^nP(-x)$ is the dual chromatic polynomial, first defined by Stanley \cite{stanley} enumerating certain coloured acyclic orientations of $\Delta$. It is a straightforward exercise to see that reciprocality is equivalent to $$\bar{P}_{\Gamma,G}(x)=F_G(x).$$
\end{remark}
\mbox{}\newline
\textbf{Problem} Find all reciprocal pairs.

\medskip

This problem is interesting because, as we will see, there are a substantial
number of such pairs, for reasons not fully understood. In the remainder of
the paper,  we present the evidence for this, and some preliminary results
on the above problem.

A basic result about reciprocal pairs is the following.

\begin{lemma}\label{lem:sumtrans}
Suppose that $(G,\Gamma)$ is a reciprocal pair. Then the number of edges of
$\Gamma$ is the sum of the number of transpositions in $G$ and the number of
transpositions $(i,j)$ in $G$ for which $\{i,j\}$ is a non-edge.
\end{lemma}

\begin{proof}
Whitney~\cite{whitney} showed that the leading terms in the chromatic
polynomial of a graph $\Gamma$ with $n$ vertices and $m$ edges are
$x^n-mx^{n-1}+ \cdots$. This follows from the inclusion-exclusion formula for the chromatic polynomial: $x^n$ is the total number of colourings of the vertices of $\Gamma$, and for each edge $\{i,j\}$, the number of colourings in which $i$ and $j$ have the same colour is $x^{n-1}$. So in the formula (\ref{e:orbchrom}) for $P_{\Gamma,G}(x)$, the identity element of $G$ contributes $x^n-mx^{n-1}+\cdots.$ The only additional contributions to the coefficient of $x^{n-1}$ in $P_{\Gamma,G}(x)$ come from elements $g$ of $G$ such that only a single edge of $\Gamma$ is contracted to obtain $\Gamma/g$ and these are transpositions. A transposition $(i,j)$ makes a non-zero contribution if and only if $\{i,j\}$
is a non-edge. So the coefficient of $x^{n-1}$ is $-m+t^0(G)$, where $t^0(G)$
is the number of transpositions with this property.

On the other hand, the coefficient of $x^{n-1}$ in $F_G(x)$ is the number of
permutations in $G$ with $n-1$ cycles, that is, the total number $t(G)$ of
transpositions. So the coefficient in $(-1)^nF_G(-x)$ is $-t(G)$.

Equating the two expressions gives $m=t(G)+t^0(G)$, as required.
\end{proof}

We remark that the converse to Lemma \ref{lem:sumtrans} does not hold: consider the group $G=S_3 \wr S_3$ acting on 3 copies of $K_3$ (see
Proposition~\ref{p:wreath}: but note that $(3K_3, S_3\wr C_3)$ is a reciprocal
pair, by Proposition~\ref{p:wreath_recip}). We also observe the following corollary to Lemma \ref{lem:sumtrans}.

\begin{corollary}
If $\Gamma$ is not a complete graph and $(\Gamma,G)$ is a reciprocal pair then $\Gamma$ has at most $\frac{(n-1)^2}{2}$ edges.
\end{corollary}

\begin{proof}
If $\Gamma$ has ${n \choose 2}-\delta$ edges then by Lemma \ref{lem:sumtrans}, $${n \choose 2}-\delta = t(G)+t^0(G) \le t(G) + \delta.$$ If $0  < \delta < \frac{n-1}{2}$ then $$t(G) \ge {n \choose 2}-2\delta > {n \choose 2} - (n-1) ={n-1 \choose 2}.$$ It is well-known that a permutation group of degree $n$ containing at least ${n-1 \choose 2}+1$ transpositions must be the full symmetric group. But this implies that $\Gamma$ is a complete graph, a contradiction. 
\end{proof}

\medskip

According to Lemma \ref{lem:sumtrans}, if $\Gamma$ is not a null graph and
$(\Gamma,G)$ is a reciprocal pair, then $G$ contains transpositions. Now
as is well-known, if a subgroup $G$ of $S_n$ contains a transposition, then
the transpositions generate a normal subgroup $N$ which is the direct product
of symmetric groups whose degrees sum to $n$. (Some degrees may be $1$, in
which case the corresponding factor is absent.) 

A $G$-invariant graph must induce a complete or null graph on each of these
sets. Moreover, between any two such sets, we have either all possible edges
or no edges.

Suppose that $n_1,\ldots,n_r$ are the sizes of the $N$-orbits carrying complete graphs and
$m_1,\ldots,m_s$ the orbits containing null graphs. Then
Lemma~\ref{lem:sumtrans} shows that the total number of edges of the graph
is
\[\sum_{i=1}^r{n_i\choose 2}+2\sum_{j=1}^s{m_j\choose 2}.\]
The first term counts edges within $N$-orbits, so the second term counts
edges between different $N$-orbits.

\subsection{Examples}

\begin{proposition}\label{p:ex1}
The following hold:
\begin{enumerate}
\item[(a)] Let $\Gamma$ be a null graph, and $G$ a subgroup of the symmetric group
$S_n$. Then $P_{\Gamma,G}(x)=F_G(x)$.
\item[(b)] Let $\Gamma$ be a complete graph, and $G$ a subgroup of the symmetric
group $S_n$. Then $P_{\Gamma,G}(x)=x(x-1)\cdots(x-n+1)$, independent of
$G$.
\end{enumerate}
\end{proposition}

\begin{proof}
\begin{enumerate}
\item[(a)] The chromatic polynomial of a null graph on $n$ vertices is $x^n$. So, if
$g\in G$ has $c(g)$ cycles, then $\Gamma/g$ is a null graph on $c(g)$ vertices.
Thus
\[P_{\Gamma,G}(x)=\sum_{g\in G}x^{c(g)}=F_G(x).\]

\item[(b)] In the formula (\ref{e:orbchrom}) for $P_{\Gamma,G}(x)$, $P_{\Gamma/g}(x)$ is $0$ unless $g$ is the identity element of $G$, when it is $x(x-1)\cdots(x-n+1)$.\qedhere  %Suppose that $\Gamma$ is a complete graph. Any proper colouring of $\Gamma$ has all colours disjoint, so any permutation group $G$ on proper $a$-colourings is semiregular, and has $a(a-1)\cdots(a-n+1)/|G|$ orbits. Thus the orbital chromatic polynomial of $G$ is $x(x-1)\cdots(x-n+1)$, independent of $G$.
\end{enumerate}
\end{proof}

\begin{corollary}
The following hold:
\begin{enumerate}
\item[(a)] If $\Gamma$ is a null graph, then $(\Gamma,G)$ is a reciprocal pair
if and only if $G$ contains no odd permutations.
\item[(b)] If $\Gamma$ is a complete graph, then $(\Gamma,G)$ is a reciprocal
pair if and only if $G$ is the symmetric group.
\end{enumerate}
\end{corollary}

\begin{proof}
\begin{enumerate}
\item[(a)] This follows from Proposition~\ref{p:parity}.
\item[(b)] We saw in the preceding section that, if $G=S_n$, then
$F_G(x)=x(x+1)\cdots(x+n-1)$. Thus, we see that $(G,\Gamma)$ is a reciprocal
pair if and only if $G$ is the symmetric group.\qedhere
\end{enumerate}
\end{proof}


\begin{proposition}\label{p:direct_recip}
Let $\Gamma$ be the disjoint union of graphs $\Gamma_1,\ldots,\Gamma_r$,
and $G$ the direct product of groups $G_1,\ldots,G_r$, where
$G_i\le\Aut(\Gamma_i)$. Then
\[P_{\Gamma,G}(x)=\prod_{i=1}^rP_{\Gamma_i,G_i}(x).\]
In particular, if $(\Gamma_i,G_i)$ is a reciprocal pair for $i=1,\ldots,r$,
then $(\Gamma,G)$ is a reciprocal pair.
\end{proposition}

The proof is straightforward; the last statement follows from
Proposition~\ref{p:prod}. The result for wreath products is similar:

\begin{proposition}\label{p:wreath_recip}
Let $\Gamma$ be the disjoint union of $m$ copies of the $n$-vertex graph
$\Delta$. Let $G\le\Aut(\Delta)$, and $H$ a group of permutations of degree
$m$. Then
\[P_{\Gamma,G\wr H}(x) = |G|^mF_H(P_{\Delta,G}(x)/|G|).\]
In particular, if $(\Delta,G)$ is a reciprocal pair and $H$ contains no odd
permutations, then $(\Gamma,G\wr H)$ is a reciprocal pair.
\end{proposition}

\begin{proof}
Given $q$ colours, there are $P_{\Delta,G}(q)/|G|$ orbits on colourings of each
copy of $\Delta$; so the overall number of orbits is the same as the number of
orbits of $H$ on an $m$-vertex null graph with $P_{\Delta,G}(q)/|G|$ colours
available.

For the last part, the hypotheses imply that $P_{\Delta,G}(q)=(-1)^nF_G(-q)$,
and that the degree of each term in $F_H$ is congruent to $m$ mod~$2$. So
the expression evaluates to $|G|^mF_H(F_G(-x)/|G|)$ if either $m$ or $n$ is
even, and the negative of this if both are odd; that is,
$(-1)^{mn}|G|^mF_H(F_G(-x)/|G|)$.
\end{proof}

We now give some more examples of reciprocal pairs.

\begin{example}
Let $\Gamma$ be a $4$-cycle, and $G$ its automorphism group, the dihedral
group of order $8$. There are $4$ edges in $\Gamma$, and $2$ transpositions in
$G$, each of which interchanges two non-adjacent points (an opposite pair of
vertices of the $4$-cycle); so the equality of the lemma holds. Direct
calculation shows that
\[P_{\Gamma,G}(x)=x(x-1)(x^2-x+2),\qquad F_G(x)=x(x+1)(x^2+x+2),\]
so $P_{\Gamma,G}(x)=(-1)^nF_G(-x)$ holds in this case.

The $4$-cycle is also the complete bipartite graph $K_{2,2}$. We note that,
for $n>2$, $(K_{n,n},S_n\wr S_2)$ is not a reciprocal pair. This can be seen
from the fact that $F_{S_n\wr S_2}(x)$ has factors $x$, $x+1,\ldots,x+n-1$, whereas $K_{n,n}$ has chromatic number $2$ and so $x-2$ is not a
factor of $P_{K_{n,n},G}(x)$ for any $G\le S_n\wr S_2$. 

We do not know whether other complete multipartite graphs support
reciprocal pairs.

%Also  $(K_{n,m},S_n \times S_m)$ is not a reciprocal pair for $n,m > 2$ since the equality $$mn=2\left({m \choose 2}+{n \choose 2}\right)=m(m-1)+n(n-1)$$ of Lemma \ref{lem:sumtrans} is easily seen to have no solutions in this range.

\end{example}

\begin{example}
Let $\Gamma$ be a path with 3 vertices and $G$ its automorphism group which is cyclic of order 2. Direct calculation shows that 
\[P_{\Gamma,G}(x)=x^2(x-1),\qquad F_G(x)=x^2(x+1).\]
This graph is an example of a star graph, for which we give a complete
analysis in the next section.
\end{example}

\begin{example} Write $N_n$ for the null graph on $n$ vertices and let $\Gamma$ be the disjoint union of $K_m$ and $N_n$ together with all edges in between and set $G=S_m \times S_n \le$ Aut$(\Gamma)$. Then $\Gamma$ has ${m \choose 2} +mn$ edges and $G$ has ${m \choose 2}+{n \choose 2}$ transpositions of which ${n \choose 2}$ correspond to non-edges in $\Gamma$. Thus, according to Lemma \ref{lem:sumtrans} we need $n=m+1$. Now the only elements $g \in G$ which give non-zero contribution to $P_{\Gamma,G}(x)$ lie in the $S_n$ component. We get: $$P_{\Gamma,G}(x)=x(x-1)\cdots (x-(m-1)) \cdot \sum_{g \in S_n} (x-m)^{c(g)}.$$ Using Proposition \ref{p:ex1}(b) and $m=n-1$ this becomes $$(-1)^m F_{S_m}(-x) F_{S_n}(-x) \cdot (-1)^n,$$ which is equal to $-F_G(-x)$ by Proposition \ref{p:prod}.
\end{example}



%\begin{lemma}\label{c:wreath}
%Let $n,m$ be natural numbers with $m$ even and $\Gamma$ consist of $n$ copies of $K_m$. Let $H \le S_n$ and $G=S_m \wr H \le$ Aut$(\Gamma)$. Then $P_{\Gamma,G}(x)=F_G(-x)$.
%\end{lemma}

%\textbf{[Recheck!]}

%\begin{proof}
%Suppose that $V(\Gamma)=\{1,2,\ldots, mn\}$ and that each copy of $K_m$ in $\Gamma$ is supported on the vertices $\{i,n+i,2n+i,\ldots, (m-1)n+i\}$ for $1 \le i \le n$. We thus identify $G$ with a subset of Sym$(\{1,2,\ldots, mn\})$. For $0 \le j \le m-1$ write $$G_j:= \mbox{Sym}(\{jn+1,jn+2,\ldots, (j+1)n\}),$$ and let $H_j \cong H$ be the copy of $H$ inside $G_j$ given by the mapping $$(1,2,\ldots, n) \mapsto (jn+1,jn+2,\ldots, (j+1)n).$$

%The only conjugacy classes of elements of $G$ which give a non-zero contribution to $P_{\Gamma,G}$ are represented by elements of the form $\alpha:=\alpha_0\alpha_1\cdots \alpha_{m-1}$ where $\alpha_j \in H_j$, and where each $\alpha_j$ has the same cycle type. Let $k_\alpha$ denote the number of cycles in $\alpha_j$ and $\ell_{\alpha}$ denote the size of the conjugacy class of $\alpha_j$ in $H_j$. We make the following two observations:
%\begin{itemize}
%\item[(i)] $|\alpha^G|=m!^{n-k_\alpha} \cdot \ell_\alpha$;
%\item[(ii)] for each $g \in \alpha^G$, $\Gamma/g$ consists of exactly $k_\alpha$ copies of $K_m$.
%\end{itemize}
%Now summing over conjugacy class representatives $\alpha$ in $H$ we have,
%$$\begin{array}{rcl}
%P_{\Gamma,G}(x) &=&\sum_{\alpha} m!^{n-k_\alpha} \ell_\alpha P_{K_m,S_m}(x)^{k_\alpha} \\
%
%&=&m!^n\sum_{\alpha} m!^{-k_\alpha} \ell_\alpha ((-1)^m F_{S_m}(-x))^{k_\alpha} \\
%
%
%&=& m!^n \cdot F_{H}((-1)^m F_{S_m}(-x)/m!) \\
%
%&=& F_G(-x)
%\end{array} $$ where the last equality follows from Proposition \ref{p:wreath} and the fact that $m$ is even. The result follows. 
%\end{proof}

%\begin{lemma}
%The converse to Lemma \ref{lem:sumtrans} does not hold.
%\end{lemma}

%\begin{proof}
%Start with the disjoint union of $K_m$ and $N_n$ with all possible edges between, and then let $\Gamma$ be the disjoint union of this graph with $K_p$ and all possible edges between. The automorphism group $G$ of $\Gamma$ is $S_m \times S_n \times S_p$. The number of edges in $\Gamma$ is ${m \choose 2} + mn+pn+mp$ and the number of transpositions in $G$ is ${m \choose 2}+{n \choose 2}+{p \choose 2}$ of which  ${n \choose 2}+{p \choose 2}$ correspond to non-edges. According to Lemma \ref{lem:sumtrans} we need $$mn+pn+mp=n(n-1)+p(p-1),$$ which is satisfied, for example, with $(m,n,p)=(n-1,n,2n)$. When $n=3$, one calculates that $P_{\Gamma,G}(x)$ is exactly: 

%$$x^{11} - 19x^{10} + 186x^9 - 1146x^8 + 4689x^7 - 12843x^6 + 22964x^5 - 25304x^4 + 15360x^3 - 3888x^2.$$

%However this is not equal to $(-1)^n F_G(-x)$ which is

%$$x^{11} - 19x^{10} + 150x^9 - 642x^8 + 1629x^7 - 2511x^6 + 2300x^5 - 1148x^4 + 240x^3.$$

%(despite the fact that the coefficients of $x^{10}$ coincide).

 
%\end{proof}


%\textbf{[Combine this with example (b) above?]}


\section{Reciprocal pairs containing a tree}

In this section we show that the only trees that can occur in a reciprocal
pair are stars, and we determine the groups that can be paired with them. 

\begin{theorem}\label{t:recpairstree}
Suppose $\Gamma$ is a tree and $(\Gamma,G)$ is a reciprocal pair. Let $n$ be the number of vertices in $\Gamma$ and assume $n \ge 3$. The following hold:
\begin{enumerate}\itemsep0pt
\item[(a)] $n$ is odd;
\item[(b)] $\Gamma$ is a star;
\item[(c)] $(C_2)^k \le G \le C_2 \wr S_k$ where $n=2k+1$.
\end{enumerate}
Conversely any pair $(\Gamma,G)$ which satisfies conditions (a)-(c) is reciprocal.
\end{theorem} 

In what follows we assume the following:
\begin{itemize}\itemsep0pt
\item $\Gamma$ is a tree;
\item $(\Gamma,G)$ is a reciprocal pair;
\item $n$ is the number of vertices in $\Gamma$ and $n \ge 3$.
\end{itemize}

Note that any two vertices interchanged by a transposition are non-adjacent.
For suppose that a transposition flips an edge $\{v,w\}$. If the tree is
central, then there are paths of the same length from the centre to $v$ and
$w$, creating a cycle. If it is bicentral, then the same argument applies
unless $\{v,w\}$ is the central edge, in which case the tree has only two 
vertices, a contradiction.

\begin{lemma}\label{lem:treeequal}
If $n$ is odd then $(\Gamma,G)$ is a reciprocal pair if and only if
\begin{equation}\label{e:eq1} x(F_G(x-1)+F_G(-x))=F_G(-x).
\end{equation}
\end{lemma}

\begin{proof}
 The chromatic polynomial of a tree with $r$ vertices is easily seen to be $x(x-1)^{r-1}$. Hence $$P(\Gamma/g)=x(x-1)^{c(g)-1}$$ for each $g \in G$ and then
 $$P_{\Gamma,G}(x)=\sum_{g \in G} P(\Gamma/g) = \sum_{g \in G} x(x-1)^{c(g)-1} = \frac{x}{x-1}F_G(x-1).$$ Rearranging (and using that $n$ is odd) yields the Lemma.
\end{proof}


 

\begin{lemma}\label{lem:nodd}
$G$ has $\frac{n-1}{2}$ transpositions; in particular, $n$ is odd.
\end{lemma}

\begin{proof}
As in Lemma \ref{lem:sumtrans}, let $t(G)$ be the number of transpositions in $G$ and $t^0(G)$ be the number of transpositions $(i,j)$ in $G$ for which $i \not\sim j$ in $\Gamma$. If $G$ fixes an edge $(u,v)$ of $\Gamma$ then $(u,v) \in G$ implies $n = 2$, a contradiction. Thus $(u,v) \notin G$, and every transposition in $G$ is a non-edge. Hence $t^0(G)=t(G)$ so $2t(G)=n-1$ by Lemma \ref{lem:sumtrans}.
\end{proof}

\begin{lemma}\label{lem:groupstruct}
$(C_2)^k \le G \le C_2 \wr S_k$ where $n=2k+1$.
\end{lemma}

\begin{proof}
The transpositions in a permutation group $G$ generate a normal subgroup $H$
which is a direct product of symmetric groups. If there are two non-disjoint
transpositions in $G$, one of the direct factors is a symmetric group with
degree at least $3$, and hence $F_H(x)$ has a root $-2$ by Propositions
\ref{p:sym} and \ref{p:prod}. Then by Proposition~\ref{p:overgroups},
$F_G(x)$ has a root $-2$. By Lemma \ref{lem:treeequal} with $x=2$,

$$0=F_G(-2)=2(F_G(1)+F_G(-2))=2F_G(1) = 2|G|,$$ a contradiction. So the transpositions are pairwise disjoint, and generate a subgroup $(C_2)^k$ with $n=2k+1$ by Lemma \ref{lem:nodd}. Thus the conclusion of the lemma holds.
\end{proof}

\begin{lemma}\label{lem:star}
$\Gamma$ is a star.
\end{lemma}

\begin{proof}
Let $v$ be the unique fixed point of $G$. By Lemma \ref{lem:groupstruct}, for each $u \neq v$ there exists a unique vertex $u'$ with $(u,u') \in G$. This is possible only if each $u$ has distance 1 from $v$. Hence $\Gamma$ is a star.
\end{proof}

\begin{proof}[Proof of Theorem \ref{t:recpairstree}]
(a),(b) and (c) follow from Lemmas \ref{lem:nodd}, \ref{lem:star} and \ref{lem:groupstruct} respectively. Conversely, suppose that (a),(b) and (c) hold. Then $G=C_2\wr K$ for some permutation
group $K$ of degree $k$. By Proposition~\ref{p:wreath}, 
$F_G(x)=x \cdot 2^kF_K(x(x+1)/2)$. Now it is clear that $$-\frac{F_G(-x)}{x}=2^k \cdot F_K\left(\frac{x(x-1)}{2}\right)=\frac{F_G(x-1)}{x-1},$$ so that
(\ref{e:eq1}) holds and we deduce from Lemma \ref{lem:treeequal} that $(\Gamma,G)$ is a reciprocal pair. Our proof is complete.
\end{proof}


Given a set of reciprocal pairs $(\Gamma_1,G_1),\ldots, (\Gamma_m,G_m)$ with each $\Gamma_i$ a star we can take direct products and wreath products (using Propositions \ref{p:direct_recip} and \ref{p:wreath_recip}) to obtain reciprocal pairs $(\Gamma,G)$ with $\Gamma$ a forest. We do not know whether all such pairs arise in this way.

\section*{Acknowledgements}

The authors would like to thank the anonymous referees for their valuable comments and suggestions.

%\section{Connection with other polynomials}



%Harden and Penman studied the \emph{fixed point polynomial} $P_G(x)$ of a permutation group $G$ (the generating function for fixed points of elements of $G$), given by $P_G(x)=Z_G(x,1,\ldots,1)$.

%Another related construction is the \emph{Parker vector} of $G$, usually presented as an $n$-tuple rather than a polynomial: its $k$th entry is obtained by differentiating $Z_G$ with respect to $s_k$, and putting all variables equal to $1$ (and dividing by $|G|$). It is relevant to computing the Galois group of an integer polynomial. See~\cite[Section 2.8]{permgps}.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


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