#  Classification of Element Systems over Finite Commutative Groups

Shouchuan Zhang a, Min Wu a,b, Hengtai Wang a
a: Department of Mathematics, Hunan University, Changsha, 410082
b: Department of Mathematics, Tsinghua University, Beijing, 100084

###### Abstract

We obtain the formula computing the number of isomorphic classes of element
systems with characters over finite commutative group $G$.

2000 Mathematics Subject Classification: 16W30, 05H99.

Keywords: Euler function, character, commutative group.

##  0 Introduction

Classification of Hopf algebras was developed and popularized in the last
decade of the twentieth century, which would have applications to a number of
other areas of mathematics, aside from its intrinsic algebraic interest. In
mathematical physics, Drinfeld’s and Jambo’s work was to provide solutions to
quantum Yang-Baxter equation. In conformal field theory, I. Frenkel and Y. Zhu
have shown how to assign a Hopf algebra to any conformal field theory
model[8]. In topology, quasi-triangular and ribbon Hopf algebras provide many
invariants of knots, links, tangles and 3-manifolds[9, 11, 14, 15]. In
operator algebras, Hopf algebras can be assigned as an invariant for certain
extensions.

Researches on the classification of Hopf algebra is in the ascendant. N.
Andruskiewitsch and H. J. Schneider have obtained interesting result in
classification of finite-dimensional pointed Hopf algebras with commutative
coradical [1, 2, 3, 4]. More recently, they have also researched this problem
in case of non-commutative coradical. Pavel Etingof and Shlomo Gelaki gave the
complete and explicit classification of finite-dimensional triangular Hopf
algebras over an algebraically closed field $k$ of characteristic $0$ [7]. The
classification of monomial Hopf algebras, which are a class of co-path Hopf
algebras, and simple-pointed sub-Hopf algebras of co-path Hopf algebras were
recently obtained in [6] and [12], respectively.

Assume that $k$ is an algebraically closed field of characteristic zero with a
primitive $\mid\\!G\\!\mid$th root of 1 and $G$ is a finite abelian group.
Element systems with characters can be applied to classify quiver Hopf
algebras, multiple Taft algebras over $G$ and Nichols algebras in
${}^{FG}_{FG}{\cal YD}$ (see [17, Theorem 3, Theorem 4] ). In this paper, we
obtained the formula computing the number of isomorphic classes of element
systems with characters over finite commutative group $G$.

In Section 2 we give the explicit formula computing the number of isomorphic
classes of element systems over finite cycle $p$-groups. In Section 3 we give
the explicit formula computing the number of isomorphic classes of element
systems over finite cycle groups. In Section 4 we give the explicit formula
computing the number of isomorphic classes of element systems of primary
commutative groups. In Section 5 we give the formula computing the number of
isomorphic classes of element systems over finite commutative groups.
Unfortunately, we have not found the explicit formula in the general case; we
only change the problem into the numbers of solutions of congruence class
equations (5.7) and (5.8).

##  1 Preliminaries

Unless specified otherwise, in the paper we have the following assumption and
notations. $G$ is a abelian group with order $m$; $F$ is a field containing a
primitive $m$th root of 1;Aut$G$ and Inn$G$ denote the automorphism and inner
automorphism group, respectively; $1$ denotes the unity element of $G;$
$\widehat{G}$ denotes the set of characters of $G$, where a character of $G$
is a group homomorphism from $G$ to $F-\\{0\\}$. $C_{m}$ denotes a cycle group
with order $m$; $\mathbb{Z}$ denotes the set of all integers; $\mathbb{N}$
denotes the set of all positive integers; $Z_{m}$ denotes the ring of integers
modulo $m$; $A^{B}$ denotes the cartesian product $\prod\limits_{i\in
B}A_{i}$, where $A_{i}=A$ for any $i\in B$. Since field $F$ contains a
primitive $|G|$th root of 1, $\widehat{G}$ has $|G|$ elements, i.e.
$|\widehat{G}|=|G|.$

$\varphi(n)$ denotes the Euler function, i.e. $\varphi(n)$ is the number of
elements in set $\\{x\mid(x,n)=1,1\leq x\leq n\\}$. For convenience, we denote
$\bar{i}\in Z_{m}$ by $i$. Obviously, $\\{i\mid(i,m)=1,1\leq i\leq m\\}$ is
the set of all invertible elements in multiplicative group $Z_{m}$, written
$Z_{m}^{*}$. It is clear that the number of elements in set $Z_{m}^{*}$ is
$\varphi(m)$.

######  Lemma 1.1.

(i) There are
$\frac{n!}{\lambda_{1}!\lambda_{2}!\cdots\lambda_{n}!1^{\lambda_{1}}2^{\lambda_{2}}\cdots
n^{\lambda_{n}}}$ permutations of type $1^{\lambda_{1}}2^{\lambda_{2}}\cdots
n^{\lambda_{n}}$ in $S_{n}$.

(ii) If $n=p^{e_{1}}_{1}p_{2}^{e_{2}}\cdots p_{s}^{e_{s}}$ and
$p_{1},p_{2},\cdots,p_{s}$ are mutually different prime numbers with positive
integer $e_{i}$ for $i=1,2,\cdots,s,$ then

$\varphi(n)=n\left(1-\frac{1}{p_{1}}\right)\left(1-\frac{1}{p_{2}}\right)\cdots\left(1-\frac{1}{p_{s}}\right).$

In particular, $\varphi(p^{e_{i}})=p^{e_{i}}-p^{e_{i}-1}$.

(iii) If $n\in\mathbb{N},$ then $\sum\limits_{d|n}\varphi(d)=n$.

(iv) If $n\in\mathbb{N}$, then
$\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{1}{\lambda_{1}!\lambda_{2}!\cdots\lambda_{n}!1^{\lambda_{1}}2^{\lambda_{2}}\cdots
n^{\lambda_{n}}}=1$.

Proof.  (i) It follows from [10, Exercise 2.7.7].

(ii) and (iii) follow from [13, Theorem 3.3.1, Theorem 3.3.1].

(iv) Considering

$\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{n!}{\lambda_{1}!\lambda_{2}!\cdots\lambda_{n}!1^{\lambda_{1}}2^{\lambda_{2}}\cdots
n^{\lambda_{n}}}=n!,$

we complete the proof. $\Box$

A group $G$ is said to act on a non-empty set $\Omega$, if there is a map
$G\times\Omega\rightarrow\Omega$, denoted by $(g,x)\mapsto g\circ x$, such
that for all $x\in\Omega$ and $g_{1},g_{2}\in G$:

$(g_{1}g_{2})\circ x=g_{1}\circ(g_{2}\circ x)\qquad and\qquad 1\circ x=x,$

where 1 denote the unity element of $G.$

For each $x\in\Omega$, let

$G_{x}:=\\{g\circ x\mid g\in G\\},$

called the orbit of $G$ on $\Omega$. For each $g\in G$, let

$F_{g}:=\\{x\in\Omega\mid g\circ x=x\\},$

called the fixed point set of $g$.

Burnside’s lemma, sometimes also called Burnside’s counting theorem, which is
useful to compute the number of orbits.

######  Theorem 1.2.

(See [10, Theorem 2.9.3.1]) ( Burnside’s Lemma ) Let $G$ be a finite group
that acts on a finite set $\Omega$, then the number $\mathcal{N}$ of orbits is
given by the following formula:

$\mathcal{N}=\frac{1}{\mid G\mid}\sum\limits_{g\in G}\mid F_{g}\mid.$

######  Definition 1.3.

$(G,\overrightarrow{g},\overrightarrow{\chi},J)$ is called an element system
with characters (simply, ESC) if $G$ is a group, $J$ is a set,
$\overrightarrow{g}=\\{g_{i}\\}_{i\in J}\in Z(G)^{J}$ and
$\overrightarrow{\chi}=\\{\chi_{i}\\}_{i\in J}\in\widehat{G}^{J}$ with
$g_{i}\in Z(G)$ and $\chi_{i}\in\widehat{G}$.
$ESC(G,\overrightarrow{g},\overrightarrow{\chi},J)$ and
$ESC(G^{\prime},\overrightarrow{g^{\prime}},\overrightarrow{\chi^{\prime}},J^{\prime})$
are said to be isomorphic if there exist a group isomorphism
$\phi:G\rightarrow G^{\prime}$ and a bijective map $\sigma:J\rightarrow
J^{\prime}$ such that $\phi(g_{i})=g^{\prime}_{\sigma(i)}$ and
$\chi^{\prime}_{\sigma(i)}\phi=\chi_{i}$ for any $i\in J$.

$ESC(G,\overrightarrow{g},\overrightarrow{\chi},J)$ can be written as
$ESC(G,g_{i},\chi_{i};i\in J)$ for convenience.

Given a finite commutative group and a positive integer $n$, define
$\Omega(G,n):=$
$\\{(G,\overrightarrow{g},\overrightarrow{\chi},J)\mid(G,\overrightarrow{g},\overrightarrow{\chi},J)$
$\hbox{ is an }ESC\hbox{ and }$ $J=\\{1,2,\cdots,n\\}\\}$. Let
$\left(\begin{array}[]{c}j\\\ g_{j}\\\ \chi_{j}\end{array}\right)$ denote
$ESC(G,g_{j},\chi_{j};j\in J)$ $\in\Omega(G,n)$ in short. Define the action
$\circ$ of group $M:=$Aut$G\times S_{n}$ on $\Omega(G,n)$ as follows:

$\displaystyle(\phi,\sigma)\circ\left(\begin{array}[]{c}j\\\ g_{j}\\\
\chi_{j}\end{array}\right)=\left(\begin{array}[]{c}\sigma(j)\\\ \phi(g_{j})\\\
\chi_{j}\phi^{-1}\end{array}\right).$ (1.7)

It is clear that it is an action. Indeed,

$(\phi,\sigma)\circ\left((\phi^{\prime},\sigma^{\prime})\circ\left(\begin{array}[]{c}j\\\
g_{j}\\\
\chi_{j}\end{array}\right)\right)=(\phi,\sigma)\circ\left(\begin{array}[]{c}\sigma^{\prime}(j)\\\
\phi^{\prime}(g_{j})\\\
\chi_{j}\phi^{\prime-1}\end{array}\right)=\left(\begin{array}[]{c}\sigma\sigma^{\prime}(j)\\\
\phi\phi^{\prime}(g_{j})\\\
\chi_{j}\phi^{\prime-1}\phi^{-1}\end{array}\right)=(\phi\phi^{\prime},\sigma\sigma^{\prime})\circ\left(\begin{array}[]{c}j\\\
g_{j}\\\ \chi_{j}\end{array}\right),$

and

$(1,1)\circ\left(\begin{array}[]{c}j\\\ g_{j}\\\
\chi_{j}\end{array}\right)=\left(\begin{array}[]{c}j\\\ g_{j}\\\
\chi_{j}\end{array}\right).$

Let ${\cal N}(G,n)$ denote the number of isomorphic classes in $\Omega(G,r)$.
This article is mostly devoted to investigate the formula of ${\cal N}(G,n)$.
It is clear that each orbit of $\Omega(G,n)$ represents an isomorphic class of
ESC’s. As a result, ${\cal N}(G,n)$ is equal to the number of orbits in
$\Omega(G,n)$.

##  2 Cycle $p$-group

In this section we give the formula computing the number of isomorphic classes
of element systems over finite cycle $p$-groups.

Let $G=\langle g\rangle$ and $\mid\\!G\\!\mid=m=p^{e}.$ It is clear that
$\hat{G}$ consists of the following maps:

$\displaystyle\chi^{(l)}:G$ $\displaystyle\rightarrow F$ $\displaystyle g$
$\displaystyle\mapsto\omega^{l},$

where $\omega$ is a primitive $m$th root of 1; Aut$G$ consists of the
following maps:

$\displaystyle\phi_{i}:G$ $\displaystyle\rightarrow G$ $\displaystyle g$
$\displaystyle\mapsto g^{i},$

where $(i,m)=1$. That is, $\mathrm{Aut}G\cong Z_{m}^{*}$ (See [18, Theorem
2.3.3]). Thus $|\mathrm{Aut}G|=|Z^{*}_{m}|=\varphi(m)$.

Since
$\chi^{(l_{j})}\phi_{i}^{-1}(g)=\chi^{(l_{j})}\phi_{i^{-1}}(g)=\chi^{(l_{j})}(g^{i^{-1}})=\omega^{l_{j}i^{-1}}$,
$\chi^{(l_{j})}\phi_{i}^{-1}=\chi^{(l_{j}i^{-1})}$. By (1.7), we have

$(\phi_{i},\sigma)\circ\left(\begin{array}[]{c}j\\\ g^{k_{j}}\\\
\chi^{(l_{j})}\end{array}\right)=\left(\begin{array}[]{c}\sigma(j)\\\
\phi_{i}(g^{k_{j}})\\\
\chi^{(l_{j})}\phi_{i}^{-1}\end{array}\right)=\left(\begin{array}[]{c}\sigma(j)\\\
g^{i\cdot k_{j}}\\\ \chi^{(l_{j}i^{-1})}\end{array}\right),$

where $i^{-1}$ is the inverse of $i$ in group $Z_{m}^{*}$.

######  Definition 2.1.

Assume $(i,p)=1$ and $p$ is a prime number. Define $\vec{\delta}_{e}(i)$ $:=$
$(\delta_{1}(i),\delta_{2}(i),$ $\ldots,\delta_{e}(i))$, where $\delta_{s}(i)$
is the order of $i$ in $Z^{*}_{p^{s}}$ for $1\leq s\leq e$.

Remark. It is easy to check $\vec{\delta}_{e}(i)=\vec{\delta}_{e}(i^{-1})$.

######  Lemma 2.2.

If $p$ is an odd prime number, then there exists $\alpha\in\mathbb{Z}$ such
that

$Z^{*}_{p^{s}}=\langle\alpha\rangle=\\{1,\alpha,\ldots,\alpha^{\varphi(p^{s})-1}\\}$

for any positive integer $s$. That is, $\alpha$ is a common generator of
$Z^{*}_{p^{s}}(s=1,2,\ldots)$.

If $p=2$, then there exist $\beta\in\mathbb{Z}$(for example, $\beta=5$ ) such
that

$Z^{*}_{p^{s}}=\langle-1\rangle\times\langle\beta\rangle=\\{1,-1\\}\times\\{1,\beta,\ldots,\beta^{2^{s-2}-1}\\}$

when $s>2$.

Proof. It follows from [13, Theorem 5.2.3; Theorem 5.2.1]. $\Box$

######  Lemma 2.3.

(i) If $d\mid m$, then the number of elements with order $d$ in cycle group
$(Z_{m},+)$ is exactly $\varphi(d)$.

(ii) If $p$ is an odd prime number or $e\leq 2$, then $\vec{\delta}_{e}(i)$ is
a case of following for any $i\in Z_{m}^{*}$:

$\vec{\delta}_{e}^{(k,d)}=(d,\ldots,\overset{k}{d},pd,p^{2}d,\ldots,p^{e-k}d),\
d|p-1,k=1,\ldots,e.$

Conversely, if $d|p-1$ and $1\leq k\leq e$, then there exactly exist
$\varphi(p^{e-k}d)$ elements in $Z^{*}_{p^{e}}$ satisfying
$\vec{\delta}_{e}(i)=\vec{\delta}^{(k,d)}_{e}$.

(iii) If $p=2$ with $e\geq 3$, then $\vec{\delta}_{e}(i)$ is a case of the
following for any $i\in Z_{m}^{*}$:

$\vec{\delta}_{e}^{(k,d)}=(1,d,\ldots,\overset{k}{d},2,2^{2},\ldots,2^{e-k}),\
d=1\mbox{ or }2,k=2,\ldots,e.$

Conversely, if $d=1$ or 2 and $2\leq k\leq e$, then there exactly exist
$\varphi(2^{e-k})$ elements in $Z^{*}_{2^{e}}$ satisfying
$\vec{\delta}_{e}(i)=\vec{\delta}_{e}^{(k,d)}$.

Remark. In (iii) above,
$\vec{\delta}_{e}^{(e-1,2)}=\vec{\delta}_{e}^{(e,2)}=(1,2,\ldots,2)$,
$\vec{\delta}_{e}^{(e,1)}=(1,1,\ldots,1)$.

Proof. (i) Let $A=\\{i\mid 1\leq i\leq d,(i,d)=1\\}$ and $B=\\{i\mid 1\leq
i\leq m,$ the order of $i$ in $(Z_{m},+)$ is $d\\}$. Define a map $\phi:\
A\rightarrow B$ such that $\phi(i)=i\frac{m}{d}$ for any $i\in A.$ It is clear
that $\phi$ is bijective.

(ii) By Lemma2.2, we can assume $i=\alpha^{\nu}$ and $\nu=p^{k-1}c$ with
$e\geq k\geq 1$ and $(p,c)=1.$ For any $1\leq s\leq e$, since the order of
$\alpha$ in $Z^{*}_{p^{s}}$ is $\varphi(p^{s})=p^{s-1}(p-1)$, we have that the
order of $i$ in $Z^{*}_{p^{s}}$ is

$\delta_{s}(i)=\frac{p^{s-1}(p-1)}{(\nu,p^{s-1}(p-1))}=\frac{p^{s-1}(p-1)}{(p^{k-1}c,p^{s-1}(p-1))}=\left\\{\begin{array}[]{ll}p^{s-k}\frac{p-1}{(c,p-1)}&\hbox{
when $s\geq k$}\\\ \frac{p-1}{(c,p-1)}&\hbox{when $s<k$}.\end{array}\right.$
(2.1)

Let $\frac{p-1}{(c,p-1)}=d$. Obviously $d|p-1$. To complete the proof of (ii),
we now show the following conclusion:

If $i,i^{\prime}\in Z_{p^{e}}$ and $(ii^{\prime},p)=1,$ then
$\vec{\delta}_{e}(i)=\vec{\delta}_{e}(i^{\prime})$ if and only if $i$ and
$i^{\prime}$ have the same orders in $Z^{*}_{p^{e}}$. Indeed, the necessity is
obvious. Now we show the sufficiency. By Lemma2.2, we can assume
$i=\alpha^{\nu}$ and $i^{\prime}=\alpha^{\nu^{\prime}}$ with $\nu=p^{k-1}c$,
$\nu^{\prime}=p^{k^{\prime}-1}c^{\prime}$, $(cc^{\prime},p)=1$. Since $i$ and
$i^{\prime}$ have the same orders in $Z^{*}_{p^{e}}$,
$k=k^{\prime},(c,p-1)=(c^{\prime},p-1)$ by (2.1). Applying (2.1), we have
$\vec{\delta}_{e}(i)=\vec{\delta}_{e}(i^{\prime})$.

Next we come back to show (ii). For any $d$ and $k$ with $d|p-1$ and $1\leq
k\leq e$, Let $i=\alpha^{p^{k-1}\frac{(p-1)}{d}}$. It is easy to check
$\vec{\delta}_{e}(i)=\vec{\delta}_{e}^{(k,d)}$. Considering that there exactly
exist $\varphi(p^{e-k}d)$ elements whose orders are $p^{e-k}d$ in
$Z^{*}_{p^{e}}$, we complete the proof of (ii).

(iii) If $i=\beta^{\nu}$, we can similarly show the first part of (iii).
Notice $\delta_{1}(i)=1.$ If $i=-\beta^{\nu}$ and $\nu=2^{k-2}c$ with
$(c,2)=1$ and $e\geq k\geq 2$, then the order of $i$ in $Z^{*}_{2^{s}}$ is

$\delta_{s}(i)=[2,\frac{2^{s-2}}{(\nu,2^{s-2})}]=[2,\frac{2^{s-2}}{(2^{k-2}c,2^{s-2})}]=\left\\{\begin{array}[]{ll}2^{s-k}&\hbox{when
}s>k\\\ 2&\hbox{when $s\leq k$}\end{array}\right.$

when $s\geq 2$. Obviously, $\delta_{1}(i)=1$ and
$\vec{\delta}_{e}(i)=\vec{\delta}_{e}^{(k,2)}$. It is easy to show the second
part of (iii) by meas of the method similar to proof of (ii). $\Box$

Next we compute the fixed point set $F_{(\phi_{i},\sigma)}$ of
$(\phi_{i},\sigma)$. Assume $\vec{\delta}_{e}(i)=\vec{\delta}_{e}^{(k,d)}$ and
that the type of $\sigma\in S_{n}$ is $1^{\lambda_{1}}2^{\lambda_{2}}\cdots
n^{\lambda_{n}}$. If $\left(\begin{array}[]{c}j\\\ g^{k_{j}}\\\
\omega^{l_{j}}\end{array}\right)\in F_{(\phi_{i},\sigma)},$ then

$\left(\begin{array}[]{c}\sigma(j)\\\ g^{ik_{j}}\\\
\omega^{i^{-1}l_{j}}\end{array}\right)=\left(\begin{array}[]{c}j\\\
g^{k_{j}}\\\
\omega^{l_{j}}\end{array}\right)=\left(\begin{array}[]{c}\sigma(j)\\\
g^{k_{\sigma(j)}}\\\ \omega^{l_{\sigma(j)}}\end{array}\right),$

which implies

$ik_{j}\equiv k_{\sigma(j)}(\mathrm{mod}\ p^{e}),$ (2.2)

and

$i^{-1}l_{j}\equiv l_{\sigma(j)}(\mathrm{mod}\ p^{e})$ (2.3)

for $1\leq j\leq n$.

It is clear that (2.2 ) and (2.3 ) are independent each other, and they hold
if and only if every cycle, such as $\tau=(j_{1}\,j_{2}\,\cdots j_{r})$, in
independent cycle decomposition of $\sigma$ satisfies the following two
formulae:

$ik_{j}\equiv k_{\tau(j)}(\mathrm{mod}\ p^{e})$ (2.2′)

and

$i^{-1}l_{j}\equiv l_{\tau(j)}(\mathrm{mod}\ p^{e}).$ (2.3′)

for $j=j_{1},\ldots,j_{r}$.

For (2.2′), we have

$\displaystyle k_{j_{2}}$ $\displaystyle\equiv$ $\displaystyle
ik_{j_{1}}\equiv i^{1}k_{j_{1}},$ $\displaystyle k_{j_{3}}$
$\displaystyle\equiv$ $\displaystyle ik_{j_{2}}\equiv i^{2}k_{j_{1}},$
$\displaystyle\cdots$ $\displaystyle k_{j_{r}}$ $\displaystyle\equiv$
$\displaystyle ik_{j_{r-1}}\equiv i^{r-1}k_{j_{1}},$ $\displaystyle k_{j_{1}}$
$\displaystyle\equiv$ $\displaystyle ik_{j_{r}}\equiv i^{r}k_{j_{1}}.$ (2.5)

This implies the numbers of solutions of both (2.2′) and (2) are the same when
we view $k_{j_{1}},\ldots,k_{j_{r}}$ as indeterminate elements. (2) is
equivalent to

$(i^{r}-1)k_{j_{1}}\equiv 0(\mathrm{mod}\ p^{e}).$ (2′)

Now we give the main result.

######  Theorem 2.4.

Assume that $G$ is a cycle group with order $p^{e}$ and positive integer $e$.

(i) If $p$ is an odd prime number or $e\leq 2$, then

$\displaystyle{\cal N}(G,n)$ $\displaystyle=$
$\displaystyle\frac{1}{p^{e-1}(p-1)}\sum_{d|p-1}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(e,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}$ (2.6)
$\displaystyle+\sum_{k=1}^{e-1}p^{-k}\sum_{d|p-1}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}$ $\displaystyle=$ $\displaystyle
1+\frac{1}{p^{e-1}(p-1)}\sum\limits_{d|p-1\atop d\leq
n}\varphi(d)\left(\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2e\sum\limits_{t=1}^{[n/d]}\lambda_{td}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}-1\right)$ (2.7) $\displaystyle+$
$\displaystyle\sum\limits_{k=1}^{e-1}p^{-k}\sum\limits_{d|p-1\atop d\leq
n}\varphi(d)\left(\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}-1\right),$

where

$f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})=\sum\limits_{s=0}^{e-k-1}(k+s)\sum\limits_{1\leq
t\leq[n/(p^{s}d)]\atop(t,p)=1}\lambda_{tp^{s}d}+e\sum\limits_{t=1}^{[n/p^{e-k}d]}\
\lambda_{tp^{e-k}d}\ .$ (2.8)

(ii) If $p=2$ and $e\geq 3$, then

$\displaystyle{\cal N}(G,n)$ $\displaystyle=$
$\displaystyle\sum_{d=1}^{2}\sum_{k=2}^{e}\frac{\varphi(2^{e-k})}{2^{e-1}}\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{4^{f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}$ (2.9) $\displaystyle=$
$\displaystyle\frac{1}{2^{e-1}}\sum\limits_{\lambda_{1}+\cdots+n\lambda_{n}=n}\frac{4^{e\sum\limits_{t=1}^{n}\lambda_{t}}+4^{\sum\limits_{(t,2)=1}\lambda_{t}+e\sum\limits_{t=1}^{[n/2]}\lambda_{2t}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}$ (2.10) $\displaystyle+$
$\displaystyle\sum\limits_{d=1}^{2}\sum\limits_{k=2}^{e-1}2^{-k}\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\frac{4^{f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}},$

where

$\displaystyle
f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,1)})=\sum\limits_{s=0}^{e-k-1}(k+s)\sum\limits_{1\leq
t\leq[n/(2^{s})]\atop(t,2)=1}\lambda_{t2^{s}}+e\sum\limits_{t=1}^{[n/2^{e-k}]}\lambda_{t2^{e-k}}\
\ (2\leq k\leq e),$ $\displaystyle
f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,2)})$
$\displaystyle=$
$\displaystyle\sum_{(t,2)=1}\lambda_{t}+\sum\limits_{s=1}^{e-k-1}(k+s)\sum\limits_{1\leq
t\leq[n/(2^{s})]\atop(t,2)=1}\lambda_{t2^{s}}$ (2.12)
$\displaystyle+e\sum\limits_{t=1}^{[n/2^{e-k}]}\lambda_{t2^{e-k}}\ \ \ (2\leq
k\leq e-1)$

and

$f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(e,2)})=\sum_{(t,2)=1}\lambda_{t}+e\sum_{t=1}^{[n/2]}\lambda_{2t}.$
(2.13)

Proof. (i) For $(\phi_{i},\sigma)\in M$, $i\in Z_{p^{e}}^{*}$, by Lemma2.2,
there exists a positive integer $k$ such that $i=\alpha^{\nu}$,
$\nu=p^{k-1}c$, $(c,p)=1$ and $1\leq k\leq e$. By Lemma2.3, there exists
positive integer $d$ such that $\vec{\delta}_{e}(i)=\vec{\delta}_{e}^{(k,d)}$
and $d\mid(p-1)$. First we compute the number
$\mid\\!F_{\phi_{i},\sigma}\\!\mid$ of fixed point set.

$(1^{\circ})$ If $d\nmid r$, then $\delta_{1}(i)\nmid r$ and
$p\nmid(i^{r}-1)$, i.e. $(i^{r}-1,p^{e})=1$. By (2′), $k_{j_{1}}=0$.

$(2^{\circ})$ If $p^{s}d|r$ and $p^{s+1}d\nmid r$ ( $0\leq s\leq e-k-1$), then
$r=tp^{s}d$, where $(t,p)=1$. By Lemma2.3, the order of $i$ is $p^{s}d$ in
$Z_{p^{k+s}}^{*}$. Consequently, $i^{r}-1=cp^{k+s}$. If $p\\!\mid\\!c$, i.e.
$c=c^{\prime}p$, then $i^{r}-1=c^{\prime}p^{k+s+1}$. Since the order of $i$ is
$p^{s+1}d$ in $Z_{p^{k+s+1}}^{*}$, $p^{s+1}d\mid r$. This is a contradiction.
Thus $(c,p)=1$. This implies that the number of solutions of (2′) is
$p^{k+s}$.

$(3^{\circ})$ If $p^{e-k}d|r$, then the number of solutions of (2′) is
$p^{e}$.

Obviously, equations (2′) about all cycles in independent cycle decomposition
of $\sigma$ are independent each other. Notice that there exactly exist
$\sum\limits_{1\leq t\leq[n/(p^{s}d)]\atop(t,p)=1}\lambda_{tp^{s}d}$ cycles,
which satisfy $(2^{\circ})$ with the length $r$, in independent cycle
decomposition of $\sigma$; there exactly exist
$\sum\limits_{t=1}^{[n/(p^{e-k}d)]}\lambda_{tp^{e-k}d}$ cycles, which satisfy
$(3^{\circ})$ with the length $r$, in independent cycle decomposition of
$\sigma$; Consequently, there exactly exist

$\displaystyle\left(\prod_{s=0}^{e-k-1}p^{(k+s)\sum\limits_{1\leq
t\leq[n/(p^{s}d)]\atop(t,p)=1}\lambda_{tp^{s}d}}\right)p^{e\sum\limits_{t=1}^{[n/p^{e-k}d]}\lambda_{tp^{e-k}d}}$
$\displaystyle=$ $\displaystyle
p^{\sum\limits_{s=0}^{e-k-1}(k+s)\sum\limits_{1\leq
t\leq[n/(p^{s}d)]\atop(t,p)=1}\lambda_{tp^{s}d}+e\sum\limits_{t=1}^{[n/p^{e-k}d]}\lambda_{tp^{e-k}d}}$
$\displaystyle=$ $\displaystyle
p^{f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}$

distinct $(k_{j})_{j\in J}$ satisfying (2.2).

Obviously,

$f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(e,d)})=e\sum\limits_{t=1}^{[n/d]}\lambda_{td}.$
(2.14)

If $d>n$, then

$f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})=0\ .$ (2.15)

Similarly, there exactly exist

$p^{f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}$

distinct $(l_{j})_{j\in J}$ satisfying (2.3).

Considering the independence between (2.2) and (2.3), we have

$|F_{(\phi_{i},\sigma)}|=\left(p^{f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}\right)^{2}=p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}.$
(2.16)

By Lemma1.1 and Lemma2.3, there exist $\varphi(p^{e-k}d)$
$\frac{n!}{\lambda_{1}!\lambda_{2}!\cdots\lambda_{n}!1^{\lambda_{1}}2^{\lambda_{2}}\cdots
n^{\lambda_{n}}}$ elements, whose the number of elements in fixed point set is
$p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}$, in $M$.

See

$\displaystyle{\cal N}(G,n)$ $\displaystyle=$
$\displaystyle\frac{1}{|M|}\sum_{(\varphi_{i},\sigma)\in
M}|F_{(\varphi_{i},\sigma)}|$ $\displaystyle=$
$\displaystyle\frac{1}{n!|\mathrm{Aut}G|}\sum_{k=1}^{e}\sum_{d|p-1}\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{\varphi(p^{e-k}d)n!}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}$
$\displaystyle=$
$\displaystyle\frac{1}{p^{e-1}(p-1)}\left(\sum_{d|p-1}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(e,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right.$
$\displaystyle\left.+\sum_{k=1}^{e-1}p^{e-k-1}(p-1)\sum_{d|p-1}\varphi(d)\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{p^{e-1}(p-1)}\sum_{d|p-1}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(e,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}(\hbox{written as (I)})$
$\displaystyle+\sum_{k=1}^{e-1}p^{-k}\sum_{d|p-1}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}(\hbox{written as (II)}).$

We now compute (I) and (II), respectively. Applying Lemma1.1, (2.14) and
(2.15), we have

(I) $\displaystyle=$
$\displaystyle\frac{1}{p^{e-1}(p-1)}\left(\sum_{d|p-1\atop d\leq
n}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2e\sum\limits_{t=1}^{[n/d]}\lambda_{td}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right.$ $\displaystyle\left.+\sum_{d|p-1\atop
d>n}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{1}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right)$ $\displaystyle=$
$\displaystyle\frac{1}{p^{e-1}(p-1)}\left(\sum_{d|p-1\atop d\leq
n}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2e\sum\limits_{t=1}^{[n/d]}\lambda_{td}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}+\sum_{d|p-1\atop d>n}\varphi(d)\right)$ $\displaystyle=$
$\displaystyle\frac{1}{p^{e-1}(p-1)}\left(\sum_{d|p-1\atop d\leq
n}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2e\sum\limits_{t=1}^{[n/d]}\lambda_{td}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}+p-1-\sum_{d|p-1\atop d\leq n}\varphi(d)\right)$
$\displaystyle=$ $\displaystyle\frac{1}{p^{e-1}(p-1)}\sum\limits_{d|p-1\atop
d\leq
n}\left(\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\varphi(d)\frac{p^{2e\sum\limits_{t=1}^{[n/d]}\lambda_{td}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}-1\right)+\frac{1}{p^{e-1}}.$

(II) $\displaystyle=$
$\displaystyle\sum_{k=1}^{e-1}p^{-k}\left(\sum_{d|p-1\atop d\leq
n}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right.$ $\displaystyle\left.+\sum_{d|p-1\atop
d>n}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{1}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right)$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{e-1}p^{-k}\left(\sum_{d|p-1\atop d\leq
n}\varphi(d)\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}+p-1-\sum_{d|p-1\atop d\leq n}\varphi(d)\right)$
$\displaystyle=$ $\displaystyle\sum_{k=1}^{e-1}p^{-k}\sum_{d|p-1\atop d\leq
n}\varphi(d)\left(\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2f_{p}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}-1\right)+1-\frac{1}{p^{e-1}}.$

Since ${\cal N}(G,n)=\textrm{I}+\textrm{II}$, (2.12) holds.

(ii) We first show

$|F_{(\phi_{i},\sigma)}|=\left(2^{f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}\right)^{2}=2^{2f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}.$
(2.17)

Case 1: $i=\beta^{\nu}$ and $\nu=2^{k-2}c$ with $(c,p)=1$ and $k>1$. By
Lemma2.3, $d=1$ and $\vec{\delta}_{e}(i)=\vec{\delta}_{e}^{(k,1)}$.

$(1^{\circ})$ If $2^{s}\mid r$ and $2^{s+1}\nmid r$ with $0\leq s\leq e-k-1)$,
then the number of solutions of (2′) is $2^{k+s}$ by means of the method
similar to proof of (i).

$(2^{\circ})$ If $2^{e-k}|r$, then the number of solutions of (2′) is $2^{e}$.

There exists $\sum\limits_{1\leq
t\leq[n/(2^{s})]\atop(t,2)=1}\lambda_{t2^{s}}$ cycles, which satisfy
$(1^{\circ})$ with the length $r$, in independent cycle decomposition of
$\sigma$; there exist $\sum\limits_{t=1}^{[n/2^{e-k}]}\lambda_{t2^{e-k}}$
cycles, which satisfy $(2^{\circ})$ with the length $r$, in independent cycle
decomposition of $\sigma$. This implies (2.17).

Remark. It is clear

$f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(e,1)})=e\sum\limits_{t=1}^{n}\lambda_{t}.$
(2.18)

Case 2: $i=-\beta^{\nu}$ and $\nu=2^{k-2}c$ with $(c,p)=1$ and $e\geq k>1$. By
Lemma2.3, $d=2$ and $\vec{\delta}_{e}(i)=\vec{\delta}_{e}^{(k,2)}$.

$(1^{\circ})$ If $2\nmid r$, i.e. $r$ is odd. It is clear $i^{r}-1=2c$. Since
$\delta_{4}(i)=2$, $c$ is odd. By (2′), $k_{j_{1}}=c^{\prime}2^{e-1}$. Thus
the number of solutions of (2′) is 2.

$(2^{\circ})$ If $2^{s}|r$ and $2^{s+1}\nmid r\ (1\leq s\leq e-k-1)$, then
$r=t2^{s}$ and $(t,2)=1$; $i^{r}-1=c2^{s+k}$ with $(c,2)=1$;
$k_{j_{1}}=c^{\prime}2^{e-s-k}$. Thus the number of solutions of (2′) is
$2^{s+k}$.

$(3^{\circ})$ If $2^{e-k}|r$, then the number of solutions of (2′) is $2^{e}$.

There exist $\sum_{(t,2)=1}\lambda_{t}$ cycles, which satisfy $(1^{\circ})$
with the length $r$, in independent cycle decomposition of $\sigma$; there
exist $\sum\limits_{1\leq t\leq[n/(2^{s})]\atop(t,2)=1}\lambda_{t2^{s}}$
cycles, which satisfy $(2^{\circ})$ with the length $r$, in independent cycle
decomposition of $\sigma$; there exist
$\sum\limits_{t=1}^{[n/2^{e-k}]}\lambda_{t2^{e-k}}$ cycles, which satisfy
$(3^{\circ})$ with the length $r$, in independent cycle decomposition of
$\sigma$. This implies (2.17).

Now we show (ii). Applying Lemma2.2(ii), Lemma2.3 and Burnside’s Lemma, we
have

$\displaystyle{\cal N}(G,n)$ $\displaystyle=$
$\displaystyle\frac{1}{|M|}\sum_{(\varphi_{i},\sigma)\in
M}|F_{(\varphi_{i},\sigma)}|$ $\displaystyle=$
$\displaystyle\frac{1}{n!|\mathrm{Aut}G|}\left(\sum_{d=1}^{2}\sum_{k=2}^{e}\sum_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\varphi(2^{e-k})\frac{n!}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}4^{f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}\right.$
$\displaystyle=$
$\displaystyle\frac{1}{2^{e-1}}\left(\sum_{d=1}^{2}\sum_{k=2}^{e-1}2^{e-k-1}\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\frac{4^{f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right.$
$\displaystyle\left.+\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\frac{4^{e\sum\limits_{t=1}^{n}\lambda_{t}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}+\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\frac{4^{\sum\limits_{(t,2)=1}\lambda_{t}+e\sum\limits_{t=1}^{[n/2]}\lambda_{2t}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right)$ $\displaystyle=$
$\displaystyle\sum_{d=1}^{2}\sum_{k=2}^{e-1}2^{-k}\sum\limits_{\lambda_{1}+\cdots+n\lambda_{n}=n}\frac{4^{f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e}^{(k,d)})}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}$
$\displaystyle+\frac{1}{2^{e-1}}\sum\limits_{\lambda_{1}+\cdots+n\lambda_{n}=n}\frac{4^{e\sum\limits_{t=1}^{n}\lambda_{t}}+4^{\sum\limits_{(t,2)=1}\lambda_{t}+e\sum\limits_{t=1}^{[n/2]}\lambda_{2t}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}.\ \ \ \ \Box$

######  Corollary 2.5.

Let $G\cong C_{p^{e}}$ and $e$ be a positive integer.

(i) If $p$ is an odd prime number or $e\leq 2$, then

${\cal N}(G,1)=p^{e}+2(p^{e-1}+p^{e-2}+\cdots+p+1);$

(ii) If $p=2$ and $e\geq 3$, then

${\cal N}(G,1)=2^{e+1}+2^{e}-2.$

Proof. (i) Notice $f(\lambda_{1},\vec{\delta}_{e}^{(k,1)})=k(1\leq k\leq e-1)$
and $f(\lambda_{1},\vec{\delta}_{e}^{(e,1)})=e$ when $n=1$. By (2.6), we have

$\displaystyle{\cal N}(G,1)$ $\displaystyle=$ $\displaystyle
1+\frac{1}{p^{e-1}(p-1)}\left(p^{2e}-1\right)+\sum_{k=1}^{e-1}p^{-k}\left(p^{2k}-1\right)$
$\displaystyle=$ $\displaystyle
1+\frac{1}{p^{e-1}(p-1)}\left(p^{2e}-1\right)+\sum_{k=1}^{e-1}p^{k}-\sum_{k=1}^{e-1}p^{-k}$
$\displaystyle=$ $\displaystyle p^{e}+2(p^{e-1}+p^{e-2}+\cdots+p+1).$

(ii) By (2.4) and (2.12), $f(\lambda_{1},\vec{\delta}_{2}^{(k,1)})=k(2\leq
k\leq e-1)$ and $f(\lambda_{1},\vec{\delta}_{2}^{(k,2)})=1(2\leq k\leq e-1)$
when $n=1$. Applying (2.10), we have

$\displaystyle{\cal N}(G,1)$ $\displaystyle=$
$\displaystyle\frac{1}{2^{e-1}}(4^{e}+4)+\sum_{k=2}^{e-1}(2^{-k}4^{k})+\sum_{k=2}^{e-1}(2^{-k}4)$
$\displaystyle=$ $\displaystyle 2^{e+1}+2^{e}-2.\Box$

######  Corollary 2.6.

Assume that $G=C_{p^{e}}$ and $p$ is a prime number with positive integer $e$.

(i) If $p$ is an odd prime number, then

${\cal
N}(G,2)=1+\frac{1}{2}\frac{p^{3e}-p^{3}}{p^{3}-1}+\frac{1}{p-1}\left(\frac{1}{2}p^{3e+1}+p^{e+1}+p^{e}-p-\frac{3}{2}\right);$

(ii) ${\cal N}(C_{4},2)=76,\ {\cal N}(C_{2},2)=10$;

(iii) If $p=2$ and $e\geq 3$, then

${\cal N}(G,2)=\frac{15}{14}2^{3e}+3\times 2^{e+1}-\frac{116}{7}.$

Proof. It is clear that $(\lambda_{1},\lambda_{2})=(0,1)$ or (2,0) when $n=2$.

(i) If $p$ is an odd prime number, by (2.8), we have $d=1,2$, and
$f_{p}(\lambda_{1},\lambda_{2},\vec{\delta}_{e}^{(k,1)})$
$=k(\lambda_{1}+\lambda_{2})$ and
$f_{p}(\lambda_{1},\lambda_{2},\vec{\delta}_{e}^{(k,2)})$ $=k\lambda_{2}(1\leq
k\leq e)$. Using (2.7), we have

$\displaystyle{\cal N}(G,2)$ $\displaystyle=$ $\displaystyle
1+\frac{1}{p^{e-1}(p-1)}\left(\sum_{\lambda_{1},\lambda_{2}}\frac{p^{2e(\lambda_{1}+\lambda_{2})}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}-1+\sum_{\lambda_{1},\lambda_{2}}\frac{p^{2e\lambda_{2}}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}-1\right)+$
$\displaystyle\sum_{k=1}^{e-1}p^{-k}\left(\sum_{\lambda_{1},\lambda_{2}}\frac{p^{2k(\lambda_{1}+\lambda_{2})}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}-1+\sum_{\lambda_{1},\lambda_{2}}\frac{p^{2k\lambda_{2}}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}-1\right)$
$\displaystyle=$ $\displaystyle
1+\frac{1}{p^{e-1}(p-1)}\left(\frac{1}{2}p^{2e}+\frac{1}{2}p^{4e}-1+\frac{1}{2}p^{2e}+\frac{1}{2}-1\right)+$
$\displaystyle\sum_{k=1}^{e-1}p^{-k}\left(\frac{1}{2}p^{2k}+\frac{1}{2}p^{4k}-1+\frac{1}{2}p^{2k}+\frac{1}{2}-1\right)$
$\displaystyle=$ $\displaystyle
1+\frac{1}{2}\frac{p^{3e}-p^{3}}{p^{3}-1}+\frac{1}{p-1}\left(\frac{1}{2}p^{3e+1}+p^{e+1}+p^{e}-p-\frac{3}{2}\right).$

(ii) If $p=2$ and $e=2$, then $d=1$ and $k=1$. By (2.7), we have

$\displaystyle{\cal N}(C_{4},2)$ $\displaystyle=$ $\displaystyle
1+\frac{1}{2}\left(\sum_{\lambda_{1},\lambda_{2}}\frac{4^{e(\lambda_{1}+\lambda_{2})}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}-1\right)+\frac{1}{2}\left(\sum_{\lambda_{1},\lambda_{2}}\frac{4^{(\lambda_{1}+2\lambda_{2})}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}-1\right)$
$\displaystyle=$ $\displaystyle 76.$

If $p=2$ and $e=1$, then $d=1$. By (2.6), we have

$\displaystyle{\cal N}(C_{2},2)$ $\displaystyle=$ $\displaystyle
1+\left(\sum_{\lambda_{1},\lambda_{2}}\frac{4^{\lambda_{1}+\lambda_{2}}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}-1\right)$
$\displaystyle=$ $\displaystyle 10.$

(iii) It follows from (2.12) and (2.13) that
$f_{2}(\lambda_{1},\lambda_{2},\vec{\delta}_{e}^{(k,1)})=k\lambda_{1}+(k+1)\lambda_{2}$
and
$f(\lambda_{1},\lambda_{2},\vec{\delta}_{e}^{(k,2)})=\lambda_{1}+(k+1)\lambda_{2}(2\leq
k\leq e-1)$. Applying (2.10), we have

$\displaystyle{\cal N}(G,2)$ $\displaystyle=$
$\displaystyle\frac{1}{2^{e-1}}\sum_{\lambda_{1},\lambda_{2}}\frac{4^{e(\lambda_{1}+\lambda_{2})}+4^{\lambda_{1}+e\lambda_{2}}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}+\sum_{k=2}^{e-1}2^{-k}\sum_{\lambda_{1},\lambda_{2}}\frac{4^{k\lambda_{1}+(k+1)\lambda_{2}}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}$
$\displaystyle+\sum_{k=2}^{e-1}2^{-k}\sum_{\lambda_{1},\lambda_{2}}\frac{4^{\lambda_{1}+k\lambda_{2}}}{\lambda_{1}!\lambda_{2}!1^{\lambda_{1}}2^{\lambda_{2}}}$
$\displaystyle=$
$\displaystyle\frac{1}{2^{e-1}}\left(\frac{4^{2e}+4^{2}}{2}+\frac{4^{e}+4^{e}}{2}\right)+\sum_{k=2}^{e-1}2^{-k}\left(\frac{4^{2k}+4^{k+1}}{2}\right)$
$\displaystyle+\sum_{k=2}^{e-1}2^{-k}\left(\frac{4^{2}+4^{k+1}}{2}\right)$
$\displaystyle=$ $\displaystyle\frac{15}{14}2^{3e}+3\times
2^{e+1}-\frac{116}{7}.\Box$

######  Corollary 2.7.

If $G\cong C_{p}$ and $p$ is a prime number, then

${\cal N}(G,n)=1+\frac{1}{p-1}\sum\limits_{d|p-1\atop d\leq
n}\varphi(d)\left(\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{p^{2\sum\limits_{t=1}^{[n/d]}\lambda_{td}}}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}-1\right).$

Proof. If follows from (2.7). $\Box$

##  3 Finite Cycle Groups

In this section we give the formula computing the number of isomorphic classes
of element systems over finite cycle groups.

######  Lemma 3.1.

Assume $G=C_{m}$ and $m=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{s}^{e_{s}}$, where
$p_{1},\ldots,p_{s}$ are mutually different prime numbers, then
$G=G_{1}\oplus\cdots\oplus G_{s}$ with $G_{i}\cong C_{m_{i}}$ and
$m_{i}=p_{i}^{e_{i}}$. Furthermore,

(i) Aut$G\cong$ Aut$G_{1}\oplus\cdots\oplus$Aut$G_{s}$;

(ii) There exists a bijective map
$\psi:\widehat{G}\rightarrow\widehat{G_{1}}\oplus\cdots\oplus\widehat{G_{s}}$.

Proof. (i) It follows from [18, Theorem 1.11.10].

(ii) Define map $\psi$:
$\widehat{G}\rightarrow\widehat{G_{1}}\oplus\cdots\oplus\widehat{G_{s}}$ by
sending $\chi\in\widehat{G}$ to

$\psi(\chi)=(\chi_{1},\ldots,\chi_{s}),$

where $\chi_{i}$ is the restriction of $\chi$ on $G_{i}$. It is clear that
$\psi$ is injective.

Conversely, for any
$(\chi_{1},\ldots,\chi_{s})\in\widehat{G_{1}}\oplus\cdots\oplus\widehat{G_{s}}$,
Define $\chi\in\hat{G}$ such that
$\chi(g)=\chi_{1}(g_{1})\cdots\chi_{s}(g_{s}),$ for any
$g=(g_{1},\ldots,g_{s})\in G,g_{i}\in G_{i}.$ Thus $\psi$ is surjective.
$\Box$

######  Theorem 3.2.

Assume that finite cycle group $G\cong C_{p_{1}^{e_{1}}}\oplus
C_{p_{2}^{e_{2}}}\oplus\cdots\oplus C_{p_{s}^{e_{s}}}$ and $p_{i}$ is a prime
number with positive integer $e_{i}$ and $p_{1}<p_{2}<\cdots<p_{s}.$

(i) If all of $p_{1},\cdots,p_{s}$ are odd prime numbers or $p_{1}=2$ with
$e_{1}\leq 2$, then

${\cal
N}(G,n)=\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\left(\frac{1}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\prod_{i=1}^{s}\frac{\sum\limits_{k_{i}=1}^{e_{i}}\sum\limits_{d_{i}|p_{i}-1}\varphi(p_{i}^{e_{i}-k_{i}}d_{i})p_{i}^{2f_{p_{i}}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e_{i}}^{(k_{i},d_{i})})}}{p_{i}^{e_{i}-1}(p_{i}-1)}\right).$
(3.1)

(ii) If $p_{1}=2$ and $e_{1}\geq 3$, then

$\displaystyle{\cal N}(G,n)$ $\displaystyle=$
$\displaystyle\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\left(\frac{1}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\frac{\sum\limits_{k_{1}=2}^{e_{1}}\sum\limits_{d_{1}=1}^{2}\varphi(2^{e_{1}-k_{1}})4^{f_{2}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e_{1}}^{(k_{1},d_{1})})}}{2^{e_{1}-1}}\right.$
$\displaystyle\left.\prod_{i=2}^{s}\frac{\sum\limits_{k_{i}=1}^{e_{i}}\sum\limits_{d_{i}|p_{i}-1}\varphi(p_{i}^{e_{i}-k_{i}}d_{i})p_{i}^{2f_{p_{i}}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e_{i}}^{(k_{i},d_{i})})}}{p_{i}^{e_{i}-1}(p_{i}-1)}\right).$

Proof. Given $(\phi,\sigma)\in M$, we first compute $|F_{(\phi,\sigma)}|$.
According to Lemma3.1, we can assume $\phi=(\phi_{1},\ldots,\phi_{s})$ with
$\phi_{i}\in\mathrm{Aut}G_{i}$. It is clear

$\phi(g)=(\phi(g_{1}),\ldots,\phi(g_{s}))$ (3.2)

and

$\chi\phi^{-1}=(\chi_{1}\phi_{1}^{-1},\ldots,\chi_{s}\phi_{s}^{-1})$ (3.3)

for any $g=(g_{1},\ldots,g_{s})\in G$,
$\chi=(\chi_{1},\ldots,\chi_{s})\in\widehat{G}$.

If $\left(\begin{array}[]{c}j\\\ g_{j}\\\ \chi_{j}\end{array}\right)\in
F_{(\phi,\sigma)}$, applying (1.7), (3.2) and (3.3), we have

$\displaystyle(\phi,\sigma)\circ\left(\begin{array}[]{c}j\\\ g_{j}\\\
\chi_{j}\end{array}\right)$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}\sigma(j)\\\ \phi(g_{j})\\\
\chi_{j}\phi^{-1}\end{array}\right)=\left(\begin{array}[]{c}\sigma(j)\\\
(\phi(g_{j1}),\ldots,\phi(g_{js}))\\\
(\chi_{j1}\phi_{1}^{-1},\ldots,\chi_{js}\phi_{s}^{-1})\end{array}\right)$
(3.13) $\displaystyle=$ $\displaystyle\left(\begin{array}[]{c}j\\\
(g_{j1},\ldots,g_{js})\\\ (\chi_{j1},\ldots,\chi_{js})\end{array}\right),$
(3.17)

where $g_{j}=(g_{j1},\ldots,g_{js}),\chi_{j}=(\chi_{j1},\ldots,\chi_{js})$.
(3.13) is equivalent to

$\displaystyle\left(\begin{array}[]{c}\sigma(j)\\\ \phi_{i}(g_{ji})\\\
(\chi_{ji}\phi_{i}^{-1})\end{array}\right)=\left(\begin{array}[]{c}j\\\
g_{ji}\\\ \chi_{ji}\end{array}\right),\ 1\leq i\leq s.$

This implies $\left(\begin{array}[]{c}j\\\ g_{ji}\\\
\chi_{ji}\end{array}\right)\in F_{(\phi_{i},\sigma)}$, where
$F_{(\phi_{i},\sigma)}$ denotes the fixed point set of $(\phi_{i},\sigma)$ in
$\Omega(G_{i},n)$. Consequently,

$|F_{(\phi,\sigma)}|=\prod_{j=1}^{s}|F_{(\phi_{i},\sigma)}|.$ (3.19)

(i) By (2.16) we have

$\displaystyle{\cal N}(G,n)$ $\displaystyle=$
$\displaystyle\frac{1}{|M|}\sum_{(\phi,\sigma)\in M}|F_{(\phi,\sigma)}|$
$\displaystyle=$
$\displaystyle\frac{1}{n!|\mathrm{Aut}G|}\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\sum_{k_{1}=1}^{e_{1}}\sum_{d_{1}|p_{1}-1}\cdots\sum_{k_{s}=1}^{e_{s}}\sum_{d_{s}|p_{s}-1}\left(\frac{n!}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right.$
$\displaystyle\left.\prod_{i=1}^{s}\varphi(p_{i}^{e_{i}-k_{i}}d_{i})p_{i}^{2f_{p_{i}}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e_{i}}^{(k_{i},d_{i})})}\right)$
$\displaystyle=$
$\displaystyle\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\left(\frac{1}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\prod_{i=1}^{s}\frac{\sum\limits_{k_{i}=1}^{e_{i}}\sum\limits_{d_{i}|p_{i}-1}\varphi(p_{i}^{e_{i}-k_{i}}d_{i})p_{i}^{2f_{p_{i}}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e_{i}}^{(k_{i},d_{i})})}}{p_{i}^{e_{i}-1}(p_{i}-1)}\right).$

(ii) By (2.16) and 2.17), we have

$\displaystyle{\cal N}(G,n)$ $\displaystyle=$
$\displaystyle\frac{1}{n!|\mathrm{Aut}G|}\sum_{\lambda_{1}+\cdots+n\lambda_{n}=n}\sum_{k_{1}=2}^{e_{1}}\sum_{d_{1}=1}^{2}\sum_{k_{2}=1}^{e_{1}}\sum_{d_{2}|p_{2}-1}\cdots\sum_{k_{s}=1}^{e_{s}}\sum_{d_{s}|p_{s}-1}\left(\frac{n!}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\right.$
$\displaystyle\left.\varphi(2^{e_{1}-k_{1}})p_{1}^{2f_{p_{1}}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e_{i}}^{(k_{1},d_{1})})}\prod_{i=2}^{s}\varphi(p_{i}^{e_{i}-k_{i}}d_{i})p_{i}^{2f_{p_{i}}(\lambda_{1},\ldots,\lambda_{n},\vec{\delta}_{e_{i}}^{(k_{i},d_{i})})}\right).\
\ \Box$

######  Corollary 3.3.

If $G\cong C_{p_{1}}\oplus\cdots\oplus C_{p_{s}}$ and $p_{1},\ldots,p_{s}$ are
mutually different prime numbers, then

$\mathcal{N}(G,1)=\prod_{i=1}^{s}(p_{i}+2).$

Proof. Obviously, $k_{i}=e_{i}=1$,
$f_{p_{i}}(\lambda_{1},\vec{\delta}_{e_{i}}^{(1,1)})=1$ and
$f_{p_{i}}(\lambda_{1},\vec{\delta}_{e_{i}}^{(1,d_{i})})=0(d_{i}>1)$. Using
(3.1), we have

$\displaystyle{\cal N}(G,1)$ $\displaystyle=$
$\displaystyle\prod_{i=1}^{s}\frac{\sum\limits_{d_{i}|p_{i}-1}\varphi(d_{i})p_{i}^{2f_{p_{i}}(\lambda_{1},,\vec{\delta}_{e_{i}}^{(1,d_{i})})}}{p_{i}-1}$
$\displaystyle=$
$\displaystyle\prod_{i=1}^{s}\frac{p_{i}^{2}+\sum\limits_{d_{i}|p_{i}-1\atop
d_{i}>1}\varphi(d_{i})}{p_{i}-1}$ $\displaystyle=$
$\displaystyle\prod_{i=1}^{s}\frac{p_{i}^{2}+p_{i}-2}{p_{i}-1}$
$\displaystyle=$ $\displaystyle\prod_{i=1}^{s}(p_{i}+2).\Box$

##  4 Primary Commutative Groups

In this section we give the formula computing the number of isomorphic classes
of element systems of primary commutative groups.

If $G\cong\overbrace{C_{p}\times\cdots\times C_{p}}^{s}$ and $p$ is a prime
number, then $G$ is called a primary commutative group.

######  Lemma 4.1.

(See [18, Theorem 5.4.2]) If $G$ is a primary commutative group with order
$p^{s}$ and positive integer $s$, then

(i) Aut$G\cong GL(s,Z_{p});$

(ii) $|$Aut$G|$=$p^{\frac{1}{2}s(s-1)}\prod\limits_{i=1}^{s}(p^{i}-1).$
Here $GL(s,Z_{p})$ is a general linear group over field $Z_{p}$. Indeed,
$GL(s,Z_{p})$ also can be viewed as $s\times s$-matrix ring over field
$Z_{p}$.

Let $g_{1},\cdots,g_{s}$ be a basis of $G$, i.e. $G\cong\langle
g_{1}\rangle\times\cdots\times\langle g_{s}\rangle$ with $o(g_{i})=p,$ where
$o(g_{i})$ denotes the order of $g_{i}$. For any $g\in G$,
$g=g_{1}^{k_{1}}\cdots g_{s}^{k_{s}},$ let
$\vec{k}:=\left(\begin{array}[]{c}k_{1}\\\ \vdots\\\
k_{s}\end{array}\right)denote\ $g.

By Lemma4.1,for any $\phi\in$Aut$G$, there exists a matrix
$A=(a_{ij})_{s\times s}\in GL(s,Z_{p})$ such that

$\displaystyle\phi:G$ $\displaystyle\rightarrow G$ $\displaystyle\vec{k}$
$\displaystyle\mapsto A\vec{k}.$

Let $A$ denote $\phi$.

$\hat{G}$ consists of the following maps:

$\displaystyle\chi:G$ $\displaystyle\rightarrow F$ $\displaystyle g_{i}$
$\displaystyle\mapsto\omega^{l_{i}}\qquad i=1,2,\cdots,s;\ 0\leq l_{i}\leq
p-1.$

Here $\omega$ is a primitive $p$th root of 1. Let
$\vec{l}=(l_{1},\cdots,l_{s})^{T}$ denote the character $\chi$ above, where
$T$ denotes transposition.

Next we compute the fixed point set $F_{(A,\sigma)}$ of $(A,\sigma)$. If
$\left(\begin{array}[]{c}j\\\ \vec{k}_{j}\\\ \vec{l}_{j}\end{array}\right)\in
F_{(A,\sigma)}$, then

$\left(\begin{array}[]{c}\sigma(j)\\\ A\vec{k}_{j}\\\ (A^{-1})^{\mbox{\tiny
T}}\vec{l}_{j}\end{array}\right)=\left(\begin{array}[]{c}j\\\ \vec{k}_{j}\\\
\vec{l}_{j}\end{array}\right)=\left(\begin{array}[]{c}\sigma(j)\\\
\vec{k}_{\sigma(j)}\\\ \vec{l}_{\sigma(j)}\end{array}\right).$

That is,

$\displaystyle A\vec{k}_{j}=\vec{k}_{\sigma(j)}$ (4.1)

and

$\displaystyle(A^{-1})^{\mbox{\tiny T}}\vec{l}_{j}=\vec{l}_{\sigma(j)},\qquad
j=1,\cdots,n,$ (4.2)

where $\vec{k_{j}},\vec{l_{j}}\in(Z_{p})^{s}$.

Assume that $\tau=(j_{1},\cdots,j_{r})$ is a cycle in the independent cycle
decomposition of $\sigma$ with the length $r$. We have

$\displaystyle\begin{array}[]{llll}\vec{k}_{j_{2}}&=A\vec{k}_{j_{1}},&\vec{l}_{j_{2}}&=B\vec{l}_{j_{1}},\\\
&\cdots&\cdots&\\\
\vec{k}_{j_{r}}&=A^{r-1}\vec{k}_{j_{1}},&\vec{l}_{j_{r}}&=B^{r-1}\vec{l}_{j_{1}},\\\
\vec{k}_{j_{1}}&=A^{r}\vec{k}_{j_{1}},&\vec{l}_{j_{1}}&=B^{r}\vec{l}_{j_{1}},\end{array}$
(4.7)

where $B=(A^{-1})^{\mbox{\tiny T}}$. Thus we only need choice
$\vec{k}_{j_{1}}$ and $\vec{l}_{j_{1}}$ such that $\vec{k}_{j_{1}}$ and
$\vec{l}_{j_{1}}$ become the solutions of $(A^{r}-I)X=0$ and $(B^{r}-I)X=0$,
respectively. Applying (4.7), we compute
$\vec{k}_{j_{2}},\cdots,\vec{k}_{j_{r}},\vec{l}_{j_{2}},\cdots,\vec{l}_{j_{r}}$.
This implies that we find a solution $\vec{k}_{j}$ ’ s , $\vec{l}_{j}{}$’s of
(4.1) for $j=j_{1},j_{2},\cdots,j_{r}$. Conversely, it is true, too.

Considering

$\displaystyle B^{r}-I$ $\displaystyle=$ $\displaystyle((A^{r})^{\mbox{\tiny
T}})^{-1}-I$ $\displaystyle=$ $\displaystyle((A^{r})^{\mbox{\tiny
T}})^{-1}-((A^{r})^{\mbox{\tiny T}})^{-1}((A^{r})^{\mbox{\tiny T}}$
$\displaystyle=$ $\displaystyle((A^{r})^{\mbox{\tiny
T}})^{-1}(I-A^{r})^{\mbox{\tiny T}},$

we have rank$(A^{r}-I)=$rank$(B^{r}-I)$, i.e. the numbers of solutions of of
$(A^{r}-I)X=0$ and $(B^{r}-I)X=0$ in $(Z_{p})^{s}$ are the same, written as
$\nu_{A,r}$. Thus

$\displaystyle\nu_{A,r}=p^{s-rank(A^{r}-I)}.$ (4.8)

Assume the type of $\sigma$ is $1^{\lambda_{1}}2^{\lambda_{2}}\cdots
n^{\lambda_{n}}$. It is clear

$\displaystyle|F_{(A,\sigma)}|=\prod_{r=1}^{n}\nu_{A,r}^{2\lambda_{r}}.$ (4.9)

######  Theorem 4.2.

If $G$ is a primary commutative group with order $p^{s}$, then

${\cal
N}(G,n)=\frac{1}{p^{\frac{1}{2}s(s-1)}\prod\limits_{i=1}^{s}(p^{i}-1)}\sum\limits_{A\in
GL(s,p)}\ \sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\
\prod\limits_{i=1}^{n}\frac{p^{2(s-rank(A^{i}-I)){\lambda_{i}}}}{\lambda_{i}!i^{\lambda_{i}}}.$

Proof. Applying Lemma 4.1 and (4.9), we have

$\displaystyle{\cal N}(G,n)$ $\displaystyle=$
$\displaystyle\frac{\sum\limits_{(A,\sigma)\in M}|F_{(A,\sigma)}|}{|M|}$
$\displaystyle=$ $\displaystyle\frac{\sum\limits_{A\in GL(s,p)}\
\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\frac{n!}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\
\prod\limits_{r=1}^{n}\nu_{A,r}^{2\lambda_{r}}}{p^{\frac{1}{2}s(s-1)}\prod\limits_{i=1}^{s}(p^{i}-1)n!}$
$\displaystyle=$
$\displaystyle\frac{1}{p^{\frac{1}{2}s(s-1)}\prod\limits_{i=1}^{s}(p^{i}-1)}\sum\limits_{A\in
GL(s,p)}\ \sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\
\prod\limits_{i=1}^{n}\frac{\nu_{A,i}^{2\lambda_{i}}}{\lambda_{i}!i^{\lambda_{i}}}.\Box$

##  5 Finite Commutative Groups

In this section we give the formula computing the number of isomorphic classes
of element systems over finite commutative groups.

Let $G$ be an additive group with order $m$ and $h_{1},\cdots,h_{s}$ be a
basis of $G$, i.e. $G=<h_{1}>\times<h_{2}>\times\cdots\times<h_{s}>$, where
the order of $h_{i}$ is $m_{i}=p_{i}^{e_{i}}$ and $p_{i}$ is a prime number
with positive integer $e_{i}$.

Let $M_{s\times s}(\mathbb{Z})$ denote the set of all $s\times s$\- matrices
over $\mathbb{Z}$. For any $\phi\in$Aut$G$, there exists $A=(a_{ij})_{s\times
s}\in M_{s\times s}(\mathbb{Z})$, such that
$\phi(h_{1},h_{2},\cdots,h_{s})=(h_{1},h_{2},\cdots,h_{s})A.$ Similarly, there
exists $B=(b_{ij})_{s\times s}\in M_{s\times s}(\mathbb{Z})$ such that
$\phi^{-1}(h_{1},h_{2},\cdots,h_{s})$ $=$ $(h_{1},h_{2},\cdots,$
$h_{s})B^{T}$.

Let $\omega$ be a primitive $m$th root of 1. For
$\vec{k}\in\left(\begin{array}[]{l}Z_{m_{1}}\\\ \vdots\\\
Z_{m_{s}}\end{array}\right)$ and $\vec{l}\in\left(\begin{array}[]{l}Z_{m}\\\
\vdots\\\ Z_{m}\end{array}\right)$, define
$g_{\vec{k}}:=h_{1}{k_{1}}+\cdots+h_{s}{k_{s}}\in G$ and
$\chi_{\vec{l}}\in\hat{G}$ such that
$\chi_{\vec{l}}(h_{1},h_{2},\cdots,h_{s}):=(\omega^{l_{1}},\omega^{l_{2}},\cdots,\omega^{l_{s}})$.

Now we compute the fixed point set $F_{(\phi,\sigma)}$ of $(\phi,\sigma)$. If
$\left(\begin{array}[]{c}j\\\ g_{\vec{k}_{j}}\\\
\chi_{\vec{l}_{j}}\end{array}\right)\in F_{(\phi,\sigma)}$, then

$\left(\begin{array}[]{c}\sigma(j)\\\ \phi(g_{\vec{k}_{j}})\\\
\chi_{\vec{l}_{j}}\phi^{-1}\end{array}\right)=\left(\begin{array}[]{c}j\\\
g_{\vec{k}_{j}}\\\
\chi_{\vec{l}_{j}}\end{array}\right)=\left(\begin{array}[]{c}\sigma(j)\\\
g_{\vec{k}_{\sigma(j)}}\\\ \chi_{\vec{l}_{\sigma(j)}}\end{array}\right).$

Thus

$\displaystyle A\vec{k}_{j}=\vec{k}_{\sigma(j)}\ \ \hbox{ and }\ \ \
B\vec{l}_{j}=\vec{l}_{\sigma(j)},\qquad j=1,\cdots,n.$ (5.1)

Assume that $\tau=(j_{1},\cdots,j_{r})$ is a cycle in independent cycle
decomposition of $\sigma$ with length $r$. We have

$\displaystyle\begin{array}[]{llll}\vec{k}_{j_{2}}&=A\vec{k}_{j_{1}},&\vec{l}_{j_{2}}&=B\vec{l}_{j_{1}},\\\
&\cdots&\cdots&\\\
\vec{k}_{j_{r}}&=A^{r-1}\vec{k}_{j_{1}},&\vec{l}_{j_{r}}&=B^{r-1}\vec{l}_{j_{1}},\\\
\vec{k}_{j_{1}}&=A^{r}\vec{k}_{j_{1}},&\vec{l}_{j_{1}}&=B^{r}\vec{l}_{j_{1}}.\end{array}$
(5.6)

Thus we only need choice $\vec{k}_{j_{1}}$ and $\vec{l}_{j_{1}}$ such that
$\vec{k}_{j_{1}}$ and $\vec{l}_{j_{1}}$ become the solutions of

$\displaystyle(A^{r}-I)X=0$ (5.7)

and

$\displaystyle(B^{r}-I)X=0,\ $ (5.8)

respectively. According to (5.6), we compute
$\vec{k}_{j_{2}},\cdots,\vec{k}_{j_{r}}$ and
$\vec{l}_{j_{2}},\cdots,\vec{l}_{j_{r}}$. That is, we get $\vec{k}_{j}$’ s and
$\vec{l}_{j}{}$’ s such that (5.1) holds. Conversely it is true, too.

Let $\nu_{\phi,r}$ and $\mu_{\phi,r}$ denote the numbers of solutions of (5.7)
and (5.8) in $\left(\begin{array}[]{l}Z_{m_{1}}\\\ \vdots\\\
Z_{m_{s}}\end{array}\right)$ and $\left(\begin{array}[]{l}Z_{m}\\\ \vdots\\\
Z_{m}\end{array}\right)$, respectively. If the type of $\sigma$ is
$1^{\lambda_{1}}2^{\lambda_{2}}\cdots n^{\lambda_{n}}$, then

$\displaystyle|F_{(A,\sigma)}|=\prod_{r=1}^{n}(\nu_{\phi,r}\mu_{\phi,r})^{\lambda_{r}}.$
(5.9)

######  Theorem 5.1.

If $G$ is a commutative group with order $m$, then

${\cal N}(G,n)=\frac{1}{\mid\\!Aut(G)\\!\mid}\sum\limits_{\phi\in Aut(G)}\
\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\
\prod\limits_{i=1}^{n}\frac{(\nu_{\phi,r}\mu_{\phi,r})^{\lambda_{i}}}{\lambda_{i}!i^{\lambda_{i}}}.$

Proof. By Lemma 4.1 and (5.9), we have

$\displaystyle{\cal N}(G,n)$ $\displaystyle=$
$\displaystyle\frac{\sum\limits_{(\phi,\sigma)\in M}|F_{(\phi,\sigma)}|}{|M|}$
$\displaystyle=$ $\displaystyle\frac{\sum\limits_{\phi\in Aut(G)}\
\sum\limits_{\lambda_{1}+\cdots+n\lambda_{n}=n}\frac{n!}{\lambda_{1}!\cdots\lambda_{n}!1^{\lambda_{1}}\cdots
n^{\lambda_{n}}}\
\prod\limits_{r=1}^{n}(\nu_{\phi,r}\mu_{\phi,r})^{\lambda_{r}}}{\mid\\!Aut(G)\\!\mid
n!}$ $\displaystyle=$
$\displaystyle\frac{1}{\mid\\!Aut(G)\\!\mid}\sum\limits_{\phi\in Aut(G)}\
\sum\limits_{\lambda_{1}+2\lambda_{2}+\cdots+n\lambda_{n}=n}\
\prod\limits_{i=1}^{n}\frac{(\nu_{\phi,i}\mu_{\phi,r})^{\lambda_{i}}}{\lambda_{i}!i^{\lambda_{i}}}.\Box$

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