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\def\Z{{\mathbb Z}}
\def\Stab{{\rm Stab}}
\def\fix{{\rm fix}}
\def\la{\langle}
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\def\AGL{{\rm AGL}}
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\def\Aut{{\rm Aut}}
\def\CI{{\rm CI}}

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\title{\bf On the Cayley isomorphism problem\\ for Cayley objects of nilpotent groups\\ of some orders}

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\author{Edward Dobson\\
\small Department of Mathematics and Statistics\\[-0.8ex]
\small Mississippi State University\\[-0.8ex]
\small PO Drawer MA Mississippi State, MS 39762 U.S.A\\
\small\tt dobson@math.msstate.edu\\
\small and\\
\small IAM\\[-0.8ex]
\small University of Primorska\\[-0.8ex]
\small 6000 Koper, Slovenia
}

% \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{Jan 1, 2012}{Jan 2, 2012}\\
\small Mathematics Subject Classifications: 05E18, 05C25, 20F18}

\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
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\begin{abstract}
We give a necessary condition to reduce the Cayley isomorphism problem for Cayley objects of a nilpotent or abelian group $G$ whose order satisfies certain arithmetic properties to the Cayley isomorphism problem of Cayley objects of the Sylow subgroups of $G$ in the case of nilpotent groups, and in the case of abelian groups to certain natural subgroups.  As an application of this result, we show that $\Z_q\times\Z_p^2\times\Z_m$ is a CI-group with respect to digraphs, where $q$ and $p$ are primes with $p^2 < q$ and $m$ is a square-free integer satisfying certain arithmetic conditions (but there are no other restrictions on $q$ and $p$).

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} Cayley object; Cayley graph; isomorphism; CI-group
\end{abstract}

\section{Introduction}

In 1967 \'Ad\'am \cite{Adam1967} conjectured that any two circulant
graphs of order $n$ are isomorphic if and only if they are
isomorphic by a group automorphism of $\Z_n$.  While \'Ad\'am's
conjecture was quickly shown to be false \cite{ElspasT1970}, the conjecture
nonetheless generated much interest in the following question:
Are two Cayley graphs of a group $G$ isomorphic if and only if
they are isomorphic by a group automorphism of $G$?  If so, we say
that $G$ is a {\bf CI-group with respect to graphs}.  This problem
naturally generalizes to any class of combinatorial objects (see \cite{Muzychuk1999} for several equivalent formulations of the precise definition of a combinatorial object).
Namely, is it true that two Cayley objects of a group $G$ in some
class ${\cal K}$ of combinatorial objects are isomorphic if and
only if they are isomorphic by a group automorphism of $G$?  If
so, we say that {\bf $G$ is a CI-group with respect to ${\cal K}$}.  If
$G$ is a CI-group with respect to {\it every} class of
combinatorial objects, we say that $G$ is a {\bf CI-group}.  In 1987,
P\'alfy \cite{Palfy1987} proved the following remarkable result:

\begin{theorem}\label{Paltool}
A group $G$ is a CI-group if and only if $\gcd(n,\varphi(n)) = 1$
or $n = 4$, where $\varphi$ is Euler's phi function.
\end{theorem}

While P\'alfy's result is quite powerful, it does not tell us
anything in general about isomorphisms between Cayley objects of a
group $G$ if $G$ is not a CI-group, other than there exists isomorphic Cayley
objects of $G$ which are not isomorphic by a group automorphism of
$G$. For such groups, we are then left with the question of if two
Cayley objects of $G$ are isomorphic, then what are the possible
isomorphisms between them?  This is sometimes known as the Cayley
isomorphism problem.  Usually, one would like the solution to this
question to be a (hopefully) short list $L$ of possible
isomorphisms.  That is, two Cayley objects of $G$ are isomorphic if
and only if they are isomorphic by a function in the list $L$. In
1999, Muzychuk \cite{Muzychuk1999} showed that if $G$ is a cyclic group of
order $n$ and for any distinct primes $p$ and $q$ dividing $n$ we
have that $q$ does not divide $p - 1$, then any two Cayley objects
of $G$ are isomorphic by an automorphism that can be found in a
natural way from isomorphisms of Cayley objects of prime-power
orders that divide $n$.  Thus Muzychuk reduced the Cayley
isomorphism problem for Cayley objects of cyclic groups of some orders to the
Cayley isomorphism problem for Cayley objects of cyclic groups of
prime-power orders.  In 2003, the author \cite{Dobson2003a}, found a
sufficient condition to extend Muzychuk's result to all abelian
groups (with the same order conditions), and showed this sufficient
condition was satisfied by some abelian groups.  In this paper, we
extend the author's earlier result to nilpotent groups, as well as to abelian groups with more general order conditions (Theorem \ref{main}).  Finally, as an application we will extend the list of CI-groups with respect to digraphs by showing that $\Z_q\times\Z_p^2\times\Z_m$ is a CI-group with respect to digraphs, where $p$ and $q$ are distinct primes with $p^2 < q$ and $m$ satisfies certain arithmetic conditions (Theorem \ref{app}).

Throughout this paper, $G$ is a finite group.  For group theoretic
terms not defined in this paper, see \cite{DixonM1996}.  We begin with
some definitions.

\begin{definition}
Let $G$ be a transitive group acting on $\Omega$.  Let $X$ be the
set of all complete block systems of $G$.  Define a partial order
on $X$ by ${\cal B}\preceq{\cal C}$ if and only if every block of
${\cal C}$ is a union of blocks of ${\cal B}$.  We define ${\cal B}\vert_C$ to be
the complete block system of $\Stab_G(C) = \{g\in G:g(C) = C\}$, the {\bf set-wise stabilizer of $C\in{\cal C}$}, consisting of all those
blocks of ${\cal B}$ that are contained in $C$, $C\in{\cal C}$, and remark that ${\cal B}\vert_C$ is a complete block system of $\Stab_G(C)$ in its action on $C$.  By $\fix_G({\cal B})$ we mean the subgroup of $G$ which fixes each block of ${\cal B}$ set-wise.  That is, $\fix_G({\cal B}) = \{g\in G:g(B) = B{\rm\ for\ all\ }B\in{\cal B}\}$.  We denote by $\Stab_G(x)$ the stabilizer of $x\in X$.  That is, $\Stab_G(x) = \{g\in G:g(x) = x\}$.  Finally, $g\in G$ induces a natural permutation $g/{\cal B}$ in $S_{\cal B}$, and we set $G/{\cal B} = \{g/{\cal B}:g\in G\}$.
\end{definition}

\begin{definition}
Let $n = \Pi_{i=1}^rp_i^{a_i}$ be the prime factorization of $n$ and define $\Omega:{\mathbb N}\mapsto{\mathbb N} $ by $\Omega(n) = \Sigma_{i=1}^ra_i$.  Setting $m = \Omega(n)$, we say a transitive group $G$
of degree $n$ is {\bf $m$-step imprimitive} if there exists a
sequence of complete block system ${\cal B}_0\prec{\cal B}_1\prec
\ldots \prec {\cal B}_m$. Note that ${\cal B}_0$ consists of singletons, while ${\cal B}_m$ consists of the entire set on which $G$ acts.  A complete block system ${\cal B}$ will
be said to be {\bf normal} if ${\cal B}$ is formed by the orbits
of a normal subgroup.  We will say that $G$ is {\bf normally
$m$-step imprimitive} if each ${\cal B}_i$, $0\le i\le m$, is
formed by the orbits of a normal subgroup of $G$.
\end{definition}

\section{The main tool}

In this section, we will give a sufficient condition that will imply that the Cayley isomorphism problem for nilpotent groups of certain orders can be reduced to the Cayley isomorphism problem for groups of prime-power order (Theorem \ref{main} and Corollary \ref{nilmain}), and for abelian groups with more general arithmetic conditions (Theorem \ref{main}).  That these results have implications for the Cayley isomorphism problem is established in Theorem \ref{bigtool}.  The following result is straightforward, and so its proof is omitted.

\begin{lemma}\label{blocktool}
Let $G_1,G_2\le S_n$ be transitive such that both $G_1$ and $G_2$
admit ${\cal B}$ as a complete block system.  Then $\la
G_1,G_2\ra$ admits ${\cal B}$ as a complete block system.
\end{lemma}

%\begin{proof}
%Let $g\in\la G_1,G_2\ra$.  Then $g = g_1g_2\cdots g_r$ where if
%$i\equiv 1\ (\mod 2)$ then $g_i\in G_1$ and if $i\equiv 0\ (\mod
%2)$ then $g_i\in G_2$.  We will show by induction on $r$ that
%$g(B)\in{\cal B}$ for every $B\in{\cal B}$.

%If $r = 1$, then this is trivial as ${\cal B}$ is a complete block
%system of $G_1$.  Assume then that $g_1\cdots g_r(B)\in{\cal B}$.
%Let $g_{r+1}\in G_i$, where $i = 1$ if $r + 1$ is odd and $i = 2$
%if $r + 1$ is even.  If $B\in{\cal B}$, then $g_{r+1}(B) = B'$ for
%some $B\in{\cal B}$.  Then
%$$g(B) = g_1\cdots g_rg_{r+1}(B) = g_1\cdots g_r(B') = B''\in{\cal B}$$
%\noindent by the induction hypothesis.  Whence $g(B)\in{\cal B}$
%for every $B\in{\cal B}$ and the result follows by induction.
%\end{proof}

%\begin{note}
%Let $G$ be a transitive group with complete block systems ${\cal
%B}\preceq{\cal C}$.  For $C\in{\cal C}$, we denote the set of all
%blocks of ${\cal B}$ that contained in $C$ by $[{\cal B}]\vert_C$.
%\end{note}

The following result is trivial after observing that the hypothesis implies that $\fix_G({\cal B}) = \Stab_G(B)$.

\begin{lemma}\label{mstep}
Let $G\le S_n$ be $m$-step imprimitive with
sequence ${\cal B}_0,\ldots,{\cal B}_m$.  If $G/{\cal B}_{m-1}$ is
cyclic of prime order $p$, then $\fix_G({\cal
B}_{m-1})\vert_B$ is $(m-1)$-step imprimitive for every
$B\in{\cal B}_{m-1}$, with $(m-1)$-step imprimitivity
sequence ${\cal B}_0\vert_B,\ldots,{\cal
B}_{m-1}\vert_B$.
\end{lemma}

%\begin{proof}
%We must show that $[{\cal B}_i]\vert_{B_{m-1}}$ is a complete
%block system of $\fix_G({\cal B}_{m-1})\vert_{B_{m-1}}$ for every
%$0\le i\le m-1$. It thus suffices to show that a block of ${\cal
%B}_i$, $0\le i\le m-1$, contained in $B_{m-1}$ is a block of
%$\fix_G({\cal B}_{m-1})\vert_{B_{m-1}}$. Let $a\in B_{m-1}$, and
%$B_i\in{\cal B}_i$ such that $a\in B_i$.  By \cite[Theorem
%1.1.5A]{DixonM1996}, there is a subgroup $H\ge\Stab_G(a)$ such that $B_i =
%\{h(a):h\in H\}$. Also by \cite[Theorem 1.1.5A]{DixonM1996}, we must show
%that there exists a subgroup $H_i\ge \Stab_{\fix_G({\cal
%B}_{m-1})\vert_{B_{m-1}}}(a)$ such that $B_i = \{h(a):h\in H_i\}$.
%Now, as $H$ fixes $B_i\subset B_{m-1}$ set-wise and $B_{m-1}$ is a
%block of $G$, $H$ fixes $B_{m-1}$ set-wise. As $G/{\cal B}_{m-1}$
%is cyclic of prime order $p$, we must have that $H\le\fix_G({\cal
%B}_{m-1})$. Similarly, as $a\in B_i\subset B_{m-1}$,
%$\Stab_G(a)\le\fix_G({\cal B}_{m-1})$. Hence $\Stab_G(a) =
%\Stab_{\fix_G({\cal B}_{m-1})}(a)$. Clearly then
%$\Stab_{\fix_G({\cal B}_{m-1})}(a)\vert_{B_{m-1}} =
%\Stab_G(a)\vert_{B_{m-1}} \le \Stab_{\fix_G({\cal
%B}_1)\vert_{B_1}}(a)$.  Conversely, if $g\in\Stab_{\fix_G({\cal
%B}_{m-1})}(a)\vert_{B_{m-1}}$, then there exists
%$g'\in\fix_G({\cal B}_1)$ such that $g'\vert_{B_{m-1}} = g$.  As
%$\Stab_G(a)\le\fix_G({\cal B}_{m-1})$, $g'\in\Stab_G(a)$. But then
%$g'\in\fix_G({\cal B}_{m-1})$ so that $g\in\Stab_{\fix_G({\cal
%B}_{m-1})\vert_{B_{m-1}}}(a)$. Hence $\Stab_{\fix_G({\cal
%B}_{m-1})}(a)\vert_{B_{m-1}} = \Stab_{\fix_G({\cal
%B}_{m-1})\vert_{B_{m-1}}}(a)$. Let $H_i = H\vert_{B_i}$.  As
%$\Stab_G(a)\le H\le\fix_G({\cal B}_{m-1})$ and $\Stab_G(a) =
%\Stab_{\fix_G({\cal B}_{m-1})}(a)$, we have that
%$\Stab_{\fix_G({\cal B}_{m-1})\vert_{B_{m-1}}}(a)\le
%H\vert_{B_{m-1}} = H_i$, and clearly $\{h_i(a):h_i\in
%H\vert_{B_{m-1}}\} = \{h(a):h\in H\}$ as ${\cal B}_i\prec {\cal
%B}_{m-1}$. The result is then proved.
%\end{proof}

We will use the following basic (and known) result implicitly throughout the
paper.

\begin{lemma}\label{regularabelian}
Let $G\le S_n$ be transitive with $H\le G$ a transitive abelian
subgroup.  Then every complete block system of $G$ is normal and
is formed by the orbits of a normal subgroup of $H$.
\end{lemma}

\begin{proof}
Let ${\cal B}$ be a complete block system of $G$ consisting of $m$
blocks of size $k$. As a transitive abelian group is regular
\cite[Proposition 1.4.4]{Wielandt1964}, we have that $H/{\cal B}$ is regular
of degree $m$, so that $\fix_H({\cal B})\not = 1$ and has order
$k$.  As $\Stab_H(B) = \fix_H({\cal B})$ for every $B\in{\cal B}$
and $\Stab_H(B)\vert_B$ is transitive \cite[Exercise 1.5.6]{DixonM1996},
we have that $\fix_H({\cal B})\vert_B$ is transitive for every
$B\in{\cal B}$.  As the blocks of ${\cal B}$ have size $k$, we
conclude that the orbits of $\fix_H({\cal B})\le\fix_G({\cal B})$
form ${\cal B}$.
\end{proof}

\begin{definition}
Let $G$ be a permutation group acting on $X$ and $H$ a permutation group acting on $Y$.  Define the {\bf wreath product of $G$ and $H$}, denoted $G\wr H$, to be the group of all permutations of $G\times H$ of the form $(x,y)\to (g(x),h_x(y))$.
\end{definition}

\begin{lemma}\label{mstepsoltool}
Let $n$ be a positive integer and $G_1,G_2$ be transitive abelian
groups of degree $n$ such that $\la G_1,G_2\ra$ is $m$-step
imprimitive.  Let $n = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$ be the
prime-power decomposition of $n$.  Then there exists $\delta\in\la
G_1,G_2\ra$ and a sequence of primes $q_1,\ldots,q_m$ such that $n
= q_1\cdots q_m$ and $\la G_1,\delta^{-1}G_2\delta\ra$ is
permutation isomorphic to a subgroup of
$\AGL(1,q_1)\wr(\AGL(1,q_2)\wr(\cdots\wr\AGL(1,q_m)))$.
Furthermore, if $\la G_1,G_2\ra$ is solvable, then we may take
$\delta = 1$.
\end{lemma}

\begin{proof}
We proceed by induction on $m$.  If $m = 1$, then $n$ is prime,
and both $G_1$ and $G_2$ are Sylow $n$-subgroups of $S_n$.  Hence
there exists $\delta\in\la G_1,G_2\ra$ such that
$\delta^{-1}G_2\delta = G_1$, and the result is trivial as $\la
G_1,\delta^{-1}G_2\delta\ra$ is cyclic of order $n$. Now assume that the result is true for all $m-1\ge 1$,
and let $G_1,G_2$ be transitive abelian groups of degree $n$,
where $\Omega(n) = m$, such that $\la G_1,G_2\ra$ is $m$-step
imprimitive.

As $\la G_1,G_2\ra$ is $m$-step imprimitive, $\la G_1,G_2\ra$ admits
a normal complete block system ${\cal B}$ consisting of $n/q_m$
blocks of size $q_m$ for some prime $q_m\vert n$, and both
$G_1/{\cal B}$ and $G_2/{\cal B}$ are transitive abelian groups of
degree $n/q_m$ and $\Omega(n/q_m) = m-1$. Furthermore, as $\la G_1,G_2\ra$
is $m$-step imprimitive, $\la G_1,G_2\ra/{\cal B}$ is $(m-1)$-step
imprimitive by \cite[Lemma 8]{Dobson2003a}, so by the induction hypothesis,
there exists $\delta_1\in\la G_1,G_2\ra$ such that $\la
G_1,\delta_1^{-1}G_2\delta\ra/{\cal B}$ is permutation isomorphic to
a subgroup of
$\AGL(1,q_1)\wr(\AGL(1,q_2)\wr(\cdots\wr\AGL(1,q_{m-1})))$ for some
sequence of primes $q_1,\ldots,q_{m-1}$ such that $n/q_m = q_1\cdots
q_{m-1}$, and if $\la G_1,G_2\ra$ is solvable, we may take $\delta_1
= 1$. Furthermore, $\fix_{G_1}({\cal B})$ is semiregular of order
$q_m$, and $\fix_{\delta_1^{-1}G_2\delta_1}({\cal B})$ is also
semiregular of order $q_m$. Hence there exists $\delta_2\in\fix_{\la
G_1,\delta_1^{-1}G_2\delta_1\ra}({\cal B})$ such that
$\delta_2^{-1}\fix_{\delta_1^{-1}G_2\delta_1}({\cal B})\delta_2$ is
contained in the same Sylow $q_m$-subgroup of $\fix_{\la
G_1,\delta_1^{-1}G_2\delta_1\ra}({\cal B})$ as $\fix_{G_1}({\cal
B})$.  If $\la G_1,G_2\ra$ is solvable, then $\fix_{\la
G_1,G_2\ra}({\cal B})$ is solvable, so $\fix_{\la G_1,G_2\ra}({\cal
B})\vert_B$ is solvable, and by \cite[Exercise 3.5.1]{DixonM1996},
$\fix_{\la G_1,G_2\ra}({\cal B})\vert_B$ has a unique Sylow
$q_m$-subgroup, so we may take $\delta_2 = 1$. Let $\delta =
\delta_1\delta_2$. As a Sylow $q_m$-subgroup of $\fix_{\la
G,\delta^{-1}G\delta\ra}({\cal B})$ is contained in
$1_{S_{n/q_m}}\wr\Z_{q_m}$ we have that both $G_1$ and
$\delta^{-1}G_2\delta$ normalize $1_{S_{n/q_m}}\wr\Z_{q_m}$. This
then implies that $\Stab_{\la
G_1,\delta^{-1}G_2\delta\ra}(B)\vert_B$ has a normal Sylow
$q_m$-subgroup, so that $\Stab_{\la
G_1,\delta^{-1}G_2\delta\ra}(B)\vert_B$ is permutation isomorphic to
a subgroup of $\AGL(1,q_m)$ for every $B\in{\cal B}$. It then
follows by the Embedding Theorem \cite[Theorem 2.6]{Meldrum1995}, that $\la
G_1,\delta^{-1}G_2\delta\ra$ is permutation isomorphic to a subgroup
of $\AGL(1,q_1)\wr(\AGL(1,q_2)\wr(\cdots\wr\AGL(1,q_m)))$, and the
result follows by induction.
\end{proof}

\begin{definition}
Let $\pi$ be a set of primes.  A {\bf $\pi$-group} $G$ is a group
such that every prime divisor of $\vert G\vert$ is contained in
$\pi$.  A subgroup $H$ of $G$ is an {\bf $S_\pi$-subgroup of $G$}
if no prime in $\pi$ divides $\vert G\vert/\vert H\vert$.  By
$\pi'$, we denote the set of primes dividing $\vert G\vert$ that
are not contained in $\pi$.
\end{definition}

We shall have need a consequence of the preceding  result.

\begin{lemma}\label{piabelian}
Let $n$ be a positive integer and $\pi$ be the set of distinct
prime numbers dividing $n$.  If $G_1,G_2$ are transitive abelian
groups of degree $n$ such that $\la G_1,G_2\ra$ is $m$-step
imprimitive then there exists $\delta\in\la G_1,G_2\ra$ such that
$\la G_1,\delta^{-1}G_2\delta\ra$ is a solvable $\pi$-group.
\end{lemma}

\begin{proof}
Let $n = p_1^{a_1}\cdots p_r^{a_r}$ be the prime-power
decomposition of $n$.  By Lemma \ref{mstepsoltool}, there exists
$\delta_1\in\la G_1,G_2\ra$ and a sequence $q_1,\ldots,q_m$ of
primes such that $n = q_1\cdots q_m$ and $\la
G_1,\delta_1^{-1}G_2\delta_1\ra\le\AGL(1,q_1)\wr(\AGL(1,q_2)\wr(\cdots\wr\AGL(1,q_m)))$.
As $\AGL(1,q_i)$ is solvable for every $1\le i\le m$, $\la
G_1,\delta_1^{-1}G_2\delta_1\ra$ is solvable.  By Hall's Theorem
\cite[Theorem 6.4.1]{Gorenstein1968}, we have that $G_1$ is contained in an
$S_\pi$-subgroup $H_1$ of $\la G_1,\delta_1^{-1}G_2\delta_1\ra$
and that $G_2$ is contained in an $S_\pi$-subgroup $H_2$ of $\la
G_1,\delta_1^{-1}G_2\delta_1\ra$.  Also by Hall's Theorem, there
exists $\delta_2\in\la G_1,\delta_1^{-1}G_2\delta_1\ra$ such that
$\delta_2^{-1}H_2\delta = H_1$.  Let $\delta = \delta_1\delta_2$.
Then $\la G_1,\delta^{-1}G_2\delta\ra\le H_1$, and $H_1$ is a
solvable $\pi$-group.
\end{proof}

Let ${\cal L} = {\cal L}_G$ be the set of all normal complete
block systems of a transitive group $G$.  Then $\preceq$ is a canonical partial order on ${\cal L}$.  Define operations $\cup$ and $\cap$ on
${\cal L}$ by ${\cal B}\cup{\cal C}$ is the normal complete block
system of $G$ formed by the orbits of $\la\fix_G({\cal
B}),\fix_G({\cal C})\ra = \fix_G({\cal B})\cdot\fix_G({\cal C})$
(as both of these groups are normal), and ${\cal B}\cap{\cal C}$
is the normal complete block system of $G$ formed by the orbits
of $\fix_G({\cal B})\cap\fix_G({\cal C})$. Notice that both of
these operation do in fact give normal complete block systems as
$\la\fix_G({\cal B}),\fix_G({\cal C})\ra\tl G$ and $\fix_G({\cal
G})\cap\fix_G({\cal C})\tl G$. Thus ${\cal L}_G$ is a lattice. See \cite{Hall1976} for terms regarding lattices not defined here.  We
also have that

\begin{lemma}\label{lattice}
If $G$ contains a transitive abelian group $H$, then ${\cal L}_G$
is a modular lattice.
\end{lemma}

\begin{proof}
We must show that if ${\cal B}\preceq{\cal A}$, then ${\cal
A}\cap({\cal B}\cup{\cal C}) = {\cal B}\cup({\cal A}\cap{\cal C})$.
By Lemma \ref{regularabelian}, there exists $A,B,C\le H$ such that ${\cal A}$ is formed by the
orbits of $A$, ${\cal B}$ is formed by the orbits of $B$, and ${\cal
C}$ is formed by the orbits of $C$.  As ${\cal B}\preceq{\cal A}$,
we have that $B\le A$.  By \cite[Theorem 8.4.1]{Hall1976}, we have that
$A\cap(B\cdot C) = B\cdot(A\cap C)$. Then the orbits of
$A\cap(B\cdot C)$ are the same as the orbits of $B\cdot(A\cap C)$.
\end{proof}

We remark that the previous result is contained in \cite[Theorem 2.10]{MuzychukP2011}

In the following three results, we will have in the hypothesis
that $\gcd(n_i,n_j\cdot\varphi(n_j)) = 1$.  Notice that this
implies that $\gcd(n_i,n_j) = 1$, and that if $p_i\vert n_i$ is
prime, then $p_i$ does not divide $p_j - 1$ for any prime
$p_j\vert n_j$.

\begin{lemma}\label{solproduct}
Let $n_1,\ldots,n_r$ be positive integers such that if $i\not =
j$, then $\gcd(n_i,n_j\cdot\varphi(n_j)) = 1$, $\pi_i$ the set of
primes dividing $n_i$, and $H_i$ be a transitive, solvable,
$\pi_i$-group of degree $n_i$, $1\le i\le r$. Let $G$ be a
transitive $m$-step imprimitive subgroup of $\Pi_{i=1}^rH_i$
acting coordinate-wise, where $\Omega(n_1\cdots n_r) = m$.
Then there exists transitive subgroups $L_i\le H_i$ such that $G =
\Pi_{i=1}^rL_i$.
\end{lemma}

%\begin{proof}
%It is not difficult to see that $\Pi_{i=1}^rH_i$ admits complete
%block systems ${\cal C}_i$ consisting of $n/n_i$ blocks of size
%$n_i$, $1\le i\le r$.  As each $H_i$ is solvable, we have that $G$
%is solvable, and so contains an $S_{\pi_i}$-subgroup $L_i$ for
%every $1\le i\le r$.  Notice that if we show $\fix_G({\cal C}_i) =
%L_i$, then we will have that $L_i\cap L_j = 1$ for every $i\not =
%j$, $L_i\tl G$, and $\la L_i:1\le i\le r\ra = G$. The result will
%then follow. Now, $G/{\cal C}_i$ is certainly solvable, and we
%have by \cite[Lemma 8]{Dobson2003a} that $G/{\cal C}_i$ is $m_i$-step
%imprimitive, where $n/n_i$ has exponent $m_i$. By Lemma
%\ref{mstepsoltool}, we have that there exists a sequence of primes
%$q_1,\ldots,q_{m_i}$ such that $q_1\cdots q_{m_i} = n/n_i$ and
%$G/{\cal C}_i\le
%\AGL(1,q_1)\wr(\AGL(1,q_2)\wr(\cdots\wr\AGL(1,q_{m_i})))$. As
%$\vert\AGL(1,p)\vert = (p-1)\cdot p$ and
%$\gcd(n_i,n_j\cdot\varphi(n_j)) = 1$ if $i\not = j$, we have that
%$G/{\cal C}_i$ has order relatively prime to $n_i$.  We conclude
%that $\fix_G({\cal C}_i) = L_i$, and the result follows.
%\end{proof}

\begin{proof}
It is not difficult to see that $\Pi_{i=1}^rH_i$ admits complete
block systems ${\cal C}_i$ consisting of $n/n_i$ blocks of size
$n_i$, $1\le i\le r$, formed by the orbits of $H_i$.  As each $H_i$ is solvable, we have that $G$
is solvable, and so contains an $S_{\pi_i}$-subgroup $L_i$ for
every $1\le i\le r$.  As $\Pi_{i=1}^rH_i/{\cal C}_i$ is a $\pi_i'$-group, $G/{\cal C}_i$ is also a $\pi_i'$-group, and so $\fix_G({\cal C}_i) = L_i$.  Then $L_i\cap L_j = 1$ for every $i\not =
j$, $L_i\tl G$, and $\la L_i:1\le i\le r\ra = G$. The result now follows.
\end{proof}

We now need only one more tool to prove the main result (Theorem \ref{main}) of this section. Before proceeding to this last tool, it will be useful to develop some terminology which will simplify the statement.  First, the proof of Theorem \ref{main} will proceed by induction on $m = \Omega(n)$.  So when proving Theorem \ref{main}, we will be assuming that the conclusion of Theorem \ref{main} holds for all integers $n/p$, where $p$ divides $n$ is prime.  In particular, with $m,n_1,\ldots,n_r$ and $n$ satisfying the hypothesis of Theorem \ref{main} and $G_1,G_2$ transitive abelian or nilpotent groups of degree $n$, then whenever $K_1,K_2$ are transitive nilpotent or abelian groups of degree $n/p$ - and to simplify our notation, there is no harm in assuming that $p\vert n_1$ - such that $\la K_1,K_2\ra$ are $(m-1)$-step imprimitive, then there exists $\delta\in\la K_1,K_2\ra$ such that $\la K_1,\delta^{-1}K_2\delta\ra\le \Pi_{i=1}^r\bar{H}_i$, where $\bar{H}_i$ is a solvable $\bar{\pi}_i$ group of degree $\bar{n}_i$.  Here $\bar{n}_i = n_i$ if $i\not = 1$ while $\bar{n}_1 = n_1/p$, and $\bar{\pi}_i$ is the set of prime divisors of $\bar{n}_i$.  In this situation, we will say that $n$ {\bf satisfies the main induction hypothesis}.

\begin{lemma}\label{bottomtool}
Let $n_1,\ldots,n_r$ be positive integers such that if $i\not =
j$, then $\gcd(n_i,n_j\cdot\varphi(n_j)) = 1$.  Let $n = n_1\cdots
n_r$ with $\Omega(n) = m$, $p\vert n$ (and without loss of generality, $p\vert n_1$), and $\bar{\pi} = \cup_{j\in
I}\bar{\pi}_j$ for some $I\subseteq[r]$. If
\begin{enumerate}
\item $n$ satisfies the main induction hypothesis,
\item $\la G_1,G_2\ra$ is $m$-step imprimitive,
\item $\la G_1,G_2\ra$ admits a complete block system ${\cal B}$ of $p$
blocks of size $n/p$, and
\item $\la G_1,G_2\ra/{\cal B}$ is a $p$-group,
\end{enumerate}
\noindent then there exists $\delta\in\la G_1,G_2\ra$ such that
$\la G_1,\delta^{-1}G_2\delta\ra$ admits a complete block system
formed by the orbits of the unique $S_{\bar{\pi}}$-subgroup of
$\fix_{G_i}({\cal B})$, $i = 1,2$.
\end{lemma}

\begin{proof}
Let $B\in{\cal B}$ and $K_i = \fix_{G_i}({\cal B})\vert_B$, $i = 1,2$. As $\la G_1,G_2\ra/{\cal B}$ has
prime order $p$, we must have that $\Stab_{\la G_1,G_2\ra}(B) =
\fix_{\la G_1,G_2\ra}({\cal B})$.  Note that $K_i$ is transitive on $B$, and if $G_i$ is nilpotent or abelian, then $K_i$ is nilpotent or abelian, $i = 1,2$.  Clearly $\fix_{G_1}({\cal
B}),\fix_{G_2}({\cal B})\le\fix_{\la G_1,G_2\ra}({\cal B})$.  By
Lemma \ref{mstep}, $\fix_{\la G_1,G_2\ra}({\cal B})\vert_B$ is
$(m-1)$-step imprimitive, so that $\la K_1,K_2\ra$ is $(m-1)$-step
imprimitive.

By hypothesis, there exists $\delta_1\in\la\fix_{G_1}({\cal
B}),\fix_{G_2}({\cal B})\ra$ such that

$$\la\fix_{G_1}({\cal
B}),\delta_1^{-1}\fix_{G_2}({\cal B})\delta_1\ra\vert_B\le
\Pi_{j=1}^r\bar{H}_{B,j},$$

\noindent where each $\bar{H}_{B,j}$ is a transitive solvable
$\bar{\pi}_j$-group of degree $\bar{n}_j$. Similarly, if $B\not = B'\in{\cal B}$, then there exists $\delta_2\in\la\fix_{G_1}({\cal
B}),\delta_1^{-1}\fix_{G_2}({\cal B})\delta_1\ra$ such that

$$\la\fix_{G_1}({\cal
B}),\delta_2^{-1}\delta_1^{-1}\fix_{G_2}({\cal
B})\delta_1\delta_2\ra\vert_{B'}\le \Pi_{j=1}^r\bar{H}_{B',j},$$ where
each $\bar{H}_{B',j}$ is a transitive solvable $\bar{\pi}_j$-group of degree
$\bar{n}_j$. Furthermore, we have that
$\delta_2\vert_B\in\la\fix_{G_1}({\cal
B}),\delta_1^{-1}\fix_{G_2}({\cal B})\delta_1\ra\vert_B$.  This
then implies that
$$\la\fix_{G_1}({\cal B}),\delta_2^{-1}\delta_1^{-1}\fix_{G_2}({\cal B})\delta_1\delta_2\ra\vert_B \le \la\fix_{G_1}({\cal B}),\delta_1^{-1}\fix_{G_2}({\cal B})\delta_1\ra\vert_B.$$
\noindent Hence $\la\fix_{G_1}({\cal
B}),\delta_2^{-1}\delta_1^{-1}\fix_{G_2}({\cal
B})\delta_1\delta_2\ra\vert_B\le \Pi_{j=1}^r\bar{H}_{B,j}$.  Continuing
inductively, we have that there exists $\delta\in\la\fix_{G_1}({\cal
B}),\fix_{G_2}({\cal B})\ra$ such that $\la\fix_{G_1}({\cal
B}),\delta^{-1}\fix_{G_2}({\cal
B})\delta\ra\vert_B\le\Pi_{j=1}^r\bar{H}_{B,j}$ for every $B\in{\cal B}$,
where each $\bar{H}_{B,j}$ is a transitive solvable $\bar{\pi}_j$-group of
degree $\bar{n}_j$. Note that $\delta^{-1}\fix_{G_2}({\cal B})\delta =
\fix_{\delta^{-1}G_2\delta}({\cal B})$ as $\delta/{\cal B} = 1$. For
ease of notation, we will replace $\delta^{-1}G_2\delta$ by $G_2$
and thus assume without loss of generality that $\la\fix_{G_1}({\cal
B}),\fix_{G_2}({\cal B})\ra\vert_B\le \Pi_{j=1}^rH_{B,j}$ for each
$B\in{\cal B}$.  By Lemma \ref{solproduct}, we may assume without
loss of generality that $\la\fix_{G_1}({\cal B}),\fix_{G_2}({\cal
B})\ra\vert_B = \Pi_{j=1}^r\bar{H}_{B,j}$ for each $B\in{\cal B}$.

As $\la\fix_{G_1}({\cal B}),\fix_{G_2}({\cal B})\ra\vert_B =
\Pi_{j=1}^rH_{B,j}$ for each $B\in{\cal B}$, $\la\fix_{G_1}({\cal
B}),\fix_{G_2}({\cal B})\ra\vert_B$ has a normal subgroup $L_B$
with orbits of size $\Pi_{j\in I}\bar{n}_j$, and so
$\la\fix_{G_1}({\cal B}),\fix_{G_2}({\cal B})\ra\vert_B$ admits a
complete block system ${\cal C}_B$ formed by the orbits of $L_B$,
$B\in{\cal B}$.  Also note that the $S_{\bar{\pi}}$-subgroup of
$\fix_{G_i}({\cal B})\vert_B$ must be contained in $L_B$, as
otherwise $(\fix_{G_i}({\cal B})\vert_B)/{\cal C}_B$ contains a
nontrivial normal subgroup with orbits of size dividing $\Pi_{j\in
I}\bar{n}_j$, $i = 1,2$. However, $(\fix_{G_i}({\cal
B})\vert_B)/{\cal C}_B$ is a solvable $\bar{\pi}'$-subgroup and
$\gcd(\Pi_{j\in I}\bar{n}_j,\Pi_{j\not\in I,j\in[r]}\bar{n}_j) = 1$, $i =
1,2$, a contradiction. Then $\fix_{G_i}({\cal B})\vert_B$, $i = 1,2$, admit
complete block systems ${\cal D}_B$ formed by the orbits of their
unique $S_{\bar{\pi}}$-subgroups, respectively, and these complete block
systems must be precisely ${\cal C}_B$, for $B\in{\cal B}$.

Let ${\cal C} = \cup_{B\in{\cal B}}{\cal C}_B$.  Clearly a normal
$S_{\bar{\pi}}$-subgroup of $\fix_{G_i}({\cal B})$ has relatively prime
order and index in $\fix_{G_k}({\cal B})$, $i = 1,2$.  Hence by
\cite[Theorem 1.1.13]{Hall1976}, an $S_{\bar{\pi}}$-subgroup of
$\fix_{G_i}({\cal B})$ is characteristic in $\fix_{G_i}({\cal
B})$, $i = 1,2$.  Whence an $S_{\bar{\pi}}$-subgroup of $\fix_{G_i}({\cal
B})$ is normal in $G_i$, $i = 1,2$.  Thus $G_i$ admits complete
block systems formed by the orbits of the $S_{\bar{\pi}}$-subgroup of
$\fix_{G_i}({\cal B})$, $i = 1,2$.  As each ${\cal C}_B$ is formed
by the orbits of the $S_{\bar{\pi}}$-subgroup of $\fix_{G_i}({\cal B})$
restricted to the block $B\in{\cal B}$, the orbits of the
$S_{\bar{\pi}}$-subgroup of $\fix_{G_i}({\cal B})$ form the complete block
system ${\cal C}$, $i = 1,2$.  Hence ${\cal C}$ is a block system
of $G_i$, $i = 1,2$, and so by Lemma \ref{blocktool}, ${\cal C}$
is a complete block system of $\la G_1, G_2\ra$.
\end{proof}

We now prove the main result of this section.

\begin{theorem}\label{main}
Let $n_1,\ldots,n_r$ be positive integers such that
$\gcd(n_i,n_j\cdot\varphi(n_j)) = 1$ if $i\not = j$, and $\pi_i$
be the set of distinct prime numbers dividing $n_i$.  Let $n =
n_1\cdots n_r$ and $m = \Omega(n)$.  If either
\begin{enumerate}
\item each $n_i$ is a prime-power, and $G_1,G_2$ are transitive nilpotent
groups of degree $n$ such that $\la G_1,G_2\ra$ is $m$-step
imprimitive, or
\item $G_1,G_2$ are transitive abelian groups of
degree $n$ such that $\la G_1,G_2\ra$ is $m$-step imprimitive,
\end{enumerate}
\noindent then there exists $\delta\in\la G_1,G_2\ra$ such that
$\la G_1,\delta^{-1}G_2\delta\ra = \Pi_{i=1}^rH_i$, where each
$H_i$ is a transitive solvable $\pi_i$-group of degree $n_i$.
\end{theorem}

\begin{proof}
Throughout the proof, if case (1) holds, we let $p_i$ be prime such that $n_i = p_i^{a_i}$, $a_i\ge 1$.  First suppose that case (1) holds and $r = 1$. Then $G_1\le\Pi_1$, $G_2\le \Pi_2$, where $\Pi_1,\Pi_2$
are Sylow $p_1$-subgroups of $\la G_1,G_2\ra$.  By a Sylow
Theorem, there exists $\delta\in\la G_1,G_2\ra$ such that
$\delta^{-1}G_2\delta\le\Pi_1$.  Then $\la
G_1,\delta^{-1}G_2\delta\ra$ is a $p_1$-group and so nilpotent.

In both cases, we proceed by induction on $m$.  Suppose that $m = 1$.  Then the only case that occurs is case (1) and $r = 1$.  The result then follows by arguments above.  Assume that the result is true for all $G_1$ and $G_2$ that satisfy the hypothesis with $\Omega(n) = m-1\ge 1$, and let $G_1,G_2\le S_n$ satisfy the hypothesis where
$\Omega(n) = m$. In case (1), by arguments above, we may assume that $r
\ge 2$. In case (2), if $r = 1$ then the result follows from Lemma \ref{piabelian}, so in any case we may assume without loss of generality that $r\ge 2$.  Let ${\cal B}_1$ be a complete block system of $\la
G_1,G_2\ra$ consisting of $p_i$ blocks of size $n/p_i$, where
$p_i\vert m_i$. Then ${\cal B}_1$ is a complete block system of
both $G_1$ and $G_2$. As both $G_1$ and $G_2$ are nilpotent,
$G_1/{\cal B}_1$ and $G_2/{\cal B}_1$ are nilpotent. We conclude
that both $G_1/{\cal B}_1$ and $G_2/{\cal B}_1$ are $p_i$-groups.
Hence there exists $\delta_1\in\la G_1,G_2\ra$ such that $\la
G_1,\delta^{-1}G_1\delta\ra/{\cal B}_1$ has order $p_i$. We thus
assume without loss of generality that $\la G_1,G_2\ra/{\cal B}_1$
has order $p_i$.

Let $\pi = \cup_{j\in J}\pi_j$ where $J = [r] - \{i\}$.  As $\vert
G_1/{\cal B}_1\vert = \vert G_2/{\cal B}_1\vert = p_i$, the
$S_\pi$-subgroups of $G_1$ and $G_2$ are contained in $\fix_{\la
G_1,G_2\ra}({\cal B}_1)$.  By Lemma \ref{bottomtool} and the
induction hypothesis, there exists $\delta_2\in\la G_1,G_2\ra$
such that $\la G_1,\delta_2^{-1}G_2\delta_2\ra$ admits a complete
block system ${\cal C}$ of $n_i$ blocks of size $n/n_i$ formed by
the $S_\pi$-subgroups of $G_1$ and $G_2$. We thus assume without
loss of generality that $\la G_1,G_2\ra$ admits ${\cal C}$ as a
complete block system.  Similarly, by Lemma \ref{bottomtool} and
the induction hypothesis, there exists $\delta_3\in\la G_1,G_2\ra$
such that $\la G_1,\delta_3^{-1}G_2\delta_3\ra$ admits a complete
block system ${\cal D}$ formed by the orbits of an
$S_{\pi_i}$-subgroup of $\fix_{G_k}({\cal B}_1)$, $k = 1,2$, (we
remark that if $n_i$ is prime, then ${\cal D}$ is trivial).  We
thus also assume without loss of generality that $\la G_1,G_2\ra$
admits ${\cal D}$ as well.

If $n_i\not = p_i$, then $\la G_1,G_2\ra/{\cal D}$ is $(m - (a_i -
1))$-step imprimitive, where $\Omega(n_i) = a_i$, and
$G_k/{\cal D}$ is nilpotent, $k = 1,2$. Hence by the induction
hypothesis there exists $\delta_4\in\la G_1,G_2\ra$ such that $\la
G_1,\delta_4^{-1}G_2\delta_4\ra/{\cal D}\le
P_i\times\Pi_{j=1,j\not = i}^rH_j'$, where $H_j'\le S_{n_j}$ is a
transitive solvable $\pi_j$-group, and $P_i$ is a $p_i$-group of
degree $p_i$. Then $\la G_1,\delta_4^{-1}G_2\delta_4\ra/{\cal D}$
admits a complete block system ${\cal E}'$ of $n/(n_i/p_i)$ blocks
of size $p_i$, so that $\la G_1,\delta_4^{-1}G_2\delta_4\ra$
admits a complete block system ${\cal E}$ of $n/n_i$ blocks of
size $n_i$.

If $n_i = p_i$, then as $\la G_1,G_2\ra$ is $m$-step imprimitive,
$\la G_1,G_2\ra$ admits a complete block system ${\cal B}_2$ such
that ${\cal B}_2\preceq{\cal B}_1$ and ${\cal B}_2$ consists of
$p_ip_j$ blocks of size $n/(p_ip_j)$ for some $p_j\vert n_j$ with
$j\not = i$. Then $G_1/{\cal B}_2$ and $G_2/{\cal B}_2$ are
nilpotent and transitive.  We conclude that $G_1/{\cal B}_2$ and
$G_2/{\cal B}_2$ are cyclic. By Theorem \ref{Paltool}, there
exists $\delta_3\in\la G_1,G_2\ra$ such that $\la
G_1,\delta_3^{-1}G_2\delta_3\ra/{\cal B}_2$ is cyclic.  We thus
assume without loss of generality that $\la G_1,G_2\ra/{\cal B}_2$
is cyclic.  Thus $\la G_1,G_2\ra/{\cal B}_2$ admits a complete
block system of $p_j$ blocks of size $p_i$, so that $\la
G_1,G_2\ra$ admits a complete block system ${\cal B}_1'$ of $p_j$
blocks of size $n/p_j$, and by Lemma \ref{mstep} $\fix_{\la
G_1,G_2\ra}({\cal B}_1')\vert_{B'}$ is $(m-1)$-step imprimitive
for every $B'\in{\cal B}_1'$.  Hence by Lemma \ref{bottomtool},
there exists $\delta_4\in\la G_1,G_2\ra$ such that $\la
G_1,\delta_4^{-1}G_2\delta_4\ra$ admits a complete block system
${\cal E}$ of $n/n_i$ blocks of size $n_i$ formed by the orbits of
an $S_{\pi_i}$-subgroup of $\fix_{G_k}({\cal B}_1')$, $k = 1,2$.
Hence regardless of the value of $n_i$, we may assume without loss
of generality that $\la G_1,G_2\ra$ admits ${\cal C}$ and ${\cal
E}$ as complete block systems.  As $\la G_1,G_2\ra$ admits a
complete block system ${\cal C}$ of $n_i$ blocks of size $n/n_i$,
$\la G_1,G_2\ra\le S_{n_i}\wr S_{n/n_i}$. As $\la G_1,G_2\ra$ also
admits ${\cal E}$ as a complete block system and $\gcd(n_i,n/n_i)
= 1$, we have that $\la G_1,G_2\ra\le S_{n/n_i}\wr S_{n_i}$. We
conclude that $\la G_1,G_2\ra\le (S_{n_i}\wr
S_{n/n_i})\cap(S_{n/n_i}\wr S_{n_i}) = S_{n_i}\times S_{n/n_i}$.
We now consider (1) and (2) separately.

(1)  By the induction hypothesis, we may, after a suitable
conjugation, assume that $\la G_1,G_2\ra/{\cal D}\le
\Pi_{j\in[r]-\{i\}}S_{m_j}$, so that $\la G_1,G_2\ra\le
\Pi_{j=1}^r S_{m_j}$.  The result then follows by inductively
applying a Sylow Theorem and then Lemma \ref{solproduct}.

(2)  By Lemma \ref{regularabelian} every complete block system is a normal
complete block system.  As ${\cal L}_{\la G_1,G_2\ra}$ is a
modular lattice by Lemma \ref{lattice}, it follows by the
Jordan-Dedekind Chain Condition \cite[pg. 119]{Hall1976} that all
finite chains between two elements have the same length.  As $\la
G_1,G_2\ra$ admits ${\cal E}$ as a complete block system, any
maximal chain between the complete block systems consisting of
singletons and the complete block system consisting of one block
that contain ${\cal E}$ must have length $m$ as $\la G_1,G_2\ra$
is normally $m$-step imprimitive. We conclude that $\la
G_1,G_2\ra/{\cal E}$ is $(m-a_i)$-step imprimitive, so by the
induction hypothesis we may assume after a suitable conjugation
that $\la G_1,G_2\ra/{\cal E}\le\Pi_{j=1,j\not = i}^rH_j$, where
$H_j\le S_{n_j}$ is a transitive solvable $\pi_j$-group.
Similarly, we may assume that $\la G_1,G_2\ra/{\cal C}\le H_i$,
where $H_i$ is a transitive solvable $\pi_i$-group.  As $C\cap E$ is a singleton, for every $C\in{\cal C}$, $E\in{\cal
E}$, we have that $\la G_1,G_2\ra\le\Pi_{j=1}^rH_j$. By Lemma
\ref{solproduct}, we may assume that $\la G_1,G_2\ra =
\Pi_{j=1}^rH_j$ as required.  The result then follows by
induction.
\end{proof}

It may be worthwhile restating Theorem \ref{main} (1) in the
following form:

\begin{corollary}\label{nilmain}
Let $n = p_1^{a_1}\cdots p_r^{a_r}$, the prime-power decomposition
of $n$, be such that $p_i\not\vert(p_j-1)$, $1\le i,j\le r$.  Let
$\Omega(n) = m$.  If $G_1,G_2$ are transitive nilpotent
groups of degree $n$ such that $\la G_1,G_2\ra$ is $m$-step
imprimitive, then there exists $\delta\in \la G_1,G_2\ra$ such
that $\la G_1,\delta^{-1}G_2\delta\ra$ is nilpotent.
\end{corollary}

\section{Solving Sets}

In this section, we further develop the terminology regarding solving sets as well as the characterizations of when a particular set is a solving set that will be needed for our main results.

\begin{definition}
Let $G$ be a group and define $g_L:G\rightarrow G$ by $g_L(x) =
gx$.  Let $G_L = \{g_L:g\in G\}$.  Then $G_L$ is the {\bf
left-regular representation of $G$}.  We define a {\bf Cayley
object of $G$} to be a combinatorial object $X$ (e.g. digraph,
graph, design, code) such that $G_L\le\Aut(X)$, where $\Aut(X)$ is
the {\bf automorphism group of $X$} (note that this implies that
the vertex set of $X$ is in fact $G$).  If $X$ is a Cayley object
of $G$ in some class ${\cal K}$ of combinatorial objects with the
property that whenever $Y$ is another Cayley object of $G$ in
${\cal K}$, then $X$ and $Y$ are isomorphic if and only if they
are isomorphic by a group automorphism of $G$, then we say that
$X$ is a {\bf CI-object of $G$ in ${\cal K}$}.  If every Cayley
object of $G$ in ${\cal K}$ is a CI-object of $G$ in ${\cal K}$,
then we say that $G$ is a {\bf CI-group with respect to ${\cal
K}$}.  If $G$ is a CI-group with respect to every class of
combinatorial objects, then $G$ is a {\bf CI-group}.
\end{definition}

\begin{definition}
Let $G$ be a finite group.  We say that $S\subseteq S_G$ is a {\bf
solving set for a Cayley object $X$} in a class of Cayley objects
${\cal K}$ if for every $X'\in {\cal K}$ such that $X\cong X'$,
there exists $s\in S$ such that $s(X) = X'$, $s(1_G) = 1_G$ for every $s\in S$, and $\Aut(G)\le S$. We
say that $S\subseteq S_G$ is a {\bf solving set for a class ${\cal
K}$ of Cayley objects of $G$} if whenever $X,X'\in {\cal K}$ are
Cayley objects of $G$ and $X\cong X'$, then $s(X) = X'$ for some
$s\in S$, and $s(1_G) = 1_G$ for every $s\in S$, and $\Aut(G)\le S$.  Finally, a set
$S$ is a {\bf solving set for $G$} if whenever $X,X'$ are isomorphic
Cayley objects of $G$ in any class ${\cal K}$ of combinatorial objects,
then $s(X) = X'$ for some $s\in S$, $s(1_G) = 1_G$ for all $s\in S$, and $\Aut(G)\le S$.
\end{definition}

\begin{remark}
Note that the definition of a solving set given above differs from
those in \cite{Muzychuk1999} and \cite{Dobson2003a}, as here, to simplify both the statements of results and their proofs, we insist that every
element of the solving set fixes $1_G$.  It is easy to see that $\alpha^{-1}G_L\alpha = G_L$ for every $\alpha\in\Aut(G)$, so the image of a Cayley object of $G$ under a group automorphism of $G$ is a Cayley object of $G$.  That is, in order to test for isomorphism, automorphisms of $G$ must be considered.  However, it is not always the case that {\it every} automorphism of $G$ needs to be considered when testing for isomorphism.  For example, Cayley graphs of cyclic groups of order $n$ each have an automorphism $x\to -x$ that is also a group automorphism of $\Z_n$, and so the image of a Cayley graph under this automorphism of $\Z_n$ is itself.  So, while our definition of a solving set is convenient for this paper, it does not always capture the idea behind a solving set (i.e. that it should be as small as possible) exactly, but will only necessarily include extra automorphism of $G$ (which could then be excluded).  Also note that in
\cite{Muzychuk1999} and \cite{Dobson2003a}, solving sets were only defined for
abelian groups.
\end{remark}

Let $X$ be a Cayley object of $G$ in ${\cal K}$.  We define a
{\bf CI-extension of $G$ with respect to $X$}, denoted by
$\CI(G,X)$, to be a set of permutations in $S_G$ that each
fix $1_G$ and whenever $\delta\in S_G$ such that
$\delta^{-1}G_L\delta\le\Aut(X)$, then there exists $v\in\Aut(X)$
such that $v^{-1}\delta^{-1}G_L\delta v = t^{-1}G_Lt$ for some
$t\in \CI(G,X)$.

\begin{lemma}\label{existence}
Let $G$ be a finite group, and $X$ a Cayley object of $G$ in some
class ${\cal K}$ of combinatorial objects.  Then $\CI(G,X)$ exists.
\end{lemma}

\begin{proof}
To show existence, we only need show that there is a set of
permutations $T$ in $S_G$ such that whenever $\delta\in S_G$ such
that $\delta^{-1}G_L\delta\le\Aut(X)$ and $v\in\Aut(X)$, then
$v^{-1}\delta^{-1}G_L\delta v = t^{-1}G_Lt$ for some $t\in
T$ and $t(1_G) = 1_G$.  This follows almost immediately.  As
$\delta v$ is a permutation, there exists $g\in G$ such that
$\delta v(1_G) = g$.  Let $t_{\delta v} = g^{-1}_L\delta v$.  Then
$t_{\delta v}(1_G) = g^{-1}_L\delta v(1_G) = g^{-1}_L(g) = g^{-1}
g = 1_G$. Furthermore, $t_{\delta v}^{-1}G_Lt_{\delta v} =
v^{-1}\delta^{-1}g_LG_Lg_L^{-1}\delta v =
v^{-1}\delta^{-1}G_L\delta v$, and existence is established with $T = \{t_{\delta v}:\delta^{-1}G_L\delta\le\Aut(X), v\in \Aut(X)\}$.
\end{proof}

Note for $X$ a Cayley object of $G$ in ${\cal K}$,
$\CI(G,X)$ is not unique as if $T$ is CI-extension of $X$ with
respect to $G$, then for $\alpha\in\Aut(G)$, $\{\alpha t:t\in T\}$
is also a CI-extension of $X$ with respect to $G$.  The following result shows the importance of
$\CI(G,X)$, as if $\CI(G,X)$ is known, then the isomorphism problem
is solved.

\begin{lemma}\label{contool1}
Let $G$ be a finite group, and $X$ a Cayley object of $G$ in some
class ${\cal K}$ of combinatorial objects.  Then the following are
equivalent:
\begin{enumerate}
\item $S = \{\alpha t:\alpha\in\Aut(G), t\in T \}$ is a solving set for $X$,
\item $T$ is a $\CI(G,X)$.
\end{enumerate}
\end{lemma}

\begin{proof}
1) implies 2).   Let $\delta\in S_G$ such that
$\delta^{-1}G_L\delta\le\Aut(X)$.  Then $\delta(X)$ is a Cayley
object of $G$ in ${\cal K}$  as $\Aut(\delta(X)) = \delta\Aut(X)\delta^{-1}\ge G_L$.  As $S$ is a solving set for $X$,
$\delta(X) = s(X)$ for some $s\in S$, and $s = \alpha t$ for some
$\alpha\in\Aut(G)$ and $t\in T$.  Thus $v =
\delta^{-1}s\in\Aut(X)$. Then
\begin{eqnarray*}
v^{-1}\delta^{-1}G_L\delta v & = & s^{-1}\delta \delta^{-1}G_L\delta\delta^{-1}s\\
  & = & s^{-1}G_Ls = t^{-1}\alpha^{-1}G_L\alpha t\\
  & = & t^{-1}G_Lt
\end{eqnarray*}

\noindent and $T$ is a $\CI(X,G)$.

2) implies 1).  Let $X$ and $X'$ be isomorphic Cayley objects of
$G$ in ${\cal K}$.  Then there exists $\delta\in S_G$ such that
$\delta(X) = X'$.  As $G_L\le\Aut(X')$,
$\delta^{-1}G_L\delta\le\Aut(X)$. As $T$ is a $\CI(X,G)$, there exists
$t\in T$ and $v\in\Aut(X)$ such that
$v^{-1}\delta^{-1}G_L\delta v = t^{-1}G_Lt$.  Hence
$tv^{-1}\delta^{-1}G_L\delta v t^{-1} = G_L$. As $G_L$ is
transitive, there exists $h\in G$ such that $h_L\delta
vt^{-1}(1_G) = 1_G$.  Clearly then $tv^{-1}\delta^{-1}h_L^{-1}G_L
h_L\delta v t^{-1} = G_L$ so that $h_L\delta vt^{-1}$ normalizes $G_L$ and fixes $1_G$.  By \cite[Corollary 4.2B]{DixonM1996}, we have that $h_L\delta vt^{-1} = \alpha\in\Aut(G)$. Then $\alpha t
v^{-1}\delta^{-1}h_L^{-1} = (1_G)_L$ and
%$$(h_L\delta v t^{-1})^{-1}g_Lh_L\delta v t^{-1} = \alpha^{-1}g_L\alpha$$
%\noindent for every $g\in G$.  Thus $\alpha t
%v^{-1}\delta^{-1}h_L^{-1}$ is contained in the center of $G_L$. As
%the centralizer of $G_L$ in $S_G$ is $G_R$ (that
%$\C_{S_G}(G_L)\cong G$ follows from \cite[Theorem 4.2A (iii)]{DixonM1996}.
%It is then a simple computation to show that $\C_{S_G}(G_L) =
%G_R$), we have that $\alpha tv^{-1}\delta^{-1}h_L^{-1}\in G_R$. As
%$\alpha(1_G) = 1_G$ and $h_L\delta vt^{-1}(1_G) = 1_G$, we have
%that $\alpha tv^{-1}\delta^{-1}h_L^{-1}(1_G) = 1_G$. As the only
%element of $G_R$ with a fixed point is $(1_G)_R = (1_G)_L$, we
%have that $\alpha tv^{-1}\delta^{-1}h_L^{-1} = (1_G)_L$.  Thus
$\alpha t = h_L\delta v$.  Then $\alpha t(X) = h_L\delta v(X) =
h_L\delta(X) = h_L(X') = X'$.
\end{proof}

The following result shows that if a solving set for $X$ has been found, then some $\CI(G,X)$ has also been found.

\begin{lemma}\label{CIX}
Let $G$ be a group, $X$ a Cayley object of $G$, and $S$ a solving set for $X$.  Define an equivalence relation $\equiv$ on $S$ by $s_1\equiv s_2$ if and only if $s_1 = \alpha s_2$ for some $\alpha\in\Aut(G)$.  Let $T$ be a set consisting of one representative from each equivalence class of $\equiv$.  Then $T$ is a $\CI(G,X)$.
\end{lemma}

\begin{proof}
It is straightforward to show that $\equiv$ is indeed an equivalence relation.  Choose a $T$ as is given in the statement.   Let $X'$ be a Cayley object of $G$ isomorphic to $X$ with $\delta:X\to X'$ an isomorphism.  Then $\delta^{-1}G_L\delta\le\Aut(X)$.  Also, as $S$ is a solving set for $X$, there exists $s\in S$ such that $s(X) = X'$ so that $v = \delta^{-1}s\in\Aut(X)$.  Let $t\in T$ such that $t\equiv s$ so that $\alpha t = s$ for some $\alpha\in\Aut(G)$.  Then $v^{-1}\delta^{-1}G_L\delta v = t^{-1}\alpha^{-1}G_L\alpha t = t^{-1}G_Lt$.
\end{proof}

Let ${\cal K}$ be a class of combinatorial objects, and $G$ a
group.  We define a {\bf CI-extension of $G$ with respect to
${\cal K}$}, denoted by $\CI(G,{\cal K})$, to be a set of
permutations in $S_G$ that each fix $1_G$ and whenever $X\in{\cal
K}$ is a Cayley object of $G$ and $\delta\in S_G$ such that
$\delta^{-1}G_L\delta\le\Aut(X)$, then there exists $t\in
\CI(G,{\cal K})$ and $v\in\Aut(X)$ such that
$v^{-1}\delta^{-1}G_L\delta v = t^{-1}G_Lt$. The proofs of the
following results are straightforward.

\begin{lemma}\label{contool2}
Let $G$ be a finite group, and ${\cal K}$ a class of combinatorial
objects.  Then the following are equivalent:
\begin{enumerate}
\item $S = \{\alpha t:\alpha\in\Aut(G), t\in T\}$ is a solving set for $G$ in ${\cal K}$,
\item $T$ is a $\CI(G,{\cal K})$.
\end{enumerate}
\end{lemma}

\begin{lemma}\label{CIK}
Let $G$ be a group, ${\cal K}$ a class of combinatorial objects, and $S$ a solving set for $G$ in ${\cal K}$.  Define an equivalence relation $\equiv$ on $S$ by $s_1\equiv s_2$ if and only if $s_1 = \alpha s_2$ for some $\alpha\in\Aut(G)$.  Let $T$ be a set consisting of one representative from each equivalence class of $\equiv$.  Then $T$ is a $\CI(G,{\cal K})$.
\end{lemma}

Let $G$ be a finite group.  We define a {\bf CI-extension of
$G$}, denoted by $\CI(G)$, to be a set of permutations in
$S_G$ that each fix $1_G$ and whenever $X\in{\cal K}$ is a Cayley
object of $G$ in some class ${\cal K}$ of combinatorial objects,
and $\delta\in S_G$ such that $\delta^{-1}G_L\delta\le\Aut(X)$,
then there exists $t\in \CI(G)$ and $v\in\la G_L,\delta^{-1}G_L\delta\ra$ such that
$v^{-1}\delta^{-1}G_L\delta v = t^{-1}G_Lt$. Repeated application of Lemma \ref{existence} for every combinatorial object $X$ in every class ${\cal K}$ of combinatorial objects shows that $\CI(G)$ exists.  The proofs of the
following results are straightforward.

\begin{lemma}\label{contool3}
Let $G$ be a finite group.  Then the following are equivalent:
\begin{enumerate}
\item $S = \{\alpha t:\alpha\in\Aut(G), t\in T\}$ is a solving set for $G$,
\item $T$ is a $\CI(G)$.
\end{enumerate}
\end{lemma}

\begin{lemma}\label{CIG}
Let $G$ be a group, and $S$ a solving set for $X$.  Define an equivalence relation $\equiv$ on $S$ by $s_1\equiv s_2$ if and only if $s_1 = \alpha s_2$ for some $\alpha\in\Aut(G)$.  Let $T$ be a set consisting of one representative from each equivalence class of $\equiv$.  Then $T$ is a $\CI(G)$.
\end{lemma}

\section{Applications}

At the present time, the isomorphism problem has not been solved for any nilpotent group that is not abelian in any class of combinatorial objects, so there are not at this time any applications for Theorem \ref{main} (1) (although as soon as the isomorphism problem has been solved for any nonabelian $p$-group in certain classes of combinatorial objects, such as color digraphs, that will change immediately).  We do though have an application of Theorem \ref{main} (2) which will not only provide new examples of CI-groups with respect to color digraphs, but also illustrate how Theorem \ref{main} (2) generalizes the main result of \cite{Dobson2003a}.

%For clarity, we now give the basic definitions concerning the CI-problem, some of which have already been stated.

%\begin{defin}
%Let $G$ be a group and define $g_L:G\rightarrow G$ by $g_L(x) =
%gx$.  Let $G_L = \{g_L:g\in G\}$.  Then $G_L$ is the {\bf
%left-regular representation of $G$}.  We define a {\bf Cayley
%object of $G$} to be a combinatorial object $X$ (e.g. digraph,
%graph, design, code) such that $G_L\le\Aut(X)$, where $\Aut(X)$ is
%the {\bf automorphism group of $X$} (note that this implies that
%the vertex set of $X$ is in fact $G$).  If $X$ is a Cayley object
%of $G$ in some class ${\cal K}$ of combinatorial objects with the
%property that whenever $Y$ is another Cayley object of $G$ in
%${\cal K}$, then $X$ and $Y$ are isomorphic if and only if they
%are isomorphic by a group automorphism of $G$, then we say that
%$X$ is a {\bf CI-object of $G$ in ${\cal K}$}.  If every Cayley
%object of $G$ in ${\cal K}$ is a CI-object of $G$ in ${\cal K}$,
%then we say that $G$ is a {\bf CI-group with respect to ${\cal
%K}$}.  If $G$ is a CI-group with respect to every class of
%combinatorial objects, then $G$ is a {\bf CI-group}.
%\end{defin}

The following result weakens the hypothesis (replacing normally
$s$-step imprimitive with $s$-step imprimitive) of \cite[Theorem
16]{Dobson2003a} and generalizes this result from abelian to nilpotent
groups.

\begin{theorem}\label{bigtool}
Let $n_1,\ldots,n_r$ be positive integers such that
$\gcd(n_i,n_j\cdot\varphi(n_j)) = 1$ if $i\not = j$, and $\pi_i$
be the set of distinct prime numbers dividing $n_i$.  Let $n =
n_1\cdots n_r$, $\Omega(n) = m$, and $G$ a nilpotent group of degree $n$.  Let $G = \Pi_{i=1}^rN_i$ where each $N_i$ is a $\pi_i$-subgroup of $G$, and $S(i)$ a solving set for $N_i$.  If
\begin{enumerate}
\item each $n_i$ is prime-power or $G$ is abelian, and
\item whenever $\delta\in S_G$ there exists $\phi\in\la G_L,\delta^{-1}G_L\delta\ra$ such that $\la G_L,\phi^{-1}\delta^{-1}G_L\phi\delta\ra$ is $m$-step imprimitive,
\end{enumerate}
\noindent then $\Pi_{i=1}^rS(i)$ is a solving set for $G$.
\end{theorem}

\begin{proof}
Let $\delta\in S_G$.  By the hypothesis, we have that there exists $\phi\in\la G_L,\delta^{-1}G_L\delta\ra$ such that $\la G_L,\phi^{-1}\delta^{-1}G_L\delta\phi\ra$ is $m$-step imprimitive.  By Theorem \ref{main}, there exists
$\omega\in\la G_L,\phi^{-1}\delta^{-1}G_L\delta\phi\ra$ such
that $L = \la G_L, \omega^{-1}\phi^{-1}\delta^{-1}G_L\delta\phi\omega\ra =
\Pi_{i=1}^rL_i$, where each $L_i$ is a transitive $\pi_i$-group of
degree $n_i$, $1\le i\le r$.  Let $\CI(N_i)$ be a CI-extension of $N_i$ as given by Lemma \ref{CIG}.  As $S(i)$ is a solving set for $N_i$, by Lemma \ref{contool3} there exists $t_i\in \CI(N_i)$ and
$v_i\in L_i$ such that $v_i^{-1}((\omega^{-1}\phi^{-1}\delta^{-1}G_L\delta\phi\omega)/{\cal
C}_i)v_i = t_i^{-1}(G_L/{\cal C}_i)t_i$, where ${\cal C}_i$ is the complete block system of $L$ formed by the orbits of $\Pi_{j=1,j\not = i}^rL_i$.  Let $t = (t_1,\ldots,t_r)$ and $v = (v_1,\ldots,v_r)$. Note that $t$ fixes $1_G$.  As $L = \Pi_{i=1}^rL_i$, we have that $v\in L$ and $t^{-1}G_Lt =
v^{-1}\omega^{-1}\phi^{-1}\delta^{-1}G_L\delta\phi\omega v$. By definition, $\Pi_{i=1}^r\CI(N_i)$ is a CI-extension of $G$, and so by Lemma \ref{contool3}, $S = \{\alpha t:t\in\Pi_{i=1}^r\CI(N_i), \alpha\in\Aut(G)\}$ is a solving set for $G$.  Finally, it is not difficult to see that $\Aut(G) = \Pi_{i=1}^r\Aut(N_i)$ as $G$ is nilpotent and if $i\not = j$ then $\gcd(n_i,n_j) = 1$, and so $\alpha = (\alpha_1,\alpha_2,\ldots,\alpha_r)$, $\alpha_i\in\Aut(N_i)$.  Then $\alpha t = (\alpha_1 t_1,\alpha_2 t_2,\ldots,\alpha_r t_r)$, $\alpha_i\in\Aut(N_i)$ and $t_i\in \CI(N_i)$.  Thus $\alpha t\in\Pi_{i=1}^rS(i)$, $S\le\Pi_{i=1}^r S(i)$, and $\Pi_{i=1}^rS(i)$ is a solving set for $G$.
%
%Now let $u\in v^{-1}\omega^{-1}\phi^{-1}\delta^{-1}G_L\delta\phi\omega v$ such that $\delta\phi\omega vu(1_G) = %1_G$, and $\gamma = \phi\omega vu$.  Then $\gamma\in\la G_L,\delta^{-1}G_L\delta\ra$, and $\delta\gamma$ fixes %$1_G$.
%Also, $t\gamma^{-1}\delta^{-1}G_L\delta\gamma t^{-1} = G_L$, and so by \cite[Corollary 4.2B]{DixonM1996} it follows %that $\delta\gamma t^{-1}\in\Aut(G)\cdot G_L$.  As both $t$ and $\delta\gamma$ fix $1_G$, we have that $\delta\gamma %t^{-1} = \alpha\in\Aut(G)$, and so $\delta\gamma = \alpha t$.  As $\delta\gamma(X) = X'$, $\alpha t(X) = X'$.  %Finally, it is not difficult to see that $\Aut(G) = \Pi_{i=1}^r\Aut(P_i)$ as $G$ is nilpotent, and so $\alpha = %(\alpha_1,\alpha_2,\ldots,\alpha_r)$, $\alpha_i\in\Aut(P_i)$.  Then $\alpha t = (\alpha_1 t_1,\alpha_2 %t_2,\ldots,\alpha_r t_r)$, $\alpha_i\in\Aut(P_i)$ and $t_i\in \CI(P_i)$.  Thus $\alpha t\in\Pi_{i=1}^rS(i)$ and the %result follows.
\end{proof}

\begin{definition}
{\rm Let $\Omega$ be a set.  A {\bf $k$-ary relational structure} on
$\Omega$ is an ordered pair $(\Omega, U)$, where $U\subseteq \Omega^k = \Pi_{i=1}^k\Omega$.  A group
$G\le S_{\Omega}$ is called {\bf $k$-closed} if $G$ is the
intersection of the automorphism groups of some set of $k$-ary
relational structures.  The {\bf $k$-closure of $G$}, denoted
$G^{(k)}$, is the intersection of all $k$-closed subgroups of
$S_{\Omega}$ that contain $G$.}
\end{definition}

Note that a $2$-closed group is the automorphism group of a color digraph.  The following result of Kalu\v znin and Klin \cite{KaluzninK1976} will prove useful.

\begin{lemma}\label{product}
Let $G\le S_X$ and $H\le S_Y$.  Let $G\times H$ act canonically on
$X\times Y$.  Then $(G\times H)^{(k)} = G^{(k)}\times H^{(k)}$ for
every $k\ge 2$.
\end{lemma}

If, in Theorem \ref{bigtool}, ${\cal K}$ is the class of $k$-ary
relational structures, and the groups $L/{\cal C}_i$ are as in the
proof of Theorem \ref{bigtool}, then by Lemma \ref{product} we may assume that each
$L/{\cal C}_i$ is $k$-closed (although there is no reason to believe that each $L/{\cal C}_i$ is a $\pi_i$-subgroup - but this fact is not used in the proof of Theorem \ref{bigtool}).  Proceeding as in Theorem
\ref{bigtool} and applying Lemma \ref{contool2} instead of Lemma
\ref{contool3}, we have the following result.

\begin{corollary}\label{kcor}
Let $n_1,\ldots,n_r$ be positive integers such that
$\gcd(n_i,n_j\cdot\varphi(n_j)) = 1$ if $i\not = j$, and $\pi_i$
be the set of distinct prime numbers dividing $n_i$.  Let $n =
n_1\cdots n_r$, $\Omega(n) = m$, and $G$ a nilpotent group of degree $n$.  Let $G = \Pi_{i=1}^rN_i$ where each $N_i$ is a $\pi_i$-subgroup of $G$, and $S(i)$ a solving set for $N_i$ in the class of $k$-ary relational structures.  If
\begin{enumerate}
\item each $n_i$ is prime or $G$ is abelian, and
\item whenever $\delta\in S_G$ there exists $\phi\in\la G_L,\delta^{-1}G_L\delta\ra$ such that $\la G_L,\phi^{-1}\delta^{-1}G_L\phi\delta\ra$ is $m$-step imprimitive,
\end{enumerate}
\noindent then $\Pi_{i=1}^rS(i)$ is a solving set for $G$ in the class of $k$-ary relational structures.
\end{corollary}

%\begin{cor}
%Let $m$ be a positive integer with $\gcd(m,\varphi(m)) = 1$ and $m
%= p_1p_2\cdots p_r$ be the prime power decomposition of $m$. Let
%$n = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$, where $a_\ell\ge 1$ is a
%positive integer.  Let $X$ be $k$-ary Cayley object of a nilpotent
%group $G = \Pi_{i=1}^rP_i$ of order $n$, where $P_i$ is a Sylow
%$p_i$-subgroup of $G$.  Let $S(i)$ be the solving set for $P_i$ in
%the class of $k$-ary relational structures. If, whenever
%$\delta\in S_G$ such that $\delta^{-1}G_L\delta\le\Aut(X)$ there
%exists $\omega\in\Aut(X)$ such that $\la
%G_L,\omega^{-1}\delta^{-1}G_L\delta\omega\ra$ is $s$-step
%imprimitive, then there exists a solving set for $X$ that is
%contained in $\Pi_{i=1}^rS(i)$.
%\end{cor}

%Babai \cite{Babai1977} characterized the CI-property in the following
%manner.

%\begin{lem}\label{CIso}
%For a Cayley object $X$ of $G$ in ${\cal K}$ the following are
%equivalent:
%\begin{enumerate}
%\item $X$ is a CI-object,
%\item given a permutation $\phi \in S_G$ such that $\phi^{-1}G_L\phi\le \Aut(X)$,
%$G_L$ and $\phi^{-1}G_L\phi$ are conjugate in $\Aut(X)$.
%\end{enumerate}
%\end{lem}

%We remark that in \cite{Babai1977}, similar characterizations of CI-groups with respect to classes of combinatorial %objects and to all classes of combinatorial objects were also give.

We now give the promised application of Theorem \ref{main} (2) which gives new CI-groups with respect to Cayley color digraphs of a particular group.  Using the results in \cite{Dobson2003a}, this result could be obtained but only in the special cases where the following additional arithmetic conditions hold: $p$ does not divide $q - 1$, and each $n_i$ is prime.  Before proceeding, we need a preliminary lemma.

\begin{lemma}\label{bigblocks}
Let $n$ be a positive integer with a prime divisor $p\vert n$ such that $n/p < p$.  If $G$ is a regular group of order $n$, and $\phi\in S_n$, then there exists $\delta\in \la G,\phi^{-1}G\phi\ra$ such that $\la G,\delta^{-1}\phi^{-1}G\phi\delta\ra$ admits a normal complete block system with blocks of size $p$.
\end{lemma}

\begin{proof}
First observe that as $n/p < p$, $G$ must have a unique cyclic Sylow $p$-subgroup $P$ of order $p$.  So $P\tl G$ and $G$ admits a normal complete block system ${\cal B}$ consisting of blocks of size $p$, formed by the orbits of $P$.  Furthermore, a Sylow $p$-subgroup of $S_n$ is of the form $1_{S_{n/p}}\wr\Z_p$, so we see that $\la P\vert_B:B\in{\cal B}\ra$ is a Sylow $p$-subgroup of $S_n$.  By a Sylow theorem, there exists $\delta\in \la G,\phi^{-1}G\phi\ra$ such that $\delta^{-1}\phi^{-1}G\phi\delta\le \la P\vert_B:B\in{\cal B}\ra$.  Let $P_1$ be the largest subgroup of $\la G,\delta^{-1}\phi^{-1}G\phi\delta\ra$ contained in $\la P\vert_B:B\in{\cal B}\ra$.  Then $G$ normalizes $P_1$ as does $\delta^{-1}\phi^{-1}G\phi\delta$.  Hence the orbits of $P_1$, which is ${\cal B}$, is a complete block system of $\la G,\delta^{-1}\phi^{-1}G\phi\delta\ra$ and the result follows.
\end{proof}

\begin{theorem}\label{app}
Let $p$ and $q$ be distinct primes with $p^2 < q$, and $q_1,\ldots,q_r$ distinct primes such that $qp^2 < q_1$ and $qp^2q_1\cdots q_i < q_{i+1}$, $1\le i\le r - 1$.  Let $m = q_1\cdots q_r$, $n_0 = p^2q$, and $n_1,\ldots,n_s$ divisors of $m$ such that $n_1\cdots n_s = m$. If $\gcd(n_i,n_j\cdot\varphi(n_j)) = 1$ then $\Z_q\times\Z_p^2\times\Z_m$ is a CI-group with respect to digraphs.
\end{theorem}

\begin{proof}
Let $\Gamma$ be a Cayley color digraph of $G = \Z_q\times\Z_p^2\times\Z_m$, and $\phi\in S_G$ such that $\phi^{-1}G_L\phi\le\Aut(\Gamma)$.  We first show by induction on $r$ that $\la G_L,\phi^{-1}G_L\phi\ra$ is $s$-step imprimitive, where $s = r + 3$.  If $r = 0$, then as $p^2 < q$ by Lemma \ref{bigblocks} we may assume after an appropriate conjugation that $\la G,\phi^{-1}G\phi\ra$ admits a (normal) complete block system ${\cal B}_q$ with blocks of size $q$. Then $G_L/{\cal B}_q$ and $\phi^{-1}G_L\phi/{\cal B}_q$ are contained in conjugate Sylow $p$-subgroups of $\la G_L,\phi^{-1}G_L\phi\ra/{\cal B}_q$, and so after another appropriate conjugation we may assume that $\la G_L,\phi^{-1}G_L\phi\ra/{\cal B}_q$ is a $p$-group.  The center of $\la G_L,\phi^{-1}G_L\phi\ra/{\cal B}_q$ contains an element of order $p$ whose orbits give a complete block system ${\cal B}_p$ consisting of blocks of size $p$.  Then ${\cal B}_p$ induces a complete block system ${\cal B}_q\prec {\cal B}_{pq}$ of $\la G_L,\phi^{-1}G_L\phi\ra$ with blocks of size $pq$.  Hence $\la G_L,\phi^{-1}G_L\phi\ra$ is $3$-step imprimitive.  Assume that $\la G_L,\phi^{-1}G_L\phi\ra$ is $s$-step imprimitive when $r\ge 0$ and let $G_L$ be such that $m$ is a product of $r + 1$ primes.  Again, after an application of Lemma \ref{bigblocks}, we may assume that $\la G_L,\phi^{-1}G_L\phi\ra$ admits a complete block system ${\cal B}_{r + 1}$ with blocks of size $q_{r + 1}$.  By induction we may assume that $\la G_L,\phi^{-1}G_L\phi\ra/{\cal B}_{q_{r+1}}$ is $(r + 3)$-step imprimitive.  It is then not difficult to see that $\la G_L,\phi^{-1}G_L\phi\ra$ is $(r + 4)$-step imprimitive.  By induction, we then have $\la G_L,\phi^{-1}G_L\phi\ra$ is $(r + 3)$-step imprimitive as required.

Write $G_L = \Pi_{i=0}^sG_i$, where $G_i\le G_L$ is of order $n_i$.  As $G_i$ is a CI-group with respect to Cayley color digraphs by \cite{KovacsM2009} if $i = 0$ and by \cite{Muzychuk1997} otherwise, by Corollary \ref{kcor}, we see that if $S$ is a solving set of $G$ in the class of color digraphs then $S \subseteq \Pi_{i=1}^r\Aut(G_i)\le\Aut(G)$.  Hence $G$ is a CI-group with respect to color digraphs.
\end{proof}

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
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\end{document}

\section{Introduction}

The reconstruction conjecture states that the multiset of unlabeled
vertex-deleted subgraphs of a graph determines the graph, provided it
has at least three vertices.  This problem was independtly introduced
by Ulam~\cite{Ulam} and Kelly~\cite{Kelly}.  The reconstruction
conjecture is widely studied
\cite{Bollobas,FGH,HHRT,KSU,RM,RR,Stockmeyer} and is very interesting
because...... See \cite{WikipediaReconstruction} for more about the
reconstruction conjecture.

\begin{definition}
  A graph is \emph{fabulous} if....
\end{definition}

\begin{theorem}
  \label{Thm:FabGraphs}
  All planar graphs are fabulous.
\end{theorem}

\begin{proof}
  Suppose on the contrary that some planar graph is not fabulous....
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Broglington Manifolds}

This section describes background information about Broglington
Manifolds.

\begin{lemma}
  \label{lem:Technical}
  Broglington manifolds are abundant.
\end{lemma}

\begin{proof}

\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proof of Theorem~\ref{Thm:FabGraphs}}

In this section we complete the proof of Theorem~\ref{Thm:FabGraphs}.

\begin{proof}[Proof of Theorem~\ref{Thm:FabGraphs}]
Let $G$ be a graph... Hence
 % use the amsmath align environment for multi-line equations
  \begin{align}
    |X| &= abcdefghijklmnopqrstuvwxyz \nonumber\\
    &= \alpha\beta\gamma
  \end{align}
  This completes the proof of Theorem~\ref{Thm:FabGraphs}.
\end{proof}

\begin{figure}[!h]
  \begin{center}
    % use \includegraphics to import figures
    % \includegraphics{filename}
  \end{center}
  \caption{\label{fig:InformativeFigure} Here is an informative
    figure.}
\end{figure}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
Thanks to Professor Querty for suggesting the proof of
Lemma~\ref{lem:Technical}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% \bibliographystyle{plain}
% \bibliography{myBibFile}
% If you use BibTeX to create a bibliography
% then copy and past the contents of your .bbl file into your .tex file

\begin{thebibliography}{10}

\bibitem{Bollobas} B{\'e}la Bollob{\'a}s.  \newblock Almost every
  graph has reconstruction number three.  \newblock {\em J. Graph
    Theory}, 14(1):1--4, 1990.

\bibitem{WikipediaReconstruction} Wikipedia contributors.  \newblock
  Reconstruction conjecture.  \newblock {\em Wikipedia, the free
    encyclopedia}, 2011.

\bibitem{FGH} J.~Fisher, R.~L. Graham, and F.~Harary.  \newblock A
  simpler counterexample to the reconstruction conjecture for
  denumerable graphs.  \newblock {\em J. Combinatorial Theory Ser. B},
  12:203--204, 1972.

\bibitem{HHRT} Edith Hemaspaandra, Lane~A. Hemaspaandra,
  Stanis{\l}aw~P. Radziszowski, and Rahul Tripathi.  \newblock
  Complexity results in graph reconstruction.  \newblock {\em Discrete
    Appl. Math.}, 155(2):103--118, 2007.

\bibitem{Kelly} Paul~J. Kelly.  \newblock A congruence theorem for
  trees.  \newblock {\em Pacific J. Math.}, 7:961--968, 1957.

\bibitem{KSU} Masashi Kiyomi, Toshiki Saitoh, and Ryuhei Uehara.
  \newblock Reconstruction of interval graphs.  \newblock In {\em
    Computing and combinatorics}, volume 5609 of {\em Lecture Notes in
    Comput. Sci.}, pages 106--115. Springer, 2009.

\bibitem{RM} S.~Ramachandran and S.~Monikandan.  \newblock Graph
  reconstruction conjecture: reductions using complement, connectivity
  and distance.  \newblock {\em Bull. Inst. Combin. Appl.},
  56:103--108, 2009.

\bibitem{RR} David Rivshin and Stanis{\l}aw~P. Radziszowski.
  \newblock The vertex and edge graph reconstruction numbers of small
  graphs.  \newblock {\em Australas. J. Combin.}, 45:175--188, 2009.

\bibitem{Stockmeyer} Paul~K. Stockmeyer.  \newblock The falsity of the
  reconstruction conjecture for tournaments.  \newblock {\em J. Graph
    Theory}, 1(1):19--25, 1977.

\bibitem{Ulam} S.~M. Ulam.  \newblock {\em A collection of
    mathematical problems}.  \newblock Interscience Tracts in Pure and
  Applied Mathematics, no. 8.  Interscience Publishers, New
  York-London, 1960.

\end{thebibliography}

\end{document}