# The Indoor Mobile Coverage Problem Using UAVs

Hazim Shakhatreh  Abdallah Khreishah  and Issa Khalil Hazim Shakhatreh and
Abdallah Khreishah are with the Department of Electrical and Computer
Engineering, New Jersey Institute of Technology (email:
<EMAIL_ADDRESS>
Issa Khalil is with Qatar Computing Research Institute (QCRI), HBKU, Doha,
Qatar (email<EMAIL_ADDRESS>Part of this work was presented in ICC 2017
[1] and ICICS 2017 [2].

###### Abstract

Unmanned aerial vehicles (UAVs) can be used as aerial wireless base stations
when cellular networks are not operational due to natural disasters. They can
also be used to supplement the ground base station in order to provide better
coverage and higher data rates for the users. Prior studies on UAV-based
wireless coverage typically consider an Air-to-Ground path loss model, which
assumes that the users are outdoor and located on a 2D plane. In this paper,
we propose using UAVs to provide wireless coverage for indoor users inside a
high-rise building. First, we present realistic Outdoor-Indoor path loss
models and describe the tradeoff introduced by these models. Then we study the
problem of efficient placement of a single UAV, where the objective is to
minimize the total transmit power required to cover the entire high-rise
building. The formulated problem is non-convex and is generally difficult to
solve. To that end, we consider three cases of practical interest and provide
efficient solutions to the formulated problem under these cases. Then we study
the problem of minimizing the number of UAVs required to provide wireless
coverage to high rise buildings and prove that this problem is NP-complete.
Due to the intractability of the problem, we use clustering to minimize the
number of UAVs required to cover the indoor users. We demonstrate through
simulations that the method that clusters the building into regular structures
and places the UAVs in each cluster requires 80% more number of UAVs relative
to our clustering algorithm.

###### Index Terms:

Unmanned aerial vehicles, Outdoor-to-Indoor path loss model, gradient descent
algorithm, particle swarm optimization, $k$-means clustering.

##  I Introduction

UAVs can be used to provide wireless coverage during emergency cases where
each UAV serves as an aerial wireless base station when the cellular network
goes down [3]. They can also be used to supplement the ground base station in
order to provide better coverage and higher data rates for the users [4].

In order to use a UAV as an aerial wireless base station, the authors in [5]
presented an Air-to-Ground path loss model that helped the academic
researchers to formulate many important UAV-based coverage problems. The
authors in [6] utilized this model to evaluate the impact of a UAV altitude on
the downlink ground coverage and to determine the optimal values for altitude
which lead to maximum coverage and minimum required transmit power. In [7],
the authors used the path loss model to propose a power-efficient deployment
for UAVs under the constraint of satisfying the rate requirement for all
ground users. The authors in [8] utilized the path loss model to study the
optimal deployment of multiple UAVs equipped with directional antennas, using
circle packing theory. The 3D locations of the UAVs are determined in a way
that the total coverage area is maximized. In [9], the authors used the path
loss model to find the minimum number of UAVs and their 3D locations so that
all outdoor ground users are served. However, it is assumed that all users are
outdoor and the location of each user can be represented by an outdoor 2D
point. These assumptions limit the applicability of this model when one needs
to consider indoor users.

Providing good wireless coverage for indoor users is very important. According
to Ericsson report [10], 90% of the time people are indoor and 80% of the
mobile Internet access traffic also happens indoors [11, 12]. To guarantee
wireless coverage, service providers are faced with several key challenges,
including providing service to a large number of indoor users and the ping
pong effect due to interference from near-by macro cells [13, 14, 15]. In this
paper, we propose using UAVs to provide wireless coverage for users inside a
high-rise building during emergency cases and special events (such as
concerts, indoor sporting events, etc.), when the cellular network service is
not available or it is unable to serve all indoor users. To the best of our
knowledge, this is the first work that proposes using UAVs to provide wireless
coverage for indoor users. We summarize our main contributions as follows:

  * •

We utilize an Outdoor-Indoor path loss model for low-SHF band (450 MHz to 6
GHz) [16], certified by ITU, and an Outdoor-Indoor path loss model for high-
SHF band (over 6 GHz) [17], then we show the tradeoff introduced by these
models.

  * •

We formulate the problem of efficient placement of a single UAV, where the
objective is to minimize the total transmit power required to cover the entire
high-rise building.

  * •

Since the formulated problem is non-convex and is generally difficult to
solve, we consider three cases of practical interest and provide efficient
solutions to the formulated problem under these cases and for different
operating frequencies (low-SHF and high-SHF bands). In the first case, we aim
to find the minimum transmit power such that an indoor user with the maximum
path loss can be covered. In the second case, we assume that the locations of
indoor users are symmetric across the dimensions of each floor (such as office
buildings or hotels), and propose a gradient descent algorithm for finding an
efficient location of a UAV. In the third case, we assume that the locations
of indoor users are uniformly distributed in each floor, and propose a
particle swarm optimization algorithm to find an efficient 3D placement of a
UAV that tries to minimize the total transmit power required to cover the
indoor users.

  * •

Due to the limited transmit power of a UAV, we formulate the problem of
minimizing the number of UAVs required to provide wireless coverage to high
rise building and prove that this problem is NP-complete.

  * •

Due to the intractability of the problem, we use clustering to minimize the
number of UAVs required to cover indoor users. We demonstrate through
simulations that the method that clusters the building into regular structures
and places the UAVs in each cluster requires 80% more number of UAVs relative
to our clustering algorithm.

##  II System Model

###  II-A System Settings

Let ($x_{UAV}$,$y_{UAV}$,$z_{UAV}$) denote the 3D location of the UAV. We
assume that all users are located inside a high-rise building as shown in
Figure 3, and use ($x_{i}$,$y_{i}$,$z_{i}$) to denote the location of user
$i$. The dimensions of the high-rise building, in the shape of a rectangular
prism, are $[0,x_{b}]$ $\times$ $[0,y_{b}]$ $\times$ $[0,z_{b}]$. Also, let
$d_{out,i}$ be the distance between the UAV and indoor user $i$, let
$\theta_{i}$ be the incident angle , and let $d_{in,i}$ be the distance
between the building wall and indoor user $i$.

###  II-B Outdoor-Indoor Path Loss Models

The Air-to-Ground path loss model presented in [5] is not appropriate when we
consider wireless coverage for indoor users, because this model assumes that
all users are outdoor and located at 2D points. In this paper, we adopt the
Outdoor-Indoor path loss model, certified by the ITU [16], for low-SHF
operating frequency. The path loss is given as follows:

$\begin{split}L_{i}=L_{F}+L_{B}+L_{I}=(w\log_{10}d_{out,i}+w\log_{10}f_{Ghz}+g_{1})\\\
+(g_{2}+g_{3}(1-\cos\theta_{i})^{2})+(g_{4}d_{in,i})\end{split}$

where $L_{F}$ is the free space path loss, $L_{B}$ is the building penetration
loss, and $L_{I}$ is the indoor loss. In this model, we also have $w$=20,
$g_{1}$=32.4, $g_{2}$=14, $g_{3}$=15,$g_{4}$=0.5 [16] and $f_{Ghz}$ is the
carrier frequency.

In [17], the authors clarify the Outdoor-to-Indoor path loss characteristics
based on the measurement for 0.8 to 37 GHz frequency band. We adopt this path
loss model for high-SHF operating frequency. The path loss is given as
follows:

$\begin{split}L_{i}=L_{F}+L_{B}+L_{I}=(\alpha_{1}+\alpha_{2}\log_{10}d_{out,i}+\alpha_{3}\log_{10}f_{Ghz})+~{}~{}~{}\\\
(\beta_{1}+\frac{\beta_{2}-\beta_{1}}{1+exp(-\beta_{3}(\theta_{i}-\beta_{4}))})+(\gamma_{1}d_{in,i})\end{split}$

In this model, we have $\alpha_{1}$=31.4, $\alpha_{2}$=20, $\alpha_{3}$=21.5,
$\beta_{1}$=6.8, $\beta_{2}$=21.8, $\beta_{3}$=0.453, $\beta_{4}$=19.7 and
$\gamma_{1}$=0.49.

Note that there is a key tradeoff in the path loss models when the horizontal
distance between the UAV and a user changes. When this horizontal distance
increases, the free space path loss (i.e., $L_{F}$) increases as $d_{out,i}$
increases, while the building penetration loss (i.e., $L_{B}$) decreases as
the incident angle (i.e., $\theta_{i}$) decreases (Figure 3 shows the
penetration loss for high-SHF band). Similarly, when this horizontal distance
decreases, the free space path loss (i.e., $L_{F}$) decreases as $d_{out,i}$
decreases, while the building penetration loss (i.e., $L_{B}$) increases as
the incident angle (i.e., $\theta_{i}$) increases.

##  III Providing Wireless coverage using a single UAV

Figure 1: Parameters of the path loss model

Figure 2: Building penetration loss for high-SHF

Figure 3: Transmit power required to cover the building

###  III-A Problem Formulation

Consider a transmission between a UAV located at
($x_{UAV}$,$y_{UAV}$,$z_{UAV}$) and an indoor user $i$ located at
($x_{i}$,$y_{i}$,$z_{i}$). The rate for user $i$ is given by:

$\begin{split}C_{i}=Blog_{2}(1+\dfrac{P_{t,i}/L_{i}}{N})\end{split}$

where $B$ is the transmission bandwidth of the UAV, $P_{t,i}$ is the UAV
transmit power to indoor user $i$, $L_{i}$ is the path loss between a UAV and
an indoor user $i$ and $N$ is the noise power. In this paper, we do not
explicitly model interference, and instead, implicitly model it as noise.

Let us assume that each indoor user has a channel with bandwidth equals $B/M$,
where $M$ is the number of users inside the building and the rate requirement
for each user is $v$. Then the minimum power required to satisfy this rate for
each user is given by:

$\begin{split}P_{t,i,min}=(2^{\frac{v.M}{B}}-1)\star N\star L_{i}\end{split}$

Our goal is to find the optimal location of UAV such that the total transmit
power required to satisfy the downlink rate requirement of each indoor user is
minimized. The objective function can be represented as:

$\begin{split}P=\sum_{i=1}^{M}(2^{\frac{v.M}{B}}-1)\star N\star L_{i},\\\
\end{split}$

where $P$ is the UAV total transmit power. Since $(2^{\frac{v.M}{B}}-1)\star
N$ is constant, our problem can be formulated as:

$\begin{split}\min_{x_{UAV},y_{UAV},z_{UAV}}L_{Total}=\sum_{i=1}^{M}L_{i}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
subject~{}to~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
x_{min}\leq x_{UAV}\leq
x_{max},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
y_{min}\leq y_{UAV}\leq
y_{max},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
z_{min}\leq z_{UAV}\leq
z_{max},~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
L_{Total}\leq
L_{max}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$

Here, the first three constraints represent the minimum and maximum allowed
values for $x_{UAV}$, $y_{UAV}$ and $z_{UAV}$. In the fourth constraint,
$L_{max}$ is the maximum allowable path loss and equals
$P_{t,max}$$/$$((2^{\frac{v.M}{B}}-1)\star N)$, where $P_{t,max}$ is the
maximum transmit power of UAV.

Finding the optimal placement of UAV is generally difficult because the
problem is non-convex. Therefore, in the next subsection, we consider three
special cases of practical interest and derive efficient solutions under these
cases.

###  III-B Efficient Placement of a Single UAV

Case 1. The worst location in building: In this case, we find the minimum
transmit power required to cover the building based on the location that has
the maximum path loss inside the building. The location that has the maximum
path loss in the building is the location that has maximum $d_{out,i}$,
maximum $\theta_{i}$, and maximum $d_{in,i}$. The locations that have the
maximum path loss are located at the corners of the highest and lowest floors.
Since the locations that have the maximum path loss inside the building are
the corners of the highest and lowest floors, we place the UAV at the middle
of the building ($y_{UAV}$= 0.5$y_{b}$ and $z_{UAV}$=0.5$z_{b}$). Then, given
Outdoor-to-Indoor path loss models for low-SHF and high-SHF bands, we need to
find an efficient horizontal point $x_{UAV}$ for the UAV such that the total
transmit power required to cover the building is minimized.

Figure 4: Transmit power required to cover the building, $f_{c}$=2 GHz

Figure 5: Transmit power required to cover the building, $f_{c}$=10 GHz

Figure 6: Transmit power required to cover 30 meter building height

Now, when the horizontal distance between the UAV and this location increases,
the free space path loss also increases as $d_{out,i}$ increases, while the
building penetration loss decreases because we decrease the incident angle
$\theta_{i}$. In Figure 3, we demonstrate the minimum transmit power required
to cover a building of different heights, where the minimum transmit power
required to cover the building is given by:

$\begin{split}P_{t,min}(dB)=P_{r,th}+L_{i}\end{split}$
$\begin{split}P_{r,th}(dB)=N+\gamma_{th}\end{split}$

Here, $P_{r,th}$ is the minimum received power, $N$ is the noise power (equals
-120dBm), $\gamma_{th}$ is the threshold SNR (equals 10dB), $y_{b}$=50 meters
, $x_{b}$=20 meters and the carrier frequency is 2Ghz. The numerical results
show that there is an optimal horizontal point that minimizes the total
transmit power required to cover a building. Also, we note that when the
height of the building increases, the optimal horizontal distance increases.
This is to compensate for the increased building penetration loss due to an
increased incident angle.

In Theorem 1, we characterize the optimal incident angle $\theta$ for low-SHF
band that minimizes the transmit power required to cover the building. This
helps us finding the optimal horizontal distance between the UAV and the
building.

######  Theorem 1.

For the low-SHF operating frequency case, when we place the UAV at the middle
of building , the optimal incident angle $\theta$ that minimizes the transmit
power required to cover the building will be equal to $48.654^{o}$ and the
optimal horizontal distance between the UAV and the building will be equal to
$((\dfrac{0.5z_{b}}{tan(48.654^{o})})^{2}-(0.5y_{b})^{2})^{0.5}-x_{b}$.

###### Proof.

In order to find the optimal horizontal point, we rewrite the equation that
represents the path loss in terms of the incident angle ($\theta_{i}$) and the
altitude difference between the UAV and the user $i$ ($\Delta h_{i}$):

$\begin{split}L_{i}(\Delta h_{i},\theta_{i})=w\log_{10}\dfrac{\Delta
h_{i}}{\sin\theta_{i}}+w\log_{10}f_{Ghz}+g_{1}\\\
+g_{2}+g_{3}(1-\cos\theta_{i})^{2}+g_{4}d_{in,i}\end{split}$

We know that the altitude difference between the UAV and the location that has
the maximum path loss is constant for a given building. Now, when we take the
first derivative with respect to $\theta$ and assign it to zero, we get:

$\begin{split}\dfrac{dL(\theta)}{d\theta}=\dfrac{w}{ln10}\dfrac{\dfrac{-\Delta
h.\cos\theta}{\sin^{2}\theta}}{\dfrac{\Delta
h}{\sin\theta}}+2g_{3}\sin\theta(1-\cos\theta)=0~{}~{}~{}~{}~{}~{}~{}~{}\\\
\dfrac{dL(\theta)}{d\theta}=\dfrac{-w}{ln10}\dfrac{\cos\theta}{\sin\theta}+2g_{3}\sin\theta(1-\cos\theta)=0~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
\dfrac{w}{ln10}\cos\theta=2g_{3}sin^{2}\theta(1-\cos\theta)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
\dfrac{w}{ln10}\cos\theta=2g_{3}(1-\cos^{2}\theta)(1-\cos\theta)~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
2g_{3}\cos^{3}\theta-2g_{3}\cos^{2}\theta-(\dfrac{w}{ln10}+2g_{3})\cos\theta+2g_{3}=0~{}~{}~{}~{}~{}~{}\end{split}$
(1)

To prove that the function is convex, we take the second derivative and we
get:

$\begin{split}\dfrac{d^{2}L}{d\theta^{2}}=\dfrac{w}{ln10}\dfrac{1}{\sin^{2}\theta}+2g_{3}\cos\theta(1-\cos\theta)+2g_{3}\sin^{2}\theta>0~{}~{}~{}for~{}0<\theta\leq
90\end{split}$

Ecrf (1) has only one valid solution which is $\cos\theta$$=$0.6606.
Therefore, the optimal incident angle between the UAV and the location that
has the maximum path loss inside the building will be $48.654^{o}$.

In order to find the optimal horizontal distance between the UAV and the
building, we apply the pythagorean’s theorem. This gives us:

$\begin{split}d_{H}=((\dfrac{0.5z_{b}}{tan(48.654^{o})})^{2}-(0.5y_{b})^{2})^{0.5}\end{split}$

Therefore, the optimal horizontal distance between the UAV and the building is
given by:

$\begin{split}d_{opt}=((\dfrac{0.5z_{b}}{tan(48.654^{o})})^{2}-(0.5y_{b})^{2})^{0.5}-x_{b}\end{split}$

∎

In Figure 6, we demonstrate the transmit power required to cover the building
as a function of the incident angle, we notice that the optimal angle that we
characterize in Theorem 1 gives us the minimum transmit power.

Now, we find an efficient incident angle $\theta$ for high-SHF band that
minimizes the transmit power required to cover the building. In order to find
an efficient angle, we rewrite the equation that represents the path loss in
terms of the incident angle ($\theta$) and the altitude difference between the
UAV and location that has the maximum path loss inside the building ($\Delta
h$), we get:

$\begin{split}L(\Delta h,\theta)=(\alpha_{1}+\alpha_{2}\log_{10}\frac{\Delta
h}{\sin\theta}+\alpha_{3}\log_{10}f_{Ghz})+~{}~{}~{}\\\
(\beta_{1}+\frac{\beta_{2}-\beta_{1}}{1+exp(-\beta_{3}(\theta_{i}-\beta_{4}))})+(\gamma_{1}d_{in,i})\end{split}$

By numerically plotting the transmit power required to cover the location that
has the maximum path loss inside the building (see Figure 6 and Figure 6),
where $y_{b}$=50 meters and $x_{b}$=20 meters, we show that for different
building heights and different operating frequencies there exists only one
global minimum value. As can be seen from the figures, to provide wireless
coverage to small buildings, the UAV transmit power must be very high, due to
the high free space path loss, this demonstrates the need for multiple UAVs to
cover the high rise building when we use high-SHF operating frequency. To find
an efficient incident angle that could give us the global minimum value, we
use the ternary search algorithm. A ternary search algorithm is a method for
finding the minimum of a unimodal function, it iteratively splits the domain
into three separate regions and discards the one where the minimum does not
belong to. The pseudo code of this algorithm is shown in Algorithm 1. From our
numerical results, we found that the angle that minimizes the power is always
$15^{o}$. This is because the building penetration loss will be minimized at
this angle (see Figure 3). The angles less than $15^{o}$ will also give us
minimum building penetration loss but the free space path loss will increase
as the incident angle $\theta_{i}$ decreases. Note that for the high-SHF case
the incident angle that results in the minimum path loss is smaller than that
for low-SHF case. This is due to the fact that the building penetration loss
at high operating frequency will be higher than that at low operating
frequency.

Input:

The interval [$a$,$b$] of unimodal function that contains the efficient
incident angle.

The absolute precision $=$$\mu$.

If $|$$b$-$a$$|$ $<$ $\mu$:

Return $\frac{(a+b)}{2}$

$l$ $=$ $a$$+$$\frac{(b-a)}{3}$

$r$ $=$ $b$$-$$\frac{(b-a)}{3}$

If $f(l)$ $>$ $f(r)$

Return ternary$\\_$search($f$, $l$, $b$, $\mu$)

Else

Return ternary$\\_$search($f$, $a$, $r$, $\mu$)

Algorithm 1 Ternary search algorithm

Case 2. The locations of indoor users are symmetric across the $xy$ and $xz$
planes: In this case, we assume that the locations of indoor users are
symmetric across the $xy$-plane ((0,0,0.5$z_{b}$),($x_{b}$,0,0.5$z_{b}$)
,($x_{b}$,$y_{b}$,0.5$z_{b}$),(0,$y_{b}$,0.5$z_{b}$))) and the $xz$-plane
((0,0.5$y_{b}$,0), ($x_{b}$,0.5$y_{b}$,0),
($x_{b}$,0.5$y_{b}$,$z_{b}$),(0,0.5$y_{b}$,$z_{b}$)). First, we prove that
$z_{UAV}$=$0.5z_{b}$ and $y_{UAV}$=$0.5y_{b}$ when the locations of indoor
users are symmetric across the $xy$ and $xz$ planes and the operating
frequency is low-SHF (Theorem 2) or high-SHF (Theorem 3). Then we use the
gradient descent algorithm to find an efficient $x_{UAV}$ that minimizes the
transmit power required to cover the building.

######  Theorem 2.

For the low-SHF operating frequency case, when the locations of indoor users
are symmetric across the $xy$ and $xz$ planes, the optimal
($y_{UAV}$,$z_{UAV}$) that minimizes the power required to cover the indoor
users will be equal ($0.5y_{b}$,$0.5z_{b}$).

The proof is presented in Appendix A. The question now is how to find an
efficient horizontal point $x_{UAV}$ that minimizes the total transmit power.
In order to find this point, we use the gradient descent algorithm [18]:

$\begin{split}x_{UAV,n+1}=x_{UAV,n}-a\dfrac{dL_{Total}}{dx_{UAV,n}}\\\
\end{split}$

Where:

$\begin{split}\dfrac{dL_{Total}}{dx_{UAV}}=\sum_{i=1}^{M}\dfrac{w}{ln10}\dfrac{-(x_{i}-x_{UAV})}{d_{out,i}^{2}}+2g_{3}.(1-\dfrac{((x_{i}-x_{UAV})^{2}+(y_{i}-y_{UAV})^{2})^{0.5}}{d_{out,i}}).~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
(\dfrac{(x_{i}-x_{UAV})d_{out,i}((x_{i}-x_{UAV})^{2}+(y_{i}-y_{UAV})^{2})^{-0.5}}{{d_{out,i}^{2}}}-~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
\dfrac{((x_{i}-x_{UAV})^{2}+(y_{i}-y_{UAV})^{2})^{0.5}(x_{i}-x_{UAV})d_{out,i}^{-1}}{d_{out,i}^{2}})~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$

$a$: the step size.
$d_{out,i}$=$((x_{i}-x_{UAV})^{2}+(y_{i}-y_{UAV})^{2}+(z_{i}-z_{UAV})^{2})^{0.5}$

The pseudo code of this algorithm is shown in Algorithm 2. Now, we prove that
$z_{UAV}=0.5z_{b}$ and $y_{UAV}=0.5y_{b}$ when the locations of indoor users
are symmetric across the xy and xz planes and the operating frequency is high-
SHF.

Input:

The 3D locations of the users inside the building.

The step size $a$, the step tolerance $\epsilon$.

The dimensions of the building $[0,x_{b}]$ $\times$ $[0,y_{b}]$ $\times$
$[0,z_{b}]$.

The maximum number of iterations $N_{max}$.

Initialize $x_{UAV}$

For $n$=1,2,…, $N_{max}$

$x_{UAV,n+1}$ $\leftarrow$ $x_{UAV,n}$$-$ $a\dfrac{dL_{Total}}{dx_{UAV,n}}$

If $\lVert$ $x_{UAV,n}$ $-$ $x_{UAV,n+1}$ $\rVert$ $<$ $\epsilon$

Return: $x_{UAV,opt}$ $=$ $x_{UAV,n+1}$

End for

Algorithm 2 Efficient $x_{UAV}$ using gradient descent algorithm

######  Theorem 3.

For the high-SHF operating frequency case, when the locations of indoor users
are symmetric across the $xy$ and $xz$ planes, the optimal ($y_{UAV}$,
$z_{UAV}$) that minimizes the power required to cover the indoor users will be
equal ($0.5y_{b}$,$0.5z_{b}$).

The proof is presented in Appendix B. To find an efficient horizontal point
$x_{UAV}$ that minimizes the total transmit power, we use the gradient descent
algorithm, where:

$\begin{split}\dfrac{dL_{Total}}{dx_{UAV}}=\sum_{i=1}^{M}\dfrac{\alpha_{2}}{ln10}\dfrac{(x_{UAV}-x_{i})}{d_{out,i}^{2}}+(\dfrac{-(\beta_{2}-\beta_{1})(\frac{-\beta_{3}}{\sqrt{1-u^{2}}})(\frac{-(z_{UAV}-z_{i})(x_{UAV}-x_{i})}{d_{out,i}^{3}})}{(1+exp(-\beta_{3}(sin^{-1}u-\beta_{4})))}.~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
\dfrac{exp(-\beta_{3}(sin^{-1}u-\beta_{4}))}{(1+exp(-\beta_{3}(sin^{-1}u-\beta_{4})))})~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$

$d_{out,i}$=$((x_{i}-x_{UAV})^{2}+(y_{i}-y_{UAV})^{2}+(z_{i}-z_{UAV})^{2})^{0.5}$
$u$=$(\dfrac{(z_{UAV}-z_{i})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})^{0.5}})$

Case 3. The locations of indoor users are uniformly distributed in each floor:
In this case, we propose the Particle Swarm Optimization (PSO) [19] to find an
efficient 3D placement of the UAV, when the locations of indoor users are
uniformly distributed in each floor.

The particle swarm optimization algorithm starts with (npop) random solutions
and iteratively tries to improve the candidate solutions based on the best
experience of each candidate (particle(i).best.location) and the best global
experience (globalbest.location). In each iteration, the best location for
each particle (particle(i).best.location) and the best global location
(globalbest.location) are updated and the velocities and locations of the
particles are calculated based on them [9]. The velocity value indicates how
much the location can be changed (see ecrf (2)). The velocity is given by:

$\begin{split}particle(i).velocity=w*particle(i).velocity+~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
c_{1}*rand(varsize)*(particle(i).best.location-particle(i).location)\\\
+c_{2}*rand(varsize)*(globalbest.location-
particle(i).location)~{}~{}~{}~{}~{}\end{split}$

where $w$ is the inertia weight, $c_{1}$ and $c_{2}$ are the personal and
global learning coefficients, and $rand(varsize)$ are random positive numbers.
Also, the location of each particle is updated as:

$\begin{split}particle(i).location=particle(i).location+particle(i).velocity\end{split}$
(2)

The pseudo code of the PSO algorithm is shown in Algorithm 3. Convergence of
the candidate solutions has been investigated for PSO [20]. This analyses has
resulted in guidelines for selecting a set of coefficients
($\kappa$,$\phi_{1}$,$\phi_{2}$) that are believed to cause convergence to a
point and prevent divergence of the swarm’s particles. We selected our
parameters according to this analysis (see Table I and Algorithm 3).

Input:

The lower and upper bounds of decision variable (varmin,varmax), Construction
coefficients ($\kappa$,$\phi_{1}$,$\phi_{2}$), Maximum number of iterations
(maxit), Population size (npop)

Initialiaztion:

$\phi$=$\phi_{1}$+$\phi_{1}$, $\chi$ =
${2\kappa}/{|2-\phi-(\phi^{2}-4\phi)^{0.5}|}$

$w$=$\chi$, $c_{1}$=$\chi$$\phi_{1}$, $c_{2}$=$\chi$$\phi_{2}$,
globalbest.cost=inf

for i=1:npop

particle(i).location=unifrnd(varmin, varmax, varsize)

particle(i).velocity=zeros(varsize)

particle(i).cost=costfunction(particle(i).location)

particle(i).best.location=particle(i).location

particle(i).best.cost=particle(i).cost

if particle(i).best.cost $<$ globalbest.cost

globalbest=particle(i).best

end if

end

PSO Loop:

for t=1:maxit

for i=1:npop

particle(i).velocity=$w$*particle(i).velocity+

$c_{1}$*rand(varsize)*(particle(i).best.location- particle(i).location)+

$c_{2}$*rand(varsize)*(globalbest.location-particle(i).location)

particle(i).location=particle(i).location+ particle(i).velocity

particle(i).cost=costfunction(particle(i).location)

if particle(i).cost $<$ particle(i).best.cost

particle(i).best.location = particle(i).location

particle(i).best.cost = particle(i).cost

if particle(i).best.cost $<$ globalbest.cost

globalbest=particle(i).best

end if

end if

end

end

Algorithm 3 Efficient UAV placement using PSO algorithm

##  IV PROVIDING WIRELESS COVERAGE USING MULTIPLE UAVs

Providing wireless coverage to High-rise building using a single UAV can be
impractical, due to the limited transmit power of a UAV. The transmit power
required to cover the building is too high. It is in the range of 50dBm to
65dBm (see Figures 3, 5 and 6), which corresponds to 100-3000 watts.

Our problem can be formulated as:

$\begin{array}[]{ll}\displaystyle\min|k|\\\ \textrm{subject to}\\\
\displaystyle\sum_{j=1}^{|k|}y_{ij}=1~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\forall
i\in m&(3.a)\\\
\displaystyle\sum_{i=1}^{|m|}(2^{\frac{v.|m|}{B}}-1).N.L_{ij}.y_{ij}\leq
P~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\forall j\in k&(3.b)\\\ x_{min}\leq
x_{j}\leq
x_{max}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\forall
j\in k&(3.c)\\\ y_{min}\leq y_{j}\leq
y_{max}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\forall
j\in k&(3.d)\\\ z_{min}\leq z_{j}\leq
z_{max}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\forall
j\in k&(3.e)\\\ \end{array}$ (3)

where $k$ is a set of fully charged UAVs, $m$ is a set of indoor users,
$\upsilon$ is the rate requirement for each user (constant), $N$ is the noise
power (constant), $B$ is the transmission bandwidth (constant), $L_{ij}$ is
the total path loss between UAV $j$ and user $i$ and $P$ is the maximum
transmit power of UAV (constant). We also introduce the binary variable
$y_{ij}$ that takes the value of 1 if the indoor user $i$ is connected to the
UAV $j$ and equals 0 otherwise. The objective is to minimize the number of
UAVs that are needed to provide a wireless coverage for indoor users.
Constraint set (3.a) ensure that each indoor user should be connected to one
UAV. Constraint set (3.b) ensure that the total power consumed by a UAV should
not exceed its maximum power consumption limit. Constraints (3.c-3.e)
represent the minimum and maximum allowed values for $x_{j}$, $y_{j}$ and
$z_{j}$.

######  Theorem 4.

The problem represented by (3) is NP-complete.

###### Proof.

The number of constraints is polynomial in terms of the number of indoor
users, UAVs and 3D locations. Given any solution for our problem, we can check
the solution’s feasibility in polynomial time, then the problem is NP.
To prove that the problem is NP-hard, we reduce the Bin Packing Problem which
is NP-hard [21] to a special case of our problem. In the Bin Packing Problem,
we have a set of items $G=\\{1,2,..,N\\}$ in which each item has volume
$z_{n}$ where $n\in G$. All items must be packed into a finite number of bins
($b_{1}$, $b$,…,$b_{B}$), each of volume $V$ in a way that minimizes the
number of bins used. The reduction steps are: 1) The $b$-th bin in the Bin
Packing Problem is mapped to the $j$-th UAV in our problem, where the volume
$V$ for each bin is mapped to the maximum transmit power of the UAV $P$. 2)
The $n$-th item is mapped to the indoor $i$-th user, where the volume for each
item $n$ is mapped to the power required to cover the $i$-th indoor user. 3)
All UAVs have the same maximum transmit power $P$. 4) The power required to
cover the $i$-th indoor user from any 3D location will be constant. If there
exists a solution to the bin packing problem with cost $C$, then the selected
bins will represent the UAVs that are selected and the items in each bin will
represent the indoor users that the UAV must cover and the total cost of our
problem is $C$. ∎

Due to the intractability of the problem, we study clustering indoor users. In
the $k$-means clustering algorithm [22], we are given a set of points $m$, and
want to group the points into a $k$ clusters such that each point belongs to
the cluster with the nearest mean. The first step in the algorithm is to
choose the number of clusters $k$. Then, randomly initialize $k$ clusters
centroids. In each iteration, the algorithm will do two things:1) Cluster
assignment step. 2) Move centroids step. In cluster assignment step, the
algorithm goes through each point and chooses the closest centroids and
assigns the point to it. In move centroids step, the algorithm calculates the
average for each group and moves the centroids there. The algorithm will
repeat these two steps until it converges. The algorithm will converge when
the assignments no longer change. To find the minimum number of UAVs required
to cover the indoor users, we utilize this algorithm to cluster the indoor
users. In our algorithm, we assume that each cluster will be covered by only
one UAV. We start the algorithm with $k=2$ and after it finishes clustering
the indoor users, it applies the particle swarm optimization [19] to find the
UAV 3D location and UAV transmit power needed to cover each cluster. Then, it
checks if the maximum transmit power is sufficient to cover each cluster, if
not, the number of clusters $k$ is incremented by one and the problem is
solved again. The pseudo code of this algorithm is shown in Algorithm 4.

Figure 7: UAV optimal placement (upper part) and convergence speed of the GD
algorithm (lower part) for different building heights, $f_{c}=2G~{}Hz$ Figure
8: UAV optimal placement (upper part) and convergence speed of the GD
algorithm (lower part) for different building heights, $f_{c}=15G~{}Hz$

Input:

The maximum transmit power of UAV $(P)$.

The 3D locations of $m$ indoor users $(x_{i},y_{i},z_{i})$.

Number of clusters $(|k|=2)$.

START:

1: Initialize cluster centroids $\gamma_{1},\gamma_{2},...,\gamma_{k}\in
R^{n}$ randomly.

2: Repeat until convergence:

For every indoor user $i\in m$, set

$c^{(i)}=arg~{}\min\limits_{j\in k}||(x_{i},y_{i},z_{i})-\gamma_{j}||^{2}$

For each cluster $j\in k$, set

$\gamma_{j}=\frac{\displaystyle\sum_{i\in
m,c^{(i)}=j}(x_{i},y_{i},z_{i})}{\displaystyle\sum_{i\in m,c^{(i)}=j}1}$

3: Using particle swarm optimization algorithm, calculate the UAV efficient 3D
location and the transmit power for each cluster $j\in k$:

$P(j)=\displaystyle\sum_{i\in m,c^{(i)}=j}(2^{\frac{v.|m|}{B}}-1)\star N\star
L_{i}$

4: For $j=1~{}to~{}|k|$

If ($P(j)>P$)

$|k|=|k|+1$

go to START

End

Output:

$|k|$ Clusters.

The transmit Power of each UAV.

The 3D locations of UAVs.

Algorithm 4 Clustering Indoor Users

##  V Numerical Results

###  V-A Simulation results for single UAV

First, we verify our results for the second case, when the locations of indoor
users are symmetric across the $xy$ and $xz$ planes, using different operating
frequencies, 2GHz for low-SHF band and 15GHz for high-SHF. We assume that each
floor contains 20 users. Then we apply the gradient descent (GD) algorithm to
find the optimal horizontal point $x_{UAV}$ that minimizes the transmit power
required to cover the indoor users. Table I lists the parameters used in the
numerical analysis for single UAV cases.

TABLE I: Parameters in numerical analysis for single UAV Vertical width of building $y_{b}$ | 50 meters
---|---
Hight of each floor | 5 meters
Step size $a$ ”GD algorithm” | 0.01
Maximum number of iterations $N_{max}$ ”GD algorithm” | 500
The carrier frequency $f_{Ghz}$, low-SHF | 2Ghz
The carrier frequency $f_{Ghz}$, high-SHF | 15Ghz
Number of users in each floor | 20 users
(varmin,varmax) ”PSO algorithm” | (0,1000)
($\kappa$,$\phi_{1}$,$\phi_{2}$) ”PSO algorithm” | (1,2.05,2.05)
Figure 9: UAV optimal placement (upper part) and convergence speed of the GD algorithm (lower part) for different building widths, $f_{c}=2G~{}Hz$ Figure 10: UAV optimal placement (upper part) and convergence speed of the GD algorithm (lower part) for different building widths, $f_{c}=15G~{}Hz$ TABLE II: Simulation Results Algorithm | Distribution | Building height | Horizontal building | Vertical building | Efficient 3D placement | Efficient total
---|---|---|---|---|---|---
|  | $z_{b}$ |  width $x_{b}$ |  width $y_{b}$ |  | path loss(dB)
GD | symmetric | 200 | 20 | 50 | (20.025, 25, 100) | $7.8825*10^{4}$
PSO | symmetric | 200 | 20 | 50 | (20.040, 25.0130, 100.0015) | $7.8825*10^{4}$
GD | symmetric | 250 | 20 | 50 | (30.809, 25, 125) | $9.9971*10^{4}$
PSO | symmetric | 250 | 20 | 50 | (30.736 , 24.960, 124.956) | $9.9971*10^{4}$
GD | symmetric | 300 | 20 | 50 | (40.746, 25, 150) | $1.2146*10^{5}$
PSO | symmetric | 300 | 20 | 50 | (40.758, 25.048, 150.054) | $1.2146*10^{5}$
GD | uniform | 200 | 20 | 50 | (24.725, 25, 100) | $7.8853*10^{4}$
PSO | uniform | 200 | 20 | 50 | (21.799, 37.389, 111.790) | $7.8645*10^{4}$
GD | uniform | 250 | 20 | 50 | (33.818, 25, 125) | $9.9855*10^{4}$
PSO | uniform | 250 | 20 | 50 | (32.921, 28.712, 124.029) | $9.9725*10^{4}$
GD | uniform | 300 | 20 | 50 | (43.117, 25, 150) | $1.2154*10^{5}$
PSO | uniform | 300 | 20 | 50 | (46.589, 31.506 ,143.858) | $1.2117*10^{5}$
GD | uniform | 250 | 10 | 50 | (38.521, 25, 125) | $9.7413*10^{4}$
PSO | uniform | 250 | 10 | 50 | (32.104, 21.017, 129.266) | $9.7252*10^{4}$
GD | uniform | 250 | 30 | 50 | (29.393, 25, 125) | $1.0275*10^{5}$
PSO | uniform | 250 | 30 | 50 | (25.529, 4.938, 138.765) | $1.0211*10^{5}$
GD | uniform | 250 | 50 | 50 | (22.711, 25, 125) | $1.0753*10^{5}$
PSO | uniform | 250 | 50 | 50 | (14.548, 17.308 ,131.8940) | $1.0696*10^{5}$
Figure 11: UAV efficient placement (upper part) and convergence speed of the
PSO algorithm (lower part) for different building heights Figure 12: UAV
efficient placement (upper part) and convergence speed of the PSO algorithm
(lower part) for different building widths

In Figures 7 and 8, we find the optimal horizontal points for a building of
different heights. In the upper part of the figures, we find the total path
loss at different locations ($x_{UAV}$,0.5$y_{b}$,$z_{UAV}$) and the optimal
horizontal point $x_{UAV}$ that results in the minimum total path loss using
the GD algorithm. In the lower part of the figures, we show the convergence
speed of the GD algorithm. As can be seen from the figures, when the height of
the building increases, the optimal horizontal point $x_{UAV}$ increases. This
is to compensate the increased building penetration loss due to an increased
incident angle.

In Figures 9 and 10, we investigate the impact of different building widths
(i.e., $x_{b}$). We fix the building height to be 250 meters for low-SHF
operating frequency and 25 meters for high-SHF, then we vary the building
width. As can be seen from the figures, when the building width increases, the
optimal horizontal distance decreases. This is to compensate for the increased
indoor path loss due to an increased building width.

Now, we validate the simulation results for low-SHF operating frequency by
using the particle swarm optimization (PSO) algorithm and verify our result
for the third case, when the locations of indoor users are uniformly
distributed in each floor, using low-SHF operating frequency. As can be seen
from the simulation results in Table II, both algorithms converge to the same
3D placement, when the locations of indoor users are symmetric across the $xy$
and $xz$ planes.

After that, we assume that each floor contains 20 users and the locations of
these users are uniformly distributed in each floor. When we apply the GD
algorithm, the 3D efficient placements and the total costs for 200 meter, 250
meter and 300 meter buildings are (24.7254, 25, 100) ($7.8853*10^{4}$),
(33.8180, 25, 125) ($9.9855*10^{4}$) and (43.1170, 25, 150)($1.2154*10^{5}$),
respectively. UAV efficient placement and the convergence speed of the PSO
algorithm for different building heights is shown in Figure 11. The 3D
efficient placements and the total costs for 200 meter, 250 meter and 300
meter buildings are (21.7995, 37.3891, 111.7901) ($7.8645*10^{4}$), (32.9212,
28.7125, 124.0291) ($9.9725*10^{4}$) and (46.5898, 31.5061
,143.8588)($1.2117*10^{5}$), respectively. As can be seen from the simulation
results, the PSO algorithm provides better results. It provides total cost
less than the cost that the GD algorithm provides by (37dB-208dB). This is
because the PSO algorithm is designed for the case in which the locations of
indoor users are uniformly distributed in each floor. On the other hand, the
GD algorithm is designed for the case in which the locations of indoor users
are symmetric across the dimensions of each floor.

We also investigate the impact of different building widths (i.e., $x_{b}$)
using the GD and PSO algorithms (see Figure 12). We fix the building height to
be 250 meters and vary the building width. As can be seen from the simulation
results, the PSO algorithm provides better results. It provides total cost
less than the cost that the GD algorithm provides by (57dB-161dB).

We can notice that the tradeoff in case three is similar to that in case two,
when the height of the building increases, the efficient horizontal point
$x_{UAV}$ computed by our algorithm increases. This is to compensate the
increased building penetration loss due to an increased incident angle. Also,
when the building width increases, the efficient horizontal distance computed
by our algorithm decreases. This is to compensate the increased indoor path
loss due to an increased building width.

###  V-B Simulation results for multiple UAVs

In this section, we verify our results for multiple UAVs scenario. First, we
assume that a building will host a special event (such as concert, conference,
etc.), the dimensions of the building are $[0,20]\times[0,50]\times[0,100]$.
The organizers of the event reserve all floors higher than 75 meters and they
expect that 200 people will attend the event. Due to interference from near-by
macro cells, the organizers decide to use UAVs to provide wireless coverage to
the indoor users. We assume that 200 indoor users are uniformly distributed in
upper part of the building (higher than 75 meters) and 200 indoor users are
uniformly distributed in the lower part (less than 75 meters). Then, we apply
the clustering indoor users algorithm to find the minimum number of UAVs
required to cover the indoor users. Table III lists the parameters used in the
numerical analysis for multiple UAVs.

The algorithm starts with $k=2$ and after it finishes clustering the indoor
users, it applies the particle swarm optimization to find the UAV 3D location
and UAV transmit power needed to cover each cluster. Then, it checks if the
maximum transmit power is sufficient to cover each cluster, if not, the number
of clusters $k$ is incremented by one and the problem is solved again. As can
be seen from the simulation results in Figure 13, we need 5 UAVs to cover the
indoor users. We can notice that an efficient horizontal point $x_{UAV}$ for
all UAVs 3D locations is the same $x_{UAV}=25$, the minimum allowed value for
$x_{UAV}$, this is because the tradeoff (shown in Figure 3) disappears when a
UAV covers small height of building.

In Figure 14, we uniformly split the building into $k$ parts and cover it by
$k$ UAVs. As can be seen from the simulation results, we need 9 UAVs to cover
the indoor users. The clustering algorithm provides better results, this is
because it utilizes the distribution of indoor users to divide them into
clusters. On the other hand, the uniformly split method is designed for the
case in which the locations of indoor users uniformly distributed in the
building.

TABLE III: Parameters in numerical analysis for multiple UAVs Maximum transmit power of UAV ($P$) | 5 Watt
---|---
Operating frequency ($f$) | 2Ghz
Transmission bandwidth ($B$) | 50M Hz
Rate requirement for each user ($\upsilon$) | 2.2Mbps
Noise power ($N$) | -150 dBm
Min and Max allowed values for $x_{j}$,$[x_{min},x_{max}]$ | [25,1000]
Min and Max allowed values for $y_{j}$, $[y_{min},y_{max}]$ | [0,50]
Min and Max allowed values for $z_{j}$, $[z_{min},z_{max}]$ | [0,1000]
Figure 13: UAVs efficient placements using clustering algorithm Figure 14:
UAVs efficient placements using uniform split method

##  VI Conclusion

In this paper, we study the problem of providing wireless coverage for users
inside a high-rise building using UAVs. First, we demonstrate why the Air-to-
Ground path loss model is not appropriate for considering indoor users with 3D
locations. Then, we present Outdoor-to-Indoor path loss models, show the
tradeoff in these models, and study the problem of minimizing the transmit
power required to cover the building. Due to the intractability of the
problem, we study an efficient placement of a single UAV under three cases.
Due to the limited transmit power of a UAV, we formulate the problem of
minimizing the number of UAVs required to provide wireless coverage to high
rise building and prove that this problem is NP-complete. Due to the
intractability of the problem, we use clustering to minimize the number of
UAVs required to cover the indoor users. In order to model more realistic
scenarios, we will study the problem of providing wireless coverage for
multiple buildings in our future work.

##  Appendix A Proof of Theorem 2

Consider that $m_{1}$ represents the users that have altitude lower than the
UAV altitude and $m_{2}$ represents the users that have altitude higher than
the UAV altitude, then:

$\begin{split}d_{out,i}=((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})^{0.5},~{}~{}\forall
z_{UAV}>z_{i}\\\
d_{out,i}=((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{i}-z_{UAV})^{2})^{0.5},~{}~{}\forall
z_{UAV}<z_{i}\\\ \end{split}$

Also,

$\begin{split}cos_{\theta_{i}}=\dfrac{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})^{0.5}},~{}~{}\forall
z_{UAV}>z_{i}\\\
cos_{\theta_{i}}=\dfrac{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{i}-z_{UAV})^{2})^{0.5}},~{}~{}\forall
z_{UAV}<z_{i}\\\ \end{split}$

Rewrite the total path loss:

$\begin{split}L_{Total}=\sum_{i=1}^{m_{1}}(wlog_{10}(d_{out,i})+g_{3}(1-\cos\theta_{i})^{2})+\sum_{i=1}^{m_{2}}(wlog_{10}(d_{out,i})+g_{3}(1-\cos\theta_{i})^{2})+K\end{split}$

Where:

$\begin{split}K=\sum_{i=1}^{M}(wlog_{10}f_{Ghz}+g_{1}+g_{2}+g_{4}d_{in,i})~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$

Now, take the derivative with respect to $z_{UAV}$, we get:

$\begin{split}\dfrac{dL_{Total}}{dz_{UAV}}=\sum_{i=1}^{m_{1}}\dfrac{w}{ln10}\frac{(z_{UAV}-z_{i})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})}+~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
2g_{3}.(1-\dfrac{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})^{0.5}}).~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
(\dfrac{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}(z_{UAV}-z_{i})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})^{\frac{3}{2}}})+~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
\sum_{i=1}^{m_{2}}\dfrac{w}{ln10}\frac{-(z_{i}-z_{UAV})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{i}-z_{UAV})^{2})}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
+2g_{3}.(1-\dfrac{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{i}-z_{UAV})^{2})^{0.5}}).~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
(\dfrac{-((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}(z_{i}-z_{UAV})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{i}-z_{UAV})^{2})^{\frac{3}{2}}})~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$

Rewrite the $\dfrac{dL_{Total}}{dz_{UAV}}$ again, we have:

$\begin{split}\dfrac{dL_{Total}}{dz_{UAV}}=\sum_{i=1}^{m_{1}}\dfrac{w}{ln10}\frac{(z_{UAV}-z_{i})}{d_{out,i}^{2}}+2g_{3}.(1-\dfrac{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}}{d_{out,i}}).~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
(\dfrac{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}(z_{UAV}-z_{i})}{d_{out,i}^{3}})+\sum_{i=1}^{m_{2}}\dfrac{w}{ln10}\frac{-(z_{i}-z_{UAV})}{d_{out,i}^{2}}+~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
2g_{3}.(1-\dfrac{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}}{d_{out,i}}).(\dfrac{-((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2})^{0.5}(z{i}-z_{UAV})}{d_{out,i}^{3}})~{}~{}~{}\\\
\end{split}$

The equation above equals zero when the UAV altitude equals the half of the
building height, where the locations of indoor users are symmetric across the
$xy$ and $xz$ planes.

##  Appendix B Proof of Theorem 3

Consider that $m_{1}$ represents the users that have altitude lower than the
UAV altitude and $m_{2}$ represents the users that have altitude higher than
the UAV altitude, then:

$\begin{split}d_{out,i}=((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})^{0.5},~{}~{}~{}\forall
z_{UAV}>z_{i}\\\
d_{out,i}=((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{i}-z_{UAV})^{2})^{0.5},~{}~{}~{}\forall
z_{UAV}<z_{i}\\\ \end{split}$

Also,

$\begin{split}\theta_{i}=sin^{-1}(\dfrac{(z_{UAV}-z_{i})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})^{0.5}}),~{}~{}~{}\forall
z_{UAV}>z_{i}\\\
\theta_{i}=sin^{-1}(\dfrac{(z_{i}-z_{UAV})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{i}-z_{UAV})^{2})^{0.5}}),~{}~{}~{}\forall
z_{UAV}<z_{i}\\\ \end{split}$

Rewrite the total path loss:

$\begin{split}L_{Total}=\sum_{i=1}^{m_{1}}\alpha_{2}log_{10}(d_{out,i})+\dfrac{(\beta_{2}-\beta_{1})}{(1+exp(-\beta_{3}(sin^{-1}(u)-\beta_{4})))}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
+\sum_{i=1}^{m_{2}}\alpha_{2}log_{10}(d_{out,i})+\dfrac{(\beta_{2}-\beta_{1})}{(1+exp(-\beta_{3}(sin^{-1}(u)-\beta_{4})))}+K~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$

Where:

$\begin{split}u=(\dfrac{(z_{UAV}-z_{i})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})^{0.5}}),~{}~{}~{}\forall
z_{UAV}>z_{i}\\\
u=(\dfrac{(z_{i}-z_{UAV})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{i}-z_{UAV})^{2})^{0.5}}),~{}~{}~{}\forall
z_{UAV}<z_{i}\\\
K=\sum_{i=1}^{M}(\alpha_{1}+\alpha_{3}log_{10}f_{Ghz}+\beta_{1}+\gamma_{1}d_{in,i})~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
\end{split}$

Now, take the derivative with respect to $z_{UAV}$, we get:

$\begin{split}\dfrac{dL_{Total}}{dz_{UAV}}=\sum_{i=1}^{m_{1}}\dfrac{\alpha_{2}}{ln10}\frac{(z_{UAV}-z_{i})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{UAV}-z_{i})^{2})}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
+(\dfrac{-(\beta_{2}-\beta_{1})(\frac{-\beta_{3}}{\sqrt{1-u^{2}}})(\frac{d_{out,i}-(z_{UAV}-z_{i})^{2}d_{out,i}^{-1}}{d_{out,i}^{2}})}{(1+exp(-\beta_{3}(sin^{-1}u-\beta_{4})))}.\dfrac{exp(-\beta_{3}(sin^{-1}u-\beta_{4}))}{(1+exp(-\beta_{3}(sin^{-1}u-\beta_{4})))})+~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
\sum_{i=1}^{m_{2}}\dfrac{\alpha_{2}}{ln10}\frac{-(z_{i}-z_{UAV})}{((x_{UAV}-x_{i})^{2}+(y_{UAV}-y_{i})^{2}+(z_{i}-z_{UAV})^{2})}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\\\
+(\dfrac{-(\beta_{2}-\beta_{1})(\frac{-\beta_{3}}{\sqrt{1-u^{2}}})(\frac{-d_{out,i}+(z_{UAV}-z_{i})^{2}d_{out,i}^{-1}}{d_{out,i}^{2}})}{(1+exp(-\beta_{3}(sin^{-1}u-\beta_{4})))}.\dfrac{exp(-\beta_{3}(sin^{-1}u-\beta_{4}))}{(1+exp(-\beta_{3}(sin^{-1}u-\beta_{4})))})~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}\end{split}$

The equation above equals zero when the UAV altitude equals the half of the
building height, where the locations of indoor users are symmetric across the
$xy$ and $xz$ planes.

## References

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# Block Form of Frobenius Groups

Jiwen Zeng  School of Mathematic Science, Xiamen University, Xiamen, Fujian,
361005, China  and  Jiping Zhang  School of Mathematic Science, Peking
University, Beijing, 100871, China<EMAIL_ADDRESS>
###### Abstract.

The aim of this paper is to apply character properties of Frobenius group to a
local block form of an group algebra. We start by establishing a block form of
Brauer permutation Lemma by using block participation of conjugate classes of
a group $G$. Then we can define a pair of Frobenius corresponding blocks
between a group $G$ and its normal subgroup $N$. A near group condition is
given to determine a pair of Frobenius corresponding blocks. With a pair of
Frobenius corresponding blocks, we study its group structure. At last we prove
connections between nilpotent properties and Frobenius corresponding blocks.

###### Key words and phrases:

Frobenius group, block algebra, characters

###### 1991 Mathematics Subject Classification:

MSC 2020: 20C20; Secondary 20C15

* Corresponding author: Jiwen Zeng<EMAIL_ADDRESS>
## Introduction

Frobenius groups are well-known in the area of finite group study. There are
at least five equivalent conditions to define Frobenius groups. Hence we can
find a lot of generalization of Frobenius groups. If you generalize Frobenius
group in one way, you could not find the other ways to define the same
generalized Frobenius group. This is one of interests to study Frobenius
groups. Our interest is the generalization of Frobenius groups in character
theory. Related paper refer to [1][2][3][4].

Suppose $N$ is a normal subgroup of a group $G$. Let Irr$(N)$ denote the set
of all irreducible ordinary characters of $N$. If any
$\varphi\in\textrm{Irr}(N)$ is induced irreducibly to
$\varphi^{G}\in\textrm{Irr}(G)$, we say $G$ is Frobenius group and $N$ is the
kernel. When you consider an block algebra $b$ of $N$, naturally we would like
to consider local character property: is any $\varphi\in\textrm{Irr}(b)$
induced irreducibly to $G$?

As usual, a property held for a block algebra is said to be a local property
for group algebras. If any character $\varphi$ from a block algebra $b$ of
normal subgroup $N$ of $G$ is induced irreducibly to $G$, we say that the
property is local Frobenius property. According to the properties of Brauer
correspondence, the induced irreducible character $\varphi^{G}$ belongs to a
block $B=b^{G}$, which is called the Brauer correspondence of $b$. Or we say
$b^{G}$ is defined. Hence it is easily to define $(b,b^{G})$ as a pair of
Frobenius corresponding blocks.

If $G$ is a Frobenius group with kernel $N$, any pair $(b,b^{G})$ will be a
pair of Frobenius corresponding blocks for a block $b$ of $N$. If there exists
a pair of Frobenius corresponding blocks $(b,b^{G})$ for a group $G$ and its
normal subgroup $N$, how much can we say the group $G$ and its normal subgroup
$N$? In this paper, we are going to study the problem in three respects.

First, we extend Brauer permutation Lemma to a local block form by using block
participation of conjugate classes[6]. Then we can give a near group condition
to decide a pair of Frobenius corresponding blocks. Second, we will study the
structure of the group if it has a pair of Frobenius corresponding blocks.
Some results of lower defect groups[9] and Brauer correspondence [11][12] will
be used for the purpose of studying these groups with a pair of Frobenius
corresponding blocks. At last, we also obtain some results about nilpotent
properties of Frobenius corresponding blocks. These results partly answer some
questions in paper[13][14].

In section 5, We have examples to show that there exists non-Frobenius finite
group with a pair of Frobenius corresponding blocks. Hence we have a good
reason to define Frobenius corresponding blocks and study them.

In this paper, most notations and basic definitions can refer to [5]. Let $G$
be a finite group. Let Irr($G$) and IBr($G$) denote the set of irreducible
ordinary characters and irreducible Brauer characters of $G$, respectively.

The following notation and terminology will be used throughout in this paper.
$G$ is a finite group. $p>0$ is a prime number. $(K,R,F)$ is a splitting
$p$-module system, that means $R$ is a complete discrete valuation ring with a
maximal ideal $\pi$ such that $F=R/\pi$ is a field of characteristic $p$ and
$K$ is the quotient field of $R$. A $p$-regular element means an element of
$G$ whose order is prime to $p$. $G^{0}$ is the set of all $p$-regular
elements in $G$. Bl($G$) denotes the set of $p$-blocks of $G$. $B_{0}$ denotes
the principal block of $G$, containing the trivial character. We use LBr($G$)
to denote the set of all linear Brauer characters of $G$.

Our paper is mostly self-contained. Although some basic results are well-known
and could be found in literatures, we like to bring convenience to readers for
the proof. Our methods of proofs in this paper most come from modular
representation of group theory.

We have six sections. Basic results and notations appear in the first two
sections and they could be found in different literatures. Section 1 is Brauer
permutation Lemma and its proof. Section 2 is block partition of conjugate
classes of group $G$. In section 3 we will give block form of Brauer
permutation Lemma which localizes the global Brauer permutation Lemma for our
use in the following sections. Lemma 3.1 is our main result in Section 3. In
section 4, we will define a pair of Frobenius corresponding blocks and our
main result is Theorem 4.6, which proves that near group condition can decide
a pair of Frobenius corresponding blocks. Section 5 will give our main
results: Theorem 5.4 and 5.5 in this paper. Here we discuss the structure of
finite groups with a pair of Frobenius corresponding blocks. In the last
section, we study the connection between nilpotent properties and Frobenius
corresponding blocks.

##  1\. Brauer Permutation Lemma

In this section we state the well-known Brauer permutation Lemma and its
application to character theory. We start with a few definitions and introduce
notation and terminology that is consistent throughout this paper.

Given a set $\Omega=\\{1,2,...,n\\},S_{\Omega}$ is the permutation group of
$\Omega.$ Let a group $G$ act on the set $\Omega=\\{1,2,...,n\\}$ by
$\rho:x\rightarrow\rho(x)\in S_{\Omega}$. we denote by:

$\rho(x)=\left(\begin{array}[]{cccc}1&2&\cdots&n\\\
1^{x}&2^{x}&\cdots&n^{x}\\\ \end{array}\right)$

This defines a permutation matrix by:

$a^{x}_{ij}=\left\\{\begin{array}[]{cc}1&j=i^{x}\\\ 0&{\rm otherwise}\\\
\end{array}\right\\}$

The following Brauer permutation Lemma should appear in many literatures. Here
for the reader’s convenience, we give a proof by using modular representation
of group theory.

######  Lemma 1.1.

(Brauer Permutation Lemma) Suppose $\Omega=\\{1,2,...,n\\},$ and $G$ acts on
$\Omega$ by $\rho$ and $\sigma$, respectively. If there exists a matrix
$A=(a_{ij})\in{\rm GL}(K)$ such that $\rho(x)$ permutes the rows of $A$ and
$\sigma(x)$ permutes the columns of $A$ satisfying

$a_{i^{\rho(x)}j^{\sigma(x)}}=a_{ij}$

for all $x,i,j$, then the following holds:

  * •

for any $x\in G,$ the number of rows fixed by $\rho(x)$ is equal to that of
columns fixed by $\sigma(x)$

  * •

the number of $G$-orbits on the set of the rows of $A$ equals to that of
$G$-orbits on the set of the columns of $A$

###### Proof.

Consider $K\Omega$ as $G-$permutation module by actions $\rho(x)$ and
$\sigma(x)$, respectively. It suffices to prove they are isomorphic as
$G-$module. Take a map

$f:K\Omega\longrightarrow K\Omega$

by matrix $A$. Note $\Omega$ is a basis of $K\Omega$. Then $f$ is a one to one
map. To check $G$-homomorphism, we need to prove

$f(i^{\rho(x)})=f(i)^{\sigma(x)}$

Using the corresponding matrix to express:

$A\sigma(x)=\rho(x)A$

This is equivalent to the following:

$\begin{array}[]{c}\rho(x^{-1})A\sigma(x)=A\\\
\rho(x^{-1})=(\rho^{x^{-1}}_{ij}),\sigma(x)=(\sigma^{x}_{ij})\\\ \end{array}$

Note that

$\rho^{x^{-1}}_{ij}=1\Leftrightarrow j=i^{\rho(x^{-1})}\Leftrightarrow
i=j^{\rho(x)},\rho^{x^{-1}}_{j^{\rho(x)}j}=1$

It means $\rho(x^{-1})=(\rho^{x^{-1}}_{ij})$ acting on the rows on the
identity matrix by:

$i\rightarrow i^{\rho(x)},i=1,2,...,n$

Hence $\rho(x^{-1})A=(a_{i^{\rho(x)}j})$. Similarly, note

$\sigma^{x}_{ij}=1\Leftrightarrow
j=i^{\sigma(x)},\sigma^{x}_{ii^{\sigma(x)}}=1$

It means $\sigma(x)=(\sigma^{x}_{ij})$ acting on the columns on the identity
matrix by:

$j\rightarrow j^{\sigma(x)},j=1,2,...,n$

Hence $\rho(x^{-1})A\sigma(x)=(a_{i^{\rho(x)}j^{\sigma(x)}})$ and we have
$\rho(x^{-1})A\sigma(x)=A$ ∎

Let us denote the set of conjugate classes of $G$ by ${\rm Cl}(G$). If a group
$A$ acts on the group $G$ by automorphism, the group $A$ naturally acts on the
set ${\rm Irr}(G)$ by $\chi^{a}(x)=\chi(x^{a^{-1}}).$ The following result is
the well-known application of Brauer permutation Lemma to character theory.

######  Corollary 1.2.

Let a group $A$ acts the set ${\rm
Irr}(G)=\\{\chi_{1},\chi_{2},...,\chi_{n}\\}$ and the set ${\rm
Cl}(G)=C_{1},C_{2},...,C_{n}$ satisfying:

$\chi_{i}^{a}(x_{j}^{a})=\chi_{i}(x_{j}),x_{j}\in C_{j},a\in
A,i,j=1,2,...,n,x_{j}^{a}\in C_{j}^{a}$

Then the following holds:

  * •

for any $a\in A$, the number of fixed irreducible characters in ${\rm Irr}(G)$
equals that of fixed classes in ${\rm Cl}(G)$.

  * •

the number of $A$-orbits in ${\rm Irr}(G)$ equals that of $A$-orbits in ${\rm
Cl}(G)$.

###### Proof.

Note matrix $X=(\chi_{i}(x_{j}))$ is nonsingular. Consider the situations when
$A$ acts on the rows and columns of the matrix $X$. Then the conditions of
Brauer permutation Lemma are satisfied. ∎

##  2\.  Block Partition of conjugate classes of group $G$

In this section we state Brauer’s results[6] of block partition of conjugate
classes of a group $G$. Some related results refer to [7, 8, 9]. Although
their proof can be found in literature, we give a proof for the reader’s
convenience.

######  Theorem 2.1.

(Brauer)Let $V$ be a free $R$-module with a basis
$\Omega=\\{v_{1},v_{2},...,v_{n}\\}$. Given a direct sum decomposition:

(2.1) $V=W_{1}\bigoplus W_{2}\bigoplus\cdots\bigoplus W_{r}$

and a Projection $\Pi_{i}$ of $V$ on $W_{i},i=1,2,...,r$, there exists a
partition:

(2.2)
$\Omega=\bigcup_{i=1,...,r}\Omega_{i},\Omega_{i}\bigcap\Omega_{j}=\emptyset,i\neq
j.$

such that $\Pi_{i}(\Omega_{i})$ is a basis of $W_{i},i=1,2,...,r.$

###### Proof.

Let $\\{w^{j}_{1},w^{j}_{2},...,w^{j}_{n_{j}}\\}$ be a basis of
$W_{j},j=1,2,...,r.$ Write

$v_{i}=\sum_{j,k}\alpha_{i,(j,k)}w_{k}^{j},i=1,2,...,n.$

Then

$\Pi_{j}(v_{i})=\sum_{k=1,...,n_{j}}\alpha_{i,(j,k)}w_{k}^{j}$

Let $A=(\alpha_{i,(j,k)})$ be the $n\times n$ matrix with $i$ as the row index
and $(j,k)$ as the columns in the order of basis of $W_{1},W_{2},...,W_{r}.$

Since the first $n\times n_{1}$ sub-matrix of $A$ has a rank $n_{1}$, we can
rearrange the rows of $A$ to get a new matrix $A_{1}$, which has first
principal $n_{1}\times n_{1}$ minor as nonsingular. Then consider lower right
corner $(n-n_{1})\times(n-n_{1})$ sub-matrix $B$ of $A_{1}$. Proceeding as
above, we can get second principal $n_{2}\times n_{2}$ minor as nonsingular.
Continue the process above, we get a matrix which has principal minors:

$n_{1}\times n_{1},n_{2}\times n_{2},...,n_{r}\times n_{r}$

all of which are nonsingular. Corresponding the new matrix, we have a
partition of

$\Omega=\bigcup_{i=1,2,...,r}\Omega_{i},\Omega_{i}\bigcap\Omega_{j}=\emptyset,i\neq
j$

with $\Pi_{j}(\Omega_{j})$ is a basis of $W_{j},j=1,2,...,r.$ ∎

The partition of $\Omega$ as in the Theorem above is said to be a partition
associated with the decomposition of $V=\bigoplus_{i=1,...,r}W_{i}.$

######  Corollary 2.2.

With the same notations and conditions above, given a partition as Theorem
2.1, if $U$ is a $R$-submodule of $V$ with a basis $\Lambda$, satisfying:

  * •

$\Pi_{j}(U)\subseteq U,j=1,2,...,r$

  * •

$\Lambda\subseteq\Omega$

Then $\Lambda=\bigcup_{i=1,...,r}(\Lambda\bigcap\Omega_{i})$ is a partition
associated with the decomposition $U=\bigoplus_{i}\Pi_{i}(U)$. In particular,
${\rm Rank}_{R}\Pi_{i}(U)=|\Lambda\bigcap\Omega_{i}|$

The following results and notations is standard as usual(refer to [5, Chapter
5]).

Block decomposition:

let

$V=Z(RG)=\bigoplus_{B\in{\rm Bl}(G)}e_{B}Z(RG),W_{B}=e_{B}Z(RG).$

The projective map from $Z(RG)$ is:$Z(RG)\rightarrow e_{B}Z(RG)$

Basis of $Z(RG)$:

the set of conjugate class sums of

$\Omega=\\{\widehat{C}|C\in{{\rm Cl}}(G),\widehat{C}=\sum_{x\in C}x\\}$

Rank of $e_{B}Z(RG)$:

Note:$K$ is the factor field of $R$. In the $KG$, we have
$e_{B}=\sum_{\chi\in{{\rm Irr}}(B)}e_{\chi}$ and
Rank${}_{R}(e_{B}Z(RG))=$Dim${}_{K}(e_{B}Z(KG))=k(B)$.

######  Theorem 2.3.

[5, Chapter 5]There is a partition of ${{\rm Cl}}(G)$, the set of conjugate
classes of $G$,

(2.3) ${{\rm Cl}}(G)=\bigcup_{B\in{\rm
Bl}(G)}\Omega_{B},\Omega_{B}\bigcap\Omega_{B^{\prime}}=\emptyset,B\neq
B^{\prime}$

such that the following holds:

  * •

$\\{e_{B}\widehat{C}|C\in\Omega_{B}\\}$ is a $R$\- basis of $e_{B}Z(RG)$,
hence $k(B)=|\Omega_{B}|$

  * •

the value of $\nu$ at det$(\omega_{\chi}(\widehat{C}))_{\chi\in
B,C\in\Omega_{B}}$ is not large than any other values of $\nu$ at the
$k(B)\times k(B)$ minors of the $k(B)\times k$ matrix
det$(\omega_{\chi}(\widehat{C}))_{\chi\in B,C\in{{\rm Cl}}(G)},k={\rm Cl}(G)$.

###### Proof.

The first case is a clear result from Theorem 2.1.

For any $C\in{\rm Cl}(G)$, we have

$e_{B}\widehat{C}=\sum_{C_{i}\in\Omega_{B}}a_{CC_{i}}e_{B}\widehat{C_{i}},a_{CC_{i}}\in
R$

Hence for $\chi\in{\rm Irr}(B),C\in{\rm Cl}(G)$, there is

$\omega_{\chi}(\widehat{C})=\omega_{\chi}(e_{B}\widehat{C})=\sum_{C_{i}\in\Omega_{B}}a_{CC_{i}}\omega_{\chi}(e_{B}\widehat{C_{i}}),a_{CC_{i}}\in
R$

This is a matrix relation, (just taking $k(B)$ conjugate classes from
Cl$(G)$):

$(\omega_{\chi}(e_{B}\widehat{C}))=(\omega_{\chi}(e_{B}\widehat{C_{i}}))(a_{CC_{i}})^{T}$

by which we have the second assertion, as
$\omega_{\chi}(e_{B}\widehat{C})=\omega_{\chi}(\widehat{C}).$ ∎

Block partition of ${\rm Cl}(G)$:

The partition of ${\rm Cl}(G)$ in the Theorem 2.3 is called Block partition of
${\rm Cl}(G)$.

##  3\.  Block form of Brauer Permutation Lemma

In this section, we are going to localize the globe Brauer permutation Lemma
for our use in the following sections. The previous two sections allows us to
attain our purpose.

Suppose that a group $A$ acts on the group $G$. A block partition of Cl$(G)$
is given as in Theorem 2.3. Naturally the action of an element $a$ of $A$ on
$G$ produces an automorphism

$RG:\sum_{g\in G}a_{g}g\mapsto\sum_{g\in G}a_{g}g^{a}$

It is also clear that $Z(RG)=Z(RG)^{a}$, that is to say, the center of $Z(RG)$
is fixed under automorphism.

An action of an element $a$ of $A$ produces a permutation on the set Bl$(G)$
and block partition $\\{\Omega_{B}|B\in G\\}$. This is because

$\begin{array}[]{c}e_{B}\mapsto e_{B}^{a}\\\
\Omega_{B}\mapsto\Omega_{B}^{a}=\\{C^{a}|C\in\Omega_{B}\\}\end{array}$

Since $e_{B}\Omega_{B}$ is a basis of $e_{B}Z(RG),e_{B}^{a}\Omega_{B}^{a}$ is
a basis of $e_{B}^{a}Z(RG)$. Hence we say:

${\rm Cl}(G)=\bigcup_{B}\Omega_{B}^{a}$

is also a block partition of Cl$(G)$.

Suppose $\\{e_{B}^{a}|a\in A\\}$ is the orbit of $e_{B}$ under action of $A$
on $G$, then $\\{\Omega_{B}^{a}|a\in A\\}$ is also an orbit of $A$-action. Let
$A_{B}$ denote the fixed subgroup of $e_{B}$ in $A$ and $A/A_{B}$ is the coset
of $A_{B}$ in $A$. Then we have the following clear facts:

Center idempotent:

we denote $f_{B}=\sum_{a\in A/A_{B}}e_{B}^{a}$, which is an center idempotent
of $RG.$

Set of Irr$f_{B}$:

Irr$(f_{B})=\bigcup_{a\in A/A_{B}}{\rm Irr}(B^{a})$.

Set of conjugate classes corresponding to Irr$f_{B}$:

$\Omega_{f_{B}}=\bigcup_{a\in A/A_{B}}\Omega_{B}^{a}.$

By the block partition of Cl$(G),$ we know that

$f_{B}Z(RG)=\bigoplus_{a\in A/A_{B}}e_{B}^{a}Z(RG)$

and it has a basis

$\bigcup_{a\in
A/A_{B}}e_{B}^{a}\Omega_{B}^{a}=\\{e_{B}^{a}\widehat{C^{a}}|a\in
A/A_{B},C^{a}\in\Omega_{B}^{a}\\}$

Suppose $\chi\in{\rm Irr}(B)$. We have

$\chi^{a}(x^{a})=\chi^{a}(e_{B}^{a}x^{a})=\chi(e_{B}x)=\chi(x),x\in C$

since $\chi^{a}\in{\rm Irr}(B^{a}).$

$k(B)$:

denote the number of characters in Irr$(B)$ for a block $B$ in $G$.

In a word, we have a summary of all arguments above:

######  Theorem 3.1.

With the same notations and conditions above, suppose that a group $A$ acts on
the group $G$. For a block $B\in{{\rm Bl}}(G)$, $A$ naturally acts on the set
${\rm Irr}(f_{B})$ and the set $\Omega_{f_{B}}$ satisfying:

$\chi_{i}^{a}(x_{j}^{a})=\chi_{i}(x_{j}),x_{j}\in C_{j},a\in
A,i,j=1,2,...,m,x_{j}^{a}\in C_{j}^{a}$

where $m=\sum_{a\in A/A_{B}}k(B^{a})=|A/A_{B}|k(B)$. Then the following holds:

  * •

for any $a\in A$, the number of fixed irreducible characters in ${{\rm
Irr}}(f_{B})$ equals that of fixed classes in $\Omega_{f_{B}}$.

  * •

the number of $A$-orbits in ${{\rm Irr}}(f_{B})$ equals that of $A$-orbits in
$\Omega_{f_{B}}$.

In particular, if $B$ is fixed under $A$-action, we have

######  Corollary 3.2.

With the same notations and conditions above, suppose that a group $A$ acts on
the group $G$. For a block $B\in{{\rm Bl}}(G)$, if $B$ is fixed under
$A$-action, then $A$ naturally acts on the set Irr($B$) and the set
$\Omega_{B}$ satisfying:

$\chi_{i}^{a}(x_{j}^{a})=\chi_{i}(x_{j}),x_{j}\in C_{j},a\in
A,i,j=1,2,...,m,x_{j}^{a}\in C_{j}^{a}$

where $m=k(B)$, and the following holds:

  * •

for any $a\in A$, the number of fixed irreducible characters in ${\rm Irr}(B)$
equals that of fixed classes in $\Omega_{B}$.

  * •

the number of $A$-orbits in ${\rm Irr}(B)$ equals that of $A$-orbits in
$\Omega_{B}$.

##  4\.  Frobenius Corresponding Block

From this section in whole paper, we suppose $N$ is a normal subgroup of $G$.
Bl$(N)$ is the set of blocks of $N$. If $b$ is a block of $N$, we use $e_{b}$
to denote the block idempotent of $b$. Consider the group $G$ acts on its
normal subgroup $N$ by conjugation. Suppose we have block partition of
Cl$(N)$, denoted by

$\Omega=\bigcup_{b\in{\rm Bl}(N)}\Omega_{b}$

We keep to use the same notations as before. For a block $b$ of $N$, let

$T(b)=\\{x\in G|b^{x}=b\\}$

denote the inertial group of $b$. Then

Center idempotent :

$f_{b}=\sum_{x\in G/T(b)}e_{b}^{x},e_{b}^{x}=e_{b^{x}}$

Set of irreducible characters of $f_{b}$:

Irr$(f_{b})=\bigcup_{x\in G/T(b)}{\rm Irr}(b^{x})$

Set of classes of $f_{b}$ by block partition:

$\Omega_{f_{b}}=\bigcup_{x\in G/T(b)}\Omega_{b}^{x}=\Omega_{b^{x}}$

Note $f_{b}$ is also a center idempotent of $RG$ and it has a block idempotent
decomposition in $Z(RG)$:

$f_{b}=\sum_{B}e_{B}$

As usually the block $b$ is said to be covered by a block $B$ if $e_{B}$
appears in the sums above.

Now we give a definition of Frobenius corresponding block.

######  Definition 4.1.

With the same notations as above, suppose $N$ is a normal subgroup of $G$. For
$b\in{\rm Bl}(N)$ and $B\in{\rm Bl}(G)$, we use $(b,B)$ denote $b$ covered by
$B$. For any nontrivial $\chi\in{\rm Irr}(b)$, if induced $\chi^{G}$ as $G$
character is irreducible, we say $(b,B)$ is a pair of Frobenius corresponding
blocks, or $b$ is Frobenius block corresponding to $B$.

Note: Since $b$ is isomorphic to $b^{x},x\in G$ by $G$-conjugation, $(b,B)$ is
a pair of Frobenius corresponding blocks if and only if $(b^{x},B)$ is a pair
of Frobenius corresponding blocks.

For example, If $G$ is a Frobenius group with kernel $N$, then all $(b,B)$ is
a pair of Frobenius corresponding blocks.

What is the relation between Irr$(b)$ and Irr$(B)$ when $b$ is covered by $B$?
The following result should be well-known(Nagao’s book, Page P337-339, Lemma
5.4, Lemma 5.7, Lemma 5.8):

######  Lemma 4.2.

Let $B$ covers $b$. Then

  * •

$B$ covers $b$, if and only if: any $\chi\in{\rm Irr}(B)$, all irreducible
constituent of $\chi_{N}$ belong to some Irr$(b^{x})$.

  * •

Let $\chi\in{\rm Irr}(B)$. Then for each $x\in G/T(b)$, there exists
$\varphi\in{\rm Irr}(b^{x})$ as an irreducible constituent of $\chi_{N}$

  * •

If $\varphi\in{\rm Irr}(b)$, there is always $\chi\in{\rm Irr}(B)$ such that
$\varphi$ is an irreducible constituent of $\chi_{N}$

  * •

if $\varphi\in{\rm Irr}(b)$, there exists $\chi\in{\rm Irr}(B)$ as an
irreducible constituent of $\varphi^{G}$

  * •

$B$ covers $b$, if and only if any $\varphi\in{\rm Irr}(b)$, there exists
$\chi\in{\rm Irr}(B)$ as an irreducible constituent of $\varphi^{G}$

Consider block cover is correspondence from a set of conjugate blocks in $N$
to more blocks in $G$.

######  Corollary 4.3.

If $(b,B)$ is a pair of Frobenius corresponding blocks, then $B$ is only one
block to cover $b$.

###### Proof.

Since $\varphi\in{\rm Irr}(b)$ is induced to an irreducible character
$\varphi^{G}$, it belongs to a unique block $B$ of $G$. ∎

Note that $f_{b}$ is a primitive idempotent of $(RN)^{G}$. Hence we also have

######  Corollary 4.4.

If $(b,B)$ is a pair of Frobenius corresponding blocks, then the block
idempotent $e_{B}$ of $B$ belongs to $(RN)^{G}$.

###### Proof.

This is because

$e_{B}=f_{b}=\sum_{x\in G/T(b)}e_{b}^{x}$

∎

By Theorem 3.1, the block form of Brauer permutation Lemma, we have a
equivalent condition for a pair of Frobenius corresponding blocks.

Inertial group $T(\chi)=$:

$\\{x\in G|\chi^{x}=\chi\\},\chi\in{\rm Irr}(b)$

######  Theorem 4.5.

With the same notations as above, $(b,B)$ is a pair of Frobenius corresponding
blocks if and only if any $1\neq C\in\Omega_{b},C$ is only fixed by $N$ under
$G$-conjugate action.

###### Proof.

Note that for any $1\neq\chi\in{\rm Irr}(b),\chi^{G}$ is irreducible if and
only if its inertial group $T(\chi)$ is $N$. Hence by Theorem 3.1, we have the
assertion. ∎

By using block partition of the set Cl$(N)$ of conjugate classes of $N$, We
give a pure group condition for a pair of Frobenius corresponding blocks.

######  Theorem 4.6.

Suppose $N$ is normal subgroup of $G$ and $b$ is a block of $N$. Let
$\Omega_{b}$ be the set of these conjugate classes decided by block partition
of ${\rm Cl}(N)$. Then $(b,B)$ is a pair of Frobenius corresponding blocks if
and only if: for any $1\neq a\in C,C\in\Omega_{b},C_{G}(a)\leq N.$

###### Proof.

By Theorem 4.5, it is suffice to prove the condition is equivalent to: for any
$1\neq C\in\Omega_{b},C$ is only fixed by $N$ under $G$-conjugate action. We
denote the fixed-point group $C$ by $C_{G}(C).$ First we claim that:

$C_{G}(a)\leq C_{G}(C),a\in C$

Let $x\in C_{G}(a)$, then

$\begin{array}[]{c}a^{x}=a\\\ \Rightarrow\forall n\in
N,a^{nx}=a^{xn^{x}}=a^{n^{x}},n^{x}\in N\\\ \Rightarrow a^{nx}\in C\\\
\Rightarrow x\in C_{G}(C)\end{array}$

Our claim is proved. Hence

$C_{G}(C)\leq N\Rightarrow C_{G}(a)\leq N,\forall a\in C.$

Conversely, suppose $C_{G}(a)\leq N,a\in C$. We want to prove $C_{G}(C)\leq
N.$ By using contradictions, suppose $x\in C_{G}(C)-N.$ Since $a^{x}=a^{n}\in
C,n\in N$, we have

$\begin{array}[]{c}a^{xn^{-1}}=a\\\ \Rightarrow xn^{-1}\in N\\\ \Rightarrow
x=xn^{-1}n\in N\end{array}$

which is a contradiction. Hence $C_{G}(C)\leq N.$ ∎

##  5\.  Some properties of Frobenius corresponding blocks

Keep to use the same notations as before. Let $N$ be a normal subgroup of $G$.
We denote the principal block of $G$ by $B_{0}$: containing trivial character
$1_{R}.$ Similarly, $b_{0}$ is the principal block of $N$.

######  Lemma 5.1.

Suppose $(b_{0},B_{0})$ is a pair of Frobenius corresponding blocks. Then
${\rm Irr}(G/N)$ belongs to principal $B_{0}$.

###### Proof.

Take any $\chi\in{\rm Irr}(G/N)$ and suppose $\chi\in{\rm
Irr}(B^{\prime}),B^{\prime}\in{\rm Bl}(G).$ Since $\chi_{N}$ have an
irreducible constituent in common with $1_{R}$, then $B^{\prime}$ and $B_{0}$
cover the same block $b_{0}$(Refer to Nagao’s book, Page 340, Theorem 5.9). By
Corollary 4.3 $B_{0}$ is the unique block to cover $b_{0}$. Hence
$B^{\prime}=B_{0}.$ Therefore Irr$(G/N)$ belongs to $B_{0}.$ ∎

Noation: For a finite group $G$, let $\pi(G)$ denote the set of all prime
number dividing $|G|$. By Lemma above, we have

######  Corollary 5.2.

With the same notation as above, Let $N$ be a normal subgroup of $G$. If
$p-$principal block $(b_{0}(p),B_{0}(p))$ is a pair of Frobenius corresponding
blocks for all $p\in\pi(N)$, then

${\rm Irr}(G/N)\subseteq\bigcap_{p\in\pi(N)}{\rm Irr}B_{0}(p).$

Now we give a relation between $k(B)$ and $k(b)$, which denote the number of
irreducible characters of $B$ and $b$, respectively.

######  Theorem 5.3.

Suppose $(b,B)$ is a pair of Frobenius corresponding blocks. Then we have

  1. (1)

$k(B_{0})=k(G/N)+\frac{k(b_{0})-1}{|G/N|}$

for principal block.

  2. (2)

$k(B)=\frac{|N|k(b)}{|T(b)|}$

for non-principal blocks.

###### Proof.

1\. Note that $f_{b_{0}}=e_{b_{0}},\Omega_{f_{b_{0}}}=\Omega_{b_{0}}$ and

$|{\rm Irr}(f_{b_{0}})|=|{\rm Irr}(b_{0})|=k(b_{0})$

Take any $\chi\in{\rm Irr}(B_{0})$, which belongs to Irr$(G/N)$ or is induced
from Irr$(b_{0})-\\{1\\}$. The set Irr$(b_{0})-\\{1\\}$ is divided into orbits
under $G$-conjugate action. Each orbit has a length $|G/N|$. Hence we have our
assertion.

2\. First we know $B$ is the unique block to cover $b$ by Corollary 4.3. As we
proved in the Lemma 5.1, any $\chi\in{\rm Irr}(G/N)$ belongs to a block
$B^{\prime}$, which covers the same principal block $b_{0}$ of $N$ as
principal block $B_{0}$ of $G$. Hence Irr$(B)\bigcap$Irr$(G/N)=\emptyset.$
This tells us all irreducible characters of Irr$(B)$ are induced from
Irr$(\Omega_{f_{b}})$, which has $|G/T(b)|k(b)$ elements and divided into
$k(B)$ orbits of $|G/N|$ elements. ∎

Defect group:

For a block $B\in{\rm Bl}(G)$, we use $D(B)$ to denote the defect group of
$B$.

For the non-principal block $b\in{\rm Bl}(N)$, if $(b,B)$ is a pair of
Frobenius corresponding blocks, we have

######  Theorem 5.4.

For a pair of Frobenius corresponding blocks $(b,B)$, we have

  1. (1)

For a non-principal block, $D(B)\leq N$.

  2. (2)

$D(b)=_{H}D(B),|T(b):N|\not\equiv 0\;{\rm mod}p$

  3. (3)

For a pair of principal Frobenius corresponding block $(b,B)$, we have that
Sylow $p-$subgroup $P\leq N$ and $G/N$ is a $p^{\prime}$-group.

###### Proof.

1\. In Theorem 5.3, we have proved the fact: every irreducible characters of
$B$ is induced from $b$ for non-principal block. Hence every irreducible
module $V$ affording irreducible character $\chi$ in $B$ is $N$-projective. By
Green Theorem, we have $D(B)\leq N$.

2\. According to results of Fong[5, P345, Lemma 5.16] and part 1, the part 2
is easy to know.

3\. Now we consider the principal block $(b,B)$. Since $1\neq\chi\in{\rm
Irr}(b)$ is induced irreducibly to $\chi^{G}\in{\rm Irr}(B)$, we know that
$b^{G}=B$ is Brauer correspondence from $b$ to $B$. Hence $B$ is the regular
cover of $b$.

By the block participation of Cl$(N)$, we can find that
$\\{e_{b}\widehat{C_{0}}|C_{0}\in\Omega_{b}\\}$ is a basis of $e_{b}Z(FN)$
over $F$. Since $T(b)=G$ and $e_{b}=e_{B}$, we claim that

$\\{e_{B}\widehat{C}|C_{0}\in\Omega_{b}\bigcap C,C\in{\rm Cl}(G)\\}$

is the basis of $e_{B}(FN)^{G}$. First we have

$e_{B}(FN)^{G}\leq e_{B}(FN)^{N}=e_{B}Z(FN)=e_{b}Z(FN)$

which implies $e_{B}(FN)^{G}$ is generated by
$\\{e_{B}\widehat{C}|C_{0}\in\Omega_{b}\bigcap C,C\in{\rm Cl}(G)\\}$ over $F$.
Second, $\\{e_{B}\widehat{C}|C_{0}\in\Omega_{b}\bigcap C,C\in{\rm Cl}(G)\\}$
is linearly independent over $F$ by the fact that
$\\{e_{b}\widehat{C_{0}}|C_{0}\in\Omega_{b}\\}$ is a basis of $e_{b}Z(FN)$
over $F$. Our claim is proved.

Suppose we have a block participation of Cl$(G)$. Let
$\\{e_{B}\widehat{C}|,C\in\Omega_{B}\\}$ is the basis of $e_{B}Z(FG).$ Then we
have a direct sum:

$e_{B}Z(FG)=e_{B}(FN)^{G}\bigoplus e_{B}(F(G-N))^{G}$

The summands in the above equality have a basis
$\\{e_{B}\widehat{C}|C\in\Omega_{B}\bigcap N\\}$ and a basis
$\\{e_{B}\widehat{C}|C\in\Omega_{B}\bigcap(G-N)\\}$, respectively. Therefore
we have a basis

$\Delta=\\{e_{B}\widehat{C}|C_{0}\in\Omega_{b}\bigcap C,C\in{\rm
Cl}(G)\\}\bigcup\\{e_{B}\widehat{C}|C\in\Omega_{B}\bigcap(G-N)\\}$

for $e_{B}Z(FG)$ by our previous claim.

According to [9], the number of classes $C$ in $\Delta$ with $D(C)=D(B)$ is
equal to the number of major subsections of $B$. Then by [11][12], the number
of major subsections of $B$ is the number of the inertial group
$T(\overline{B})$-conjugate classes in $Z(D)$, where $\overline{B}$ is the
root of $B$ in $DC_{G}(D),D=D(B)$. From $D>1,$ it follows that the number is
bigger than $1.$ Hence we can find at least one $1\neq C$ in $\Delta$, such
that $D(C)=D(B).$

Now compute the linear $F-$representation $\omega^{*}_{B}:Z(FG)\rightarrow
F(\widehat{C}\rightarrow\omega(\widehat{C})^{*})$. Since $B$ is the principal
block, we have

$\omega(\widehat{C})^{*}=(\frac{|G|}{|C_{G}(x)|})^{*},x\in C$

where $*:R\rightarrow F=R/(\pi),p\in(\pi).$ Hence we have the following:

$\omega(\widehat{C})^{*}=0\Leftrightarrow D(C)<|G|_{p}=|D(B)|.$

Take $C\in\Delta.$ If $C\in G-N$, then $\omega(\widehat{C})^{*}=0$ by the fact
that $B$ is the regular cover of $b$. Therefore we can find that
$C\in\Delta,1\neq C_{0}\in\Omega_{b}\bigcap C$ such that
$\omega(\widehat{C})^{*}\neq 0.$ Hence there exist $x\in C_{0}$ such that
$D(B)\leq C_{G}(x)\leq N$ by Theorem 4.6.

∎

######  Theorem 5.5.

Suppose $N$ is a normal subgroup of $G$. For each prime number $p||N|$, if the
$p-$ principal block $(b_{0},B_{0})$ is a pair of Frobenius corresponding
block, then we have $G=NH,N\bigcap H=1.$

###### Proof.

From Theorem 5.4, $G/N$ is a $p^{\prime}-$group for each $p||N|$. Hence $N$ is
a normal Hall-subgroup of $G$. By the Schur-Zassenhaus theorem, we conclude
that there exists $H\leq G$ such that $G=NH,N\bigcap H=1.$ ∎

######  Example 5.6.

Take $G=A_{4}$, the subgroup of symmetric group $S_{4}$ on the set with $4$
elements. It has $12$ elements and has a normal subgroup

$K_{4}=\\{(1),(12)(34),(13)(24),(14)(23)\\}.$

Suppose $K_{4}$ has linear irreducible characters:
$1,\lambda_{1},\lambda_{2},\lambda_{3}.$ It is easy to check that $A_{4}$ has
three linear irreducible characters from $A_{4}/K_{4}:1,\delta,\delta^{2}$.
Another is $\chi=\lambda_{1}^{G}=\lambda_{2}^{G}=\lambda_{3}^{G}$, induced
from $K_{4}.$ Hence $A_{4}$ is a Frobenius group with $K_{4}$ as kernel.

When $p=2,$ both $A_{4}$ and $K_{4}$ have only one block, principal block:
$B_{0},b_{0}.$ Then we have

$k(B_{0})=k(G/N)+\frac{k(b_{0})-1}{|G/N|}=3+1=4$

When $p=3$, the kernel $N$ has $4$ blocks:

${{\rm Irr}}(b_{0})=\\{1\\},{{\rm
Irr}}(b_{i})=\\{\lambda_{i}\\},T(b_{i})=N,i=1,2,3.$

Hence $A_{4}$ has $2$ blocks:

${{\rm Irr}}(B_{0})=\\{1,\delta,\delta^{2}\\},{{\rm
Irr}}(B_{1})=\\{\chi=\lambda_{1}^{G}\\}$

satisfying

$k(B_{0})=k(G/N)=3,k(B_{1})=\frac{|N|k(b_{1})}{|T(b_{1})|}=1$

######  Example 5.7.

Take $G=S_{4}$, the symmetric group over $4$ elements. With the same notation
as Example 5.6, $A_{4}$ has one irreducible character $\chi$ of dimension $3$
and three linear characters:$1,\delta,\delta^{2}.$ Since $A_{4}$ is the
commutator subgroup of $S_{4}$, there are two linear characters: $1,\rho$ in
Irr$(S_{4}/A_{4})\subseteq$ Irr$(S_{4})$. Consider inertial group $T(\chi)$ in
$G$, it must be equal to $S_{4}$, because of $|S_{4}/A_{4}|=2.$ If a character
$\varphi$ in $G$ covers $\chi$, it must be a extension of $\chi$ to $G$. Hence
we have two extensions: $\varphi,\varphi\rho,$ such that
$\varphi_{A_{4}}=\chi,(\varphi\rho)_{A_{4}}=\chi$. Considering the inertial
group $T(\delta)$ in $G,$ it must be $A_{4},$ as the linear character $\rho$
can not be its extension to $G$. Hence $\delta^{G}=(\delta^{2})^{G}=\zeta$ is
an irreducible character of $S_{4}$. Now we have

${\rm Irr}(S_{4})=\\{1,\rho,\zeta,\varphi,\varphi\rho,\\}$

with dimension $\\{1,1,2,3,3\\}.$

When $p=2$, the block in $H=A_{4}$ is covered by a unique block in $G=S_{4},$
as $G/H$ is a $2-$group. Since $H$ has only one $2$-block, the principal block
$b_{0}$, the principal $b_{0}$ is only covered by the principal block $B_{0}$
in $G$. Hence $G$ has only one $2$-block, the principal block $B_{0}.$ By
arguments above, the character $\chi$ in Irr$(b_{0})$ is not induced
irreducibly to $B_{0}$. Hence $(b_{0},B_{0})$ is not the Frobenius
corresponding block.

When $p=3,H=A_{4}$ has two blocks:

  * •

$b_{0}$ with irreducible characters $1,\delta,\delta^{2}$;

  * •

$b_{1}$ with irreducible characters $\chi$.

Since $\delta^{G}=\zeta$ is irreducible in $G$, $B_{0}$ is the unique block to
cover $b_{0}$. The irreducible characters in $G/H$ must belong to blocks to
cover principal block $b_{0}$, because their restriction to $H$ only has the
trivial character $1$ as their irreducible constituents [5, Page 340]. So
$\rho\in$ Irr$(B_{0})$. Both $\varphi$ and $\varphi\rho$ are irreducible
characters of dimension $3$, hence each of them constitutes a block of defect
zero in $G$, denoted by $B_{1}$ and $B_{2}$, respectively. Since $\chi$ is
only covered by its extension $\varphi,\varphi\rho$ in $G$, both block $B_{1}$
and $B_{2}$ are the cover block of $b_{1}$.

Therefore, $(b_{0},B_{0})$ is Frobenius corresponding block. Neither
$(b_{1},B_{1})$ nor $(b_{1},B_{2})$ is Frobenius corresponding block. We also
have

$k(B_{0})=k(G/H)+\frac{k(b_{0})-1}{|G/H|}=3$

##  6\. Frobenius corresponding block and nilpotent property

Since Frobenius group has a nilpotent kernel, we naturally need to study the
connection between Frobenius corresponding blocks and nilpotent properties.
According article [13][14], we will obtain some results in this section.

Notation: $\pi(G)$ denote the set of prime number dividing $|G|$. We state the
following result

######  Proposition 6.1.

For $p,q\in\pi(N)$, suppose that both $(b_{0}(p),B_{0}(p))$ and
$(b_{0}(q),B_{0}(q))$ are pairs of Frobenius corresponding blocks. Then

$\displaystyle{\rm Irr}(G/N)={\rm Irr}(B_{0}(p))\bigcap{\rm Irr}(B_{0}(q))$
$\displaystyle\Leftrightarrow\\{1\\}={\rm Irr}(b_{0}(p))\bigcap{\rm
Irr}(b_{0}(q))$

###### Proof.

First by Lemma 5.1, we know that

${\rm Irr}(G/N)\subseteq{\rm Irr}(B_{0}(p))\bigcap{\rm Irr}(B_{0}(q)).$

If there exists $1\neq\varphi\in{\rm Irr}(b_{0}(p))\bigcap{\rm
Irr}(b_{0}(q))$, then we have that

$\varphi^{G}\in{\rm Irr}(B_{0}(p))\bigcap{\rm Irr}(B_{0}(q))$

and $\varphi^{G}\not\in{\rm Irr}(G/N).$

Conversely, if there exists a $\chi\in{\rm Irr}(B_{0}(p))\bigcap{\rm
Irr}(B_{0}(q))$ and $\chi\not\in{\rm Irr}(G/N),$ then we have that
$1\neq\varphi\in{\rm Irr}(b_{0}(p))$ such that $\varphi^{G}=\chi$ and
$1\neq\psi\in{\rm Irr}(b_{0}(q))$ such that $\psi^{G}=\chi.$ Hence we have
$\psi^{g}=\varphi,g\in G$ by Clifford Theorem. It means that $\varphi\in{\rm
Irr}(b_{0}(q))^{g}$. but ${\rm Irr}(b_{0}(q))^{g}={\rm Irr}(b_{0}(q)).$ Hence
we get

$1\neq\varphi\in{\rm Irr}(b_{0}(p))\bigcap{\rm Irr}(b_{0}(q)).$

∎

In [13][14], authors define

$B_{0}(G)_{\pi}=\bigcap_{p\in\pi}\textrm{Irr}(B_{0}(p)),\pi\subseteq\pi(G)$

and denote $B_{0}(G)$ for $\pi=\pi(G).$ If $B_{0}(G)_{\pi}=\\{1_{G}\\}$, they
say that $\textrm{Irr}(G)$ is principally $\pi$-separated. If
$B_{0}(G)=\\{1_{G}\\}$, it is said that $\textrm{Irr}(G)$ is principally
separated. According to result in [13][14], we have the following result by
Proposition 6.1 above:

######  Theorem 6.2.

Suppose that $(b_{0}(p),B_{0}(p))$ is a pair of Frobenius corresponding blocks
for any $p\in\pi(N)$. Then $N$ is nilpotent if and only if

${\rm Irr}(G/N)={\rm Irr}(B_{0}(p))\bigcap{\rm Irr}(B_{0}(q)),p,q\in\pi(N)$

###### Proof.

This is because $N$ is nilpotent if and only if ${\rm Irr}(N)$ is principally
$\\{p,q\\}$-separated for any two different prime numbers $p,q\in\pi(N).$ Then
by Proposition 6.1, we have the result. ∎

Theorem 6.2 induces the following result easily:

######  Corollary 6.3.

With the same conditions as in Theorem 6.2, If $N$ is nilpotent, then

${\rm Irr}(G/N)=\bigcap_{p\in\pi(N)}{\rm Irr}(B_{0}(p))$

###### Proof.

This is because

$\textrm{Irr}(G/N)\subseteq\bigcap_{p\in\pi(N)}{\rm
Irr}(B_{0}(p))\subseteq\textrm{Irr}(B_{0}(p))\bigcap\textrm{Irr}(B_{0}(q))$

for any $p,q\in\pi(N).$ ∎

If $\textrm{Irr}(G)$ is not principal separated, one of questions in[14] is:
How big can $B_{0}(G)$ be?

The following result is related to their question. According to results in
[15, X, Theorem 1.5, P416] and [14], we are going to prove the following
result:

######  Theorem 6.4.

Let $N\unlhd G$ and $G/N$ is solvable. Suppose $(b_{0}(p),B_{0}(p))$ is a pair
of Frobenius corresponding blocks for all $p\in\pi(N)$. Let
$H=\prod_{p\in\pi(G/N)}O_{p^{\prime}}(G)$. Then

${\rm Irr}(G/H)=B_{0}(G).$

In particular, ${\rm Irr}(G)$ is principally separated if and only if
$G=\prod_{p\in\pi(G/N)}O_{p^{\prime}}(G).$

###### Proof.

First we claim a general case for any $\pi\leq\pi(G)$:

$\textrm{Irr}(G/H)=\bigcap_{p\in\pi}\textrm{Irr}(G/O_{p^{\prime}}(G)).$

Since $O_{p^{\prime}}(G)\leq H,$ we have

$\textrm{Irr}(G/H)\subseteq\bigcap_{p\in\pi}\textrm{Irr}(G/O_{p^{\prime}}(G)).$

If $\chi\in\bigcap_{p\in\pi}\textrm{Irr}(G/O_{p^{\prime}}(G))$, it means that:
$H\leq\textrm{ker}\chi$ and

$\bigcap_{p\in\pi}\textrm{Irr}(G/O_{p^{\prime}}(G))\subseteq\textrm{Irr}(G/H).$

Hence our claim is proved.

If $G$ is $p-$solvable, we have
$\textrm{Irr}(G/O_{p^{\prime}}(G))=\textrm{Irr}(B_{0}(p))$.

Now take $\pi=\pi(G/N)$. By arguments above and the given conditions, we have

${\rm Irr}(G/H)=\bigcap_{p\in\pi(G/N)}\textrm{Irr}(B_{0}(p)).$

Since $(b_{0}(p),B_{0}(p))$ is a pair of Frobenius corresponding blocks for
all $p\in\pi(N)$, we know that $N\unlhd H$ by Theorem 5.4. Hence
$\textrm{Irr}(G/H)\subseteq\textrm{Irr}(G/N)$.

By corollary 5.2, we have

$\textrm{Irr}(G/N)\subseteq\textrm{Irr}(B_{0}(p)),p\in\pi(N).$

Note

$\pi(G)=\pi(N)\bigcup\pi(G/N),\pi(N)\bigcap\pi(G/N)=\emptyset$

by Theorem 5.4. Then we have

$\displaystyle B_{0}(G)$
$\displaystyle=\bigcap_{p\in\pi(G)}\textrm{Irr}(B_{0}(p))$
$\displaystyle=\bigcap_{p\in\pi(G/N)}\textrm{Irr}(B_{0}(p))\bigcap_{p\in\pi(N)}\textrm{Irr}(B_{0}(p))$
$\displaystyle=\textrm{Irr}(G/H)\bigcap_{p\in\pi(N)}\textrm{Irr}(B_{0}(p))$
$\displaystyle=\textrm{Irr}(G/H)$

Our conclusion is proved. ∎

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