# Statistical Channel Modeling for Long-Range Ground-to-Air FSO Links

Hossein Safi, and Akbar Dargahi Hossein Safi is with the Department of
Electrical Engineering, Shahid Beheshti University G. C., 1983969411, Tehran,
Iran, (e-mail: h_safi@sbu.ac.ir). Akbar Dargahi is with the Department of
Electrical Engineering, Shahid Beheshti University G. C., 1983969411, Tehran,
Iran (e-mail<EMAIL_ADDRESS>
###### Abstract

To provide high data rate aerial links for 5G and beyond wireless networks,
the integration of free-space optical (FSO) communications and aerial
platforms has been recently suggested as a practical solution. To fully reap
the benefit of aerial-based FSO systems, in this paper, an analytical channel
model for a long-range ground-to-air FSO link under the assumption of plane
wave optical beam profile at the receiver is derived. Particularly, the model
includes the combined effects of transmitter divergence angle, random wobbling
of the receiver, jitter due to beam wander, attenuation loss, and atmospheric
turbulence. Furthermore, a closed-form expression for the outage probability
of the considered link is derived which makes it possible to evaluate the
performance of such systems. Numerical results are then provided to
corroborate the accuracy of the proposed analytical expressions, and to prove
the superiority of the proposed channel model over the previous models in
long-range aerial FSO links.

###### Index Terms:

Aerial platforms, beam profile, channel model, free-space optics, plane wave.

##  I Introduction

More recently, the idea of integrating free space optical (FSO) communications
with aerial platforms, i.e., balloons, airships, drones, etc. has been the
subject of many studies in academia and industry to establish high capacity
and easy-to-deploy low power consumption links [1, 2]. Such systems seem
capable of carrying out multiple tasks, especially they can be utilized as
aerial stations to enable a ubiquitous connectivity for the next generation of
wireless networks [3].

However, to fully obtain the benefits of airborne FSO links, the communication
channel should be explicitly characterized. There has recently been an
increasing amount of literature on this topic [4, 5, 6, 7]. For instance, a
theoretical model for a link between an unmanned aerial vehicle (UAV) and a
satellite is developed in [4] when the effects of Doppler frequency shift,
pointing error, and atmospheric turbulence are taken into account. The work in
[5] proposes a complex channel model for a short link airborne FSO system with
which the effect of transceivers’ wobbling (also termed vibrations) on the
link performance is investigated in detail. Meanwhile, for the considered
system model in [5], the authors further develop their results and derive a
more tractable channel model in [6]. Najafi et al., propose a statistical
model for the geometric loss of a UAV-based FSO link in [7], and also analyze
the performance of such link in terms of outage probability and ergodic rate
under different turbulence conditions in [8].

Nevertheless, these prior works mainly concern short-range aerial FSO links,
and thus, their proposed pointing error models have been developed under the
assumption of the Gaussian profile for the received optical beam. Since plane
wave and spherical wave models are more accurate optical wave models for
characterizing the beam profile in long-range FSO links [9, 10], these
proposed channel models, however, may not be directly applicable for long-
range FSO links. In addition, prior studies [5, 6, 7, 8, 11] have not been
able to account for all aspects of link parameters, e.g, the effect of beam
wander on the link performance has not been investigated.

To fill this gap, in this paper, we consider a long-range ground-to-air FSO
link111For example, a link to an aerial platform above the stratosphere layer
(more than 20 km above the ground level) and derive a new model for pointing
error under the assumption of plane wave for the optical beam profile at the
receiver. In addition to the pointing error that includes the effects of
receiver’s wobbling and beam wander, we take the effects of atmospheric
turbulence and attenuation loss into account and derive an analytical channel
model for the considered system. Moreover, we derive a closed-form expression
for the outage probability of the considered link under strong atmospheric
turbulence conditions. We report simulation results to corroborate the
accuracy of the derived analytical expressions. We then show that, unlike the
proposed model in [5], our proposed analytical-based model can achieve a
perfect match with the exact simulation-based results. The developed results
of this paper can thus be applied for studying and designing long-range FSO
links without resorting to time-consuming simulations.

The rest of this paper is organized as follows. In Section II, we present the
system model and the main assumptions considered in this paper. Next, in
Section III, we provide analytical formulations for the channel model and the
outage probability of the considered ground-to-air FSO link. We presents
numerical results to verify our analytical expressions and demonstrate the
need of an accurate channel model for our considered setup in Section IV.
Finally, Section V concludes the paper.

##  II System Model

The positions of the receiver and transmitter are determined by
$H_{Tx}=(0,0,0)$ and $H_{Rx}=(l_{x},l_{y},Z+l_{z})$, respectively. It is
assumed that the receiver node is hovering at a fixed altitude, and thus,
there is no Doppler spreading effect. The independent random variables (RVs)
$l_{x}$, $l_{y}$, and $l_{z}$, which are zero-mean Gaussian distributed having
variance $\sigma^{2}_{l}$, denote the random deviations due to the receiver
wobbling along the axes of coordinates [5, 12]. In comparison with the link
length $Z$, the variation of $l_{z}$ is negligible, thus the effect of
pointing error due to receiver wobbling around the $z$-axis can be reasonably
neglected. However, for long-range FSO links, the effect of optical beam
wander induced pointing errors should be taken into account [10]. Accordingly,
at the receiver aperture, the displacements of the optical beam because of
beam wander effects along the $x$ and $y$ coordinates, denoted by $b_{x}$ and
$b_{y}$, are RVs with Gaussian distribution having zero-mean and variance [4]

$\displaystyle\sigma_{b}^{2}=2.07\int_{h_{0}}^{H}C_{n}^{2}\left(h\right)\left(Z-h\right)^{2}w_{h}^{-\frac{1}{3}}dh,$
(1)

where $w_{h}=\frac{h\theta_{\textrm{div}}}{2}$, and $C_{n}^{2}\left(h\right)$
is the refractive-index structure parameter as a function of the altitude $h$.

Employing intensity modulation and direct detection (IM/DD) techniques, at the
$k$th symbol interval, the output photo-current can be written as

$\displaystyle r[k]=\eta\,h\,s[k]+n[k],$ (2)

where $\eta$, $h$, $s[k]$, and $n[k]$ denote photo-detector (PD) responsivity,
the channel gain, the transmitted symbol with average optical power $P_{t}$,
and the Gaussian noise having zero mean and variance $\sigma_{n}^{2}$,
respectively. Moreover, in an FSO system, the instantaneous electrical signal-
to-noise ratio (SNR) can be written as

$\displaystyle\gamma=\dfrac{\eta^{2}P_{t}^{2}h^{2}}{\sigma_{n}^{2}}.$ (3)

In practical FSO systems the coherence time of the channel seems quite long
relative to the bit time [13]. Thus, for such systems with slow fading
property, outage probability (i.e., the probability with which the
instantaneous SNR falls bellow a certain threshold $\gamma_{\textrm{th}}$) is
an appropriate parameter for evaluating the link performance [14]. Thereby,
the outage probability can be written as

$\displaystyle\mathbb{P}_{\textrm{out}}=\int_{0}^{\gamma_{\textrm{th}}}f_{\gamma}(\gamma)d\gamma=\int_{0}^{\frac{\sqrt{\gamma_{\textrm{th}}\sigma_{n}^{2}}}{\eta
P_{t}}}f_{h}(h)dh,$ (4)

where $f_{\gamma}(\gamma)$ and $f_{h}(h)$ denote the probability density
function (PDF) of $\gamma$ and $h$, respectively.

##  III  Channel Modeling

The channel gain $h$ can be written as

$\displaystyle h={h_{a}h_{t}h_{p}},$ (5)

where the parameters $h_{a}$, $h_{t}$, and $h_{p}$, denote the attenuation
loss, the atmospheric turbulence, and the effective pointing error induced
geometrical loss, respectively. In the sequel, we discuss about these channel
parameters in more detail.

###  III-1 Attenuation loss and Atmospheric Turbulence

The attenuation loss can be modeled by the Beers-Lambert law as
$h_{a}=\exp\left(-Z\xi\right)$, where $\xi$ denotes the attenuation
coefficient which is a function of the visibility [15]. The optical turbulence
can be modeled as Gamma-Gamma distribution which is the dominant statistical
fading model for FSO links because of its excellent agreement with measurement
data for weak to strong turbulence conditions[10]. Thus, the PDF of $h_{t}$
can be written as

$\displaystyle
f_{\rm{GG}}(h_{t})=\frac{2(\alpha\beta)^{\frac{\alpha+\beta}{2}}}{\Gamma(\alpha)\Gamma(\beta)}h_{t}^{\frac{\alpha+\beta}{2}-1}K_{\alpha-\beta}(2\sqrt{\alpha\beta
h_{t}})$ (6)

where $\Gamma(\cdot)$ denotes the Gamma function, and $K_{\delta}(\cdot)$
represents the modified Bessel function of the second kind of order $\delta$.
Furthermore, $\alpha$ and $\beta$ are the parameters related to the number of
large-scale and small-scale eddies, respectively.

Figure 1:  Beam footprint (red circle) at the receiver aperture (blue circle).
The center of the received beam is deviated from the center of receiver lens
due to beam wander effects and the random wobbling of hovering receiver node.

###  III-2 Effective Pointing Error

The receiver aperture with area $A_{r}=\pi r_{a}^{2}$ has been located at a
large distance $Z$ from the transmitter, i.e., $Z\gg r_{a}$. Thus, at the
receiver the phase difference over the optical wavefront can be reasonably
omitted and the paraboloidal phase factor of the optical beam approaches zero.
According to [16, (1.2.5) and (1.2.6)], when the optical transmitter radiates
all its power through the two-dimensional solid angle $\Omega_{s}$, the field
intensity at distance $Z$ is

$\displaystyle I(Z)=\frac{P_{t}}{\Omega_{s}Z^{2}}.$ (7)

Considering $\theta_{\textrm{div}}$ as the divergence angle of the transmitted
beam, the radius of received optical beam footprint at the receiver aperture
can be approximated as $w_{z}=\dfrac{\theta_{\textrm{div}}Z}{2}$, we also have
$\Omega_{s}=2\pi\left(1-\cos\left(\frac{\theta_{\textrm{div}}}{2}\right)\right)$.
Accordingly, when incident optical beam is orthogonal to the PD plane, the
amount of collected power by the receiver with a collecting aperture area of
$A_{r}=\pi r_{a}^{2}$ is

$\displaystyle
P_{r}=\frac{r_{a}^{2}P_{t}}{2\left(1-\cos(\frac{\theta_{\textrm{div}}}{2})\right)Z^{2}}.$
(8)

Let $s_{d}$ represents the separation distance between the center of optical
beam footprint and the center of the receiver aperture. As a result, the
effective pointing error as a function of $s_{d}$ is formulated as

$\displaystyle h_{p}\left(s_{d}\right)=$
$\displaystyle\frac{r_{a}^{2}\cos\left(\theta_{d}\right)u(w_{z}-r_{a}-s_{d})}{2\left(1-\cos\left(\frac{\theta_{div}}{2}\right)\right)Z^{2}}$
(9)
$\displaystyle+\frac{A_{i}(s_{d},\theta_{d})\left(u(w_{z}-r_{a}-s_{d})-u(w_{z}+r_{a}-s_{d})\right)}{2\pi\left(1-\cos\left(\frac{\theta_{div}}{2}\right)\right)Z^{2}},$

where $u(\cdot)$ is the well-known step function, and
$A_{i}\left(s_{d}\right)$ is the overlapping area between the receiver lens
and received beam when $w_{z}-r_{a}\leq s_{d}\leq w_{z}+r_{a}$ (i.e., circle-
circle intersection) and it can be obtained as

$\displaystyle
A_{i}\left(s_{d}\right)=-\frac{1}{2}\sqrt{\left(-s_{d}+r_{a}+w_{z}\right)\left(s_{d}+r_{a}-w_{z}\right)}$
(10)
$\displaystyle\times\sqrt{\left(s_{d}-r_{a}+w_{z}\right)\left(s_{d}+r_{a}+w_{z}\right)}$
$\displaystyle+r_{a}^{2}\cos^{-1}\left(\frac{s_{d}^{2}+r_{a}^{2}-w_{z}^{2}}{2s_{d}r_{a}}\right)+w_{z}^{2}\cos^{-1}\left(\frac{s_{d}^{2}+w_{z}^{2}-r_{a}^{2}}{2s_{d}w_{z}}\right).$

To find the PDF of the effective pointing error, $f_{h_{p}(h_{p})}$, we resort
to the following theorem.

Theorem 1. _The PDF of $h_{p}$ is analytically derived as (III-2), where
$\delta(\cdot)$ is the Dirac delta function and $\Pi(\cdot)$ is the gate
function defined as_

$\Pi(x)=\left\\{\begin{array}[]{rl}1&\leavevmode\nobreak\ \text{if
}\leavevmode\nobreak\ x<1\\\ 0&\leavevmode\nobreak\ \text{if
}\leavevmode\nobreak\ x>1\end{array}\right..$ (11)

_Moreover, in ( III-2), $\mathcal{C}_{3}=\mathcal{C}_{2}\left(\pi
w_{z}r_{a}-r_{a}^{2}+w_{z}^{2}\right)$,
$\mathcal{C}_{2}=\left(\frac{\mathcal{C}_{1}}{w_{z}r_{a}}\right)$,
$\mathcal{C}_{1}=\frac{r_{a}^{2}}{4\pi\left(1-\cos\left(\frac{\theta_{div}}{2}\right)\right)Z^{2}}$,
$h_{pg1}=\mathcal{C}_{1}\left(\pi-\frac{2r_{a}^{2}+2r_{a}w_{z}}{w_{z}r_{a}}\right)$,
and
$h_{pg2}=\mathcal{C}_{1}\left(\pi-\frac{2r_{a}^{2}-2r_{a}w_{z}}{w_{z}r_{a}}\right)$.
_

$\displaystyle f_{h_{p}}(h_{p})\simeq$
$\displaystyle\left[1-\exp\left(-\frac{(w_{z}-r_{a})^{2}}{2\sigma_{r}^{2}}\right)\right]\times\delta\left(h_{p}-2\pi\mathcal{C}_{1}\right)+\exp\left(-\frac{(w_{z}+r_{a})^{2}}{2\sigma_{r}^{2}}\right)\times\delta\left(h_{p}\right)$
$\displaystyle+\frac{1}{2\mathcal{C}_{2}\sigma_{r}^{2}}\exp\left(\frac{h_{p}-\mathcal{C}_{3}}{2\mathcal{C}_{2}\sigma_{r}^{2}}\right)\times\left[\Pi\left(\frac{h_{p}}{h_{pg2}}\right)-\Pi\left(\frac{h_{p}}{h_{pg1}}\right)\right].$
(12)

###### Proof:

We have $s_{d}=\sqrt{(l_{x}+b_{x})^{2}+(l_{y}+b_{y})^{2}}$. Since $l_{x}$ and
$l_{y}$ Gaussian RVs having zero mean and variance $\sigma_{l}^{2}$, the RV
$s_{d}$ can be modeled by a Rayleigh distribution shown as

$\displaystyle
f_{s_{d}}(s_{d})=\frac{s_{d}}{\sigma_{r}^{2}}\exp\left(-\frac{s_{d}^{2}}{2\sigma_{r}^{2}}\right),\leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ s_{d}\geq 0$
(13)

where $\sigma_{r}^{2}=\sigma_{l}^{2}+\sigma_{b}^{2}$. From (9), the PDF of
$h_{p}$ can be formulated as

$\displaystyle
f_{h_{p}}(h_{p})=f_{1h_{p}}(h_{p})\\!+\\!f_{2h_{p}}(h_{p})\\!+\\!f_{3h_{p}}(h_{p}),$
(14)

where $f_{1h_{p}}(h_{p})$, $f_{2h_{p}}(h_{p})$, and $f_{3h_{p}}(h_{p})$ are
the distribution of $h_{p}$ when $(s_{d}\leq w_{z}-r_{a})$, $(w_{z}-r_{a}\leq
s_{d}\leq w_{z}+r_{a})$, and $(s_{d}\geq w_{z}+r_{a})$, respectively. From (9)
and (14), $f_{1h_{p}}(h_{p})$ and $f_{3h_{p}}(h_{p})$ can be obtained
respectively as

$\displaystyle f_{1h_{p}}(h_{p})=$ $\displaystyle\leavevmode\nobreak\
\left(1-\exp\left(-\frac{(w_{z}-r_{a})^{2}}{2\sigma_{r}^{2}}\right)\right)$
$\displaystyle\times\delta\left(h_{p}-\frac{r_{a}^{2}}{2\left(1-\cos\left(\frac{\theta_{div}}{2}\right)\right)Z^{2}}\right),$
(15)

and

$\displaystyle
f_{3h_{p}}(h_{p})=\exp\left(-\frac{(w_{z}+r_{a})^{2}}{2\sigma_{r}^{2}}\right)\times\delta\left(h_{p}\right).$
(16)

Calculating $f_{2h_{p}}(h_{p})$ by substituting $A_{i}\left(s_{d}\right)$ from
(10) is mathematically cumbersome. In practical situations, $r_{a}$ is on the
order of cm (2-10 cm), and, as shown in the numerical result section, to
reduce the effect of geometrical spreading loss via adjusting the divergence
angle of transmitted beam, $w_{z}$ must be set on the order of several meters.
Under these conditions, eq. (10) can be approximated as

$\displaystyle A_{i}\left(s_{d}\right)\simeq
r_{a}^{2}\left(\frac{\pi}{2}-\frac{s_{d}^{2}+r_{a}^{2}-w_{z}^{2}}{2w_{z}r_{a}}\right).$
(17)

Substituting (17) in (9), for $w_{z}-r_{a}\leq s_{d}\leq w_{z}+r_{a}$, $h_{p}$
can be approximated as

$\displaystyle h_{p}\simeq$ $\displaystyle\leavevmode\nobreak\
\frac{r_{a}^{2}\left(\frac{\pi}{2}-\frac{s_{d}^{2}+r_{a}^{2}-w_{z}^{2}}{2w_{z}r_{a}}\right)}{2\pi\left(1-\cos\left(\frac{\theta_{div}}{2}\right)\right)Z^{2}}.$
(18)

Now, from (18) and (13) and after some manipulations, ${\rm
for}\leavevmode\nobreak\ h_{pg1}<h_{p}<h_{pg2}$ we have

$\displaystyle
f_{2h_{p}|\theta_{d}}(h_{p})\simeq\frac{1}{2\mathcal{C}_{2}\sigma_{r}^{2}}\exp\left(\frac{h_{p}-\mathcal{C}_{3}}{2\mathcal{C}_{2}\sigma_{r}^{2}}\right),$
(19)

where $\mathcal{C}_{3}=\mathcal{C}_{2}\left(\pi
w_{z}r_{a}-r_{a}^{2}+w_{z}^{2}\right)$,
$\mathcal{C}_{2}=\left(\frac{\mathcal{C}_{1}}{w_{z}r_{a}}\right)$,
$\mathcal{C}_{1}=\frac{r_{a}^{2}}{4\pi\left(1-\cos\left(\frac{\theta_{div}}{2}\right)\right)Z^{2}}$.
Subsequently, by substituting (III-2), (16) and (19) in (14),
$f_{h_{p}}(h_{p})$ is obtained as (III-2) where
$h_{pg1}=\mathcal{C}_{1}\left(\pi-\frac{2r_{a}^{2}+2r_{a}w_{z}}{w_{z}r_{a}}\right)$,
and
$h_{pg2}=\mathcal{C}_{1}\left(\pi-\frac{2r_{a}^{2}-2r_{a}w_{z}}{w_{z}r_{a}}\right)$.

∎

Finally, by taking the effects of atmospheric turbulence and attenuation loss
into account, the PDF of $h=h_{a}h_{t}h_{p}$ can be obtained as

$\displaystyle f_{h}(h)=$ $\displaystyle\int
f_{h|h_{t}}(h)f_{h_{t}}(h_{t})dh_{t}$ (20) $\displaystyle=$
$\displaystyle\int\frac{1}{h_{a}h_{t}}f_{h_{p}}\left(\dfrac{h}{h_{a}h_{t}}\right)f_{h_{t}}(h_{t})dh_{t}.$

Substituting (6) and (III-2) in (20) and after some algebra, the analytical
expressions of $f_{h}(h)$ is obtained in (III-2).

$\displaystyle f_{h}(h)=$ $\displaystyle\leavevmode\nobreak\
e^{-\frac{(w_{z}+r_{a})^{2}}{2\sigma_{r}^{2}}}\delta(h)+\frac{2}{\Gamma(\alpha)\Gamma(\beta)}\left(\frac{\alpha\beta}{2\mathcal{C}_{1}h_{a}}\right)^{\frac{\alpha+\beta}{2}}\left[1-e^{-\frac{(w_{z}-r_{a})^{2}}{2\sigma_{r}^{2}}}\right]h^{\frac{\alpha+\beta}{2}-1}k_{\alpha-\beta}\left(2\sqrt{\alpha\beta\left(\frac{h}{2\mathcal{C}_{1}h_{a}}\right)}\right)$
$\displaystyle+\frac{(\alpha\beta)^{\frac{\alpha+\beta}{2}}}{h_{a}\mathcal{C}_{2}\sigma_{r}^{2}\Gamma(\alpha)\Gamma(\beta)}\int_{\frac{h}{h_{a}h_{pg2}}}^{\frac{h}{h_{a}h_{pg1}}}\exp\left(\frac{\frac{h}{h_{a}h_{t}}-\mathcal{C}_{3}}{2\mathcal{C}_{2}\sigma_{r}^{2}}\right)h_{t}^{\frac{\alpha+\beta}{2}-2}k_{\alpha-\beta}(2\sqrt{\alpha\beta
h_{t}})dh_{t}.$ (21)

(a)

(b)

Figure 2: Channel distribution $f_{h}(h>0)$, for $r_{a}=5$ cm, $Z=$ 20 km, and
a) $\theta_{\textrm{div}}=0.1\leavevmode\nobreak\ \textrm{mrad}$ &
$w_{z}/r_{a}=20$ and, b) $\theta_{\textrm{div}}=0.2\leavevmode\nobreak\
\textrm{mrad}$ & $w_{z}/r_{a}=40$. Figure 3: Outage probability versus $P_{t}$
for $r_{a}=10\textrm{\leavevmode\nobreak\ cm}$, $\theta_{\textrm{div}}=0.2$
mrad, and different values of link length $Z$.

Finally, to derive a mathematically tractable closed-form expression for the
outage probability of the considered link, we employ the expansion of the
modified Bessel function. Accordingly, from [17], the modified Bessel function
$K_{\delta}(x)$ can be expanded as

$\displaystyle
K_{\delta}(x)\\!=\\!\dfrac{\pi}{2\sin(\pi\delta)}\\!\sum_{n=0}^{\infty}\\!\left[\\!\dfrac{(x/2)^{2n-\delta}}{\Gamma(n-\delta+1)n!}\\!-\\!\dfrac{(x/2)^{2n+\delta}}{\Gamma(n+\delta+1)n!}\\!\right]\\!.$
(22)

According to (III-2) and (22), and after some manipulations, the outage
probability can be written as

$\displaystyle\mathbb{P}_{\textrm{out}}=$
$\displaystyle\sum_{n=0}^{\infty}\dfrac{{\mathcal{C}_{5}}}{2\sin\big{(}\pi(\alpha-\beta)\big{)}}\left(\dfrac{h_{\textrm{th}}^{n+\beta}}{\mathcal{C}_{n1}}-\dfrac{h_{\textrm{th}}^{n+\alpha}}{\mathcal{C}_{n2}}\right),$
(23)

where $\mathcal{C}_{n1}=(n+\beta)\Gamma(n-\alpha+\beta+1)n!$,
$\mathcal{C}_{n2}=(n+\alpha)\Gamma(n+\alpha-\beta+1)n!$, and
$\mathcal{C}_{5}=\frac{2\pi}{\Gamma(\alpha)\Gamma(\beta)}\mathcal{C}_{4}^{\frac{\alpha+\beta+1}{2}}\left[1-e^{-\frac{(w_{z}-r_{a})^{2}}{2\sigma_{r}^{2}}}\right]$.

##  IV Simulation Results and Analysis

We verify the accuracy of the provided analytical expressions by performing
Monte Carlo simulations with over $50\times 10^{6}$ independent runs. For
simulations, we consider typical system parameters as follows. The optical
wavelength $\lambda=1550$ nm, transmit bit-rate 1 Gbps, PD responsibility
$\eta$ = 0.9, aperture radius $r_{a}$ = 5 cm, the link length $Z$ = 20 km,
$\alpha=4$ and $\beta=1.7$.

In Figs. 2a and 2b, we plot the channel distribution for the different values
of the optical beam-width $w_{z}$ at the receiver aperture. We also plot the
proposed channel model in [5] as a benchmark. According to these two figures,
we find that the accuracy of the derived analytical channel model depends on
the values of $w_{z}$ and, for sufficiently large values of $w_{z}$, a perfect
match between simulations and theory can be realized. For the considered long-
range FSO systems, $w_{z}$ holds the value of several meters, hence, our
derived analytical expression for the channel model of such links meets an
acceptable degree of accuracy. Accordingly, relevant performance evaluation
such as outage probability can be analytically carried out without performing
Monte-Carlo simulations. Moreover, from Figs. 2a and 2b, one can readily
observe that the proposed channel model in [5], which is based on Gaussian
beam profile for short-range UAV-based link, is not able to achieve acceptable
accuracy for long-range links, i.e., when plane wave regime is assumed for the
optical beam at the receiver.

Finally, the curves of outage probability versus $P_{t}$ under different
values of link length in Fig. 3. As expected, for the same transmit power,
$\mathbb{P}_{\textrm{out}}$ increases when the link length is increased.
Meanwhile, the results of this figure verify the accuracy of the derived
analytical expression for the outage probability that makes it easier
designing such FSO links without resorting to time-consuming simulations.

##  V CONCLUSION

A statistical channel model for a long-range ground-to-air FSO link under the
plane wave regime was proposed in this paper. Different link parameters, i.e.,
transmitter divergence angle, receiver’s wobbling, jitter due to beam wander,
attenuation loss, and atmospheric turbulence were included in the proposed
model. Subsequently, a closed-form expression for the outage probability was
derived to evaluate the performance of the system. Our analytical results,
have made it possible to study and design such FSO systems without resorting
to time-consuming simulations.

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# On finite groups whose power graph is claw-free

Pallabi Manna  Pallabi Manna, Harish Chandra Research Institute, Prayagraj -
211019, India<EMAIL_ADDRESS>,  Santanu Mandal  Santanu
Mandal, School of Computing Science and Engineering, VIT Bhopal University,
Bhopal-466114, India<EMAIL_ADDRESS>and  Andrea Lucchini  A.
Lucchini, Università di Padova, Dipartimento di Matematica “Tullio Levi-
Civita”, Via Trieste 63, 35121 Padova, Italy<EMAIL_ADDRESS>
###### Abstract.

A graph is called claw-free if it contains no induced subgraph isomorphic to
the complete bipartite graph $K_{1,3}$. The undirected power graph of a group
$G$ has vertices the elements of $G$, with an edge between $g_{1}$ and $g_{2}$
if one of the two cyclic subgroups $\langle g_{1}\rangle,\langle g_{2}\rangle$
is contained in the other. It is denoted by $P(G)$. The reduced power graph,
denoted by $P^{*}(G),$ is the subgraph of $P(G)$ induced by the non-identity
elements. The main purpose of this paper is to explore the finite groups whose
reduced power graph is claw-free. In particular we prove that if $P^{*}(G)$ is
claw-free, then either $G$ is solvable or $G$ is an almost simple group. In
the second case the socle of $G$ is isomorphic to $\mathrm{PSL}(2,q)$ for
suitable choices of $q$. Finally we prove that if $P^{*}(G)$ is claw-free,
then the order of $G$ is divisible by at most 5 different primes.

###### Key words and phrases:

Power graph, claw-free graph, nilpotent groups, solvable groups, simple
groups.

###### 1991 Mathematics Subject Classification:

05C25, 20D10, 20D15, 20E32

##  1\. Introduction

There are numerous significant and well-known families of graphs that are
likewise claw-free. This family includes line graphs, complement of triangle-
free graphs, the graph of the icosahedron, the comparability graphs, and so
on. The initial incentive to investigate the characteristics of claw-free
graphs seems to have emerged from Beineke’s characterization of line graphs in
[3]. However, it was primarily in the late 1970s and early 1980s that the
graph theory community began to pay attention to the class of claw-free
graphs. In this time frame, the matching number of these graphs was noticed,
and initial findings on Hamiltonian properties were validated. Still, perhaps
more significant were the findings that the independence number can be found
using a polynomial method and that Berge’s Perfect Graph Conjecture holds true
for the class of graphs that are free of claws.

In algebraic graph theory we enjoy fantastic applications of abstract algebra
in graph theory. In this context, numerous graph structures to algebraic
structures, such as groups, semigroups, rings, etc., were defined by various
academics. This type of research aids in the characterization of the resultant
graphs, the exploration of the algebraic structure from the isomorphic graph,
and the recognition of a relationship between the related graphs and algebraic
structures. Graph defined on groups has lot of collections such as the
generating graph, enhanced power graph, commuting graph, soluble graph, and so
on. From ancient to the latest collection, interested readers may follow the
articles [2, 6, 7, 8, 10, 25]. The power graph is one of them. The concept of
power graph was initially presented by Kelarev and Quinn in 2002 ([22]) as a
directed graph.

######  Definition 1.1.

The directed power graph of a group $G$, denoted by $\overrightarrow{P}(G)$,
is a graph whose vertices are the elements of the group $G$, and there is an
arc $u\rightarrow v$ ($u\neq v$) if and only if $v=u^{i}$, for some positive
integer $i$.

Later in 2009, Chakrabarty et al. defined the undirected power graph [11].

######  Definition 1.2.

The undirected power graph (or simply power graph) of a group $G$, denoted by
$P(G)$, is a graph whose vertices are the elements of the group $G$, in which
two vertices $u$ and $v$ are connected by an edge if and only if either
$u=v^{i}$ or $v=u^{j}$ for some positive integers $i$, $j$.

The reduced power graph, denoted by $P^{*}(G),$ is the graph
$P(G)\setminus\\{e\\}$, where $e$ is the identity of the group. In a variety
of literature sources, power graphs have investigated in several ways. We
refer to two survey publications [1, 24] for intriguing results regarding
power graphs.

In terms of forbidden induced subgraphs, several important classes of graphs
can be specified. Among other graph types, these classes include triangle-free
graphs, claw-free graphs, chordal graphs, cograph, split, and threshold
graphs. This topic was introduced for the different forbidden induced
subgraphs of power graphs in an earlier study [15, 26]. In [9], the authors
answered the question: “Which power graphs are cograph?”. Bera discussed the
line graph characterization of the power graph in [4].

In the power graph, the identity element is adjacent to the other vertices of
the graph. Therefore the condition ‘$P(G)$ is claw-free’ can occur only if $G$
is either a cyclic group of prime power order or a cyclic group of the form
$C_{p^{m}q}$, where $p,q$ are distinct primes and $m\geq 1$. However, the
problem gets more intriguing if we remove the identity from the power graph.

This paper addresses the problem: “Identify the groups whose reduced power
graph is claw-free”.

In Section 2 we introduce some preliminary considerations which allow, in
particular, to obtain some information on finite nilpotent groups whose
reduced power graph is claw-free. We prove that if $G$ is a $p$-group with $p$
odd and $P^{*}(G)$ is claw-free, then either $G$ is cyclic or $\exp(G)=p$. The
situation is more complicate when $p=2:$ if $G$ is a non-cyclic 2-group and
$P^{*}(G)$ is claw-free, then either $G$ is a dihedral group or $\exp(G)\leq
4$. Moreover a nilpotent group $G$ whose reduce power-graph is claw-free is
either a $p$-group or a cyclic group whose order is divisible by at most 2
different primes. Finally we prove that if $G$ is not nilpotent and $P^{*}(G)$
is claw-free, then either $G$ is indecomposable or there exist two distinct
primes $p$ and $q$ such that $G=H\times K$ where $H\cong C_{p}$ and $K$ is the
semidirect product of a cyclic $q$-group $Q$ and a cyclic $p$-subgroup $P$ of
$\operatorname{Aut}(Q).$

Our main result in the solvable case, proved in Section 3, is the following:

######  Theorem 1.3.

Let $G$ be a finite solvable group. If $P^{*}(G)$ is claw-free, then
$\pi(G)\leq 4,$ denoting by $\pi(G)$ the number of distinct prime divisors of
the order of $G.$

The previous result is best possible. For example let $C=\langle
x_{1},x_{2}\rangle,$ with $|x_{1}|=31,$ and $|x_{2}|=61;$ then $\langle
x_{1}\rangle$ has an automorphism $\alpha_{1}$ of order 15 and $\langle
x_{2}\rangle$ has an automorphism $\alpha_{2}$ of order 15. Let $\langle
y\rangle\cong{C_{15}}.$ Then we may define a semidirect product
$G=C\rtimes\langle y\rangle\cong C_{31\cdot 61}\rtimes C_{15},$ in which
$(x_{1}x_{2})^{y}=x_{1}^{\alpha_{1}}x_{2}^{\alpha_{2}}.$ It follows from
Proposition 3.4 that $P^{*}(G)$ is claw-free.

In Section 4 we investigate the structure of finite groups $G$ with claw-free
reduced power graph, when the solvability assumption is removed. Our main
result is:

######  Theorem 1.4.

Let $G$ be a finite group. If $G$ is not solvable and $P^{*}(G)$ is claw-free,
then $G$ is an almost simple group.

Section 5 contains a complete classification of finite simple groups whose
reduced power graph is claw-free. It turn out that if $G$ is a finite non-
abelian simple group and $P^{*}(G)$ is claw free, then
$G\cong\mathrm{PSL}(2,q)$. More precisely:

######  Theorem 1.5.

Let $G$ be a finite non-abelian simple groups. Then $P^{*}(G)$ is claw-free if
and only if $G\cong\mathrm{PSL}(2,q)$ and each of the two numbers $\frac{q\pm
1}{(2,q-1)}$ is either a prime-power or the product of a prime and a prime-
power.

Combining Theorems 1.4 and 1.5 we finally prove the following:

######  Theorem 1.6.

Let $G$ be a finite group. If $P^{*}(G)$ is claw-free, then $\pi(G)\leq 5.$

The bound given in the previous statement is best possible. For example the
order of $\mathrm{PSL}(2,64)$ is divisible by the five primes 2, 3, 5, 7, 13,
and $P^{*}(\mathrm{PSL}(2,64))$ is claw-free (see Theorem 5.10).

This paper uses standard notations. To learn the fundamental concepts and
symbols of group theory that are not covered here, the reader can consult any
of [20, 28].

##  2\. Some preliminary results

Assume that $P^{*}(G)$ contains a claw made by $3$ pendent vertices
$a_{1},a_{2},a_{3}$ and a central vertex $b$. For $1\leq i\leq 3,$ the
directed graph $\overrightarrow{P}(G)$ contains at least one of the two edges
$a_{i}\to b$ and $b\to a_{i}.$ Moreover it cannot be that
$\overrightarrow{P}(G)$ contains $a_{i_{1}}\to b$ and $b\to a_{i_{2}},$ for
$a_{i_{1}}\neq a_{i_{2}}$, otherwise it would also contain the edge
$a_{i_{1}}\to a_{i_{2}}.$ It follows that there are two possible types of
claws in $P^{*}(G):$

  1. (1)

the subgraph of $\overrightarrow{P}(G)$ induced by the four vertices
$a_{1},a_{2},a_{3}$ and $b$ consists of the three edges: $b\to a_{1},b\to
a_{2},b\to a_{3}.$ In this case we will say that $a_{1},a_{2},a_{3},b$ induces
a claw of first type and we will use the symbol $(b\to a_{1},a_{2},a_{3}).$

  2. (2)

the subgraph of $\overrightarrow{P}(G)$ induced by the four vertices
$a_{1},a_{2},a_{3}$ and $b$ consists of the three edges: $a_{1}\to b,a_{2}\to
b,a_{3}\to b.$ In this case we will say that $a_{1},a_{2},a_{3},b$ induces a
claw of second type and we will use the symbol $(a_{1},a_{2},a_{3}\to b).$

Notice that if $(b\to a_{1},a_{2},a_{3})$ is a claw of first type, then
$\langle a_{1}\rangle,\langle a_{2}\rangle,\langle a_{3}\rangle$ are
incomparable subgroups of the cyclic group $\langle b\rangle$: this implies
that $|b|$ cannot be a prime power, since the subgroup lattice of a cyclic
$p$-group is a chain. In particular if every non-trivial element of $G$ has
prime-power order, then $P^{*}(G)$ does not contain claws of first type.
Recall that a finite group is called an EPPO group if every non-identity
element is of prime-power order. EPPO groups were first introduced in 1950 by
Higman [18]; in 1960 Suzuki classified the simple EPPO groups [30] and a
complete classification can be found in [17]. Assume that $G$ is an EPPO
group. Since $P^{*}(G)$ cannot contain claws of the first type, we deduce that
$P^{*}(G)$ is claw-free if and only if $G$ does not contain three distinct
cyclic subgroups with non-trivial intersection. In particular $P^{*}(G)$ is
claw-free if all the non-trivial elements of $G$ have prime order (this occurs
if and only if $G$ has exponent $p$, or $G$ is a Frobenius group of order
$p^{a}q$ and the Sylow $p$-subgroup of $G$ has exponent $p$, or $G=A_{5}$
[12]).

######  Remark 2.1.

Let $g$ be a non-trivial element of a finite group $G.$

  1. (1)

If $g$ has ordine $p_{1}p_{2}p_{3}$ where $p_{1},p_{2},p_{3}$ are distinct
primes, then $(g\to g^{p_{1}},g^{p_{2}},g^{p_{3}})$ is a claw of first type in
$P^{*}(G).$

  2. (2)

If $g$ has ordine $p_{1}^{2}p_{2}^{2}$ where $p_{1},p_{2}$ are distinct
primes, then $(g\to g^{p_{1}^{2}},g^{p_{2}^{2}},g^{p_{1}p_{2}})$ is a claw of
first type in $P^{*}(G).$

######  Remark 2.2.

If $G=\langle g\rangle$ is a cyclic group of order $p^{a}q,$ where $p$ and $q$
are distinct primes, then $P^{*}(G)$ is claw-free.

###### Proof.

Assume that there exists a claw in $P^{*}(G)$ with pendants
$a_{1},a_{2},a_{3}.$ Since the subgroup lattice of $\langle g^{q}\rangle$ is a
chain, at most one of the three elements $a_{1},a_{2},a_{3}$ can belong to
$\langle g^{q}\rangle.$ So we may assume $a_{1}=g^{p^{n_{1}}}$ and
$a_{2}=g^{p^{n_{2}}}$ with $n_{1}\leq n_{2}$. But in this case $P^{*}(G)$
contains the edge $a_{1}\to a_{2}.$ ∎

######  Remark 2.3.

Let $p$ be an odd prime and $G=C_{p^{2}}\times C_{p}$ or let $p=2$ and
$G=C_{4}\times C_{2}\times C_{2}.$ Then $P^{*}(G)$ contains a claw of the
second type.

###### Proof.

The group $G$ contains an element $g$ of order $p^{2}$ and three pairwise
distinct subgroups $\langle a_{1}\rangle,\langle a_{2}\rangle,\langle
a_{3}\rangle$ of order $p$, not contained in $\langle g\rangle$. It can be
easily seen that $(a_{1}g,a_{2}g,a_{3}g\to g^{p})$ is a claw in $P^{*}(G).$ ∎

Notice that the first part of the previous statement is not true when $p=2.$
Indeed if $G=C_{4}\times C_{4}$, then every element of order 2 in $G$ is
contained in exactly two cyclic subgroups of order 4 and therefore $P^{*}(G)$
is a claw-free graph. As a consequence, $P^{*}(C_{4}\times C_{2})$ is also
claw-free. However, when $p=2,$ the following weaker version of Remark 2.3
holds.

######  Remark 2.4.

Let $G=C_{2}\times C_{8}.$ Then $P^{*}(G)$ contains a claw of the second type.

###### Proof.

If $a$ in an element of $G$ of order 8 and $b$ is an element of order 2 with
$b\neq a^{4},$ then $(a,ab,a^{2}b\to a^{4})$ is a claw in $P^{*}(G).$ ∎

######  Corollary 2.5.

Let $G$ be an abelian $p$-group. If $G$ is cyclic or has exponent $p,$ then
$P^{*}(G)$ is claw-free. Moreover, if $G$ is neither cyclic nor elementary
abelian, then $P^{*}(G)$ is claw-free if and only if $p=2$ and either $G\cong
C_{4}\times C_{2}$ or $G\cong C_{4}\times C_{4}$.

######  Proposition 2.6.

Let $p$ be an odd prime. If $G$ is a finite $p$-group, then $P^{*}(G)$ is
claw-free if and only if either $G$ is cyclic or $G$ has exponent $p.$

###### Proof.

Assume that $G$ is not cyclic and contains an element $g$ of order $p^{2}$ and
that $P^{*}(G)$ is claw-free. By the previous corollary, $G$ is non-abelian.
First assume that $G$ is not of maximal class. By [5, Exercise 5, §10], there
exists a non-cyclic abelian subgroup $H$ of $G$ that contains $g.$ By Remark
2.3, $P^{*}(H)$ contains a claw, so we get a contradiction. So we may assume
that $G$ is of maximal class. But then $G$ contains a normal subgroup $N$ with
$N\cong C_{p}\times C_{p}.$ Let $K=NC,$ with $C=\langle g\rangle$. It cannot
be $N\cap C=1$. Indeed in this case $(Z(K)\cap N)C$ is a non-cyclic abelian
group of exponent $p^{2}$ and its reduce power graph contains a claw. We
remains with the case when $K$ is a non-abelian $p$-group of exponent $p^{2}.$
But then $K\cong C_{p^{2}}\rtimes C_{p}=\langle
a,b|a^{p^{2}}=b^{p}=1,bab^{-1}=a^{p+1}\rangle$ and $P^{*}(K)$ contains the
claw $(a,ab,a^{2}b\to a^{p})$. ∎

It is possible to prove a weaker result when $p=2.$

######  Proposition 2.7.

Let $G$ be a non-cyclic $2$-group. If $P^{*}(G)$ is claw-free then either $G$
has exponent at most 4 or $G$ is a dihedral group.

###### Proof.

Assume that $G$ is a non-cyclic $2$ group containing an element $x$ of order
$8.$ By [5, Proposition 10.30] either $G$ is a dihedral group (in which case
it can be easily checked that $P^{*}(G)$ is claw-free) or one of the following
holds:

1) $G\cong{\rm{Q}}_{2^{n+1}}$ is a generalized quaternion group of order
$2^{n+1},$ with $n\geq 3.$ In this case $G$ contains a cyclic normal subgroup
$N$ of order $2^{n}$, a unique involution and all the elements in $G\setminus
N$ have order 4. Thus $P^{*}(G)$ contains a claw.

2) $G\cong{\rm{SD}}_{2^{n+1}}$ is a semidihedral group of order $2^{n+1},$
with $n\geq 3.$ In this case $G$ contains a subgroup $H$ isomorphic to
${\rm{Q}}_{2^{n}}$, so $P^{*}(H)$ contains a claw.

3) $G$ contains an elements $y$ of order $8$ such that $\langle
x\rangle\cap\langle y\rangle$ has order 4. Let $K=\langle x,y\rangle.$ Then
$K$ is a group of order 16, containing two different cyclic subgroups of order
8. There are two possibilities: $K\cong C_{8}\times C_{2}$ or $K\cong\langle
a,b\mid a^{8}=1,b^{2}=1,a^{b}=a^{5}\rangle.$ In the first case we already know
that $P^{*}(K)$ is not-claw free. In the second case, it can be easily checked
that $(a,ab,a^{2}b\to a^{4})$ is a claw in $P^{*}(K).$ ∎

We need the following lemma to obtain more information on the 2-groups of
exponent 4 whose reduced power graph is claw-free.

######  Lemma 2.8.

Suppose that $G$ is a finite 2-group of exponent 4. If $P^{*}(G)$ is claw-
free, then $G$ is $Q_{8}$-free.

###### Proof.

We prove by induction on $|G|$ that if $G$ has exponent 4 and a section $X/Y$
of $G$ is isomorphic to the quaternion group $Q_{8},$ then $P^{*}(G)$ contains
a claw. We may assume $Y\neq 1,$ otherwise $X\cong Q_{8}$ and $P^{*}(Y)$
contains a claw. Let $Z=\langle z\rangle$ be a subgroup of order 2 contained
in $Y\cap Z(X).$ By induction $P^{*}(X/Z)$ contains a claw
$(a_{1}Z,a_{2}Z,a_{3}Z\to bZ)$. For $1\leq i\leq 3,$ since
$4=|a_{i}Z|\leq|a_{i}|\leq 4,$ we have $|a_{i}|=4$, which implies in
particular $\langle a_{i}\rangle\cap Z=1.$ Moreover $b$ has order 2, $\langle
b\rangle Z\cong C_{2}\times C_{2}$ and $\\{a_{1}^{2},a_{2}^{2},a_{3}^{2}\\}\
\subseteq\\{b,bz\\}.$ Thus we may assume $a_{1}^{2}=a_{2}^{2}$, but then
$(a_{1},a_{2},a_{1}z\to a_{1}^{2})$ is a claw in $P^{*}(G).$ ∎

Combining the previous lemma with the classification of the quaternion-free
2-group (see [21] and [31]), we obtain the following result:

######  Proposition 2.9.

Let $G$ be a non-abelian 2-group of exponent 4. If $P^{*}(G)$ is claw-free,
then $G=N\langle x\rangle$, where $N$ is an abelian normal subgroup of $G$.
Moreover $|x|\leq 4$ and either $N$ is elementary abelian or $N\leq
C_{4}\times C_{4}.$

###### Proof.

Assume that $G$ is a non-abelian 2-group of exponent 4 and that $P^{*}(G)$ is
claw-free. By the previous lemma, $G$ is $Q_{8}$-free. By [21, Proposition
1.6], if $G$ were modular, then the condition ${\rm{exp}}(G)=4$ would imply
that $G$ is abelian. So $G$ is non-modular and one of the three conditions
$(a),(b),(c)$ described in [21, Theorem 1.7] occurs. We may exclude case (c),
since it implies the existence of an element in $G$ of order at least 16. In
cases (a) and (b) $G=N\langle x\rangle,$ where $N$ is a normal abelian
subgroup. Since $P^{*}(N)$ is claw-free, either $N\leq C_{4}\times C_{4}$ or
$N$ is an elementary abelian 2-group. ∎

Although it is difficult to give a complete description of the finite
$p$-groups whose reduced power graph is claw-free, it is easy to describe the
nilpotent groups that have this property and are not $p$-groups. For this
purpose, the following easy observation is needed.

######  Remark 2.10.

If $G\cong C_{p}\times C_{p}\times C_{q}$, with $p$ and $q$ different primes,
then $P^{*}(G)$ contains a claw of second type.

###### Proof.

The group $G$ contains three different subgroups $\langle a_{1}\rangle,\langle
a_{2}\rangle,\langle a_{3}\rangle$ of order $p$ and a subgroup $\langle
b\rangle$ of order $q$. It suffices to notice that $(a_{1}b,a_{2}b,a_{3}b\to
b)$ is a claw in $P^{*}(G).$ ∎

######  Proposition 2.11.

Suppose that $G$ is a finite nilpotent group and that $P^{*}(G)$ is claw-free.
Then either $G$ is a $p$-group or $G$ is cyclic. In the second case
$|G|=p^{n}q$ where $n\in\mathbb{N}$ and $p,q$ are distinct primes.

###### Proof.

If $\pi(G)\geq 3,$ then $\pi(\langle g\rangle)=3$ for some $g\in G,$ and it
follows from Remark 2.1 (1) that $P^{*}(G)$ contains a claw. Hence we may
assume that $G\cong P\times Q,$ where $P$ is a $p$-group and $Q$ is a
$q$-group. Suppose that $Q$ is not cyclic. Then either $Q$ contains a subgroup
isomorphic to $C_{q}\times C_{q}$ or $Q$ is a generalized quaternion group. It
can be easily seen that if $Q$ is a generalized quaternion group, then
$P^{*}(Q)$ contains a claw. On the other hand if $Q$ contains $C_{q}\times
C_{q}$ then $G$ contains $C_{q}\times C_{q}\times C_{p}$ and $P^{*}(G)$
contains a claw by the previous remark. So if $G$ is claw-free then $Q$ (and
$P$ for the same reason) is cyclic. Hence $G$ itself is cyclic and the
conclusion follows from Remark 2.1 (2). ∎

######  Proposition 2.12.

Let $G$ be a finite group, $g$ a non-trivial element of $G$ and $C=C_{G}(g).$
If $P^{*}(C)$ is claw-free then $\pi(C)\leq 2.$ Moreover if $\pi(C)=2,$ then
one of the Sylow subgroups of $C$ is cyclic and normal.

###### Proof.

It is not restrictive to assume $|g|=p,$ with $p$ a prime. Assume that $q$ is
a different prime divisor of $|C|$ and let $Q$ be a Sylow $p$-sybgroup of $C.$
Since $\langle g\rangle Q$ is nilpotent, by Proposition 2.11, $Q$ is cyclic.
Moreover $Q$ is normal in $C,$ otherwise $C$ would contain three different
Sylow $q$-subgroups, say $\langle y_{1}\rangle,\langle y_{2}\rangle,\langle
y_{3}\rangle$ and $P^{*}(C)$ would contain the claw $(gy_{1},gy_{2},gy_{3}\to
g).$ If $q_{1}$ and $q_{2}$ were two different prime divisors of $|C|,$ with
$p\notin\\{q_{1},q_{2}\\},$ then the $q_{1}$-Sylow subgroup and the
$q_{2}$-Sylow subgroup would both be normal and $C$ would contain an element
$y$ of order $q_{1}q_{2}.$ But then $gy$ would have order $pq_{1}q_{2}$ and
$C$ would contain a claw by Remark 2.1. ∎

######  Theorem 2.13.

Assume that $G$ is a finite non-nilpotent group that can be decomposed as
direct product of non-trivial factors. If $P^{*}(G)$ is claw free, then there
exist two distinct primes $p$ and $q$ such that $G=H\times K$ where $H\cong
C_{p}$ and $K$ is the semidirect product of a cyclic $q$-group $Q$ and a
cyclic $p$-subgroup $P$ of $\operatorname{Aut}(Q).$ Moreover either $P\cong
C_{p}$ or $p=2$ and $P\cong C_{4}.$

###### Proof.

Suppose that $G=H\times K,$ with $H\neq 1$ and $K$ not nilpotent. Then, for a
suitable prime divisor $p$ of $|K|,$ $K$ contains a Sylow $p$-subgroup, say
$P$, which is not normal in $K$.

We claim that $H$ is a $p$-group. Indeed, assume that $H$ contains an element
$\langle x\rangle$ of prime order $q,$ with $q\neq p.$ Then the reduced power
graph $\langle x\rangle\times P$ is claw-free, and by Proposition 2.11, $P$ is
cyclic. Since $P$ is not normal in $K$, $K$ contains three distict Sylow
$p$-subgroups $P_{1}=\langle y_{1}\rangle,P_{2}=\langle
y_{2}\rangle,P_{3}=\langle y_{3}\rangle.$ But then $P^{*}(G)$ contains the
claw $(xy_{1},xy_{2},xy_{3}\to x).$

Since $G$ is not nilpotent, there exists a prime $q\neq p$ dividing $|K|.$
Since $K$ centralizes the elements of $H$, it follows from the Proposition
2.12 and its proof that $q$ is the unique prime divisor of $|K|$ different
from $p$ and that the Sylow $q$-subgroup $Q$ of $G$ is cyclic and normal in
$C.$

Since $H\times Q$ is nilpotent and its reduced power graph is claw-free, $H$
is a cyclic $p$-group. Moreover we have proved that $K=QP$ with $P$ a
$p$-group and $Q$ a normal cyclic $q$-subgroup. We claim that $|H|=p.$
Otherwise $H$ contains an element $x$ of order $p^{2}$ and $K$ an element $y$
of order $p$ and an element $z$ of order $q$. In this case $(x,xy,x^{p}z\to
x^{p})$ is a claw in $P^{*}(G).$ We have also $C_{P}(Q)=1,$ otherwise $K$
contains an element of order $pq$ and $C_{p}\times C_{p}\times C_{q}$ is
isomorphic to a subgroup of $G.$ Thus $P\leq\operatorname{Aut}(Q)$ is cyclic.
Finally, since $P^{*}(H\times P)$ is claw-free, either $P\cong C_{p}$ or $p=2$
and $P\cong C_{4}.$ ∎

##  3\. Solvable groups

The aim of this section is to prove Theorem 1.3. A key ingredient in the proof
is the following property of Frobenius complements.

######  Lemma 3.1.

[27, Lemma 2.1] Suppose $X$ is a Frobenius complement. Then every Sylow
subgroup of $X$ is cyclic or generalized quaternion. If $X$ has an odd order,
then any two elements of prime order commute, and $X$ is metacyclic. If $X$
has an even order, then $X$ contains a unique involution.

The proof will also use the following remark.

######  Remark 3.2.

Let $x$ be an element of $G$ of prime order $p.$ If $C_{G}(x)$ contains three
cyclic subgroups $\langle a_{1}\rangle,\langle a_{2}\rangle,\langle
a_{3}\rangle$ of prime orders, say $p_{1},p_{2},p_{3},$ and
$p\notin\\{p_{1},p_{2},p_{3}\\}$, then $(a_{1}x,a_{2}x,a_{3}x\to x)$ is a claw
in $P^{*}(G).$

######  Lemma 3.3.

Let $G=P\rtimes Q$, where $P$ is a non-cyclic elementary abelian $p$-group and
$Q$ is a $q$-group, with $p$ and $q$ different primes. If $P^{*}(G)$ is claw-
free, then $C_{P}(y)=1$ for every $1\neq y\in Q.$

###### Proof.

Assume that, for some $1\neq y\in Q$, $C:=C_{P}(y)\neq 1.$ If $C$ is not
cyclic, then it contains at least $p+1$ subgroups of order $p$, but then, by
Remark 3.2, $P^{*}(G)$ contains a claw. So $C=\langle x\rangle$, with $|x|=p.$
Moreover $1=[x^{z},y^{z}]=[x,y^{z}]$ for every $z\in P$: this implies that
$|P:C|\leq 2,$ otherwise we have at least $3$ different conjugates of $\langle
y\rangle,$ all commuting with $x.$ If $p\neq 2,$ we would have $C=P=\langle
x\rangle,$ against the assumption that $P$ is not cyclic. We reach the same
conclusion if $p=2.$ Indeed if $C\neq P$, then $P\cong C_{2}\times C_{2}$ and
$y$ centralizes a subgroup of $C$ of order 2. This would imply that $y$
centralizes $P$ and $C=P,$ a contradiction. ∎

###### Proof of Theorem 1.3.

Suppose that $G$ is a minimal counterexample. Since $P^{*}(G)$ is claw-free,
$P^{*}(H)$ is claw-free for every subgroup $H$ of $G$. In particular
$\pi(H)\leq 4$ for every proper subgroup $H$ of $G$. Let $N$ be a maximal
normal subgroup of $G$. Then $|G:N|=p_{1}$ is a prime and $p_{1}$ does not
divide $N$, otherwise $\pi(G)=\pi(N)\leq 4.$ In particular $G=N\rtimes
C_{p_{1}}$ and $\pi(N)=4.$ Now let $N/M$ be a chief factor of $G.$ Then $N/M$
is a $p_{2}$-group for some prime $p_{2}.$ Moreover, if $P_{1}$ is a Sylow
$p_{1}$-subgroup of $G$, then $MP_{1}<G$ and therefore $\pi(MP_{1})\leq 4.$
Since $\pi(G)=\pi(MP_{1})\cup\\{p_{2}\\}$, we must have that
$p_{2}\notin\pi(MP_{1})$ and therefore $G=(M\rtimes P_{2})\rtimes P_{1},$
being $P_{2}$ a Sylow $p_{2}$-subgroup of $N.$ Repeating the argument, we may
conclude that $G=(((P_{5}\rtimes P_{4})\rtimes P_{3})\rtimes P_{2})\rtimes
P_{1}$ where $P_{i}$ is a Sylow $p_{i}$-subgroup of $G$ and its is elementary
abelian; more precisely $P_{i}$ is isomorphic to a chief factor of $G.$

We distinguish two cases:

1) There exists $1\neq x\in P_{5}$ such that $C_{G}(x)>P_{5}$. Then there
exists $y$ of order $q$ (with $q\in\\{p_{1},p_{2},p_{3},p_{4}\\}$) such that
$[x,y]=1.$ By Lemma 3.3, $P_{5}=\langle x\rangle.$ So
$X:=C_{G}(x)=C_{G}(P_{5})$ is a normal subgroup of $G$. By Remark 3.2, $X$
must contain a unique subgroup of order $q$, so $\langle y\rangle$ is
characteristic in $X$ (and therefore normal in $G$). But then be may assume
$P_{5}P_{4}=\langle xy\rangle.$ Now consider $Y=C_{G}(xy)$: it must be
$Y=\langle x,y\rangle$ (otherwise we would have in $G$ an element of order
divisible by 3 different primes). This $G/C_{G}(xy)\cong P_{3}P_{2}P_{1}$ is
abelian, a contradiction.

2) $H=P_{4}P_{3}P_{2}P_{1}$ acts fixed point freely on $P_{5}$. But then $H$
is a Frobenius complement, and we may apply Lemma 3.1. If $|H|$ is even, then
$H$ contains a unique involution and this involution commutes with all the
elements of $H$ of prime order In this case we may assume $p_{4}=2$. If $|H|$
is odd, then every two elements of $H$ of prime order commute. In any case,
$G$ would contain cyclic subgroups of order $p_{4}p_{3},$ $p_{4}p_{2},$
$p_{4}p_{1}$, containing the same subgroup of order $p_{4}.$ ∎

In order to see that the bound $\pi(G)\leq 4$ in Theorem 1.3 is best possible,
we analyze what happens if $|G|=p_{1}p_{2}p_{3}p_{4}$, with
$p_{1},p_{2},p_{3},p_{4}$ four distinct primes.

######  Proposition 3.4.

Suppose that the order of $G$ is the product of four distinct primes. Then
$P^{*}(G)$ is claw-free if and only if $G\cong C_{p_{1}p_{2}}\rtimes
C_{p_{3}p_{4}}$, where $p_{1},p_{2},p_{3},p_{4}$ are distinc primes, and the
centralizer in $G$ of every non-trivial element of $C_{p_{1}p_{2}}$ coincides
with $C_{p_{1}p_{2}}.$

###### Proof.

Since all the Sylow subgroups of $G$ are cyclic, $G$ is solvable (see [20,
Corollary 5.15]). Let $F$ be the Fitting subgroup of $G$ and assume that
$P^{*}(G)$ is claw-free. Then $F$ is cyclic, so, since $P^{*}(F)$ is claw-
free, we must have $\pi(F)\leq 2.$ Thus we may assume that either $|F|=p_{1}$
or $|F|=p_{1}p_{2}.$

First assume $|F|=p_{1}.$ By [28, 5.4.4], $C_{G}(F)=F$, so $G/F$ is isomorphic
to a subgroup of $\operatorname{Aut}(F)$. But $\operatorname{Aut}(C_{p_{1}})$
is cyclic, so $G/F$ is cyclic of order $p_{2}p_{3}p_{4}$. Hence $G$ contains
an element of order $p_{2}p_{3}p_{4}$, in contradiction with Remark 2.1 (1).

Assume now that $|F|=\langle x\rangle$ is cyclic of order $p_{1}p_{2}.$ Again
we use the fact that $C_{G}(F)=F$ to deduce that $G/F$ is an abelian group.
This implies that $G=F\rtimes K$, with $K\cong C_{p_{3}}C_{p_{4}}.$ Now let
$1\neq y\in F.$ By Proposition 2.12, $\pi(C_{G}(y))\leq 2$. Since $F\leq
C_{G}(y),$ we deduce $F=C_{G}(y).$

Conversely assume $C_{G}(P_{1})=C_{G}(P_{2})=F$. Then the nontrivial subgroups
of $G$ are precisely the following: $F,P_{1},P_{2},$ $p_{1}p_{2}$ subgroups of
order $p_{3}$, $p_{1}p_{2}$ subgroups of order $p_{4}$ and $p_{1}p_{2}$
subgroups of order $p_{3}p_{4}.$ In particular every subgroup of prime order
in $G$ is property contained in a unique cyclic subgroup of $G$ and therefore
$P^{*}(G)$ is claw-free. ∎

##  4\. Non solvable groups

Before to prove Theorem 1.4 we need some preliminary results. The following is
an immediate consequence of Proposition 2.12.

######  Proposition 4.1.

Let $G$ be a finite group. If $P^{*}(G)$ is claw-free and $\pi(G)\geq 3$, then
$Z(G)=1.$ In particular if $P^{*}(G)$ is claw-free and $G$ is not solvable,
then $Z(G)=1.$

######  Proposition 4.2.

If $P^{*}(G)$ is claw-free, then either $G$ is almost simple or
$F(G)=F^{*}(G)$, where $F(G)$ and $F^{*}(G)$ are the Fitting subgroup and the
generalized Fitting subgroup of $G$ respectively.

###### Proof.

Suppose $F^{*}(G)\neq F(G).$ Then $G$ contains a subnormal quasisimple
subgroup $X.$ By Proposition 4.1 $Z(X)=1$ and therefore $X$ is a non-abelian
simple group. Moreover $F^{*}(X)=XC,$ with $C=C_{F^{*}(G)}(X).$ It follows
from Theorem 2.13 that $C=1.$ So we conclude that $X=F^{*}(G)$ is a non-
abelian simple group and therefore $G$ is almost simple. ∎

###### Proof of Theorem 1.4.

Let $G$ be a minimal counterexample. In particular $G$ is not almost simple,
so by Proposition 4.2 $F^{*}(G)=F(G).$ We will use several times in the proof
that this implies $F(G)\neq 1$ and $C_{G}(F(G))=Z(F(G)).$

Let $R(G)$ be the solvable radical of $G$ and $Q$ a Sylow subgroup of $R(G).$
By the Frattini argument, $G=R(G)N_{G}(Q).$ Assume that $N_{G}(Q)<G.$ Then
$N_{G}(Q)$ is not solvable, and $P^{*}(N_{G}(Q))$ is claw-free, hence by the
minimality of $G$, it follows that $N_{G}(Q)$ is almost simple. But this is
not possible, since $Q$ is normal in $N_{G}(Q).$ Thus all the Sylow subgroups
of $R(G)$ are normal in $G$. But then $R(G)$ is nilpotent and therefore
$F(G)=R(G).$ If $F(G)$ is not a $p$-group, then by Proposition 2.11, $F(G)$ is
cyclic. Since $C_{G}(F(G))\leq F(G),$ we would have that
$G/F(G)\leq\operatorname{Aut}(F(G))$ is abelian. Thus $G$ is solvable, a
contradiction. So $F(G)$ is a $p$-group, for a suitable prime $p.$

Let $N/F(G)$ be a minimal normal subgroup of $G/F(G).$ Since $F(G)=R(G),$
$N/F(G)$ is non-solvable. If $N\neq G,$ then, by the minimality of $G$, $N$ is
almost simple, in contradiction with the fact that $F(G)$ is a non-trivial
nilpotent normal subgroup of $N$. Thus $G=N$ and $G/F(G)$ is isomorphic to a
non-abelian simple group $S.$

We claim that $p=2.$ Indeed if $p\neq 2,$ then a Sylow 2-subgroup $P$ of $G$
is isomorphic to a Sylow 2-subgroup of $S.$ In particular $P$ contains a
subgroup isomorphic to $C_{2}\times C_{2}.$ By Lemma 3.1, the action of $P$ on
$F(G)$ is not fixed-point-free, hence $F(G)P$ is not claw-free since it
contains a subgroup isomorphic to $C_{p}\times C_{2}\times C_{2}.$

Suppose that $F(G)$ contains an element $g$ of order 4. Then the subgroup $X$
generated by the elements of order 2 in $Z(F(G))$ has order at most 4
(otherwise $C_{F}(g)$ would contain a subgroup isomorphic to $C_{4}\times
C_{2}\times C_{2}))$. If follows from Proposition 4.1, that $C_{G}(X)=F$. But
then $G/F$ is isomorphic to a subgroup of $\operatorname{Aut}(X),$ which is
solvable, a contradiction. So $F$ is an elementary abelian 2-group.

Let $P$ be a Sylow 2-subgroup of $G$. Notice that $P$ is non-abelian,
otherwise $P\leq C_{G}(F(G))=F(G),$ in contradiction with the fact that
$P/F(G)$ is isomorphic to a Sylow 2-subgroup of a non-abelian simple group.
Moreover $P$ cannot be a dihedral group; otherwise $F$ would be a normal
subgroup of index at least 4 of a dihedral group, and therefore $F$ would be
cyclic and $G/F\leq\operatorname{Aut}(F)$ would be abelian. It follows from
Proposition 2.7 that $P$ has exponent 4, so we may apply Proposition 2.9 to
deduce that $P=N\langle x\rangle,$ with $N$ an abelian normal subgroup of $P.$
Morevor, since $P/F(G)$ is isomorphic to a Sylow 2-subgroup of a non-abelian
simple group, $P/F(G)$ is not cyclic.

First assume that $N\leq C_{4}\times C_{4}.$ This implies $F(G)\leq
C_{2}\times C_{2}\times C_{2}.$ Since $G/F(G)$ is a non-abelian simple group,
the unique possibily is that $F(G)\cong C_{2}\times C_{2}\times C_{2}$ and
$G/F(G)\cong{\rm{SL}}(3,2).$ In particular $G$ is perfect of order 1344. Using
the library of perfect groups available in GAP [16], it can be checked that
there are two possibilities. Either $G\cong\rm{{AGL}}(3,2)$ or $F(G)$ is the
Frattini subgroup of $G.$ In the first case in $G$ there are 16 subgroups of
order 6 containing the same involution, in the second case $G$ contains an
element of order 8. So in both the cases, $P^{*}(G)$ is not claw-free.

Now assume that $N$ is an elementary abelian 2-group. Since $P/F(G)$ is not
cyclic, $N$ is not contained in $F(G).$ This implies $N\not\leq C_{P}(F(G)).$
Let $y\in F(G)\setminus C_{P}(N).$ Then $\langle y\rangle N$ is non-abelian
and therefore there exists $n\in N$ such that $z=yn$ has order 4. Since $P$
contains no subgroup isomorphic to $C_{4}\times C_{2}\times C_{2},$ we deduce
$|C_{N}(z)|\leq 4.$ Since $F(G)$ is abelian, $F(G)\cap N\leq
C_{N}(z)=C_{N}(y)\leq C_{2}\times C_{2}$. This would imply $F(G)\leq
C_{2}\times C_{2}\times C_{2}$. We can conclude as in the previous paragraph.
∎

##  5\. Non-abelian simple groups

It is well known that a finite non-abelian simple group contains a four-
subgroup, since no simple group has a cyclic group or a quaternion group for
its Sylow 2-subgroup. Moreover if $G$ is a finite non-abelian simple group and
$P^{*}(G)$ is claw-free then no subgroup of $G$ can be isomorphic to
$C_{2}\times C_{2}\times C_{p}$ with $p$ an odd prime. This implies that the
centralizer of every four-subgroup is a $2$-group. Thus a key role in the
study of finite simple groups whose reduced-power graph is claw-free is played
by the following result:

######  Theorem 5.1.

[29, Theorem 1] A finite simple group in which the centralizer of every four-
subgroup is a $2$-group is isomorphic with one of the following:

  1. (1)

$\mathrm{PSL}(2,q),$ ${\rm{Sz}}(q),$ ${\rm{J}}_{1},$ ${\rm{M}}_{11},$
$\mathrm{PSL}(3,4),$ ${}^{2}{\rm{F}}_{4}(2)^{\prime},$

  2. (2)

$\mathrm{PSL}(3,q),$ $\mathrm{PSU}(3,q)$ or $\mathrm{G}_{2}(q)$, for suitable
odd $q\geq 3.$

In order to prove Theorem 1.5 it suffices to investigate the groups listed in
the statement of the previous theorem.

######  Lemma 5.2.

$P^{*}(J_{1})$ is not claw-free.

###### Proof.

It follows from Lemma 4.1, noticing that $J_{1}$ contains a subgroup
isomorphic to $C_{2}\times A_{5}$ [14]. ∎

######  Lemma 5.3.

$P^{*}({\rm{M}}_{11})$ is not claw-free.

###### Proof.

It suffice to notice that any involution of ${\rm{M}}_{11}$ is contained in
three different cyclic subgroups of order 4. ∎

######  Lemma 5.4.

Let $G={\rm{Sz}}(q)$, where $q=2^{2k+1}$, where $k\geq 1$ is a positive
integer. Then $P^{*}(G)$ is never claw-free.

###### Proof.

According to [30], a Sylow $2$-subgroup of $G$ has order $q^{2}$; it contains
$q-1$ elements of order $2$, and $q^{2}-q$ elements of order $4$. So, at least
$\dfrac{q}{2}$ subgroups of order $4$ share a common subgroup of order $2$.
Since $q/2\geq 3,$ this implies that the reduced power graph of a Sylow
$2$-subgroup of $G$ always contains a claw. ∎

######  Lemma 5.5.

$P^{*}(\mathrm{PSL}(3,4))$ is not claw-free.

###### Proof.

It suffices to prove that in $\mathrm{PSL}(3,4)$ every involution is the
square of precisely 12 elements of order 4. ∎

######  Lemma 5.6.

If $q$ is odd, then $P^{*}(\mathrm{PSL}(3,q))$ is not claw-free.

###### Proof.

$\mathrm{PSL}(3,q)$ contains a subgroup $H\cong{\rm{SL}}(2,q).$ If $q\neq 3$,
then $H$ is not solvable and $Z(H)$ has order 2. But then $P^{*}(H)$ contains
a claw by Proposition 4.1. It is not difficult to check that
$\mathrm{PSL}(3,3)$ contains 3 different cyclic subgroups of order 4
containing the same subgroup of order 2. ∎

######  Lemma 5.7.

If $q$ is odd, then $P^{*}(\mathrm{G}_{2}(q))$ is not claw-free.

###### Proof.

When $q$ is odd, $\mathrm{G}_{2}(q)$ contains a subgroup $H$ isomorphic to
${\rm{SL}}(3,q)$ [23]. If $Z(H)=1,$ then $H\cong\mathrm{PSL}(3,q)$ and we
conclude from the previous lemma that $P^{*}(H)$ contains a claw. It $Z(H)\neq
1,$ then the same conclusion is ensured by Proposition 4.1. ∎

######  Lemma 5.8.

$P^{*}(^{2}{\rm{F}}_{4}(2)^{\prime})$ is not claw-free.

###### Proof.

It suffices to notice that ${}^{2}{\rm{F}}_{4}(2)^{\prime}$ contains a
subgroup $H\cong\mathrm{PSL}(3,3)$ [14] and that $P^{*}(H)$ is claw-free by
Lemma 5.6. ∎

######  Lemma 5.9.

If $q$ is odd, then $P^{*}(\mathrm{PSU}(3,q))$ is not claw-free.

###### Proof.

$\mathrm{PSU}(3,q)$ contains a subgroup $H\cong{\rm{SL}}(2,q).$ If $q\neq 3$,
then $H$ is not solvable and $Z(H)$ has order 2. But then $P^{*}(H)$ contains
a claw by Proposition 4.1. If $q=3$ the conclusion follows from the fact that
every element of $\mathrm{PSU}(3,3)$ of order 2 is contained in 4 different
cyclic subgroups of order 4. ∎

######  Theorem 5.10.

Let $G=\mathrm{PSL}(2,q)$ and let $d=(q-1,2).$ Then $P^{*}(G)$ is claw-free if
and only if each of the two numbers $(q\pm 1)/d$ is either a prime power or
the product of some prime with a prime-power.

###### Proof.

Let $\Omega$ be the set of positive integers that are either prime-power or of
the form $p^{n}q,$ where $p$ and $q$ are distinct primes and $n\in\mathbb{N}.$
Moreover let $z_{1}=(q-1)/d$ and $z_{2}=(q+1)/d.$ Assume that $q=p^{t},$ with
$p$ a prime, and let $K$ be a Sylow $p$-subgroup of $G.$ By [19, II, Satz 8.5]
$G$ contains a cyclic subgroup $H_{1}$ of order $z_{1}$ and a cyclic subgroup
$H_{2}$ of order $z_{2}$ and the conjugates of $K,H_{1},H_{2}$ give rise to a
partition of $G.$ Thus $P^{*}(G)$ is claw-free if and only
$P^{*}(K),P^{*}(H_{1})$ and $P^{*}(H_{2})$ are claw-free. On the other hand
$P^{*}(K)$ is claw-free since $K$ is an elementary abelian $p$-group, while,
by Remark 2.1, for $i\in\\{1,2\\},$ $P^{*}(H_{i})$ is claw-free if and only if
$z_{i}\in\Omega.$ ∎

##  6\. Proof of Theorem 1.6

###### Proof of Theorem 1.6.

Assume that $P^{*}(G)$ is claw-free. If $G$ is solvable, then $\pi(G)\leq 4$
by Theorem 1.3. So assume that $G$ is almost simple. Let
$S=\operatorname{soc}(G).$ By Theorem 1.5, $S=\mathrm{PSL}(2,q),$ with
$q=p^{f}$ for a suitable prime $p.$ Let $d=(q-1,2).$ Since
$|S|=q(q-1)(q+1)/d,$ it follows from Theorem 1.5 that $\pi(S)\leq 5.$ Recall
that the order of $G/S$ divides $2f.$ Assume that $r$ is a prime dividing
$|G|$ but not $|S|$. Then $r$ divides $f$. Let $u=f/r$ and let $g$ be an
element of order $r$ in $f$. Then $C_{S}(g)\cong\mathrm{PSL}(2,p^{u})$ and $G$
contains a subgroup $H$ isomorphic to $\mathrm{PSL}(2,p^{u})\times C_{r}.$
However $|\mathrm{PSL}(2,p^{u})|$ is divisible by at least two different
primes and is coprime with $r$, so $P^{*}(H)$ contains a claw by Theorem 2.13.
∎

##  7\. Acknowledgements

Pallabi Manna expresses gratitude to the Department of Atomic Energy (DAE),
India for providing funding (in the form of post-doctoral fellowship) during
this work. Santanu Mandal acknowledges the infrastructure provided by VIT
Bhopal University, India.

##  8\. Statements and Declarations

Conflict of interest: On behalf of all authors, the corresponding author
states that there is no conflict of interest.

##  9\. Data Availability

Data sharing is not applicable to this article as no data were created or
analyzed in this study.

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