# On LinDistFlow Model Congestion Pricing:
Bounding the Changes in Power Tariffs

Shourya Bose, Kejun Chen, and Yu Zhang  ECE Department, University of
California, Santa Cruz  Emails: {shbose, kchen158<EMAIL_ADDRESS>
###### Abstract

The optimal power flow (OPF) problem is an important mathematical program that
aims at obtaining the best operating point of an electric power grid. The
optimization problem typically minimizes the total generation cost subject to
certain physical constraints of the system. The so-called linearized
distribution flow (LinDistFlow) model leverages a set of linear equations to
approximate the nonlinear AC power flows. In this paper, we consider an OPF
problem based on the LinDistFlow model for a single-phase radial power
network. We derive closed-form solutions to the marginal values of both real
and reactive power demands. We also derive upper bounds on the congestion
price (a.k.a. ‘shadow price’), which denotes the change in marginal demand
prices when the apparent power flow limits of certain lines are binding at
optimum. Various cases of our result are discussed while simulations are
carried out on a $141$-bus radial power network.

##  I Introduction

Radially energized power networks are prevalent in grid-scale power systems
such as utility distribution networks and microgrids [1]. They are defined as
power distribution networks wherein the energized section assumes a topology
of a _connected tree_. Traditionally, the only generation source in a radial
network is a sub-station as the upstream to the network, which is interfaced
with some high-voltage transmission network using power conversion devices.
With the advent of consumer-level generation devices such as photovoltaic
panels, wind turbines, and microturbines, radial networks can now accommodate
_prosumers_ , i.e. agents on the network which can either inject/withdraw
power in/from the network. Desirable set points of generation power can be
determined by solving for set points which are optimal with respect to some
generation cost function, subject to physical and operational constraints.
This optimization problem, well known as _Optimal Power Flow_ (OPF) was first
introduced in literature by Carpentier [2, 3]. Depending on the nature of
constraints and the cost function, OPF can be variously categorized as DC-OPF,
AC-OPF, security constrained OPF, etc [4]. The former two are different ways
of modeling physics of the network, while the latter adds security or
contingency constraints meant to ensure robust operation of the network. It is
important to note that in this article we only consider OPF problems wherein
the network operator and prosumers seek to optimize the same cost function.
This is as opposed to scenarios wherein the prosumers may seek to optimize a
cost function different from that of the network operator [5, 6].

The DC-OPF, which models the physics of the network using a set of linear
equations, has been widely researched and used in practice for transmission
networks, where the topology may be meshed. Combined with administrative
constraints, DC-OPF provides reasonably accurate set points _vis-a-vis_ the
optimal AC-OPF solution [7]. However, the DC-OPF linearization does not model
reactive power injection, which poses a problem in analyzing radial
distribution networks with devices such as PVs and WTs having controllable
inverters. To counter this drawback, there has been significant recent
research on the _LinDistFlow_ equation [8]. First introduced by Baran and Wu
[9], this is a set of linearized equations which describes the physics of the
network (possibly multi-phase with high R/X ratios). The underlying condition
of LinDistFlow is that there are no power line losses, which allows for
linearization of the non-convex _DistFlow_ equations from which LinDistFlow is
derived [10].

#### Related Work

For a review of marginal demand costs in traditional OPF models including DC-
OPF, the reader may consult the textbook [11]. Khatami et al. provide a
detailed description of various components constituting nodal prices [12].
Biegel et al. consider congestion management through shadow prices [13]. Bai
et al. consider marginal pricing of real and reactive power demand under
various markets for the nonlinear DistFlow model [14]. Xu et al. design a
deregulated power market mechanism, which uses the idea of marginal pricing at
its core [15]. A comprehensive review of pricing mechanisms in transmission
and distribution markets, including reserves, may be found in [16]. Finally,
similar in nature to the current paper, Winnicki et al. consider marginal
pricing in the DistFlow model, but without consideration of congestion. [17]

#### Contribution

In this paper, we formulate an OPF problem with the LinDistFlow model, named
as LDF-OPF. We consider load satisfaction, generation bounds, voltage bounds,
and conic branch flows. Our proposed framework can handle more generalized
formulations with linear and conic constraints. We first show a closed-form
expression for the marginal price when there is no flow congestion at
optimality of LDF-OPF. Then, we derive an upper bound on the variation of
demand marginal costs. The proposed upper bound is function of terms involving
marginal costs associated with binding of aforementioned conic constrains
representing branch flow limits, and network topology factors. This result
provides a useful tool for network operators to estimate the change in demand
marginal prices as a function of their choice of branch flow marginal costs.

#### Notation

$\mathbb{R}$ and $\mathbb{N}$ denote the set of real numbers and integers,
respectively. Vectors and matrices are denoted with boldface. For a vector
$\mathbf{a}\in\mathbb{R}^{n}$, $\mathbf{a}(j)$ is its $j^{\text{th}}$ element
while $\left\lVert\mathbf{a}\right\rVert_{2}$ denotes its 2-norm.
$\mathbf{1}_{n}\in\mathbb{R}^{n}$ is the all-ones vector. For a positive
$n\in\mathbb{N}$, $[n]$ denotes the set $\\{1,\cdots,n\\}$. For a finite set
$\mathcal{S}$, $|\mathcal{S}|$ denotes its cardinality. For a directed graph
$\mathcal{G}=(\mathcal{N},\mathcal{L})$ where $\mathcal{N}$ is the set of
nodes and $\mathcal{L}$ the set of _directed_ edges, and for any node
$i\in\mathcal{N}$, $\mathcal{H}_{i}$ denotes the _inclusive downstream set_ of
$i$, i.e.
$\mathcal{H}_{i}\stackrel{{\scriptstyle\Delta}}{{=}}\\{j\in\mathcal{N}|\exists\text{
directed path from $i$ to $j$ in }\mathcal{L}\\}\cup\\{i\\}$.

##  II Problem Formulation

###  II-A Background

Consider a radial power network with a slack bus. The buses are labeled as
$\mathcal{N}\stackrel{{\scriptstyle\Delta}}{{=}}\\{0,1,\cdots,n\\}$, with $0$
denoting the slack bus. The network topology is represented by a directed
graph
$\mathcal{G}\stackrel{{\scriptstyle\Delta}}{{=}}(\mathcal{N},\mathcal{L})$,
where $\mathcal{L}$ denotes the set of branches. Without loss of generality,
$\mathcal{L}$ can be constructed such that the directed branches point _away_
from the slack bus. All non-slack buses are classified as the set of _load_
buses $\mathcal{N}^{l}$ and the set of _generator_ buses $\mathcal{N}^{g}$
such that $\mathcal{N}=\\{0\\}\cup\mathcal{N}^{g}\cup\mathcal{N}^{l}$. Let
$n_{g}\stackrel{{\scriptstyle\Delta}}{{=}}|\mathcal{N}^{g}|$ and
$n_{l}\stackrel{{\scriptstyle\Delta}}{{=}}|\mathcal{N}^{l}|$ denote the number
of generator and load buses, respectively. Noting that the number of branches
equals that of non-slack buses, each branch may be uniquely assigned the index
of the bus to which it is upstream.

A concrete way of analyzing the physics of power flows is through the
_DistFlow_ equations. For $i\in\mathcal{N}$, let
$s_{i}\stackrel{{\scriptstyle\Delta}}{{=}}p_{i}+\mathfrak{i}q_{i}$ be the
complex power injection at bus $i$,
$S_{i}\stackrel{{\scriptstyle\Delta}}{{=}}P_{i}+\mathfrak{i}Q_{i}$ be the
complex power flow on branch $i$, and $v_{i}$ and $l_{i}$ denote the squared
voltage magnitude at bus $i$, and squared current magnitude of branch $i$,
respectively. The DistFlow equations that hold for all $i\in\mathcal{N}$ are
given as [18]

$\displaystyle s_{i}$
$\displaystyle=\sum_{j\in\text{child}(i)}S_{j}-S_{i}+l_{i}z_{i},$ (1a)
$\displaystyle v_{i}$
$\displaystyle=v_{\text{parent}(i)}-2\text{Re}[z_{i}^{*}S_{i}]+l_{i}|z_{i}|^{2},$
(1b) $\displaystyle|S_{i}|^{2}$ $\displaystyle=v_{\text{parent}(i)}l_{i}.$
(1c)

Assuming no power line losses in the network the DistFlow equations 1 can be
linearized into the so-called _LinDistFlow_ model whose compact form is given
as [18]

$\displaystyle\mathbf{v}=\mathbf{Rp}+\mathbf{Xq}+v_{0}\mathbf{1}_{n},$ (2)

where
$\mathbf{p}\stackrel{{\scriptstyle\Delta}}{{=}}[\mathbf{p}(1),\cdots,\mathbf{p}(n)]$,
$\mathbf{q}\stackrel{{\scriptstyle\Delta}}{{=}}[\mathbf{q}(1),\cdots,\mathbf{q}(n)]$,
and
$\mathbf{v}\stackrel{{\scriptstyle\Delta}}{{=}}[\mathbf{v}(1),\cdots,\mathbf{v}(n)]$
denote the real and reactive power injections and squared voltage magnitude of
all non-slack buses, respectively. $v_{0}$ is the fixed voltage of the slack
bus. Positive semi-definite matrices $\mathbf{R},\mathbf{X}\in n\times n$
encode branch resistance and reactance, as well as the topology of
$\mathcal{G}$. Since the vector $\mathbf{p}$ (similarly $\mathbf{q}$ and
$\mathbf{v}$) contains various indices corresponding to generators and loads,
we define the matrices $\mathbf{A}_{g}\in\\{0,1\\}^{n_{g}\times n}$ and
$\mathbf{A}_{l}\in\\{0,1\\}^{n_{l}\times n}$ which help us separate generation
and load indices from $p$ as

$\displaystyle\mathbf{p}_{g}=\mathbf{A}_{g}\mathbf{p},\quad\mathbf{q}_{g}=\mathbf{A}_{g}\mathbf{q},\quad\mathbf{p}_{l}=\mathbf{A}_{l}\mathbf{p},\quad\mathbf{q}_{l}=\mathbf{A}_{l}\mathbf{q}.$

Thanks to zero line losses, the slack bus real and reactive injections become
$p_{s}=-\mathbf{1}_{n}^{\top}\mathbf{p}$ and
$q_{s}=-\mathbf{1}_{n}^{\top}\mathbf{q}$, respectively.

Let
$\mathbf{f}^{p}\stackrel{{\scriptstyle\Delta}}{{=}}[\mathbf{f}^{\mathbf{p}}(1),\cdots,\mathbf{f}^{p}(n)]$
and
$\mathbf{f}^{q}\stackrel{{\scriptstyle\Delta}}{{=}}[\mathbf{f}^{\mathbf{p}}(1),\cdots,\mathbf{f}^{q}(n)]$.
The branch flows are given as $\mathbf{f}^{p}=\mathbf{Fp}$ and
$\mathbf{f}^{q}=\mathbf{Fq}$, where $\mathbf{F}$ is derived from the _signed
branch-bus incidence matrix_ $\tilde{\mathbf{A}}\in n\times n+1$ by deleting
its first column, and invert-transposing it.

######  Lemma 1 (Properties of matrix $\tilde{\mathbf{A}}$).

The matrix $\tilde{\mathbf{A}}$ is defined as

$\displaystyle\tilde{\mathbf{A}}(i,j)=\begin{cases}1,&\text{if branch $i$
starts at bus $j-1$}\\\ -1,&\text{if branch $i$ terminates at bus $j-1$}\\\
0,&\text{otherwise.}\end{cases}$

Let $\mathbf{A}$ be the square matrix derived by deleting the first column of
$\tilde{\mathbf{A}}$. $\mathbf{A}^{-1}$ exists [18], and
$\mathbf{F}\stackrel{{\scriptstyle\Delta}}{{=}}\mathbf{A}^{-\top}$.

###  II-B Optimal Power Flow Problem

A general OPF problem based on the described power network characteristics is
given as follows.

$\displaystyle\min\limits_{p,q,v,p_{s},q_{s}}\quad$
$\displaystyle\mathbf{c}_{g}^{\top}(\mathbf{A}_{g}\mathbf{p})+c_{s}p_{s}$ (3)
s.t. $\displaystyle\mathbf{v}=\mathbf{Rp}+\mathbf{Xq}+v_{0}\mathbf{1}_{n}$
(3a) $\displaystyle\underline{\mathbf{v}}\leq\mathbf{v}\leq\bar{\mathbf{v}}$
(3b)
$\displaystyle\mathbf{A}_{l}\mathbf{p}=\hat{\mathbf{p}}_{l},\quad\mathbf{A}_{l}\mathbf{q}=\hat{\mathbf{q}}_{l}$
(3c)
$\displaystyle\underline{\mathbf{p}}_{g}\leq\mathbf{A}_{g}\mathbf{p}\leq\bar{\mathbf{p}}_{g},\quad\underline{\mathbf{q}}_{g}\leq\mathbf{A}_{g}\mathbf{q}\leq\bar{\mathbf{q}}_{g}$
(3d) $\displaystyle p_{s}=-\mathbf{1}_{n}^{\top}\mathbf{p},\quad
q_{s}=-\mathbf{1}_{n}^{\top}\mathbf{q}$ (3e)
$\displaystyle\underline{p}_{s}\leq
p_{s}\leq\bar{p}_{s},\quad\underline{q}_{s}\leq q_{s}\leq\bar{q}_{s}$ (3f)
$\displaystyle\mathbf{f}^{p}=\mathbf{Fp},\quad\mathbf{f}^{q}=\mathbf{Fq}$ (3g)
$\displaystyle\left\lVert[\mathbf{f}^{p}(i),\mathbf{f}^{q}(i)]^{\top}\right\rVert_{2}\leq\bar{f}_{i},\quad\forall
i\in[n]$ (3h)

The objective function calculates the cost of generation at generator buses
(given by
$\mathbf{c}_{g}^{\top}(\mathbf{A}_{g}\mathbf{p})=\mathbf{c}_{g}^{\top}\mathbf{p}_{g}$,
where $\mathbf{c}_{g}\in n_{g}$ is the generation cost vector) and at the
slack bus (given by $c_{s}p_{s}$, where $c_{s}$ is the slack generation cost).
(3a) is the LinDistFlow equation, while (3b) limits voltage values to within
operational limits. (3c) stipulates that the demanded real and reactive power
amounts are $\hat{\mathbf{p}}_{l}$ and $\hat{\mathbf{q}}_{l}$ respectively.
(3d), (3e), and (3f) place limits on generation. Finally, (3g) and (3h)
together describe the conic line flow for each branch.

The OPF (3) contains redundancy in its decision variables and constraints. To
that end, a much simplified and operatorized version of (3) may be written as
follows.

$\displaystyle\mathcal{J}(\boldsymbol{\ell},\bar{\mathbf{f}})\stackrel{{\scriptstyle\Delta}}{{=}}\min\limits_{\mathbf{p}_{g},\mathbf{q}_{g}}\quad$
$\displaystyle\tilde{\mathbf{c}}_{g}^{\top}\mathbf{p}_{g}$ (4)
$\displaystyle\mathbf{M}_{p}\mathbf{p}_{g}+\mathbf{M}_{q}\mathbf{q}_{g}\leq\mathbf{G}\boldsymbol{\ell}+\mathbf{h}$
(4a)
$\displaystyle\underline{\mathbf{p}}_{g}\leq\mathbf{p}_{g}\leq\bar{\mathbf{p}}_{g},\quad\underline{\mathbf{q}}_{g}\leq\mathbf{q}_{g}\leq\bar{\mathbf{q}}_{g}$
(4b)
$\displaystyle\left\lVert\begin{bmatrix}\mathbf{r}_{i}^{\top}\mathbf{p}_{g}+\mathbf{s}_{i}^{\top}\boldsymbol{\ell}\\\
\mathbf{r}_{i}^{\top}\mathbf{q}_{g}+\mathbf{t}_{i}^{\top}\boldsymbol{\ell}\end{bmatrix}\right\rVert_{2}\leq\bar{f}_{i},\quad\forall
i\in[n]$ (4c)

In the above,
$\tilde{\mathbf{c}}_{g}\stackrel{{\scriptstyle\Delta}}{{=}}\mathbf{c}_{g}-c_{s}\mathbf{1}_{n_{g}}$,
$\boldsymbol{\ell}\stackrel{{\scriptstyle\Delta}}{{=}}[\hat{\mathbf{p}}_{l}^{\top},\hat{\mathbf{q}}_{l}^{\top}]^{\top}$,
and
$\bar{\mathbf{f}}\stackrel{{\scriptstyle\Delta}}{{=}}[\bar{f}_{1},\cdots,\bar{f}_{n}]^{\top}$.
The OPF operator $\mathcal{J}(\boldsymbol{\ell},\bar{\mathbf{f}})$ maps real &
reactive power demands and line flow limits to the optimum generation cost,
which is a scalar. It can be shown that
$\mathcal{J}(\boldsymbol{\ell},\bar{\mathbf{f}})$ is jointly convex in
$\boldsymbol{\ell}$ and $\bar{\mathbf{f}}$; see [19].

######  Remark 1 (Equivalence of (3) and (4)).

To establish the equivalence of (3) and (4), first $\mathbf{v}$, $p_{s}$, and
$q_{s}$ can be eliminated as decision variables from (3) by replacing all of
their occurrences with $\mathbf{Rp}+\mathbf{Xq}+v_{0}\mathbf{1}_{n}$,
$-\mathbf{1}_{n}^{\top}\mathbf{p}$, and $-\mathbf{1}_{n}^{\top}\mathbf{q}$
respectively. Further, the indices of $\mathbf{p}$ and $\mathbf{q}$
corresponding to load buses may be written as linear combinations of elements
of $\boldsymbol{\ell}$ using (3c). Thus, the only effective decision variables
in (3) are $\mathbf{p}_{g},\mathbf{q}_{g}\in n_{g}$. Further, any expression
in (3) which contains any of the decision variables $\mathbf{v,p,q},p_{s}$, or
$q_{s}$ is linear in $\mathbf{p}_{g}$, $\mathbf{q}_{g}$ or
$\boldsymbol{\ell}$. Therefore, the objective of (3) is equivalent to the
objective of (4). Constraints (3a)-(3e) are equivalent to (4a), (3f) and (4b),
and finally (3g)-(3h) are equivalent to (4c).

###  II-C Illustrative Example

$\mathbf{\sim}$$\mathbf{\sim}$Slack$\text{Bus }3$$\text{Bus
}2$$v(2)\in[0.95,1.05]\;pu$$\text{Gen
}1$$v_{0}=1.05\;pu$$v(1)=1.05\;pu$$v(3)\in[0.95,1.05]\;pu$ Figure 1:
Illustrative example of a 4-bus radial network.

Consider a 4-bus system shown in Figure 1, which is derived from case4ba in
MATPOWER [20]. The buses are indexed such that the slack bus has index 0, the
generator bus has index 1, and the two load buses have indices 2 and 3,
respectively. Each branch is uniquely numbered based on its downstream bus.
The impedance (in p.u.) of each branch is $0.003+j0.006$ $\Omega$. Thus,
equation (3a) then becomes

$\displaystyle\begin{bmatrix}\mathbf{v}(1)\\\ \mathbf{v}(2)\\\
\mathbf{v}(3)\end{bmatrix}$ $\displaystyle=\begin{bmatrix}0.012&0.006&0\\\
0.006&0.006&0\\\ 0&0&0.006\end{bmatrix}\begin{bmatrix}\mathbf{p}(1)\\\
\mathbf{p}(2)\\\ \mathbf{p}(3)\end{bmatrix}$
$\displaystyle+\begin{bmatrix}0.024&0.012&0\\\ 0.012&0.012&0\\\
0&0&0.012\end{bmatrix}\begin{bmatrix}\mathbf{q}(1)\\\ \mathbf{q}(2)\\\
\mathbf{q}(3)\end{bmatrix}+\begin{bmatrix}v_{0}\\\ v_{0}\\\
v_{0}\end{bmatrix}.$

The voltage of the generator bus and slack bus is fixed at $1.05$ p.u., while
the voltage levels of the load buses vary in $[0.95,1.05]$ p.u.. Recognizing
that
$\boldsymbol{\ell}=[\mathbf{p}(2),\mathbf{p}(3),\mathbf{q}(2),\mathbf{q}(3)]$,
$p_{g}=\mathbf{p}(1)$ and $q_{g}=\mathbf{q}(1)$, the above equation may be
written as

$\displaystyle\mathbf{v}(1)=0.012p_{g}+0.024q_{g}+\begin{bmatrix}0.006&0&0.012&0\end{bmatrix}\boldsymbol{\ell}+v_{0}$
$\displaystyle\begin{bmatrix}\mathbf{v}(2)\\\
\mathbf{v}(3)\end{bmatrix}=\begin{bmatrix}0.006\\\
0\end{bmatrix}p_{g}+\begin{bmatrix}0.012\\\ 0\end{bmatrix}q_{g}+$
$\displaystyle\begin{bmatrix}0.006&0&0.012&0\\\
0&0.006&0&0.012\end{bmatrix}\boldsymbol{\ell}+\begin{bmatrix}v_{0}\\\
v_{0}\end{bmatrix}$

Letting $\mathbf{v}(1)=1.05$ (or equivalently, $1.05\leq\mathbf{v}(1)\leq
1.05$) and $0.95\leq\mathbf{v}(i)\leq 1.05$ for $i=2,3$ recovers (4a). Now
suppose the slack bus provides a maximum of 1 p.u. real power to the system;
i.e., $p_{s}\leq 1$. This can be equivalently written in the form of (4a) as
$-\mathbf{p}_{g}-\begin{bmatrix}1&1&0&0\end{bmatrix}\boldsymbol{\ell}\leq 1$
Finally, we demonstrate a branch flow. The matrix $\mathbf{F}$ is given as

$\displaystyle\mathbf{F}=\begin{bmatrix}-1&0&0\\\ -1&-1&0\\\
0&0&-1\end{bmatrix},$

and correspondingly

$\displaystyle\mathbf{f}^{\mathbf{p}}(1)=-\mathbf{p}(1),\quad\mathbf{f}^{\mathbf{p}}(2)=-\mathbf{p}(1)-\mathbf{p}(2),\quad\mathbf{f}^{\mathbf{p}}(3)=-\mathbf{p}(3)$
$\displaystyle\mathbf{f}^{\mathbf{q}}(1)=-\mathbf{q}(1),\quad\mathbf{f}^{\mathbf{q}}(2)=-\mathbf{q}(1)-\mathbf{q}(2),\quad\mathbf{f}^{\mathbf{q}}(3)=-\mathbf{q}(3).$

Suppose the branch flow limit of branch 2 is 3 p.u. It can be expressed as
$\left\lVert\begin{bmatrix}\mathbf{f}^{\mathbf{p}}(2)&\mathbf{f}^{\mathbf{q}}(2)\end{bmatrix}^{\top}\right\rVert_{2}\leq
3$, equivalently,

$\displaystyle\left\lVert\begin{bmatrix}(-p_{g}+\begin{bmatrix}-1&0&0&0\end{bmatrix}\boldsymbol{\ell})\\\
(-q_{g}+\begin{bmatrix}0&0&-1&0\end{bmatrix}\boldsymbol{\ell})\end{bmatrix}\right\rVert_{2}\leq
3.$

In other words, in the form of (4c) we have

$\displaystyle\mathbf{r}_{2}$ $\displaystyle=[-1],\mathbf{s}_{2}^{\top}$
$\displaystyle=\begin{bmatrix}-1&0&0&0\end{bmatrix},\mathbf{t}_{2}^{\top}$
$\displaystyle=\begin{bmatrix}0&0&-1&0\end{bmatrix}.$

For a generalized $j$, we provide the following closed form expressions on the
norm values of $\mathbf{r}_{j}$, $\mathbf{s}_{j}$ and $\mathbf{t}_{j}$.

######  Lemma 2 (Norm values of $\mathbf{r}_{j}$, $\mathbf{s}_{j}$ and
$\mathbf{t}_{j}$).

We have, for all branches $j$,

$\displaystyle\left\lVert\mathbf{r}_{j}\right\rVert_{2}=\sqrt{|\mathcal{H}_{j}\cap\mathcal{N}_{g}|},\left\lVert\mathbf{s}_{j}\right\rVert_{2}=\sqrt{|\mathcal{H}_{j}\cap\mathcal{N}_{l}|},$
$\displaystyle\left\lVert\mathbf{t}_{j}\right\rVert_{2}=\sqrt{|\mathcal{H}_{j}\cap\mathcal{N}_{l}|}.$

###### Proof.

Recall that branch $j$ shares its index with its downstream bus $j$. We use
the properties of matrix $\mathbf{F}$ in this proof. From Lemma 1,
$\mathbf{F}=\mathbf{A}^{-\top}$. It is known [21] that the
${(i,j)}^{\text{th}}$ element of matrix $\mathbf{A}^{-1}$ is +1 if branch $j$
is directed _along_ path from $i$ to slack bus, -1 if it is directed _against_
, and 0 otherwise. By the construction of $\mathcal{G}$, $\mathbf{A}^{-1}$
only has entries -1 and 0. Thus, $\mathbf{A}^{-\top}$ has
${(i,j)}^{\text{th}}$ element -1 if branch $i$ is falls in the path from bus
$j$ to the origin and 0 otherwise. Therefore, the ${i}^{\text{th}}$ row of
$\mathbf{A}^{-\top}$ collects all buses in $\mathcal{H}_{i}$. The result
follows by separating generator buses and placing their coefficients in
$\mathbf{r}_{j}$, and load buses in $\mathbf{s}_{j}$ and $\mathbf{t}_{j}$. ∎

##  III Marginal Analysis

The analysis of marginal pricing starts from the dual problem of (4) that is
given as follows.

######  Lemma 3.

The dual problem of (4) is given as

$\displaystyle\max\limits_{\begin{subarray}{c}\boldsymbol{\lambda},\boldsymbol{\alpha}_{lb},\boldsymbol{\alpha}_{ub},\boldsymbol{\beta}_{lb},\\\
\boldsymbol{\beta}_{ub},\theta_{i},\phi_{i},\mu_{i}\end{subarray}}\quad$
$\displaystyle-(\mathbf{G}\boldsymbol{\ell}+\mathbf{h})^{\top}\boldsymbol{\lambda}+\boldsymbol{\alpha}_{lb}^{\top}\underline{\mathbf{p}}_{g}-\boldsymbol{\alpha}_{ub}^{\top}\bar{\mathbf{p}}_{g}+\boldsymbol{\beta}_{lb}^{\top}\underline{\mathbf{q}}_{g}$
$\displaystyle-\boldsymbol{\beta}_{ub}^{\top}\bar{\mathbf{q}}_{g}+\sum_{i=1}^{n}\left[-(\theta_{i}\mathbf{s}_{i}+\phi_{i}\mathbf{t}_{i})^{\top}\boldsymbol{\ell}-\mu_{i}\bar{f}_{i}\right]$
(5) $\displaystyle\text{s.t.}\quad\tilde{\mathbf{c}}_{g}+$
$\displaystyle\mathbf{M}_{p}^{\top}\boldsymbol{\lambda}+\boldsymbol{\alpha}_{ub}-\boldsymbol{\alpha}_{lb}-\sum_{i=1}^{n}\theta_{i}\mathbf{r}_{i}=0$
(5a) $\displaystyle\mathbf{M}_{q}^{\top}\boldsymbol{\lambda}$
$\displaystyle+\boldsymbol{\beta}_{ub}-\boldsymbol{\beta}_{lb}-\sum_{i=1}^{n}\phi_{i}\mathbf{t}_{i}=0$
(5b)
$\displaystyle\left\lVert[\theta_{i},\phi_{i}]^{\top}\right\rVert_{2}\leq\mu_{i},\quad\forall
i\in[n]$ (5c)
$\displaystyle\boldsymbol{\lambda},\boldsymbol{\alpha}_{lb},\boldsymbol{\alpha}_{ub},\boldsymbol{\beta}_{lb},\boldsymbol{\beta}_{ub},\\{\mu_{i}\\}\geq
0$ (5d)

###### Proof.

We augment problem (4) by adding auxiliary variables
$y_{i}\stackrel{{\scriptstyle\Delta}}{{=}}\mathbf{r}_{i}^{\top}\mathbf{p}_{g}+\mathbf{s}_{i}^{\top}\boldsymbol{\ell}$
and
$z_{i}\stackrel{{\scriptstyle\Delta}}{{=}}\mathbf{r}_{i}^{\top}\mathbf{q}_{g}+\mathbf{t}_{i}^{\top}\boldsymbol{\ell}$,
and letting $\theta_{i}$ and $\phi_{i}$ denote the dual variables for the
same. Constraint (5c) can then be written as
$\left\lVert[y_{i},z_{i}]^{\top}\right\rVert_{2}\leq\bar{f}_{i},\forall
i\in[n]$. The Lagrangian of the augmented problem is given as

$\displaystyle\mathcal{L}\stackrel{{\scriptstyle\Delta}}{{=}}$
$\displaystyle\tilde{\mathbf{c}}_{g}^{\top}\mathbf{p}_{g}+\boldsymbol{\lambda}^{\top}(\mathbf{M}_{p}\mathbf{p}_{g}+\mathbf{M}_{q}\mathbf{q}_{g}-\mathbf{G}\boldsymbol{\ell}-\mathbf{h})+\boldsymbol{\alpha}_{lb}^{\top}(\underline{\mathbf{p}}_{g}-\mathbf{p}_{g})$
$\displaystyle+\boldsymbol{\alpha}_{ub}^{\top}(\mathbf{p}_{g}-\bar{\mathbf{p}}_{g})+\boldsymbol{\beta}_{lb}^{\top}(\underline{\mathbf{q}}_{g}-\mathbf{q}_{g})+\boldsymbol{\beta}_{ub}^{\top}(\mathbf{q}_{g}-\bar{\mathbf{q}}_{g})+$
$\displaystyle+\sum_{i=1}^{n}\bigg{[}\theta_{i}(y_{i}-\mathbf{r}_{i}^{\top}\mathbf{p}_{g}-\mathbf{s}_{i}^{\top}\boldsymbol{\ell})+\phi_{i}(z_{i}-\mathbf{r}_{i}^{\top}\mathbf{q}_{g}-\mathbf{t}_{i}^{\top}\boldsymbol{\ell})$
$\displaystyle+\mu_{i}\left(\left\lVert
y_{i},z_{i}]^{\top}\right\rVert_{2}-\bar{f}_{i}\right)\bigg{]}.$

The dual objective function consists of all the terms in $\mathcal{L}$ which
are not functions of any primal variables. Since $\mathcal{L}$ is linear in
$\mathbf{p}_{g}$ and $\mathbf{q}_{g}$, their respective coefficients must be
zero such that $\mathcal{L}$ is bounded from below. This gives rise to (5a)
and (5b). Finally, that for any $y,z,\theta,\phi\in$ and $\mu>0$, it holds
that

$\displaystyle\inf\limits_{y,z}\;\;[\theta,\phi]\begin{bmatrix}y\\\
z\end{bmatrix}+\mu\left\lVert\begin{bmatrix}y\\\
z\end{bmatrix}\right\rVert_{2}=\begin{cases}0,&\text{if
}\left\lVert[\theta,\phi]^{\top}\right\rVert_{2}\leq\mu,\\\ -\infty,&\text{if
}\left\lVert[\theta,\phi]^{\top}\right\rVert_{2}>\mu.\end{cases}$

Applying this property to find the infimum of all terms consisting of $y_{i}$
and $z_{i}$ in $\mathcal{L}$, we recover (5c). Finally, non-negativity of dual
variables yields (5d). ∎

We assume that there exist optimal primal and dual solutions for (4) and (3)
such that Kahrush-Kuhn-Tucker (KKT) conditions hold [22, Section 5.5.3].

######  Assumption 1 (KKT conditions).

Let
$\Gamma\stackrel{{\scriptstyle\Delta}}{{=}}\\{\boldsymbol{\lambda},\boldsymbol{\alpha}_{lb},\boldsymbol{\alpha}_{ub},\boldsymbol{\beta}_{lb},\boldsymbol{\beta}_{ub},\theta_{i},\phi_{i},\mu_{i}\\}$
denote the set of dual variables in (3). Then, there exist optimal solutions
$\mathbf{p}_{g}^{*}$ and $\mathbf{q}_{g}^{*}$ for primal problem (4) and
$\Gamma^{*}$ for dual problem (3) which satisfy the following KKT conditions:

  1. 1.

stationarity of $\mathcal{L}$ in primal variables $\mathbf{p}_{g}$ and
$\mathbf{q}_{g}$.

  2. 2.

constraints (4a)–(4c) hold for primal variables.

  3. 3.

constraints (5a)–(5d) hold for dual variables.

$\displaystyle\boldsymbol{\lambda}^{\top}(\mathbf{M}_{p}\mathbf{p}_{g}+\mathbf{M}_{q}\mathbf{q}_{g}-\mathbf{G}\boldsymbol{\ell}-\mathbf{h})=0$
$\displaystyle\boldsymbol{\alpha}_{lb}^{\top}(\underline{\mathbf{p}}_{g}-\mathbf{p}_{g})=0,\quad\boldsymbol{\alpha}_{ub}^{\top}(\mathbf{p}_{g}-\bar{\mathbf{p}}_{g})=0$
$\displaystyle\boldsymbol{\beta}_{lb}^{\top}(\underline{\mathbf{q}}_{g}-\mathbf{q}_{g})=0,\quad\boldsymbol{\beta}_{ub}^{\top}({\mathbf{q}}_{g}-\bar{\mathbf{q}}_{g})=0.$

We now introduce the _dual value function_
$\mathcal{D}(\Gamma,\boldsymbol{\ell},\bar{\mathbf{f}})$, which is defined as
the objective function of dual problem (3) as a function of _any_ values of
dual variables $\Gamma$, demands $\boldsymbol{\ell}$, and flow limits
$\bar{\mathbf{f}}$. Assumption 1 allows us to exploit duality theory in order
to equate the dual value function to the operator
$\mathcal{J}(\boldsymbol{\ell},\bar{\mathbf{f}})$.

######  Remark 2 (Strong duality).

Since the primal problem (4) is convex in the decision variables, Assumption 1
ensures that strong duality holds [22, Section 5.5.3], i.e.
$\mathcal{J}(\boldsymbol{\ell},\bar{\mathbf{f}})=\mathcal{D}(\Gamma^{*},\boldsymbol{\ell},\bar{\mathbf{f}})$.

We now provide a closed form of the _flow marginal costs_ as a function of the
optimum dual variables.

######  Lemma 4.

The flow marginal cost for line flow limits $\bar{\mathbf{f}}$, denoted as
$C^{flow}_{\bar{\mathbf{f}}}$ is given as

$\displaystyle
C^{\text{flow}}_{\bar{\mathbf{f}}}(j)\stackrel{{\scriptstyle\Delta}}{{=}}\nabla_{\bar{\mathbf{f}}(j)}\mathcal{J}(\boldsymbol{\ell},\bar{\mathbf{f}})=-\mu^{*}_{j},\quad\forall
j\in[n]$ (6)

Moreover, if $\mathbf{p}_{g1}^{*}$ is the optimum real generation with flow
limits $\bar{\mathbf{f}}_{1}$ and $\mathbf{p}_{g2}^{*}$ with
$\bar{\mathbf{f}}_{2}$ where
$\bar{\mathbf{f}}_{1}\succeq\bar{\mathbf{f}}_{2}$, then
$\tilde{\mathbf{c}_{g}}^{\top}\mathbf{p}_{g1}^{*}\leq\tilde{\mathbf{c}_{g}}^{\top}\mathbf{p}_{g2}^{*}$.

###### Proof.

Due to strong duality, it follows that
$\mathcal{J}(\boldsymbol{\ell},\bar{\mathbf{f}})=\mathcal{D}(\Gamma^{*},\boldsymbol{\ell},\bar{\mathbf{f}})$.
Taking derivative of
$\mathcal{D}(\Gamma^{*},\boldsymbol{\ell},\bar{\mathbf{f}})$ with respect to
each of the elements in $\bar{\mathbf{f}}$ recovers the closed form of
$C^{flow}_{\bar{\mathbf{f}}}$. The second part of the result follows from
observing that the feasible space of (4) is smaller under
$\bar{\mathbf{f}}_{2}$ than $\bar{\mathbf{f}}_{1}$, leading to same or higher
objective value. ∎

To this end, we present the main result as follows.

######  Theorem 1 (Bounding marginal prices).

Suppose at optimality of problem (4),
$\mathcal{I}(\bar{\mathbf{f}}_{1})\subseteq[n]$ is the nonempty index set
collecting the binding constraints in (4c) for flow limits
$\bar{\mathbf{f}}_{1}$. Then, the congested marginal cost of load demand is
given as

$\displaystyle
C^{\text{load}}_{\bar{\mathbf{f}}_{1}}\stackrel{{\scriptstyle\Delta}}{{=}}\nabla_{\boldsymbol{\ell}}\mathcal{J}(\boldsymbol{\ell},\bar{\mathbf{f}}_{1})=-\boldsymbol{\lambda}^{*\top}\mathbf{G}-\sum_{i\in\mathcal{I}(\bar{\mathbf{f}}_{1})}(\theta_{i}^{*}\mathbf{s}_{i}+\phi_{i}^{*}\mathbf{t}_{i})^{\top}.$
(7)

Figure 2: 141-bus distribution system, adapted from MATPOWER case141. We
retain the topology and real/reactive load demands from case141, but randomly
add 25 distributed generation buses. The real power generation of each
distributed generator is limited to $p_{g}\in[0,0.0654]pu$, while the reactive
power generation follows limits $q_{g}\in[-0.0270,0.0270]pu$. Figure 3:
Results from simultaneously perturbing two branch flow limits from Figure 2.
We choose branches 16 and 18 to perturb since branch 18 is downstream to
branch 16, making for easier interpretability of results. The first two
figures indicate marginal costs of real and reactive power respectively, while
the third figure is the proposed upper bound on the quantities derived in both
the first and second figure. As we see, the upper bound is valid for both the
quantities.

Furthermore, let $\mathcal{I}(\bar{\mathbf{f}}_{2})$ be empty for flow limits
$\bar{\mathbf{f}}_{2}$, denoting the case wherein the network is uncongested.
We have,

$\displaystyle|C^{\text{load}}_{\bar{\mathbf{f}}_{1}}(i)-C^{\text{load}}_{\bar{\mathbf{f}}_{2}}(i)|\leq
K_{\bar{\mathbf{f}}_{1}}\sum_{j\in\mathcal{I}(\bar{\mathbf{f}}_{1})}|C^{\text{flow}}_{\bar{\mathbf{f}}_{1}}(j)|,$
(real)
$\displaystyle|C^{\text{load}}_{\bar{\mathbf{f}}_{1}}(n_{l}+i)-C^{\text{load}}_{\bar{\mathbf{f}}_{2}}(n_{l}+i)|\leq
K_{\bar{\mathbf{f}}_{1}}\sum_{j\in\mathcal{I}(\bar{\mathbf{f}}_{1})}|C^{\text{flow}}_{\bar{\mathbf{f}}_{1}}(j)|$
(reactive)

for all $i\in[n]$, $K_{\bar{\mathbf{f}}_{1}}>0$ is a constant defined as
$K_{\bar{\mathbf{f}}_{1}}\stackrel{{\scriptstyle\Delta}}{{=}}{(n_{nz}^{(j)})}^{\frac{1}{2}}\max\limits_{j\in\mathcal{I}(\bar{\mathbf{f}}_{1})}~{}\left\lVert[\mathbf{s}_{j}^{\top},\mathbf{t}_{j}^{\top}]^{\top}\right\rVert_{2}$,
where $n_{nz}^{(j)}$ is the number of nonzeros in $\mathbf{s}_{j}$ (same as
$\mathbf{t}_{j}$).

###### Proof.

The closed form of $C^{\text{load}}_{\bar{\mathbf{f}}_{1}}$ may be derived
similar to the proof of Lemma 4. Note that
$C^{\text{load}}_{\mathbf{\bar{f}}_{1}}-C^{\text{load}}_{\mathbf{\bar{f}}_{2}}=-\sum_{i\in\mathcal{I}(\bar{\mathbf{f}}_{1})}(\theta_{i}^{*}\mathbf{s}_{i}+\phi_{i}^{*}\mathbf{t}_{i})^{\top}$.
Thus,

$\displaystyle\left|C^{\text{load}}_{\mathbf{\bar{f}}_{1}}(i)-C^{\text{load}}_{\mathbf{\bar{f}}_{2}}(i)\right|$
$\displaystyle=\left|-\sum_{j\in\mathcal{I}(\bar{\mathbf{f}}_{1})}\theta^{*}_{j}\mathbf{s}_{j}+\phi^{*}_{j}\mathbf{t}_{j}\right|\leq\sum_{j\in\mathcal{I}(\bar{\mathbf{f}}_{1})}\left|\begin{bmatrix}\mathbf{s}_{j}\\\
\mathbf{t}_{j}\end{bmatrix}^{\top}\begin{bmatrix}\theta_{j}^{*}\mathbf{z}^{s}_{j}\\\
\phi^{*}_{j}\mathbf{z}^{t}_{j}\end{bmatrix}\right|$
$\displaystyle\stackrel{{\scriptstyle[a]}}{{\leq}}\sum_{j\in\mathcal{I}(\bar{\mathbf{f}}_{1})}\left(\left\lVert\begin{bmatrix}\mathbf{s}_{j}\\\
\mathbf{t}_{j}\end{bmatrix}\right\rVert_{2}\times\left\lVert\begin{bmatrix}\theta_{j}^{*}\mathbf{z}^{s}_{j}\\\
\phi^{*}_{j}\mathbf{z}^{t}_{j}\end{bmatrix}\right\rVert_{2}\right)\quad\stackrel{{\scriptstyle[b]}}{{\leq}}\sum_{j\in\mathcal{I}(\bar{\mathbf{f}}_{1})}\mu_{j}^{*}K_{\bar{\mathbf{f}}_{1}},$

where $\mathbf{z}^{s}_{j}$ and $\mathbf{z}^{t}_{j}$ are vectors which have
value $1$ at indices where $\mathbf{s}_{j}$ and $\mathbf{t}_{j}$ are nonzero
respectively, and $0$ otherwise. $[a]$ is due to the Cauchy-Schwartz
inequality while $[b]$ follows from dual (5c) and uniformly upper bounding the
term
$\left\lVert[\mathbf{s}_{j}^{\top},\mathbf{t}_{j}^{\top}]^{\top}\right\rVert_{2}$
for all $j\in\mathcal{I}(\bar{\mathbf{f}}_{1})$. Observing that
$\nabla_{\bar{f}_{j}}\mathcal{J}(\boldsymbol{\ell},\bar{\mathbf{f}})=-\mu_{j}^{*}$
and $\mu_{j}^{*}\geq 0$ concludes the proof. ∎

Lemma (2) may be used to find the closed form of $K_{\mathbf{\bar{f}}}$ for
different $\bar{\mathbf{f}}$. In the following remark, we highlight some use
cases of Theorem 1.

######  Remark 3 (Use cases of Theorem 1).

Note that system operators often possess datasets of the form
$\\{{C^{\text{flow}}_{\mathbf{\bar{f}}}}_{k},\bar{\mathbf{f}}_{k},p_{k}\\}$,
where $p_{k}\in[0,1]$ is the probability of sample $k$ being representative of
a desired scenario. Such a probability may be derived empirically by observing
frequencies of similar data points in the dataset

  * •

The bounds may be used to quickly approximate the worst-case changes in power
tariffs as a function of implemented line flow limits $\bar{\mathbf{f}}$ using
existing datasets, without re-running the full OPF problem for different
values of $\bar{\mathbf{f}}$.

  * •

The capability to compute the marginal prices $C^{\text{flow}}$ and
$C^{\text{load}}$ as a function of $\boldsymbol{\ell}$ and $\bar{\mathbf{f}}$
may be used to train novel sensitivity informed deep learning architectures
[23]. Bounds based on sensitivity of the OPF solution to input parameters can
also be used to construct triggering conditions in distributed optimization
using the principle of event-triggered communication [24].

  * •

If the power demand $\boldsymbol{\ell}$ is a random variable drawn from a
known distribution, then Theorem 1 can provide a method to quantify the
statistical properties of demand marginal costs. Such analyses generally fall
under the domain of probabilistic (optimal) power flow, which is a topic of
significant research due to its possible applications in deep-learning based
architectures for power systems [25].

##  IV Simulation Result

In this section, we experimentally validate the bounds derived in Theorem 1 by
carrying out simulations on a 141-bus single-phase radial power network
derived from case141 in MATPOWER [20], as shown in Figure 2. Due to case141
originally containing only a single generator at the slack bus, we modify the
same to add distributed generation. This is done according to the following
steps.

  * •

First, 25 buses are randomly selected to host distributed generation. Then per
generator bounds are chosen such that the total upper bound of real power
generation approximately equals the total real power demand at the load buses.
The real power cost of distributed generator buses are chosen uniformly-at-
random in the range $[0,1]$, while the slack bus generation is chosen to be
more expensive than any distributed generator.

  * •

Following introduction of distributed generators, we run the OPF without any
branch flow limits to generate optimal generation and branch flow setpoints.
This is done using CVX [26] and MATLAB. Once the optimal flow setpoints are
generated, the branch flow limit of the line upstream to any distributed
generator is set to its corresponding flow setpoint.

  * •

Two branches, _viz._ 16 and 18 are selected, and their branch flow limits are
simultaneously reduced upto 75% of their original values. For each step in the
reduction, the perturbation of real and reactive marginal cost of load demand
at bus 20 (a load bus) is recorded, and presented in Figure 3. The proposed
upper bound is also presented in the same figure.

As seen in Figure 3, the proposed bounds are respected by the actual
perturbation in marginal costs. Here, both the marginal cost perturbations and
the upper bound are derived from the formulations presented in Theorem 1. It
should be noted that the results in Theorem 1 use optimal dual variables in
its formulation, which are generated in most modern optimization solvers.
Another interesting observation in Figure 3 is that the marginal cost of
reactive power is within some numerical error of zero. This is because the
objective function of OPF 3 does not penalize generation/consumption of
reactive power. Were an objective term concering reactive power to be added to
the OPF, we would observe non-zero marginal costs for reactive power alongside
real power demand.

##  V Conclusion

In this paper, we formulated the LDF-OPF problem and derived upper bounds on
the marginal prices of real and reactive power demand for all load buses. We
also presented certain approaches showing how the main result in Theorem 1 can
be utilized for OPF-based planning problems. Future work will develop results
in this paper to accelerate solve times of LDF-OPF.

## Acknowledgements

The authors would like to thank Professor Yihsu Chen at the University of
California, Santa Cruz for helpful discussions on this work and its future
directions.

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$\displaystyle w(t+1)\equiv 0(\hbox{\rm mod }\frac{m}{2})\quad{\rm{and}}\quad
sr^{w+1}+s\sum_{l=1}^{w+t-1}r^{l}+sr-1\equiv 0(\hbox{\rm mod }\frac{n}{2}),$
(104)

which implies

$\displaystyle r^{2w}\equiv r^{w(t+1)}\equiv 1(\hbox{\rm mod }\frac{n}{2}).$
(105)

(ii) $\hbox{\rm o}(\pi(a))=n$:

$(bc^{2x})^{n}=(c^{2w(v+1)}a^{2u\sum_{l=w+1}^{2w}r^{l}})^{\frac{n}{2}}=c^{nw(v+1)}a^{un\sum_{l=w+1}^{2w}r^{l}}=1,$

that is

$\displaystyle\frac{n}{2}w(v+1)\equiv 0(\hbox{\rm mod }\frac{m}{2}).$ (106)

(iii) $\pi$ preserves $(a^{2})^{c_{1}}=a^{2r}$:

$((a^{2})^{c^{2}})^{c}=c^{2w(v+1)}a^{2ur\sum_{l=w+1}^{2w}r^{l}}\quad{\rm{and}}\quad(a^{2r})^{c}=c^{2wr(v+1)}a^{2ur\sum_{l=w+1}^{2w}r^{l}},$

that is

$\displaystyle w(t+1)(r-1)\equiv 0(\hbox{\rm mod }\frac{m}{2}).$ (107)

(iv) $\pi$ preserves $c_{1}^{a}=a^{2s}c_{1}^{t}$:

$((c^{2})^{a})^{c}=c^{2v}a^{2ur^{w+1}}\quad{\rm{and}}\quad(a^{2s}c^{2t})^{c}=c^{2ws(v+1)+2t}a^{2sru\sum_{l=w+1}^{2w}r^{l}},$

that is

$\displaystyle v\equiv ws(v+1)+t(\hbox{\rm mod
}\frac{m}{2})\quad{\rm{and}}\quad u\equiv su\sum_{l=1}^{w}r^{l}(\hbox{\rm mod
}\frac{n}{2}).$ (108)

(v) $\pi$ preserves $c_{1}^{b}=a^{2u}c_{1}^{v}$:

$\begin{array}[]{lcl}((c^{2})^{b})^{c}&=&(c^{2})^{a^{1-2sr}c^{2-2t-2w}}=c^{2t}a^{2sr^{2-w}},\\\
(a^{2u}c^{2v})^{c}&=&(c^{2w(v+1)}a^{2u\sum_{l=w+1}^{2w}r^{l}})^{u}c^{2v}=c^{2wu(v+1)+2v}a^{2u^{2}\sum_{l=w+2}^{2w+1}r^{l}},\end{array}$

that is,

$\displaystyle t\equiv wu(v+1)+v(\hbox{\rm mod
}\frac{m}{2})\quad{\rm{and}}\quad s\equiv u^{2}\sum_{l=0}^{w-1}r^{l}(\hbox{\rm
mod }\frac{n}{2}).$ (109)

(vi) $\pi^{2}=\hbox{\rm Inn}(c_{1})$: Recall $\hbox{\rm
Inn}(c_{1})(a)=a^{1-2sr}c_{1}^{1-t},\,\hbox{\rm Inn}(c_{1})(a^{2})=a^{2r}$ and
$\hbox{\rm Inn}(c_{1})(b)=c_{1}^{v-1}a^{2ur}b$.

$a^{1-2sr}c^{2-2t}=\hbox{\rm
Inn}(c_{1})(a)=\pi^{2}(a)=b^{c}c^{2w}=a^{1-2sr}c^{2-2t-2w+2w},$

as desired;

$a^{2r}=\hbox{\rm
Inn}(c^{2})(a_{1})=\pi^{2}(a^{2})=(c^{2w(v+1)}a^{2u\sum_{l=w+1}^{2w}r^{l}})^{c}=c^{2w(v+1)(1+uw)}a^{2(u\sum_{l=1}^{w}r^{l})^{2}},$

that is

$\displaystyle w(v+1)(1+uw)\equiv 0(\hbox{\rm mod
}\frac{m}{2})\quad{\rm{and}}\quad r\equiv(u\sum_{l=1}^{w}r^{l})^{2}(\hbox{\rm
mod }\frac{n}{2}),$ (110)

which implies $(u,\frac{n}{2})=1$ and $(\sum_{l=1}^{w}r^{l},\frac{n}{2})=1$ as
$(r,\frac{n}{2})=1$; and

$c^{2(v-1)}a^{2ur}b=\hbox{\rm
Inn}(c_{1})(b)=\pi^{2}(b)=c^{2(w+t-1)+2w(v+1)(sr-1)+2vw}a^{2sur\sum_{l=1}^{w}r^{l}}b,$

that is,

$\displaystyle v\equiv t+wvsr+wsr(\hbox{\rm mod
}\frac{m}{2})\quad{\rm{and}}\quad s\sum_{l=1}^{w}r^{l}\equiv 1(\hbox{\rm mod
}\frac{n}{2}),$ (111)

which implies $(u,\frac{n}{2})=1$.

(vii) Insure $\langle c\rangle_{X}=1$: Since $\langle c\rangle_{X}\leq M$, we
get $\langle c\rangle_{X}\leq M_{X}=\langle a^{2}\rangle\langle c^{2}\rangle$.
Since $\langle c\rangle_{X}=\cap_{x\in X}C^{x}=\cap_{x\in G}C^{x}=\cap_{x\in
G}\langle c^{2}\rangle^{x}=\langle c^{2}\rangle_{X_{1}}=1$ and
$u\sum_{l=1}^{i}r^{l}\equiv 0\equiv s\sum_{l=1}^{i}r^{l}(\hbox{\rm mod
}\frac{n}{2})\Leftrightarrow i\equiv 0(\hbox{\rm mod }\frac{m}{2})$, noting
$(s,\frac{n}{2})=(u,\frac{n}{2})=1$, we have $\sum_{l=1}^{i}r^{l}\equiv
0(\hbox{\rm mod }\frac{n}{2})\Leftrightarrow i\equiv 0(\hbox{\rm mod
}\frac{m}{2})$. Now we are ready to determine the parameters by summarizing
Eq(102)-Eq(111). Since $(r,\frac{n}{2})=(u,\frac{n}{2})=1$ (after Eq(110)), we
get from Eq(102) that $\sum_{l=1}^{v-1}r^{l}\equiv 0(\hbox{\rm mod
}\frac{n}{2}).$ By (vii), $\sum_{l=1}^{i}r^{l}\equiv 0(\hbox{\rm mod
}\frac{n}{2})\Leftrightarrow i\equiv 0(\hbox{\rm mod }\frac{m}{2})$, which
means $v\equiv 1(\hbox{\rm mod }\frac{m}{2})$. Inserting $v=1$ in
Eq(102)-Eq(111), we get that $s=u^{2}\sum_{l=0}^{w-1}r^{l}$ and $t=2wu+1$ in
Eq(109); $(r^{w}+1)(1+s\sum_{l=0}^{w-1}r^{l})\equiv 0(\hbox{\rm mod
}\frac{n}{2})$ in Eq (104) $nw\equiv 0(\hbox{\rm mod }\frac{m}{2})$ in
Eq(106); and $2w(r-1)\equiv 2w(1+uw)\equiv 0(\hbox{\rm mod }\frac{m}{2})$ in
Eq(107) and (110). All these are summarized in the lemma. $\Box$

###  6.3 $M=\langle a^{2}\rangle\langle c\rangle$ and $X/M_{X}\cong A_{4}$

###### Lemma 6.3

Suppose that $X=X(D)$, $M=\langle a^{2}\rangle\langle c\rangle,\,X/M_{X}\cong
A_{4}$ and $\langle c\rangle_{X}=1$. Then

$X=\langle
a,b,c|R,a^{c^{3}}=a^{r},(c^{3})^{b}=a^{2u}c^{3},a^{c}=bc^{\frac{im}{2}},b^{c}=a^{x}b\rangle,$

where $n\equiv 2(\hbox{\rm mod }4)$ and either $i=u=0$ and $r=x=1$; or $i=1$,
$6\bigm{|}m$, $l^{\frac{m}{2}}\equiv-1(\hbox{\rm mod }\frac{n}{2})$ with
$\hbox{\rm o}(l)=m$, $r=l^{3}$, $u=\frac{l^{3}-1}{2l^{2}}$ and
$x\equiv-l+l^{2}+\frac{n}{2}(\hbox{\rm mod }n)$.

Proof Under the hypothesis, $M_{X}=\langle a^{2}\rangle\mathchar
9583\relax\langle c^{3}\rangle$. Set $n=\hbox{\rm o}(a)$ and $m=\hbox{\rm
o}(c)$. Then $n$ is even and $3\bigm{|}m$. Since
$X/M_{X}=\langle\overline{a},\overline{b}\rangle\langle\overline{c}\rangle\cong
A_{4}$, we can choose $\overline{b}$ such that the form of $X/M_{X}$ is the
following: $\overline{a}^{\overline{c}}=\overline{b}$ and
$\overline{b}^{\overline{c}}=\overline{a}\overline{b}.$ Set $c_{1}:=c^{3}$ and
$X_{1}=GM_{X}=\langle a,b\rangle\langle c_{1}\rangle$. By Lemma 6.1, we get

$X_{1}=\langle
a,b,c^{3}|R,(a^{2})^{c_{1}}=a^{2r},(c_{1})^{a}=a^{2s}c_{1}^{t},(c_{1})^{b}=a^{2u}c_{1}^{v}\rangle$

whose

$\displaystyle\begin{array}[]{ll}&r^{t-1}-1\equiv r^{v-1}-1\equiv
u(\sum_{l=0}^{v-1}r^{l}-1)\equiv 0(\hbox{\rm mod }\frac{n}{2}),\,t^{2}\equiv
v^{2}\equiv 1(\hbox{\rm mod }\frac{m}{3}),\\\ &s\sum_{l=1}^{t}r^{l}+sr\equiv
sr+s\sum_{l=1}^{v}r^{l}-u\sum_{l=1}^{t}r^{l}+ur\equiv 1-r(\hbox{\rm mod
}\frac{n}{2}),\\\ &s\sum_{l=1}^{i}r^{l}\equiv u\sum_{l=1}^{i}r^{l}\equiv
0(\hbox{\rm mod }n)\Leftrightarrow i\equiv 0(\hbox{\rm mod
}\frac{m}{3}).\end{array}$ (115)

Now $X=X.\langle c\rangle$. Set $a^{c}=bc_{1}^{w}$. Then $X$ may be defined by
$R$ and

$\displaystyle(a^{2})^{c_{1}}=a^{2r},(c_{1})^{a}=a^{2s}c_{1}^{t},(c_{1})^{b}=a^{2u}c_{1}^{v},\,a^{c}=bc_{1}^{w},b^{c}=a^{1+2x}bc_{1}^{y}.$
(116)

If $w\equiv 0(\hbox{\rm mod }\frac{m}{3})$, then $\hbox{\rm o}(a)=\hbox{\rm
o}(a^{c})=\hbox{\rm o}(b)=2$, which implies $X\cong A_{4}$. Then
$n=2,\,i=u=s=0$ and $r=x=t=v=1$, as desired. So in that follows, we assume
$w\not\equiv 0(\hbox{\rm mod }\frac{m}{3})$.

To determine the parameters $r,s,t,u,v,w,x$ and $y$, we only to consider the
last extension $X_{1}.\langle c\rangle$ in Eq(116), where $a^{c}=bc_{1}^{w}$
and $b^{c}=a^{1+2x}bc_{1}^{y}$. Set $\pi\in\hbox{\rm Aut\,}(X_{1}):a\to
bc_{1}^{w},\,b\to a^{1+2x}bc_{1}^{y},\quad c_{1}\to c_{1}.$ We need to carry
out the following eight steps:

(i) $\hbox{\rm o}(\pi(b))=2$:

$(a^{1+2x}bc^{3y})^{2}=ba^{-(1+2x)}c^{3y}a^{1+2x}bc^{3y}=c^{3y(tv+1)}a^{2r^{y}(u\sum_{l=1}^{ty}r^{l}+xr^{y}-x-s\sum_{l=1}^{y}r^{l})}=1,$

that is

$\displaystyle y(tv+1)\equiv 0(\hbox{\rm mod }\frac{m}{3})\quad{\rm{and}}\quad
u\sum_{l=1}^{ty}r^{l}+xr^{y}-x-s\sum_{l=1}^{y}r^{l}\equiv 0(\hbox{\rm mod
}\frac{n}{2}),$ (117)

which implies $r^{2y}\equiv r^{y(tv+1)}\equiv 1(\hbox{\rm mod }\frac{n}{2});$

(ii) $\hbox{\rm o}(\pi(ab))=2$:

$\begin{array}[]{lcl}1&=&(c^{3(vw+yt)}a^{2ur^{y}\sum_{l=1}^{w}r^{l}-2r^{y}x+2s\sum_{l=1}^{y}r^{l}-1})^{2}\\\
&=&c^{3(vw+yt+tvw+y)}a^{2((r^{w+y}+1)(ur^{y}\sum_{l=1}^{w}r^{l}-r^{y}x+s\sum_{l=1}^{y}r^{l})+s\sum_{l=1}^{vw+yt}r^{l}-1)},\end{array}$

that is

$\displaystyle\begin{array}[]{ll}&vw+yt+tvw+y\equiv 0(\hbox{\rm mod
}\frac{m}{3});\\\
&(r^{w+y}+1)(ur^{y}\sum_{l=1}^{w}r^{l}-r^{y}x+s\sum_{l=1}^{y}r^{l})+s\sum_{l=1}^{vw+yt}r^{l}\equiv
1(\hbox{\rm mod }\frac{n}{2}),\end{array}$ (120)

which implies $r^{2w}\equiv 1(\hbox{\rm mod }\frac{n}{2});$

(iii) $\hbox{\rm o}(\pi(a))=n$:

$(bc^{3w})^{n}=(bc^{3w}bc^{3w})^{\frac{n}{2}}=(c^{3w(v+1)}a^{2u\sum_{l=w+1}^{2w}r^{l}})^{\frac{n}{2}}=c^{3\frac{n}{2}w(v+1)}a^{nur^{w}\sum_{l=1}^{w}r^{l}}=1,$

that is

$\displaystyle\frac{n}{2}w(v+1)\equiv 0(\hbox{\rm mod }\frac{m}{3}).$ (121)

(iv) $\pi$ preserves $(a^{2})^{c_{1}}=a^{2r}$:

$\begin{array}[]{ll}&((a^{2})^{c^{3}})^{c}=(c^{3w(v+1)}a^{2u\sum_{l=w+1}^{2w}r^{l}})^{c^{3}}=c^{3w(v+1)}a^{2ur\sum_{l=w+1}^{2w}r^{l}},\\\
&(a^{2r})^{c}=(c^{3w(v+1)}a^{2u\sum_{l=w+1}^{2w}r^{l}})^{r}=c^{3wr(v+1)}a^{2ur\sum_{l=w+1}^{2w}r^{l}},\end{array}$

that is

$\displaystyle w(v+1)(r-1)\equiv 0(\hbox{\rm mod }\frac{m}{3}).$ (122)

(v) $\pi$ preserves $c_{1}^{a}=a^{2s}c_{1}^{t}$:

$\begin{array}[]{ll}&((c^{3})^{a})^{c}=(c^{3})^{bc^{3w}}=(a^{2u}c^{3v})^{c^{3w}}=c^{3v}a^{2ur^{w+1}},\\\
&(a^{2s}c^{3t})^{c}=(c^{3w(v+1)}a^{2u\sum_{l=w+1}^{2w}r^{l}})^{s}c^{3t}=c^{3ws(v+1)+3v}a^{2sur^{w+1}\sum_{l=1}^{w}r^{l}},\end{array}$

that is

$\displaystyle v\equiv ws(v+1)+t(\hbox{\rm mod
}\frac{m}{3})\quad{\rm{and}}\quad u\equiv su\sum_{l=1}^{w}r^{l}(\hbox{\rm mod
}\frac{n}{2}).$ (123)

(vi) $\pi$ preserves $c_{1}^{b}=a^{2u}c_{1}^{v}$:

$\begin{array}[]{ll}&((c^{3})^{b})^{c}=(c^{3})^{a^{1+2x}b}=(c^{3}a^{2x(1-r)})^{ab}=c^{3tv}a^{2(u\sum_{l=1}^{t}r^{l}-sr+x(r-1))},\\\
&(a^{2u}c^{3v})^{c}=(c^{3w(v+1)}a^{2u\sum_{l=w+1}^{2w}r^{l}})^{u}c^{3v}=c^{3wu(v+1)+3v}a^{2u^{2}r\sum_{l=w+1}^{2w}r^{l}},\end{array}$

that is

$\displaystyle\begin{array}[]{ll}&t\equiv wu(v+1)+1(\hbox{\rm mod
}\frac{m}{3});\\\ &u\sum_{l=1}^{t}r^{l}-sr+x(r-1)\equiv
u^{2}r^{w+1}\sum_{l=1}^{w}r^{l}(\hbox{\rm mod }\frac{n}{2}).\end{array}$ (126)

(vii) $\pi^{3}=\hbox{\rm Inn}(c^{3})$: Recall $\hbox{\rm
Inn}(c_{1})(a)=a^{1-2sr}c_{1}^{1-t},\,\hbox{\rm Inn}(c_{1})(a^{2})=a^{2r}$ and
$\hbox{\rm Inn}(c_{1})(b)=c_{1}^{v-1}a^{2ur}b$.

$\begin{array}[]{ll}&a^{1-2sr}c^{3-3t}=\hbox{\rm
Inn}(c_{1})(a)=\pi^{3}(a)=(a^{1+2x}bc^{3(w+y)})^{c}\\\
&a^{-1}c^{3vt(w+wxv+wx)+3(w+2y)}a^{2r^{w}(u\sum_{l=1}^{t(w+wxv+wx)}r^{l}-s\sum_{l=1}^{w+wxv+wx}r^{l}-ux\sum_{l=w+1}^{2w}r^{l}-x)},\end{array}$

that is

$\displaystyle\begin{array}[]{ll}&1-t\equiv vt(w+wxv+wx)+w+2y(\hbox{\rm mod
}\frac{m}{3})\\\ &1-sr\equiv
r^{w}(u\sum_{l=1}^{t(w+wxv+wx)}r^{l}-s\sum_{l=1}^{w+wxv+wx}r^{l}-ux\sum_{l=w+1}^{2w}r^{l}-x)(\hbox{\rm
mod }\frac{n}{2});\end{array}$ (129) $a^{2r}=\hbox{\rm
Inn}(c_{1})(a^{2})=\pi^{3}(a^{2})=(a^{2})^{c^{3}}=c^{3w(v+1)+3uw^{2}(v+1)(uw+1)}a^{2r^{w}(u\sum_{l=1}^{w}r^{l})^{3}},$

that is

$\displaystyle w(v+1)+uw^{2}(v+1)(uw+1)\equiv 0(\hbox{\rm mod
}\frac{m}{3})\quad{\rm{and}}\quad r\equiv
r^{w}(u\sum_{l=1}^{w}r^{l})^{3}(\hbox{\rm mod }\frac{n}{2}),$ (130)

which implies $(u,\frac{n}{2})=(\sum_{l=1}^{w}r^{l},\frac{n}{2})=1$ as
$(r,\frac{n}{2})=1$, and

$c^{3(v-1)}a^{2ur}b=\hbox{\rm
Inn}(c_{1})(b)=\pi^{3}(b)=(c^{3(w+t-1)}a^{2sr-1})^{c}=c^{3(t-1)+3wsr(v+1)}a^{2sur\sum_{l=1}^{w}r^{l}}b,$

that is

$\displaystyle v\equiv t+wsr(v+1)(\hbox{\rm mod
}\frac{m}{3})\quad{\rm{and}}\quad 1\equiv s\sum_{l=1}^{w}r^{l}(\hbox{\rm mod
}\frac{n}{2}),$ (131)

which implies $(s,\frac{n}{2})=1$.

(viii) Insure $\langle c\rangle_{X}=1$: Since $\langle c\rangle_{X}\leq M$, we
get $\langle c\rangle_{X}\leq M_{X}=\langle a^{2}\rangle\langle c^{3}\rangle$.
Since $\langle c\rangle_{X}=\cap_{x\in X}C^{x}=\cap_{x\in G}C^{x}=\cap_{x\in
G}\langle c^{3}\rangle^{x}=\langle c^{3}\rangle_{X_{1}}=1$ and
$u\sum_{l=1}^{i}r^{l}\equiv 0\equiv s\sum_{l=1}^{i}r^{l}(\hbox{\rm mod
}\frac{n}{2})\Leftrightarrow i\equiv 0(\hbox{\rm mod }\frac{m}{3})$, noting
$(s,\frac{n}{2})=(u,\frac{n}{2})=1$, we have $\sum_{l=1}^{i}r^{l}\equiv
0(\hbox{\rm mod }\frac{n}{2})\Leftrightarrow i\equiv 0(\hbox{\rm mod
}\frac{m}{3})$. Now we are ready to determine the parameters by summarizing
Eq(115)-Eq(131).

Since $(r,\frac{n}{2})=(u,\frac{n}{2})=1$ (after Eq(130)), we get from Eq(115)
that $\sum_{l=1}^{v-1}r^{l}\equiv 0(\hbox{\rm mod }\frac{n}{2}).$ By (viii),
$\sum_{l=1}^{i}r^{l}\equiv 0(\hbox{\rm mod }\frac{n}{2})\Leftrightarrow
i\equiv 0(\hbox{\rm mod }\frac{m}{3})$, which means $v\equiv 1(\hbox{\rm mod
}\frac{m}{3})$. Inserting $v=1$ in Eq(115)-Eq(131), we get $2w(wu+1)\equiv
0(\hbox{\rm mod }\frac{m}{3})$ and $v\equiv 1+2wu(\hbox{\rm mod }\frac{m}{3})$
in Eq(120), (126) and (131), which implies $2wu\equiv 0(\hbox{\rm mod
}\frac{m}{3})$. Then $2w\equiv 0(\hbox{\rm mod }\frac{m}{3})$ as $nw\equiv
0(\hbox{\rm mod }\frac{m}{3})$ and $(u,\frac{n}{2})=1$, which implies
$w=\frac{m}{6}$ as $w\not\equiv 0(\hbox{\rm mod }\frac{m}{3})$. Inserting
$w=\frac{m}{6}$ in Eq(115)-Eq(131) again, we get that $r^{w}\equiv-1(\hbox{\rm
mod }\frac{n}{2})$ in Eq(115) and (131) and $t\equiv 1(\hbox{\rm mod
}\frac{m}{3})$ in Eq(126).

Since $2y\equiv 0(\hbox{\rm mod }\frac{m}{3})$ in Eq(117), we know $y$ is
either $0$ or $\frac{m}{6}$. If $y=\frac{m}{6}=w$, then with Eq(117) and
(120), we get $s\equiv u(\hbox{\rm mod }\frac{n}{2})$. Then $2sr\equiv
1-r(\hbox{\rm mod }\frac{n}{2})$ in Eq(115), which implies
$(r-1,\frac{n}{2})=1$, and then $r=1$, contradicting with
$r^{w}\equiv-1(\hbox{\rm mod }\frac{n}{2})$. So $y=0$.

By Eq(126), we get $2x\equiv
u\sum_{l=1}^{w}r^{l}+(u\sum_{l=1}^{w}r^{l})^{2}-1(\hbox{\rm mod
}\frac{n}{2}),$ which implies $\frac{n}{2}$ is odd, then
$s\equiv\frac{r^{-1}-1}{2}(\hbox{\rm mod }\frac{n}{2})$ in Eq(115). And by
Eq(130), we get $-r\equiv(u\sum_{l=1}^{w}r^{l})^{3}(\hbox{\rm mod
}\frac{n}{2}).$ Take $l=-\frac{2ru}{1-r}$, then $r=l^{3}$,
$u=\frac{l(r-1)}{2r}$ and $1+2x\equiv-l+l^{2}(\hbox{\rm mod }\frac{n}{2})$.
Since $1+2x$ is odd, we get $1+2x\equiv-l+l^{2}+\frac{n}{2}(\hbox{\rm mod
}n)$. $\Box$In fact, if we add the conditions $t=1$ and $w\neq 0$ and delete
$\langle c\rangle_{X}=1$ in the above calculation, then we can get the
following:

###### Lemma 6.4

With the notation, suppose that $t=1$ and $w\neq 0$. Then

$X=\langle
a,b,c|R,a^{c^{3}}=a^{r},(c^{3})^{b}=a^{2u}c^{3},a^{c}=bc^{\frac{m}{2}},b^{c}=a^{x}b\rangle,$

where $n\equiv 2(\hbox{\rm mod }4)$, $m\equiv 0(\hbox{\rm mod }6)$,
$l^{\frac{m}{2}}\equiv-1(\hbox{\rm mod }\frac{n}{2})$, $r=l^{3}$,
$u=\frac{l^{3}-1}{2l^{2}}$ and $x\equiv-l+l^{2}+\frac{n}{2}(\hbox{\rm mod
}n)$.

###  6.4 $M=\langle a^{4}\rangle\langle c\rangle$, $X/M_{X}\cong S_{4}$ and
$\langle c\rangle_{X}=1$

###### Lemma 6.5

Suppose that $X=X(D)$, $M=\langle a^{4}\rangle\langle c\rangle,\,X/M_{X}\cong
S_{4}$ and $\langle c\rangle_{X}=1$. Then $X=\langle
a,b,c|R,(a^{2})^{c^{3}}=a^{2r},(c^{3})^{b}=a^{\frac{2(l^{3}-1)}{l^{2}}}c^{3},(a^{2})^{c}=bc^{\frac{im}{2}},b^{c}=a^{2(-l+l^{2}+\frac{n}{4})}b,c^{a}=a^{2+4z}c^{1+\frac{km}{3}}\rangle,$
where either $i=z=0$ and $k=l=1$; or $i=1$, $n\equiv 4(\hbox{\rm mod
}8),\,m\equiv 0(\hbox{\rm mod }6)$, $l^{\frac{m}{2}}\equiv-1(\hbox{\rm mod
}\frac{n}{4})$ with $\hbox{\rm o}(l)=m$, $r=l^{3}$, $z=\frac{1-3l}{4l}$,
$k\in\\{1,2\\}$ and $\sum_{i=1}^{j}r^{i}\equiv 0(\hbox{\rm mod
}\frac{n}{2})\Leftrightarrow j\equiv 0(\hbox{\rm mod }\frac{m}{3})$.

Proof Under the hypothesis, $M_{X}=\langle a^{4}\rangle\mathchar
9583\relax\langle c^{3}\rangle$. Set $n=\hbox{\rm o}(a)$ and $m=\hbox{\rm
o}(c)$. Then $4\bigm{|}n$ and $3\bigm{|}m$. Since
$X/M_{X}=\langle\overline{a},\overline{b}\rangle\langle\overline{c}\rangle\cong
S_{4}$, we can choose $\overline{b}$ such that the form of $X/M_{X}$ is the
following:
$(\overline{a}^{2})^{\overline{c}}=\overline{b},\,\overline{b}^{\overline{c}}=\overline{a}^{2}\overline{b}$
and $(\overline{c})^{\overline{a}}=\overline{a}^{2}\overline{c}^{2}.$ Take
$a_{1}=a^{2}$ and $c_{1}=c^{3}$. Then we set
$a_{1}^{c}=bc_{1}^{w},b^{c}=a_{1}^{x}bc_{1}^{y},c^{a}=a_{1}^{1+2z}c^{2+3d}$,
where $x$ is odd.

Suppose $w\equiv 0(\hbox{\rm mod }\frac{m}{3})$. Note $\hbox{\rm
o}(a^{2})=\hbox{\rm o}((a^{2})^{c})=\hbox{\rm o}(b)=2$, then $X\cong S_{4}$.
Then $l=1,i=z=d=0$ as desired. So in what follows, we assume $w\not\equiv
0(\hbox{\rm mod }\frac{m}{3})$.

Then consider $X_{1}=GM_{X}=\langle a,b\rangle\langle c_{1}\rangle$. Noting
$\langle a\rangle\langle c_{1}\rangle\leq X_{1}$ and $\langle
c_{1}\rangle_{X_{1}}=1$, by Lemma 4.2, we know $\langle a_{1}\rangle\lhd
X_{1}$, which implies that $c_{1}$ normalises $\langle a_{1}\rangle$. Take
$X_{2}=\langle a_{1},b\rangle\langle c\rangle$. Then we get $X_{2}=(\langle
a_{1},b\rangle\langle c_{1}\rangle).\langle c\rangle$. Note that $c_{1}$
normalises $\langle a_{1}\rangle$ in $X_{2}$. Then by Lemma 6.4, we get

$X=\langle
a_{1},b,c|R,(a_{1}^{2})^{c}=a_{1}^{2r},c_{1}^{a_{1}}=a_{1}^{2s}c_{1},c_{1}^{b}=a_{1}^{2u}c_{1},a_{1}^{c}=bc^{\frac{m}{2}},b^{c}=a^{x}b\rangle,$

where

$\displaystyle\begin{array}[]{ll}&n\equiv 4(\hbox{\rm mod }8),\,m\equiv
0(\hbox{\rm mod }6),\,v=1,\,w=\frac{m}{6},\,y=0,\\\
&r=l^{3},\,s=\frac{1-l^{3}}{2l^{3}},\,u=\frac{l^{3}-1}{2l^{2}},\,1+2x=-l+l^{2}+\frac{n}{4},\,l^{\frac{m}{2}}\equiv-1(\hbox{\rm
mod }\frac{n}{4}).\end{array}$ (134)

Note $X=X_{2}.\langle a\rangle$. So $X$ may be defined by $R$ and

$\displaystyle(a_{1}^{2})^{c}=a_{1}^{2r},c_{1}^{a_{1}}=a_{1}^{2s}c_{1},c_{1}^{b}=a_{1}^{2u}c_{1},a_{1}^{c}=bc^{\frac{m}{2}},b^{c}=a_{1}^{x}b,c^{a}=a_{1}^{1+2z}c^{2+3d}.$
(135)

What we should to determine the parameters $r,z$ and $d$ by analyse the last
one extension. $X_{2}.\langle a\rangle$, where $c^{a}=a^{2+4z}c^{2+3d}$. Set
$\pi\in\hbox{\rm Aut\,}(X_{1}):$ $a^{2}\to a^{2}$, $b\to a^{-2}b$ and $c\to
a^{2+4z}c^{2+3d}$. We need to check the following seven equalities:

(i) $\pi$ preserves $(a^{2})^{c^{3}}=a^{2r}$:

$a^{2r}=((a^{2})^{c^{3}})^{a}=(a^{2})^{c^{3(2+3d)}}=a^{2r^{2+3d}},$

that is

$\displaystyle r^{1+3d}-1\equiv 0(\hbox{\rm mod }\frac{n}{2}).$ (136)

(ii) $\pi$ preserves $(c^{3})^{b}=a^{\frac{2(l^{3}-1)}{l^{2}}}c^{3}$:

$\begin{array}[]{lcl}((c^{3})^{b})^{a}&=&(c^{3(2+3d)}a^{2r(1+2z)+4r^{1+2d}zl+r^{d}(2lr^{1+d}+2l^{2}+\frac{n}{2}+4l^{2}z)})^{ba^{2}}\\\
&=&c^{3(2+3d)}a^{2l(l^{3}-1)+2r(1+2z)+4r^{1+2d}zl+r^{d}(2lr^{1+d}+2l^{2}+\frac{n}{2}+4l^{2}z)},\end{array}$

that is

$\displaystyle 1-r\equiv
2r(1+2z)+4r^{1+2d}zl+r^{d}(2lr^{1+d}+2l^{2}+4l^{2}z)(\hbox{\rm mod
}\frac{n}{2}).$ (137)

(iii) $\pi$ preserves $(a^{2})^{c}=bc^{\frac{m}{2}}$:

$\begin{array}[]{lcl}((a^{2})^{c})^{a}&=&(a^{2})^{c^{2+3d}}=(a^{2(-l+l^{2}+\frac{n}{4})}bc^{\frac{m}{2}})^{c^{3d}}=a^{2r^{d}(l^{2}+\frac{n}{4})-2l}bc^{\frac{m}{2}}\\\
(bc^{\frac{m}{2}})^{a}&=&ba^{2}(c^{3(2+3d)}a^{2r(1+2z)+4r^{1+2d}zl+r^{d}(2lr^{1+d}+2l^{2}+\frac{n}{2}+4l^{2}z)})^{\frac{m}{6}}\\\
&=&a^{\frac{2}{1-r}(2r(1+2z)+4r^{1+2d}zl+r^{d}(2lr^{1+d}+2l^{2}+\frac{n}{2}+4l^{2}z))-2}bc^{\frac{m}{2}},\end{array}$

that is,

$\displaystyle\begin{array}[]{ll}&\frac{2}{1-r}(2r(1+2z)+4r^{1+2d}zl+r^{d}(2lr^{1+d}+2l^{2}+\frac{n}{2}+4l^{2}z))-2\equiv\\\
&2r^{d}(l^{2}+\frac{n}{4})-2l(\hbox{\rm mod }n).\end{array}$ (140)

With Eq(137), we get

$r^{d}l\equiv 1(\hbox{\rm mod }\frac{n}{4}).$

(iv) $\pi$ preserves $b^{c}=a^{2(-l+l^{2}+\frac{n}{4})}b$:

$(b^{c})^{a}=(ba^{2})^{a^{2+4z}c^{2}}=a^{2l^{2}-8zl-8l+\frac{n}{2}}b\quad{\rm{and}}\quad(a^{2(-l+l^{2}+\frac{n}{4})}b)^{a}=a^{2(-l+l^{2})-2+\frac{n}{2}}b,$

that is

$\displaystyle z\equiv\frac{1-3l}{4l}(\hbox{\rm mod }\frac{n}{4}).$ (141)

(v) $\hbox{\rm o}(\pi(c))=m$:

$1=(a^{2+4z}c^{2+3d})^{m}=c^{m(2+3d)}a^{\sum_{i=0}^{\frac{m}{3}-1}r^{i}(2r(1+2z)+4r^{1+2d}zl+r^{d}(2lr^{1+d}+2l^{2}+\frac{n}{2}+4l^{2}z))}.$

Note that $r^{\frac{m}{6}}\equiv-1(\hbox{\rm mod }\frac{n}{4})$. Then
$\sum_{i=0}^{\frac{m}{3}-1}r^{i}\equiv 0(\hbox{\rm mod }\frac{n}{4})$. Note
that $2r+\frac{n}{2}+2r^{d}l^{2}(l^{2+3d}+1)\equiv 0(\hbox{\rm mod }4)$ which
implies
$4|(2r(1+2z)+4r^{1+2d}zl+r^{d}(2lr^{1+d}+2l^{2}+\frac{n}{2}+4l^{2}z))$. Then

$\sum_{i=0}^{\frac{m}{3}-1}r^{i}(2r(1+2z)+4r^{1+2d}zl+r^{d}(2lr^{1+d}+2l^{2}+\frac{n}{2}+4l^{2}z))\equiv
0(\hbox{\rm mod }n),$

as desired.

(vi) $\pi^{2}=\hbox{\rm Inn}(a^{2})$: Recall $(a^{2})^{c^{3}}=a^{2r}$ and
$(a^{2})^{c}=bc^{\frac{m}{2}}$, then we know $\hbox{\rm
Inn}(a^{2})(c)=c^{1+\frac{m}{2}}a^{-2}b$ and $\hbox{\rm
Inn}(a^{2})(c^{3})=c^{3}a^{2-2r}$.

$c^{1+\frac{m}{2}}ba^{2}=\hbox{\rm
Inn}(a^{2})(c)=\pi^{2}(c)=(a^{2+4z}c^{2+3d})^{a}=c^{1+3(1+d)(1+3d)+\frac{m}{2}}ba^{2+l(1-r^{d}l)+\frac{n}{2}\sum_{i=0}^{d}r^{i}},$

that is,

$\displaystyle\begin{array}[]{ll}&(1+d)(1+3d)\equiv 0(\hbox{\rm mod
}\frac{m}{3}),\\\ &0\equiv
l(1-r^{d}l)+\sum_{i=0}^{d}r^{i}\frac{n}{2}(\hbox{\rm mod }n);\end{array}$
(144)

and

$c^{3}a^{2-2r}=\hbox{\rm
Inn}(a^{2})(c^{3})=\pi^{2}(c^{3})=c^{3(2+3d)^{2}}a^{(1+\sum_{i=0}^{1+3d}r^{i})(1-2l^{2}-l^{3}+2r^{1+d}+\frac{n}{2})},$

that is,

$\displaystyle 1\equiv(2+3d)^{2}(\hbox{\rm mod
}\frac{m}{3})\quad{\rm{and}}\quad
0\equiv(\sum_{i=0}^{1+3d}r^{i}-1)(1-r)(\hbox{\rm mod }n).$ (145)

(vii) Insure $\langle c\rangle_{X}=1$: Since $\langle c\rangle_{X}\leq M$, we
get $\langle c\rangle_{X}\leq M_{X}=\langle a^{4}\rangle\langle c^{3}\rangle$,
which implies $\langle c\rangle_{X}=\cap_{x\in X}C^{x}=\cap_{x\in
G}C^{x}=\cap_{x\in G}\langle c^{3}\rangle^{x}=\langle c^{3}\rangle_{X_{1}}$.
Then it is suffer to insure $\langle c^{3}\rangle_{X_{1}}=1$, where
$X_{1}=\langle
a,b,c|R,(a^{2})^{c^{3}}=a^{2l^{3}},(c^{3})^{b}=a^{\frac{2(l^{3}-1)}{l^{2}}}c^{3},(c^{3})^{a}=c^{3(2+3d)}a^{1-l^{3}+\frac{n}{2}}\rangle$.
Then by Lemma 6.1, we get $\sum_{i=1}^{j}r^{i}\equiv 0(\hbox{\rm mod
}\frac{n}{2})\Leftrightarrow j\equiv 0(\hbox{\rm mod }\frac{m}{3})$. By
Eq(136) and $r^{\frac{m}{6}}\equiv-1(\hbox{\rm mod }\frac{n}{4})$, we get
$\sum_{i=1}^{1+3d}r^{i}\equiv 0(\hbox{\rm mod }\frac{n}{2})$. Then $1+3d\equiv
0(\hbox{\rm mod }\frac{m}{3})$. Then $0\equiv l(1-r^{d}l)(\hbox{\rm mod }n)$
with Eq(144). $\Box$

###  6.5 $M=\langle a^{3}\rangle\langle c\rangle$ and $X/M_{X}\cong S_{4}$

###### Lemma 6.6

Suppose that $X=X(D)$, $M=\langle a^{3}\rangle\langle c\rangle,\,X/M_{X}\cong
S_{4}$ and $\langle c\rangle_{X}=1$. Then

$\displaystyle X=\langle
a,b,c|R,a^{c^{4}}=a^{r},b^{c^{4}}=a^{1-r}b,(a^{3})^{c^{\frac{m}{4}}}=a^{-3},a^{c^{\frac{m}{4}}}=bc^{\frac{3m}{4}}\rangle,$
(146)

where $m\equiv 4(\hbox{\rm mod }8)$ and $r$ is of order $\frac{m}{4}$ in
$\mathbb{Z}_{2n}^{*}$.

In this case, $M_{X}=\langle a^{3}\rangle\langle c^{4}\rangle$. Set
$a^{3}=a_{1}$ and $c^{4}=c_{1}$ so that $M_{X}=\langle a_{1}\rangle\langle
c_{1}\rangle$. Set $\hbox{\rm o}(a)=n$ and $\hbox{\rm o}(c)=m$, where $n\equiv
0(\hbox{\rm mod }3)$ and $m\equiv 0(\hbox{\rm mod }4)$. Then in Lemma 6.7, we
shall show $\langle a_{1}\rangle\lhd X$ and in Lemma 6.9, we shall get the
classification of $X$.

###### Lemma 6.7

$\langle a_{1}\rangle\lhd X$.

Proof Let $X_{1}=M_{X}G$. Since $\langle a\rangle\langle c_{1}\rangle\leq
X_{1}$ and $\langle c_{1}\rangle_{X_{1}}=1$, the subgroup $X_{1}$ has been
given in Lemma 6.1:

$\displaystyle X_{1}=\langle a,b,c_{1}|\langle
a,b\rangle,c_{1}^{\frac{m}{4}}=1,(a^{2})^{c_{1}}=a^{2r},(c_{1})^{a}=a^{6s}c_{1}^{t},(c_{1})^{b}=a_{1}^{u}c_{1}^{v}\rangle,$
(147)

where

  1. (1)

if $t=1$, then $v=1$, $6s\sum_{l=1}^{\frac{m}{4}}r^{l}\equiv 2(1-6sr)-2r\equiv
0(\hbox{\rm mod }n)$ and $6s\sum_{l=1}^{w}r^{l}\equiv 0\equiv
3u\sum_{l=1}^{w}(1-6sr)^{l}(\hbox{\rm mod }n)$ is if and only if $w\equiv
0(\hbox{\rm mod }\frac{m}{4})$;

  2. (2)

if $t\neq 1$, then both $n$ and $u$ are even, and $r,s,t,u,v$ are given by

$\begin{array}[]{ll}&r^{t-1}-1\equiv
r^{v-1}-1\equiv\frac{3u}{2}(\sum_{l=0}^{v-1}r^{l}-1)\equiv 0(\hbox{\rm mod
}\frac{n}{2}),\\\ &3s\sum_{l=1}^{t}r^{l}+3sr\equiv
3sr+3s\sum_{l=1}^{v}r^{l}-\frac{3u}{2}\sum_{l=1}^{t}r^{l}+\frac{3ur}{2}\equiv
1-r(\hbox{\rm mod }\frac{n}{2}),\\\ &t^{2}-1\equiv v^{2}-1\equiv 0(\hbox{\rm
mod }\frac{m}{4}),\\\
&3s\sum_{l=1}^{i}r^{l}\equiv\frac{3u}{2}\sum_{l=1}^{i}r^{l}\equiv 0(\hbox{\rm
mod }\frac{n}{2})\Leftrightarrow i\equiv 0(\hbox{\rm mod
}\frac{m}{4}).\end{array}$

Now $X=\langle X_{1},c\rangle$ and we need to write the relation of $c$ with
$X_{1}$. Since $X/M_{X}=\langle\overline{a},\overline{b}\rangle\mathchar
9583\relax\langle\overline{c}\rangle\cong S_{4}$, checked by Magma, under our
condition, the only possibilities are the following:

$\displaystyle\overline{a}^{3}=\overline{c}^{4}=\overline{b}^{2}=1,\overline{a}^{\overline{b}}=\overline{a}^{-1},\overline{a}^{\overline{c}}=\overline{a}^{i}\overline{b}\overline{c}^{3},$
(148)

where $i\in\mathbb{Z}_{3}$. Observing Eq(147) and Eq(148), we may relabel
$a^{i}b$ by $b$. Then in $X$, Eq(148) corresponds to

$\displaystyle a^{3}=a_{1},b^{2}=1,c^{4}=c_{1},a^{c}=bc^{3+4w}.$ (149)

Moreover, the conjugacy of $\overline{c}$ on $M_{X}$ is needed:

$\displaystyle(a_{1})^{c}=a_{1}^{z}c_{1}^{d}.$ (150)

Then the group $X$ is uniquely determined by Eq(147), Eq(149) and Eq(150). For
the contrary, we assume $d\neq 0$. Then we need to deal with two cases,
according to the parameter $t$ of $X_{1}$. Case 1: $t=1$. In this case, $v=1$
and $2(1-6sr)-2r\equiv 0(\hbox{\rm mod }n)$ by $X_{1}$. Set $r_{1}=1-6sr$.
Then $r_{1}\equiv 1(\hbox{\rm mod }3),\,a^{c_{1}}=a^{r_{1}}$ and
$b^{c_{1}}=a_{1}^{ur_{1}}b$. By Eq(147), Eq(149) and Eq(150), one can check
$b^{c}=a_{1}^{x}bc^{2+4y}$ for some $x$ and $y$.

Since $c$ preserves $a_{1}^{c_{1}}=a_{1}^{r_{1}}$, we get

$((a_{1})^{c_{1}})^{c}=a_{1}^{zr_{1}}c_{1}^{d}=c_{1}^{d}a_{1}^{zr_{1}^{1+d}}\,{\rm{and}}\,(a_{1}^{r_{1}})^{c}=(a_{1}^{z}c_{1}^{d})^{r_{1}}=c_{1}^{dr_{1}}a_{1}^{3z\sum_{l=1}^{r_{1}}r_{1}^{dl}},$

which gives

$\displaystyle d\equiv dr_{1}(\hbox{\rm mod }\frac{m}{4}).$ (151)

Since $c$ preserves $b^{c_{1}}=a_{1}^{ur_{1}}b$, we get

$(b^{c_{1}})^{c}=a^{{2+3x}r_{1}+3ur_{1}}bc^{4y}\quad{\rm{and}}\quad(a_{1}^{ur_{1}}b)^{c}=c_{1}^{dur_{1}}a_{1}^{z\sum_{l=1}^{ur_{1}}r_{1}^{dl}}a^{2+3x}bc^{4y},$

which gives

$\displaystyle du\equiv 0(\hbox{\rm mod }\frac{m}{4}).$ (152)

Since $a_{1}^{c_{1}}=a_{1}^{c^{4}}=a_{1}^{x_{1}}c_{1}^{d(z^{3}+z^{2}+z+1)}$
for some $x_{1}$, we get

$\displaystyle d(z^{3}+z^{2}+z+1)\equiv 0(\hbox{\rm mod }\frac{m}{4});$ (153)

By last equation of Eq(149), we get $ac=cbc^{3+4w}$. Then

$a_{1}^{ac}=a_{1}^{z}c_{1}^{d}\,{\rm{and}}\,a_{1}^{cbc^{3+4w}}=a_{1}^{x_{1}}c_{1}^{d-dz(z^{2}+z+1)},$

which gives

$\displaystyle dz(z^{2}+z+1)\equiv 0(\hbox{\rm mod }\frac{m}{4}).$ (154)

With Eq(153), we know $d\equiv 0(\hbox{\rm mod }\frac{m}{4}),$ a
contradiction. Case 2: $t\neq 1$. Suppose that $t\neq 1$. Then if $\langle
a_{1}^{2}\rangle\lhd X$, consider $\overline{X}=X/\langle a_{1}^{2}\rangle$
and $\langle\overline{c_{1}}\rangle\lhd\overline{X}$. Note $\overline{C}\leq
C_{\overline{X}}(\langle\overline{c_{1}}\rangle)=\overline{X}$, then $t=1$,
contradicting with $t\neq 1$. So in what follows, we shall show $\langle
a_{1}^{2}\rangle\lhd X$.

By Eq(147), both $u$ and $n$ are even and $r\equiv 1(\hbox{\rm mod }3)$. Then
$z$ is odd as $(a,\frac{n}{3})=1$.

By Eq(147), Eq(149) and Eq(150), one can check $b^{c}=a^{2+6x}bc^{4y}$ for
some $x$ and $y$.

Since $c$ preserves $(a_{1}^{2})^{c_{1}}=a_{1}^{2r}$, we get

$((a_{1}^{2})^{c_{1}})^{c}=(a_{1}^{z}c_{1}^{d}a_{1}^{z}c_{1}^{d})^{c_{1}}=a_{1}^{x_{1}}c_{1}^{d(t+1)},\,(a_{1}^{2r})^{c}=(a_{1}^{z}c_{1}^{d})^{2r}=a_{1}^{x_{2}}c_{1}^{d(t+1)r},$

which gives

$\displaystyle d(t+1)(r-1)\equiv 0(\hbox{\rm mod }\frac{m}{4}).$ (155)

Since $c$ preserves $(c_{1})^{b}=a_{1}^{u}c_{1}^{v}$, we get

$(c_{1}^{b})^{c}=c_{1}^{a^{2+6x}bc^{4y}}=a_{1}^{x_{3}}c_{1}^{v},\,(a_{1}^{u}c_{1}^{v})^{c}=(a_{1}^{z}c_{1}^{d})^{u}c_{1}^{v}=a_{1}^{x_{4}}c_{1}^{\frac{du(t+1)}{2}+v},$

which gives

$\displaystyle\frac{du(t+1)}{2}\equiv 0(\hbox{\rm mod }\frac{m}{4}).$ (156)

Since $c$ preserves $c_{1}^{a}=a_{1}^{2s}c_{1}^{t}$, which implies
$c_{1}^{bc^{2+4w}}=a_{1}^{2s}c_{1}^{t}$, we get

$c_{1}^{bc^{2+4w}}=(a_{1}^{u}c_{1}^{v})^{c^{2+4w}}=((a_{1}^{z}c_{1}^{d})^{z}c_{1}^{d})^{ur^{w}}c_{1}^{v}=a_{1}^{x}c_{1}^{v},$

which gives

$\displaystyle v\equiv t(\hbox{\rm mod }\frac{m}{4}).$ (157)

By last equation of Eq(149) again, we get $ac^{2}=cbc^{4(w+1)}$. Then

$(a_{1}^{2})^{ac^{2}}=a_{1}^{x}c_{1}^{d(t+1)(z+1)}\,{\rm{and}}\,(a_{1}^{2})^{cbc_{1}^{w+1}}=a_{1}^{x}c_{1}^{d(t+1)},$

which gives

$\displaystyle dz(t+1)\equiv 0(\hbox{\rm mod }\frac{m}{4}).$ (158)

Since $\hbox{\rm o}(a_{1}^{c})=\frac{n}{3}$, we get $\frac{dn(t+1)}{6}\equiv
0(\hbox{\rm mod }\frac{m}{4})$. With $(\frac{n}{6},z)=1$ and Eq(158), we get
$d(t+1)\equiv 0(\hbox{\rm mod }\frac{m}{4})$. Then
$(a_{1}^{2})^{c}=(a_{1}^{z}c_{1}^{d})^{2}=a_{1}^{x}c_{1}^{d(t+1)}=a_{1}^{x}$,
which implies $\langle a_{1}^{2}\rangle\lhd X$, as desired. $\Box$

###### Lemma 6.8

With the notation, suppose that $\langle a_{1}\rangle\lhd X$, $\langle
c\rangle_{X}=1$ and $\langle c\rangle$ is $2-$group. Then $\hbox{\rm o}(c)=4$.

Proof Consider $\overline{X}=X/\langle
a_{1}\rangle=\langle\overline{a},\overline{b}\rangle\langle\overline{c}\rangle$.
Note $\overline{M_{X}}=\langle\overline{c}^{4}\rangle$, Then one can check
$C_{\overline{X}}(\langle\overline{c}_{1}\rangle)=\overline{X}$, which implies
$\overline{X}$ is the central expansion of $S_{4}$. Since the Schur multiplier
of $S_{4}$ is $\mathbb{Z}_{2}$, we know $\hbox{\rm o}(c)$ is either $4$ or
$8$. For the contrary, we assume $\hbox{\rm o}(c)=8$. Then consider
$X_{1}=\langle a,b\rangle\mathchar 9583\relax\langle c_{1}\rangle$, where
$c_{1}=c^{4}$ and $\hbox{\rm o}(c_{1})=2$, again. By Lemma 6.1, we have

$X_{1}=\langle
a,b,c_{1}\bigm{|}R,a^{c_{1}}=a^{r},b^{c_{1}}=a_{1}^{u}b\rangle,$

where $r\equiv 1(\hbox{\rm mod }3).$ Note that $\langle a\rangle\leq\langle
c\rangle_{X}(\langle a_{1}\rangle)\lhd X$. Thus $\langle
a,bc,c^{2}\rangle\leq\langle c\rangle_{X}(\langle a_{1}\rangle),$ which
implies $\langle a_{1}\rangle\times\langle c_{1}\rangle\lhd X$. One can check
$r=1$(as $3r\equiv 3(\hbox{\rm mod }\frac{n}{3})$ and $1\equiv
r^{2}=2r-1(\hbox{\rm mod }\frac{n}{3})$). Then $\langle
a,c^{2}\rangle\leq\langle c\rangle_{X}(\langle a_{1}\rangle\times\langle
c_{1}\rangle)\lhd X$, which implies $bc\in\langle c\rangle_{X}(\langle
a_{1}\rangle\times\langle c_{1}\rangle)$. So $c_{1}^{b}=c_{1}$. Then
$c_{1}\in\langle c\rangle_{X}=1$, a contradiction. $\Box$

###### Lemma 6.9

The group $X$ is given by Eq(146).

Proof By lemma, $\langle a_{1}\rangle\lhd X$, that is $(a_{1})^{c}=a_{1}^{z}$
by Eq(150). Since $\langle a^{2}\rangle\lhd X_{1}$, we get $\langle
a\rangle\lhd X_{1}$ and so $G\lhd X_{1}$, that is $t=v=1$ in Eq(147). Then by
Eq(147), (149) and (150), we can set

$X=\langle
a,b,c\bigm{|}R,a^{c_{1}}=a^{r_{1}},b^{c_{1}}=a_{1}^{ur_{1}}b,(a_{1})^{c}=a_{1}^{z},a^{c}=bc^{3+4w}\rangle,$

where $r_{1}=1-6sr$, $r_{1}^{\frac{m}{4}}-1\equiv 2(r_{1}-r)\equiv 0(\hbox{\rm
mod }n)$ and $r_{1}^{i}-1\equiv 0\equiv 3u\sum_{l=1}^{i}r_{1}^{l}(\hbox{\rm
mod }n)$ is if and only if $i\equiv 0(\hbox{\rm mod }\frac{m}{4})$.

Set $\langle c\rangle=\langle c_{2}\rangle\times\langle c_{3}\rangle$, where
$\langle c_{2}\rangle$ is $2-$group and $\langle c_{3}\rangle$ is
$2^{\prime}-$Hall subgroup of $\langle c\rangle$. Then $\langle
c_{1}\rangle=\langle c_{2}^{4}\rangle\times\langle c_{3}\rangle$. And we shall
show $c_{2}^{4}=1$.

Consider $\overline{X}=X/\langle
a_{1}\rangle=\langle\overline{a},\overline{b}\rangle\langle\overline{c}\rangle$.
Then one can check
$C_{\overline{X}}(\langle\overline{c}_{1}\rangle)=\overline{X}$, which implies
$\langle\overline{c_{1}}\rangle\leq Z(\overline{X})$ and
$\overline{X}/\langle\overline{c_{1}}\rangle\cong S_{4}$. Note
$\langle\overline{c}_{3}\rangle\leq\langle\overline{c}_{3}\rangle(\langle\overline{c}_{2}^{4}\rangle\langle\overline{a}\rangle)\leq\overline{X}$
where
$(|\overline{X}:\langle\overline{c}_{3}\rangle(\langle\overline{c}_{2}^{4}\rangle\langle\overline{a}\rangle)|,|\langle\overline{c}_{3}\rangle|)=1$,
then by Proportion 2.5, $\langle\overline{c}_{3}\rangle$ has a complement in
$\overline{X}$, which implies $X=(\langle a,b\rangle\langle
c_{2}\rangle)\mathchar 9583\relax\langle c_{3}\rangle.$ Consider
$X_{2}=\langle a,b\rangle\langle c_{2}\rangle$, where $\langle
c_{2}\rangle_{X_{2}}=1$ and $\langle a_{1}\rangle\lhd X_{2}$. Then by Lemma
6.8, we get $|\langle c_{2}\rangle|=4$, which implies $\langle
c_{1}\rangle=\langle c_{3}\rangle$. Then

$X=(\langle a,b\rangle\langle c_{2}\rangle)\mathchar 9583\relax\langle
c_{1}\rangle.$

In what follows, we shall determine $X$. In $X_{2}=\langle a,b\rangle\langle
c_{2}\rangle$, we know $a^{c_{2}}=bc_{2}^{3}$ by Eq(149). Consider $\langle
a\rangle\leq C_{X_{2}}(\langle a_{1}\rangle)\lhd X_{2}$. Then
$C_{X_{2}}(\langle a_{1}\rangle)$ is either $\langle
a,bc_{2},c_{2}^{2}\rangle$ or $X_{2}$. Suppose that $\langle
c\rangle_{X}(\langle a_{1}\rangle)=X$. Then we know $a_{1}^{2}=1$ as
$[a_{1},b]=1$, that is $n=3$ or $6$. Then one can check

$\begin{array}[]{ll}&X=X_{2}\cong S_{4},\,{\rm{if}}\,n=3;\\\ &X=X_{2}=\langle
a,b,c|a^{6}=b^{2}=c^{4}=1,a^{b}=a^{-1},a^{c}=bc^{3},b^{c}=a^{2}b\rangle,\,{\rm{if}}\,n=6.\end{array}$

Suppose that $\langle c\rangle_{X}(\langle a_{1}\rangle)=\langle
a,bc_{2},c_{2}^{2}\rangle$. Then $a_{1}=a_{1}^{bc_{2}}=(a_{1}^{-1})^{c_{2}}$,
which implies $z=-1$. In $X_{1}=\langle a,b\rangle\mathchar 9583\relax\langle
c_{1}\rangle$, we know

$X_{1}=\langle
a,b,c_{1}\bigm{|}R,a^{c_{1}}=a^{r_{1}},b^{c_{1}}=a_{1}^{ur_{1}}b\rangle.$

Since $c_{2}$ preserves $a^{c_{1}}=a^{r_{1}}$, we get

$(bc_{2}^{3})^{c_{1}}=a_{1}^{ur_{1}}bc^{3}\,{\rm{and}}\,(a^{r_{1}})^{c_{2}}=a_{1}^{\frac{1-r_{1}}{3}}bc^{3},$

which gives

$ur_{1}\equiv\frac{1-r_{1}}{3}(\hbox{\rm mod }\frac{n}{3}).$

Then

$X=\langle
a,b,c\bigm{|}R,a^{c_{1}}=a^{r_{1}},b^{c_{1}}=a_{1}^{\frac{1-r_{1}}{3}}b,(a_{1})^{c_{2}}=a_{1}^{-1},a^{c_{2}}=bc_{2}^{3}\rangle,$

where $c_{2}=c^{\frac{m}{4}}$ and $\hbox{\rm o}(r_{1})=\frac{m}{4}$. $\Box$

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