# Quantum Key Distribution With an Integrated Photonic Receiver

Giulia Guarda  1,2 Domenico Ribezzo  2,3 Tommaso Occhipinti  4 Alessandro
Zavatta  2,4 and Davide Bacco 4, 5  1European Laboratory for Non-Linear
Spectroscopy (LENS), 50019 Sesto Fiorentino, Italy
2Istituto Nazionale di Ottica, Consiglio Nazionale delle Ricerche (CNR-INO),
50125 Firenze, Italy
3University of Naples Federico II, Department of Physics, 80125 Napoli, Italy
4QTI S.r.l., 50125 Firenze, Italy
5University of Florence, Department of Physics and Astronomy, 50019 Sesto
Fiorentino, Italy<EMAIL_ADDRESS>
††journal: opticajournal††articletype: Research Article

Photonic integrated circuits (PICs) are key in advancing quantum technologies
for secure communications. They offer inherent stability, low losses and
compactness compared to standard fiber-based and free-space systems.

Our reasearch demonstrates PIC’s effectivness in enhancing quantum
communications, implementing a three-state BB84 protocol with decoy-state
method. We employ an integrated receiver and superconducting nanowire single
photon detectors (SNSPDs) to achieve technological advancements.

One of the most notable results is the extraction of a secret key over a
record-breaking 45 dB channel attenuation. Our results demonstrate a
remarkable 220% boost in key rate compared to our prototype fiber-based
receiver over a 10 dB channel attenuation. This improvement in the secret key
rate (SKR) signifies the potential of integrated photonics to advance the
field of quantum communication.

##  1 Introduction

Our society heavily relies on secure communication links for the exchange of
information. However, recent advancements in quantum computing pose a
significant threat to the security of our well-established network
infrastructure [1]. Consequently, there is a pressing need to develop
innovative cryptographic techniques, robust against quantum computational
advances. One prominent example of this is Quantum Key Distribution (QKD) [2],
where the security of the protocols is grounded in the laws of physics and not
related to the limited computational power of an eavesdropper [3]. QKD enables
the secure distribution of a random secret key among two or more users by
harnessing the unique properties of quantum physics [4]. The most studied and
widely used protocol to perform QKD is the BB84 protocol, where a prepare and
measure scheme ensures the exchange of a secret key exploiting different
degrees of freedom of photons, e.g., polarization [5], time of arrival and
phase, [6, 7, 8], path [9], etc.

In this work, we implement an efficient BB84 protocol with time-bin and phase
encoding. In this scheme, an imbalanced Mach-Zender Interferometer (iMZI) is
required to decode the phase-encoded states. We examine the advantages of an
integrated iMZI at the receiver’s end. Compared to standard free-space [10]
and fiber-based [6, 7, 8, 11] interferometers, our integrated solution is more
compact, requires no active stabilization, and is more efficient. Free-space
devices are bulky and challenging to maintain stable over extended periods of
time, while fiber-based interferometers require additional feedback loops to
mitigate drifts in the system, often introducing noise in the detection system
and making more complex the hardware implementation. In contrast, the photonic
interferometer overcomes these limitations, offering a compact device with low
losses and long-term stability [12, 13, 14].

Integrated platforms represent a practical and scalable choice for increasing
the hardware’s complexity [15] and pave the way towards a more mature
deployment of quantum technologies. By confining the propagation modes within
a solid-state device, PICs significantly reduce noise caused by vibrations and
minimize temperature fluctuations [14, 16, 17], making the integrated devices
a more stable solution compared to fiber-based systems [18, 19]. Moreover, the
lithographic fabrication process can be precisely replicated, facilitating
scalable mass production of reproducible devices [16].

Currently, state-of-the-art integrated systems are limited by losses and form
factors [20]. Although stability performances are enhanced when using a PIC,
there is room for improvements. In this work, we analyze the performances of a
PIC-based iMZI at the receiver’s end of a QKD scheme. Hence, we focus on the
stability of the interferometer and the propagation losses of the entire
integrated device, testing it for the decoding of quantum states in an
efficient three-state BB84 protocol.

##  2 Methods

The best-known example of quantum communication is QKD, which enables the
distribution of a secret random key between two or more users by exploiting
quantum states made by photons. In our work, we implement an efficient three-
state BB84 protocol using time-bin encoding and one-decoy state method [6, 7,
8, 21, 22]. In this protocol, the transmitter randomly generates quantum
states encoded in two bases, with each quantum state consisting of two-time
bins: early and late. In contrast to the standard BB84 protocol where both
states in the X-basis - $\ket{0}_{x}$ and $\ket{1}_{x}$ \- are generated, in
our scheme only the state $\ket{0}_{x}$ is produced. Importantly, it has been
proved that this improvement does not compromise the security of the protocol
[23, 24]. However, X-basis states can exclusively be used for security checks
and not for generating the final key. This modification brings advantages in
terms of reduced complexity in the state generation system and a decreased
need for detectors in the decoding system. Moreover, by optimizing the Z-basis
choice beyond the balanced probability $p_{Z}=50\%$, Alice and Bob can reduce
the amount of bits to be discarded during the key sifting stage. The three
distinct states can be distinguished by considering both the temporal
component and the intensity of the pulses. Mathematically, they are
represented as:

$\ket{0}\textsubscript{z}=\ket{\gamma}\textsubscript{early}\ket{0}\textsubscript{late};\
\
\ket{1}\textsubscript{z}=\ket{0}\textsubscript{early}\ket{\gamma}\textsubscript{late};\
\
\ket{0}\textsubscript{x}=\frac{\ket{0}\textsubscript{z}+\ket{1}\textsubscript{z}}{\sqrt{2}}\
.$ (1)

The pulses encoding the states are generated by carving and attenuating a
continuous wave C-band laser down to the single-photon level. Therefore, they
are weak coherent states and not ideal single photon Fock states. Weak
coherent pulses are advantageous for QKD purposes since they can be generated
at higher rates than any existing single photon source [25]. However, although
they are strongly attenuated, the probability of getting a multi-photon event
is not zero, which makes the protocol vulnerable to photon number splitting
(PNS) attacks [3]. To address this issue, a technique known as decoy-state
method is implemented [26, 27, 28], where the intensity of the pulses is
randomly varied at the transmitter’s end. By doing so, if a PNS occurs, it
becomes possible to detect it by looking at the losses of the different
intensities pulses, and the quantum security of the protocol is preserved [3,
4].

Figure 1: States generation. A quantum state exhibits a temporal duration of
1.68 ns and is composed of two time-bins, each separated by a 800 ps time
delay. The interval between two consecutive time bins of two consecutive
quantum states is 880 ps. The wave function of each photon consists of one or
two pulses occurring at the center of either one or both of the time-bins. In
the Z-basis encoding, the photons are confined to a single time-bin, while the
X-basis states represent the superposition of the Z-basis states. This results
in the photon being spread across both time-bins, with a relative phase of
either zero or $\pi$. The established protocol permits the generation of only
the phase 0 state. Decoy method A three-level signal sets the different
intensities for the decoy.

The estimation of the final secret key is done in the finite block-size
regime. Here, the upper bound for the secret key rate length ($l$) is bounded
to [22]:

$l\leq
s_{Z,0}^{l}+s_{Z,1}^{l}(1-H_{2}(\phi_{Z}^{u}))-\lambda_{EC}-\lambda_{sec}-\lambda_{corr}.$
(2)

In Eq.(2), $s_{Z,0}^{l}$ and $s_{Z,1}^{l}$ are the lower bounds for vacuum and
single photon events. $H_{2}(x)$ is the binary entropy, defined as
$H_{2}(x)=-x\ log_{2}(x)-(1-x)\ log_{2}(1-x)$. $\phi_{Z}^{u}$ is the upper
bound for the phase error rate, estimated based on the interferometer
visibility as $vis_{X}=1-2QBER_{X}$. As only one state is generated in the
X-basis, $\phi_{Z}^{u}$ can only be estimated and not directly measured.
Furthermore, $\lambda_{EC}$ is the amount of disclosed bits during the error
correction stage. $\lambda_{sec}$ and $\lambda_{corr}$ are functions of the
security ($\epsilon_{sec}=10^{-12}$) and correctness
($\epsilon_{corr}=10^{-12}$) parameters in the form of:
$\lambda_{sec}=6\log_{2}(19/\epsilon_{sec})$ and
$\lambda_{corr}=\log_{2}(2/\epsilon_{corr})$. The first quantifies as
$\epsilon_{sec}=10^{-12},$ the probability of the correlation between Eve’s
and Alice’s keys being stronger than the correlation between Alice’s and Bob’s
keys. The second assesses the probability of Alice’s and Bob’s sifted keys
($S_{A}$ and $S_{B}$) differing [29]. These can be defined as:

$\mathds{1}(S_{A},S_{B};Z,C)<\epsilon_{sec};\ \ P[S_{A}\neq
S_{B}]<\epsilon_{corr}.$ (3)

Here $\mathds{1}(\cdot)$ is an information measure, $P[\cdot]$ is a
probability function, $Z$ is the eavesdropper’s bit sequence and $C$
represents the exchanged information.

##  3 Experimental setup

Figure 2 shows the architecture of the setup used in this work. In this
configuration, the transmitter (Alice) is embedded in a compact 2U rack box.
The receiver (Bob) includes a beam splitter for random detection basis choice,
and two different interferometers. The first is an integrated interferometer
enclosed in a copper case, and temperature stabilized with a thermoelectric
temperature controller. The second is a prototype fiber-based interferometer,
specifically realized for this experiment. The two parties communicate both
via a quantum channel (QC) and a classical channel (CC). The QC is composed by
a single-mode fiber at telecom wavelength (SMF-28, ITU G-652) and is used for
the transmission of the quantum states. The CC, made of electrical cables and
it is employed as the service channel, serving for clock synchronization. The
Bob’s detection system includes high-efficient superconducting nanowire
single-photon detectors (SNSPDs), with 93% efficiency, 400 Hz dark count rate
and 40 ps of jitter.

Figure 2: Setup. ALICE. A continuous wave laser passes through two intensity
modulators (IMs). A field programmable gate array (FPGA) drives the IMs. IM1:
states generation, IM2: decoy method. A variable optical attenuator (VOA) is
used to set different intensities. A beam splitter adds additional attenuation
and enables to monitor the intensities with a photon detector. An additional
fixed attenuation stage (ATT) brings the intensities down to the single photon
level. The quantum channel (QC) is made of a single-mode optical fiber and
different channel lengths are simulated using fixed optical attenuators. The
classical channel (CC) is used for sharing a synchronization signal (SYNC).
BOB. A 50/50 beam splitter makes the random detection basis choice. The states
measured in the Z-basis go straight to the SNSPD. For the phase measurement in
the X basis, the photons need to pass through an imbalanced Mach Zehnder
interferometer before being detected. The performances of two different
interferometers are compared. PIC-based iMZI. Polarization controllers adjust
the polarization of the input light on the PIC, whose performances are
polarization-dependent. The fibers are edge coupled. The PIC temperature is
stabilized with a thermoelectric temperature controller (TED). Prototype
fiber-based iMZI The 800 ps delay line interferometer (DLI) is obtained with
careful fiber splicing. A phase-lock loop (PLL) controls the fluctuations and
locks the interferometer on a desired phase difference by monitoring the
drifts of a counter propagating laser, slightly detuned from the quantum
channel wavelength. Finally, the PLL implements a
proportional–integral–derivative (PID) controller that acts on a phase shifter
in one of the interferometer arms. SNSPDs used for detection are connected to
a time-to-digital converter (TDC), whose data are sent to Bob’s
computer.Finally, Alice and Bob communicate over the CC for the final
generation of a shared secret key.

On the transmitter side, the states are generated at a repetition rate of 595
MHz by modulating a continuous-wave laser at 1545.32 nm wavelength. In a real-
world scenario, the quantum states need to be generated with different random
phases to avoid security issues [30, 31]. It does not happen in our
transmitter, however, with a view to future field trials, our setup is capable
of implementing phase randomization for each state [32]. We employ a field
programmable gate array (FPGA) to generate the electrical sequences encoding
the states. These signals drive two intensity modulators. One IM shapes the
laser signal to generate the time-bin encoding, and the second modifies the
states intensities to create the decoy. The FPGA generates a sequence of 8190
pulses, corresponding to 4095 states, repeated 145358 times per second. The
states are generated in both Z and X bases, with probabilities $p_{Z}^{A}=0.5$
and $p_{X}^{A}=(1-p_{Z}^{A})=0.5$, respectively. After the generation, these
time-bin-encoded quantum states are attenuated down to the single photon level
with a variable optical attenuator (VOA) and additional fixed optical
attenuation. Finally, they are transmitted to Bob through the QC. Different
channel lengths are simulated with several fixed optical attenuators.

The decoding scheme at Bob’s end involves the use of a 50/50 fiber-based beam
splitter for random detection basis selection. Here, probabilities
$p_{Z}^{B}=0.5$ and $p_{X}^{B}(1-p_{Z}^{B})=0.5$ are configured. When the
quantum states are measured in the Z-basis, they are directed straight to a
SNSPD channel, and the arrival time of the photons is recorded. In the case of
X-basis measurement, an imbalanced Mach Zehnder interferometer (iMZI) is
necessary to properly measure the phase. In this work, we compare the
performances of a lab-made fiber-based iMZI with a PIC-based one. The fiber-
based iMZI has been constructed specifically to conduct this experiment. It
requires active stabilization, done with a phase lock loop (PLL). This
consists of a monitor signal at 1550.00 nm, entering from the iMZ output and
counter-propagating with respect to the quantum states path, a piezo-
controlled phase shifter, and an InGaAs photodetector. A
proportional–integral–derivative (PID) controller, acting on the phase
shifter, is in charge for keeping the phase difference of the two
interferometer arms constant. As the PLL wavelength is just slightly different
from the quantum states one, locking the phase of the monitoring signal
enables concurrent phase locking for the quantum signal. Both the input and
output port of the interferometer are equipped with a wavelength division
multiplexer (WDM) in charge for combining and dividing the two signals. The
hardware is supported by an auto-tuning software that hinders drifts in the
setup. The PIC-based iMZI, on the other hand, does not require an active
stabilization system. The quantum signal is coupled into the PIC via edge
coupling method. As the waveguides are polarization-dependent, polarization
controllers are used to control the incoming photons polarization. Once
coupled into the waveguides, the two pulses composing a time-bin are
overlapped using an 800 ps delay line into one of the arms of the
interferometer. The phase between the two arms is adjusted and kept stable
using a thermoelectric cooler (TEC) controlled by a temperature controller
(TED) using a PID loop. Subsequently, the signal is collected into an optical
fiber, edge-coupled, and glued to the PIC before being directed to the SNSPD.

To ensure accurate data collection, the signals are recorded using a time-to-
digital converter (TDC), a qutag time-tagging unit produced by qutools GmBH,
and sent to Bob’s computer. Finally, Alice and Bob communicate using the CC to
obtain a quantum criptographically secure secret key. The block size, $n_{Z}$,
which determines the number of bits used for the finate key analysis, is set
to $10^{7}$. The data are collected for a full block time, i.e. the time
required to acquire a full block size, $\tau_{Z}$. The longest acquisition
time was done for the longest simulated channel, 45 dB. At this attenuation
level, it was required a block time of 3 hours and 15 minutes of continuous
acquisitions.

###  3.1 Integrated interferometer

The PCI iMZI utilized in this work is made of passive components. The platform
is composed of a borosilicate glass matrix, specifically designed for making
ion exchange waveguides, with an effective refractive index of 1.5212 at
1545.32 nm [33].

By inducing a localized alteration in the glass matrix composition, it is
possible to produce changes in the refractive index, allowing for the creation
of waveguides. This process involves thin-film deposition, photolithography,
and a chemical ion-exchange bath. The resulting waveguides exhibit minimal
losses and seamless compatibility with conventional optical fibers, making
them well-suited for telecommunication applications [33].

Once the temperature of the PIC is stabilized, using a PID loop, it remains
constant for the whole measurement time, as it is enclosed in a copper case
ensuring thermal stability. This long-term stability enables us to conduct
continuous measurements lasting for 2 days.

##  4 Results and Discussion

We compare the stability and secret key rate of two receivers. The first
receiver uses a fiber-based interferometer to measure the X-basis [11]. In
contrast, the second receiver employs a fully photonic integrated
interferometer. The advantages of the integrated solution are evident in both
stability comparisons and amount of secret key extracted from the two setups
under various attenuation channels.

The advantages of the PIC stem from its ultra-low losses and inherent
stability. On the other hand, the fiber-based interferometer relies on a PLL
to maintain stability [7]. This mitigates the interferometer fluctuations but
introduces noise due to the counter-propagating monitor signal, which hinders
secret key extraction at longer distances. Furthermore, the enhanced
integrated device stability and noise reduction, lead to increased key
extraction rates for different channel lengths, as illustrated in Fig.3a).

Figure 3: a) Secret key rate as a function of the channel losses. The dotted
green line represents the extracted key rate for different channel
attenuations for the PIC-based interferometer according to our simulation
model. The red circles are the measured SKR values for the same system. The
dashed blue curve is the simulation for the fully fiber-based interferometer
and the orange squares are the measured values for the same system. b)
Stability. Trend of QBERx during the key exchange over a channel showing 30 dB
of attenuation for over 50 hours of continuous measurements. The green circles
are the data relative to the PIC-based iMZI. QBERx is connected to the
interferometer visibility vis by $QBER_{X}=(1-vis)/2$. c) Stability
comparison. The trend of the QBERx during the key exchange at 30 dB for both
the fiber-based and the PIC-based interferometer. The purple circles are the
date relative to the PIC-based iMZI and the green cross shaped points are
relative to the fully fiber-based iMZI. All the experimental points have been
acquired for a full block time.

The PIC-based interferometer demonstrates superior stability. Even when the
fiber-based setup achieves good stability, the PIC-based interferometer
registers lower Quantum Bit Error Rate (QBER) values, due to its higher
extinction ratio. Fig.3b) illustrates the long-term stability achieved with
the integrated iMZI, demonstrating over 50 hours of continuous measurements.
This extended stability allows for extended data acquisition over a period of
more than two days.

Fig.3c) illustrates the stability of the fiber-based iMZI during a 5-hour
continuous data acquisition at 30 dB attenuation. As the system drifts, the
PLL requires an internal re-calibration, which takes some time, leading to
uneven time intervals between data points. A comparison with the integrated
interferometer highlights the significant benefits of long-term stability.

Different materials for photonic integration have been studied. The most
common include silicon-based platforms [34] such as silicon and silicon
nitride [35, 36]; nonlinear materials such as lithium niobate [37]; and glass
waveguides, such as aluminium borosilicate glass [20]. In a standard quantum
communication protocol, quantum states undergo manipulation through a
combination of active and passive components. Common discrete variables
encoding methods include polarization, path, and time-bin and phase degrees of
freedom [16], and they generally exploit active components. When it comes to
decoding the quantum information, passive components come into play. Among
these, a widely used configuration involves interferometers, which
incorporates elements like phase shifters, delay lines, and beam splitters
[38].

Tab.1 reports the most impactful solution adopted for a BB84 time-bin
receiver. Different loss features, material choices, and form factors
(FFactors) are reported.

Table 1: State-of-the-art.
($\ast$): aluminum borosilicate glass, ($\dagger$): borosilicate glass. Ref | Loss | FFactor | Material
---|---|---|---
[35] | 4.5 dB | 2.5 mm | SiOxNy
[36] | 9 dB | 32 mm | SiOxNy
[20] | 3.50 dB | 150 mm | ($\ast$)
Ours | 2.75 dB | 10 mm | ($\dagger$)

Finally, Tab.2 shows a comparison with state-of-the-art devices.

Table 2: Results. Ref | Length | Att | SKR
---|---|---|---
[35] | 20.0 km |  | 916.0 kbps
[36] | 20.0 km |  | 568.0 kbps
[20] |  | 39.5 dB | 940.0 bps
Ours |  | 45.0 dB | 12.2 bps

Overall it is possible to extract a secret key from the fully prototype fiber-
based system up to 40 dB of attenuation channel, equivalent to 205 km in low-
loss fibers, at a rate of 110 bps. On the other hand, the chip-based device
extracts a positive key at a rate of 637.8 bps at 40 dB of attenuation
channel, and 12 bps with 45 dB attenuation. This is equivalent to a distance
of 225 km in standard low-loss fibers and represents, to the best of our
knowledge [20, 39], the longest distance covered with a fully-integrated
receiver system.

##  5 Conclusion

In summary, our work showcases the pivotal role of photonic integrated
circuits in propelling the boundaries of quantum communication.

The system, supported with a PIC-based iMZI, shows a QBER as low as 2%. This
value is below what we have previously observed for standard fully fiber-based
receivers [6, 7, 11, 8], thanks to low losses in the setup and higher
stability in the interferometer. These features (stability and low-losses)
allow for over 2 days of continuous measurements with excellent stability.
Finally, the integrated receiver enables the extraction of a secret key up to
45 dB, equivalent to a distance of 225.0 km in standard low-loss optical
fibers, the furthest distance achieved nowadays with a fully integrated device
[20, 39].

Acknowledgments This work was funded by the European Union (ERC, QOMUNE,
101077917, by the Project EQUO (European QUantum ecOsystems) which is funded
by the European Commission in the Digital Europe Programme under the grant
agreement No 101091561, the Project SERICS (PE00000014) under the MUR National
Recovery and Resilience Plan funded by the European Union - NextGenerationEU,
the Project QuONTENT under the it Progetti di Ricerca, CNR program funded by
the Consiglio Nazionale delle Ricerche (CNR) and by the European Union - PON
Ricerca e Innovazione 2014-2020 FESR - Project ARS01/00734 QUANCOM, the
Project QUID (Quantum Italy Deployment) funded by the European Commission in
the Digital Europe Programme under the grant agreement No 101091408.

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# On finite CG-groups

Sekhar Jyoti Baishya  S. J. Baishya, Department of Mathematics, Pandit
Deendayal Upadhyaya Adarsha Mahavidyalaya, Behali, Biswanath-784184, Assam,
India<EMAIL_ADDRESS>
###### Abstract.

A finite group $G$ is a CG-group if $\mid\operatorname{Cent}(G)\mid=\mid
G^{\prime}\mid+2$, where $G^{\prime}$ is the commutator subgroup and
$\operatorname{Cent}(G)$ is the set of distinct element centralizers of $G$.
In this paper we give some results on CG-groups. We also give a negative
answer to [29, Conjecture 2.3] given by K. Khoramshahi and M. Zarrin.

###### Key words and phrases:

Finite group, Minimal non-abelian group, Minimal non-nilpotent group

###### 2010 Mathematics Subject Classification:

20D60, 20D99

##  1\. Introduction

Given a group $G$, let $\operatorname{Cent}(G)$ denote the set of centralizers
of $G$, i.e., $\operatorname{Cent}(G)=\\{C(x)\mid x\in G\\}$, where $C(x)$ is
the centralizer of the element $x$ in $G$. The study of finite groups in terms
of $|\operatorname{Cent}(G)|$, becomes an interesting research topic in last
few years. Starting with Belcastro and Sherman [17] in 1994 many authors have
been studied and characterised finite groups $G$ in terms of
$\mid\operatorname{Cent}(G)\mid$. More information on this and related
concepts may be found in [1, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 40, 41,
42].

While studying the number of distinct element centralizers of a finite group
$G$, we observe that there are lots of examples of finite groups (including
family of finite groups) for which $|\operatorname{Cent}(G)|=\mid
G^{\prime}\mid+2$, where $G^{\prime}$ is the commutator subgroup of $G$ (the
smallest example of such group being $S_{3}$). We have also observed that such
groups have many interesting properties (for example, in most of the cases
$G^{\prime}$ is abelian, proper element centralizers are abelian etc). This
motivates us to introduce the notion of CG-group with a hope that this might
give some new solvability criterian for a finite group. We call a finite group
$G$ to be a CG-group if $\mid\operatorname{Cent}(G)\mid=\mid
G^{\prime}\mid+2$. In this paper, we give some examples of CG-groups and some
necessary and sufficient conditions for a finite group to be a CG-group. Apart
from these, we also give a negative answer to the conjecture [29, Conjecture
2.3] given by K. Khoramshahi and M. Zarrin.

In this paper, all groups are finite (however Lemma 3.1 and Remark 3.2 holds
for any group) and all notations are usual. For example $G^{\prime}$, $Z(G)$
denotes the commutator subgroup and the center of a group $G$ respectively,
$C_{n}$ denotes the cyclic group of order $n$, $D_{2n}$ denotes the dihedral
group of order $2n$, $A_{n}$ denotes the alternating group of degree $n$,
$S_{n}$ denotes the symmetric group of degree $n$ and
$C_{n}{\rtimes}_{\theta}C_{p}$ denotes semidirect product of $C_{n}$ and
$C_{p}$, where $\theta:C_{p}\longrightarrow\operatorname{Aut}(C_{n})$ is a
homomorphism.

##  2\. Some Examples and basic results

We begin with some examples and counter examples of finite CG-groups. As we
have already mentioned, the smallest CG-group is $S_{3}$ (in fact in view of
[13, Corollary 2.5], any group with central quotient of order $pq$ is a CG-
group, where $p,q$ are primes, not necessarily distinct). The next higher
order CG-groups are $D_{8}$ and $Q_{8}$. We will see that any non-abelian
group order $pqr$ or $p^{4}$, where $p,q,r$ are primes not necessarily
distinct is a CG-group. On the other hand, in view of [17, Theorem 7], no
finite perfect group is a CG-group (recall that a group $G$ is said to be
perfect if $G^{\prime}=G$). Consequently, no finite simple, semisimple group
(a group is said to be semisimple if it is a direct product of non-abelian
simple groups) is a CG-group and $A_{4}$ is the only CG-group among the
alternating groups $A_{n},n\geq 3$. We now give some more examples of finite
CG-groups:

######  Proposition 2.1.

The generalized dihedral group $D(m,n)=\langle a,b\mid
a^{m}=b^{n}=1,bab^{-1}=a^{-1}\rangle$; $m\geq 3,n\geq 2$ and $n$ even is a CG-
group.

###### Proof.

By [4, Fact 2], we have $Z(D(m,n))=\langle a^{\frac{m}{2}},b^{2}\rangle$ or
$\langle b^{2}\rangle$ according as $m$ is even or odd. Therefore
$\langle\langle a\rangle,Z(D(m,n))\rangle$ is an abelian normal subgroup of
$D(m,n)$ of prime index. Now, the result follows using [13, Theorem 2.3]. ∎

As an immediate corollary we have the following result:

######  Corollary 2.2.

The dihedral group $D_{2n},n\geq 3$ is a CG-group.

######  Proposition 2.3.

Let $\frac{G}{Z(G)}\cong C_{n}{\rtimes}_{\theta}C_{p}$ be non-abelian, $n$ be
an integer, $p$ be a prime and
$\theta:C_{p}\longrightarrow\operatorname{Aut}(C_{n})$ is a non-trivial
homomorphism. Then $G$ is a CG-group.

###### Proof.

See [13, Proposition 2.9]. ∎

######  Proposition 2.4.

The generalized quaternion group $Q_{4m}=\langle a,b\mid
a^{2m}=1,b^{2}=a^{m},bab^{-1}=a^{-1}\rangle,m\geq 2$ is a CG-group.

###### Proof.

It is well known that $\frac{Q_{4m}}{Z(Q_{4m})}\cong D_{2m}$, for any $m\geq
2$. Therefore by Proposition 2.3, $G$ is a CG-group. ∎

Let $I(G)$ be the set of all solutions of the equation $x^{2}=1$ in $G$.
Define $\alpha(G)=\frac{\mid I(G)\mid}{\mid G\mid}$. Then we have the
following result:

######  Proposition 2.5.

Let $G$ be a finite group of order $2^{n}m$, where $n$ is any integer and
$m>1$ is an odd integer. If $\alpha(G)>\frac{1}{2}$, then $G$ is a CG-group.

###### Proof.

See [13, Corollary 2.6]. ∎

######  Proposition 2.6.

The semidihedral or quasidihedral group $SD_{2^{n}}=\langle x,y\mid
x^{2^{n-1}}=y^{2}=1,yxy^{-1}=x^{2^{n-2}-1}\rangle$ is a CG-group for any
$n\geq 3$.

###### Proof.

It follows from [13, Theorem 2.3], noting that $SD_{2^{n}},n\geq 3$ has an
abelian normal subgroup of prime index. ∎

######  Proposition 2.7.

The modular $p$-group $Mod_{n}(p)=\langle x,y\mid
x^{p^{n-1}}=y^{p}=1,yxy^{-1}=x^{1+p^{n-2}}\rangle$ is a CG-group for any
$n\geq 3$.

###### Proof.

It follows from [13, Theorem 2.3], noting that $Mod_{n}(p),n\geq 3$ has an
abelian normal subgroup of prime index. ∎

######  Proposition 2.8.

The group $U_{6n}=\langle x,y\mid x^{2n}=y^{3}=1,x^{-1}yx=y^{-1}\rangle$ is a
CG-group for any $n$.

###### Proof.

It follows from Proposition 2.3, noting that $\frac{U_{6n}}{Z(U_{6n})}\cong
S_{3}$ (see [10, Lemma 2.7]). ∎

The linear groups plays a significant role in the theory of finite groups. We
now study the CG-groups among the linear groups of degree two. In the
following results, the groups $GL(2,q),SL(2,q),PGL(2,q)$ and $PSL(2,q)$ denote
the general linear, special linear, projective general linear and projective
special linear groups respectively of degree $2$ over the field of size $q$,
where $q$ is a prime power.

######  Proposition 2.9.

The group $G=GL(2,q)$ is a CG-group if and only if $q=2$.

###### Proof.

If $q=2$, then $G=S_{3}$ is a CG-group. Next, suppose $q>2$. In view of [2,
Proposition 3.26]), the proper centralizers of $G$ are precisely the members
of the family $\\{xDx^{-1},xIx^{-1},xPZ(G)x^{-1}\mid x\in G\\}$, where

  1. (a)

$D$ is the subgroup of all diagonal matrices in $G$, and the number of
conjugates of $D$ in $G$ is $\frac{q(q+1)}{2}$,

  2. (b)

$I$ is a cyclic subgroup of $G$, and the number of conjugates of $I$ in $G$ is
$\frac{q(q-1)}{2}$,

  3. (c)

$P$ is the Sylow $p$-subgroup of $G$ consisting of all upper triangular
matrices with $1$ in the diagonal, and the number of conjugates of $PZ(G)$ in
$G$ is $q+1$.

Therefore $\mid
Cent(G)\mid=\frac{q(q+1)}{2}+\frac{q(q-1)}{2}+(q+1)+1=q^{2}+q+2$. Also by [16,
Theorem 3.1.19], for $q>2$, we have $G^{\prime}=SL(2,q)$ having order
$q(q-1)(q+1)$. In the present scenario, one can easily verify that $G$ is not
a CG-group. ∎

######  Proposition 2.10.

The group $G=SL(2,q)$ is a CG-group if and only if $q=2$ or $3$.

###### Proof.

It is easy to verify that $SL(2,2)$ and $SL(2,3)$ are CG-groups. On the
otherhand, if $q>3$, then by [32], we have $G=G^{\prime}$ and consequently
using [17, Theorem 7], $G$ is not a CG-group. ∎

We now compute the number of distinct element centralizers of $PGL(2,q)$.

######  Proposition 2.11.

Let $G=PGL(2,q)$. Then

$\mid\operatorname{Cent}(G)\mid=\begin{cases}2q^{2}+q+2&\;\text{if \, $q$ \,
is odd, $q>3$}\\\ q^{2}+q+2&\;\text{if \, $q$ \, is even, $q>4$}\\\
5&\;\text{if \, $q=2$}\\\ 14&\;\text{if \, $q=3$}\\\ 22&\;\text{if \,
$q=4$}\par\end{cases}$

###### Proof.

If $q$ is even, then $PGL(2,q)=PSL(2,q)$ (see [35]). Now, the result follows
from [39, Theorem 1.1].

Now, suppose $q$ is odd. We have $G=PGL(2,3)=S_{4}$ and
$\mid\operatorname{Cent}(G)\mid=14$. Next, suppose $q\geq 5$. In view of [24,
Page 5], it follows that the centralizer of any involution (element of order
$2$) in $G$ is either $D_{2(q-1)}$ or $D_{2(q+1)}$. Hence by [3, Proposition
2.4], $\mid\operatorname{Cent}(PGL(2,5))\mid=57$, noting that the size of
element centralizers in $PGL(2,5)=S_{5}$ are $4,5,6,8$ and $12$. On the other
hand, if $q>5$, then using [21, Table 4], it follows that the size of the
proper element centralizers of $G$ are $q,q-1,q+1,2(q-1)$ and $2(q+1)$. Again,
let $a$ and $b$ be any two distinct involutions in $G$. Suppose $C(a)=C(b)$.
Since $C(a),C(b)\in\\{D_{2(q-1)},D_{2(q+1)}\\}$, therefore $a=b$ noting that
$Z(C(a))=Z(C(b))$, which is a contradiction. Therefore $C(a)\neq C(b)$. In the
present scenario, in view of [3, Proposition 2.4], $\mid
Cent(G)\mid=(q+1)+q(q+1)+q(q-1)+1=2q^{2}+q+2$. ∎

As an immediate corollary, we have the following results, noting that
$(PGL(2,q))^{\prime}=PSL(2,q)$ for $q\geq 3$ (see [35]):

######  Corollary 2.12.

$PGL(2,q)$ is a CG-group if and only if $q=2$ or $3$.

######  Corollary 2.13.

$S_{5}$ is not a CG-group.

For $PSL(2,q)$, we have the following result:

######  Proposition 2.14.

$PSL(2,q)$ is a CG-group if and only if $q=2$ or $3$.

###### Proof.

For $q>3$ we have $PSL(2,q)$ is a simple group (see [35]) and hence by [17,
Theorem 7], $G$ is not a CG-group. On the otherhand $PSL(2,2)=S_{3}$ and
$PSL(2,3)=A_{4}$ are CG-groups. ∎

We conclude the section with the following lemmas:

######  Lemma 2.15.

Let $G$ be a finite group such that $\mid G^{\prime}\cap Z(G)\mid=1$. Then $G$
is a CG-group if and only if $\frac{G}{Z(G)}$ is a CG-group.

###### Proof.

The proof follows using [9, Lemma 3.1], noting that if $\mid G^{\prime}\cap
Z(G)\mid=1$, then $\mid(\frac{G}{Z(G)})^{\prime}\mid=\mid G^{\prime}\mid$. ∎

######  Lemma 2.16.

Let $G$ be a finite group such that all sylow subgroups are abelian. Then $G$
is a CG-group if and only if $\frac{G}{Z(G)}$ is a CG-group.

###### Proof.

In the present scenario, we have $\mid G^{\prime}\cap Z(G)\mid=1$ (see [36, p.
118]). Therefore the result follows from the previous lemma. ∎

##  3\. The main results

In this section, we prove the main results of the paper. However, we begin
with the following lemma.

######  Lemma 3.1.

Let $G$ be any group. If $\mid\frac{G}{Z(G)}\mid=pqr$ where $p,q,r$ are primes
(not necessarily distinct), then $C(x)$ is abelian for any $x\in G\setminus
Z(G)$.

###### Proof.

Let $x\in G\setminus Z(G)$. If $\frac{C(x)}{Z(G)}$ is cyclic, then $C(x)$ is
abelian. Now, suppose $\mid\frac{C(x)}{Z(G)}\mid=pq$. Then $o(xZ(G))=p,q$ or
$pq$. If $o(xZ(G))=pq$, then $C(x)$ is abelian. Next suppose $o(xZ(G))\neq
pq$. Then there exists some $y\in C(x)$ such that $\frac{C(x)}{Z(G)}=\langle
xZ(G),yZ(G)\rangle$. Consequently, $C(x)=\langle x,y,Z(G)\rangle$ and hence
$C(x)$ is abelian. If $\mid\frac{C(x)}{Z(G)}\mid=pr$ or $qr$, then using
similar arguments we can show that $C(x)$ is abelian. ∎

######  Remark 3.2.

Recall that a group $G$ is said to be a CA-group if $C(x)$ is abelian for any
$x\in G\setminus Z(G)$. It is easy to see that for such groups $C(x)\cap
C(y)=Z(G)$ for any two distinct proper centralizers $C(x)$ and $C(y)$. Also we
have seen that if $\mid\frac{G}{Z(G)}\mid=pqr$ where $p,q,r$ are primes (not
necessarily distinct), then $G$ is a CA-group.

The authors in [29, Proposition 3.6] showed that if $G$ is a finite group such
that $\frac{G}{Z(G)}$ is isomorphic to a simple group, then then $G$ and
$G^{\prime}$ are isoclinic groups. In the following we generalize this result
as follows:

######  Proposition 3.3.

Let $G$ be a finite group such that $\frac{G}{Z(G)}$ is perfect. Then $G$ and
$G^{\prime}$ are isoclinic groups.

###### Proof.

Suppose $\frac{G}{Z(G)}$ is perfect. Then
$\frac{G}{Z(G)}=(\frac{G}{Z(G)})^{\prime}=\frac{G^{\prime}Z(G)}{Z(G)}$ and
consequently, $G^{\prime}Z(G)=G$. Therefore in view of [31, Lemma 2.7], $G$ is
isoclinic to $G^{\prime}$. ∎

As an immediate consequence we have the following necessary condition for a
finite CG-group.

######  Proposition 3.4.

Let $G$ be a finite group such that $\frac{G}{Z(G)}$ is perfect. Then $G$ is
not a CG-group.

###### Proof.

Suppose $\frac{G}{Z(G)}$ is perfect. Then by Proposition 3.3, $G$ is isoclinic
to $G^{\prime}$ and consequently, using [42, Lemma 2.3], we have
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(G^{\prime})\mid$. Now,
if $G$ is a CG-group, then
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(G^{\prime})\mid=\mid
G^{\prime}\mid+2$, which is impossible by [17, Theorem 7]. ∎

The following result follows using technique similar to [9, Lemma 2.7].

######  Proposition 3.5.

Let $p$ be the smallest prime divisor of the order of a group $G$ and $\mid
G^{\prime}\mid=p$. Then $G$ is a CG-group if and only if $\frac{G}{Z(G)}\cong
C_{p}\times C_{p}$.

As an application to this we have the following result. Given a group $G$,
$\omega(G)$ denotes the size of a maximal set of pairwise non-commuting
elements of $G$.

######  Proposition 3.6.

Let $G$ be a finite non-abelian metacyclic $p$-group, where $p>2$ is a prime.
Then $G$ is a CG-group if and only if $\frac{G}{Z(G)}\cong C_{p}\times C_{p}$.

###### Proof.

In view of [23], we have $\omega(G)=\frac{\mid G^{\prime}\mid}{p}(1+p)$. Now,
suppose $G$ is a CG-group. Then $\mid
G^{\prime}\mid+2\geq\omega(G)+1=\frac{\mid G^{\prime}\mid}{p}(1+p)+1$.
Consequently, we have $\mid G^{\prime}\mid=p$. Now, the result follows using
Proposition 3.5. ∎

The following proposition gives a sufficient condition for a finite group to
be a CG-group.

######  Proposition 3.7.

Let $G$ be a finite non-abelian group with an abelian normal subgroup of prime
index. Then

  1. (a)

$G$ is a CG-group.

  2. (b)

If $\frac{G}{Z(G)}$ is abelian, then $\frac{G}{Z(G)}$ is elementary abelian.

  3. (c)

If $\frac{G}{Z(G)}$ is non-abelian, then $\frac{G}{Z(G)}$ a CG-group. In
particular, if $\frac{G}{Z(G)}$ is of order $p^{r}$ for some prime $p$, then
$\mid Cent(G)\mid=p^{r-1}+2$.

.

###### Proof.

a) See [13, Theorem 2.3]. It may be mentioned here that [22, Theorem 3.4]) is
a particular case of this result, where the author obtained the result for
$p$-groups ($p$ a prime )only, noting that $\mid Cent(G)\mid=\omega(G)+1$ if
and only if $G$ is a CA-group ([1, Lemma 2.6]).

b) By [20, Theorem A], $G$ is a CA-group and consequently,
$\\{\frac{C(x)}{Z(G)}/x\in G\setminus Z(G)\\}$ is a partition of
$\frac{G}{Z(G)}$ (see [1, Remark 2.1]). Therefore if $\frac{G}{Z(G)}$ is
abelian, by [43, p. 571], we have $\frac{G}{Z(G)}$ is elementary abelian.

c) Suppose $\frac{G}{Z(G)}$ is non-abelian. Let $N$ be an abelian normal
subgroup of $G$ of prime index. Then $N=C(x)$ for some $x\in G\setminus Z(G)$.
In the present scenerio, we have $\frac{C(x)}{Z(G)}=C(xZ(G))$, and
consequently, $C(xZ(G))$ is an abelian normal subgroup of $\frac{G}{Z(G)}$ of
prime index. Therefore by [13, Theorem 2.3], $\frac{G}{Z(G)}$ is a CG-group.

Again, suppose $\frac{G}{Z(G)}$ is non-abelian of order $p^{r}$ for some prime
$p$. By [20, Theorem A], $G$ is a CA-group. Therefore $C(x)\cap C(y)=Z(G)$ for
any $x,y\in G\setminus Z(G),xy\neq yx$. Now, $N=C(x)$ is a centralizer of $G$
of index $p$. Clearly $C(x)$ will contain exactly $p^{r-1}$ distinct right
cosets of $Z(G)$. Therefore the number of right cosets of $Z(G)$ (other than
$Z(G)$) left for the remaining proper centralizers is
$p^{r}-p^{r-1}=p^{r-1}(p-1)$. In the present scenario, one can verify that any
proper centralizer other than $C(x)$ will contain exactly $p$ distinct right
cosets of $Z(G)$. Hence $\mid Cent(G)\mid=p^{r-1}+2$. ∎

As an application of the above proposition, we have the following result for
minimal non-abelian group. Recall that a minimal non-abelian group is a non-
abelian group all of whose proper subgroups are abelian. By [26, Aufgaben III.
5.14], we have if $G$ is a finite minimal non-abelian group, then $G$ can have
at the most two distinct prime divisors and if $G$ is not a prime power group,
then $G=PQ$, where $P$ is a cyclic $p$-Sylow subgroup of $G$ and $Q$ is the
elementary abelian minimal normal $q$-Sylow subgroup of $G$.

######  Proposition 3.8.

Let $G$ be a finite minimal non-abelian group. Then

  1. (a)

$G$ is a CG-group. In particular, (see [8, Proposition 2.2]) if $G$ is a
$p$-group ($p$ a prime), then we have $\mid Cent(G)\mid=p+2$. Otherwise, $G$
is a primitive $\mid Q\mid+2$ centralizer group, where $Q$ is the normal Sylow
subgroup of $G$.

  2. (b)

If $G$ is a $p$-group ($p$ a prime), then $\frac{G}{Z(G)}\cong C_{p}\times
C_{p}$. Otherwise, $\frac{G}{Z(G)}$ is a CG-group.

.

###### Proof.

a) Let $G$ be a finite minimal non-abelian group. If $G$ is a $p$-group, for
some prime $p$, then $G$ has an abelian normal subgroup of prime index and
hence by Proposition 3.7, $G$ is a CG-group.

Next, suppose $G$ is not a $p$-group. Let $P$ be a cyclic Sylow subgroup and
$Q$ be the elementary abelian Sylow subgroup of $G$ respectively. Then $P$ has
a normal subgroup $H$ of prime index and consequently, $QH$ is an abelian
normal subgroup of $G$ of prime index. Therefore by Proposition 3.7, $G$ is a
CG-group.

Now, if $G$ is a $p$-group, then $\mid Cent(G)\mid=p+2$, noting that we have
$\mid G^{\prime}\mid=p$ by [19, Lemma 2.5]. On the otherhand, in view of [9,
Lemma 3.1], we have $G$ is a primitive $\mid Q\mid+2$ centralizer group, where
$Q$ is the normal Sylow subgroup of $G$, niting that in the present scenario,
we have $G^{\prime}=Q$ and $\mid G^{\prime}\cap Z(G)\mid=1$.

b) If $G$ is a $p$-group, then the result follows from Peoposition 3.5, noting
that we have $\mid G^{\prime}\mid=p$ by [19, Lemma 2.5]. On the other hand, if
$G$ is not a prime power group, then $\frac{G}{Z(G)}$ is not a prime power
group, noting that in the present scenario, we have $Z(G)\subsetneq P$, where
$P$ is a cyclic Sylow subgroup of $G$. Now, the result follows from
Proposition 3.7. ∎

Recall that a Frobenius group $G$ is said to be minimal if no proper subgroup
of $G$ is Frobenius. In this connection we have the following result:

######  Proposition 3.9.

Let $G$ be a Frobenius group with kernel $K$ and complement $H$.

  1. (a)

If $G$ is minimal Frobenius, then $G$ is a CG-group,

  2. (b)

If $K$ is cyclic, then $G$ is a CG-group,

  3. (c)

If $H$ is abelian, then $G$ is a CG-group if and only if $G^{\prime}$ is
abelian.

.

###### Proof.

a) If $G$ is a minimal Frobenius group, then by [33, Theorem 3.2.14], $K$ is
elementary abelian and $H$ has prime order. Again, by [25, Theorem 2.2],
$G^{\prime}=K$. Therefore from the definition of Frobenius group, $G$ is a CG-
group.

b) If $K$ is cyclic, then in view of [18], $H$ is cyclic. In the present
scenario, by [25, Theorem 2.2], we have $G^{\prime}=K$. Therefore from the
definition of Frobenius group, $G$ is a CG-group.

c) If $H$ is abelian, then by [25, Theorem 2.2], we have $G^{\prime}=K$. Now,
if $G$ is a CG-group, then $G^{\prime}$ must be abelian. Again, if
$G^{\prime}$ is abelian, then from the definition of Frobenius group, $G$ is a
CG-group. ∎

######  Remark 3.10.

If $G$ is a finite solvable group in which centralizers of non-identity
elements are abelian, then $G$ is a Frobenius group with abelian kernel and
cyclic complement (see [38, Proposition 3.1.1, Proposition 1.2.4]). In this
connection we have the following result:

######  Proposition 3.11.

Let $G$ be a finite group such that $C(x)$ is abelian for every non-identity
element $x\in G$. Then $G$ is a CG-group if and only if $G$ is a Frobenius
group with abelian kernel and cyclic complement.

###### Proof.

Suppose $G$ is a CG-group. If $G$ is non-solvable, then by [2, Lemma 3.9], $G$
is simple and consequently using [17, Theorem 7], $G$ is not a CG-group.
Therefore $G$ is solvable. Now, the result follows from Remars 3.10.

Conversely, suppose $G$ is a Frobenius group with abelian kernel $K$ and
cyclic complement. By [25, Theorem 2.2], we have $G^{\prime}=K$. Therefore
from the definition of Frobenius group, $G$ is a CG-group. ∎

As an immediate consequence we have the following result:

######  Proposition 3.12.

If $G$ is a non-abelian group of order $pqr$, $p,q,r$ being primes (not
necessarily distinct), then $G$ is a CG-group.

###### Proof.

If $\mid Z(G)\mid=1$, then in view of Remark 3.2, $G$ is a solvable CA-group
and therefore by Proposition 3.11, $G$ is a CG-group. Again, if $\mid
Z(G)\mid\neq 1$, then $\mid\frac{G}{Z(G)}\mid$ is a product of two primes and
hence by [13, Corollary 2.5], $G$ is a CG-group. ∎

######  Proposition 3.13.

If $G$ is a non-abelian group of order $p^{4}$, $p$ being prime, then $G$ is a
CG-group.

###### Proof.

It is well known that if $G$ is a non-abelian group of order $p^{4}$, $p$
being prime, then $G$ has an abelian normal subgroup of prime index and hence
by Proposition 3.7, $G$ is a CG-group. ∎

We need the following result to prove our next proposition:

######  Proposition 3.14.

(Lemma 12.12 [27]) Let $G$ be a finite group. If $A\vartriangleleft G$ with
$A$ abelian and $G/A$ cyclic, then $\mid A\mid=\mid G^{\prime}\mid\mid A\cap
Z(G)\mid$.

We now give another sufficient condition for a finite group to be a CG-group.

######  Proposition 3.15.

Let $G$ be a finite group such that
$\frac{G}{Z(G)}=\frac{K}{Z(G)}\rtimes\frac{H}{Z(G)}$ is a Frobenius group with
$K$ and $H$ abelian. Then $G$ is a CG-group.

###### Proof.

Using the third isomorphic theorem, we get $\frac{G}{K}\cong\frac{H}{Z(G)}$.
Consequently, we have $K$ is an abelian normal subgroup of $G$ such that
$\frac{G}{K}$ is cyclic. In the present scenario, in view of Proposition 3.14,
$\mid K\mid=\mid G^{\prime}\mid\mid K\cap Z(G)\mid$ which forces
$\mid\frac{K}{Z(G)}\mid=\mid G^{\prime}\mid$. Therefore by [5, Proposition
3.1], $G$ is a CG-group. ∎

As a consequence we have the following corollaries:

######  Corollary 3.16.

Let $G$ be a finite group such that
$\frac{G}{Z(G)}=\frac{K}{Z(G)}\rtimes\frac{H}{Z(G)}$ is a Frobenius group with
$H$ abelian. If $G^{\prime}$ is abelian, then $G$ is a CG-group.

###### Proof.

Note that $H$ is abelian implies $\frac{H}{Z(G)}$ is cyclic and consequently,
using [25, Theorem 2.2], we have
$(\frac{G}{Z(G)})^{\prime}=\frac{G^{\prime}Z(G)}{Z(G)}=\frac{K}{Z(G)}$. Now,
if $G^{\prime}$ is abelian, then $G^{\prime}Z(G)$ is abelian and therefore, by
Proposition 3.15, $G$ is a CG-group. ∎

######  Corollary 3.17.

Let $G$ be a finite group such that $\frac{G}{Z(G)}$ is minimal Frobenius
group. If $G^{\prime}$ is abelian, then $G$ is a CG-group.

###### Proof.

In view of [33, Theorem 3.2.14], $\frac{G}{Z(G)}$ is a Frobenius group with
cyclic complement and therefore, the result follows from Corollary 3.16. ∎

######  Corollary 3.18.

Let $G$ be a finite group such that $\frac{G}{Z(G)}$ is a Frobenius group with
cyclic kernel. Then $G$ is a CG-group.

###### Proof.

In view of [18], $\frac{G}{Z(G)}$ has cyclic Frobenius complement and
therefore the result follows from Proposition 3.15. ∎

In this connection we would like to mention the following three lemmas:

######  Lemma 3.19.

Let $G$ be a finite group such that
$\frac{G}{Z(G)}=\frac{K}{Z(G)}\rtimes\frac{H}{Z(G)}$ is a Frobenius group with
$K$ and $H$ abelian. Then $\mid G^{\prime}\cap Z(G)\mid=1$.

###### Proof.

In the present scenario, from Proposition 3.15 and [25, Theorem 2.2], we have
$\mid G^{\prime}\mid=\mid\frac{K}{Z(G)}\mid=\mid(\frac{G}{Z(G)})^{\prime}\mid$
and hence the result follows. ∎

######  Lemma 3.20.

Let $G$ be a finite group such that $\frac{G}{Z(G)}$ is a Frobenius group with
cyclic kernel. Then $\mid G^{\prime}\cap Z(G)\mid=1$.

###### Proof.

In view of [18], $\frac{G}{Z(G)}$ has cyclic Frobenius complement and
therefore the result follows from Lemma 3.19. ∎

######  Lemma 3.21.

Let $G$ be a finite group such that
$\frac{G}{Z(G)}=\frac{K}{Z(G)}\rtimes\frac{H}{Z(G)}$ is a Frobenius group with
$H$ abelian. If $G^{\prime}$ is abelian, then $\mid G^{\prime}\cap
Z(G)\mid=1$.

###### Proof.

Using arguments similar to Corolary 3.16 and Lemma 3.19 we get the result. ∎

As a consequence of Proposition 3.15, we have the following three results:

######  Proposition 3.22.

Let $G$ be a finite group such that $G^{\prime}$ is of prime order $p$ and
$G^{\prime}\cap Z(G)=\\{1\\}$. Then $G$ is a CG-group.

###### Proof.

In the present scenario, we have $(\frac{G}{Z(G)})^{\prime}$ is of prime order
$p$ and $Z(\frac{G}{Z(G)})$ is of order $1$. Therefore in view of [34,
Proposition 5], $\frac{G}{Z(G)}$ is a Frobenius group with cyclic kernel and
cyclic complement. Therefore $G$ is a CG-group by Corollary 3.18. ∎

######  Proposition 3.23.

If $G$ is a finite group such that $\mid\frac{G}{Z(G)}\mid=pqr$ or $p^{2}q$
for any primes $p<q<r$, then $G$ is a CG-group.

###### Proof.

Suppose $\mid\frac{G}{Z(G)}\mid=pqr$. By [15, Proposition 2.5], we have $\mid
Z(\frac{G}{Z(G)})\mid=1$ and therefore in view of Remark 3.2, $\frac{G}{Z(G)}$
is a solvable CA-group. But then using Remark 3.10, we have $\frac{G}{Z(G)}$
is a Frobenius group with cyclic kernel and cyclic complement. Now, the result
follows using Proposition 3.15.

Next, suppose $\mid\frac{G}{Z(G)}\mid=p^{2}q$. For $q=3$, we have
$\frac{G}{Z(G)}\cong A_{4}$ or $D_{12}$. If $\frac{G}{Z(G)}\cong A_{4}$, then
$\frac{G}{Z(G)}\cong\frac{K}{Z(G)}\rtimes\frac{H}{Z(G)}$ is a Frobenius group
with kernel of order $4$ and complement of order $3$. In the present scenario,
we have $K$ and $H$ both are abelian, and therefore by Proposition 3.15, $G$
is a CG-group. On the otherhand if $\frac{G}{Z(G)}\cong D_{12}$, then by [13,
Proposition 2.8, Proposition 2.9], $G$ is a CG-group.

Now, we consider the case when $q>3$. Using [15, Proposition 2.11], we have
$\mid Z(\frac{G}{Z(G)})\mid=1$ or $p$. Now, If $\mid Z(\frac{G}{Z(G)})\mid=p$,
then by [15, Proposition 2.11], $\frac{G}{Z(G)}\cong C_{pq}\rtimes C_{p}$ and
so $G$ is a CG-group by [13, Proposition 2.9].

Again, if $\mid Z(\frac{G}{Z(G)})\mid=1$, then in view of Remark 3.2 and
Remark 3.10, $\frac{G}{Z(G)}$ is a Frobenius group with cyclic kernel and
cyclic complement of order $p^{2}$. Now, the result follows using Corollary
3.18. ∎

######  Proposition 3.24.

If $G$ is a finite group such that $\mid\frac{G}{Z(G)}\mid=pqr$ or $pq^{2}$,
$p<q<r$ be primes, then $\mid G^{\prime}\cap Z(G)\mid=1$.

###### Proof.

By [15, Proposition 2.5, Proposition 2.11], we have $\mid
Z(\frac{G}{Z(G)})\mid=1$ and hence in view of Remark 3.2 and Remark 3.10,
$\frac{G}{Z(G)}\cong\frac{K}{Z(G)}\rtimes\frac{H}{Z(G)}$ is a Frobenius group
with abelian kernel and cyclic complement. In the present scenario, one can
verify that $K$ and $H$ both are abelian. Therefore the result follows using
Lemma 3.19. ∎

For finite groups with central quotient of order $p^{3}$, $p$ being prime, we
have the following result:

######  Proposition 3.25.

Let $G$ be a finite group such that $\mid\frac{G}{Z(G)}\mid=p^{3}$, $p$ being
prime. Then $G$ is a CG-group if and only if $G$ has an abelian normal
subgroup of prime index.

###### Proof.

Let $G$ be a CG-group. In view of Remark 3.2, we have $G$ is a CA-group and
$C(x)\cap C(y)=Z(G)$ for any $x,y\in G\setminus Z(G),xy\neq yx$. Now, suppose
$G$ has no centralizer of index $p$. Then $\mid C(x)\mid=\frac{\mid
G\mid}{p^{2}}$ for all $x\in G\setminus Z(G)$ and consequently by [28],
$G=A\times P$, where $A$ is an abelian group and $P$ is a $p$-group. In the
present scenario, each proper centralizer of $G$ will contain exactly $p$
distinct right coset of $Z(G)$. Therefore $\mid Cent(G)\mid=p^{2}+p+2$. But
then $\mid G^{\prime}\mid=p(p+1)$, which is impossible. Therefore $G$ must
have a centralizer of prime index $p$, say, $C(y)$ for some $y\in G\setminus
Z(G)$. Clearly, in view of [30, Exercise 3 (p. 9)], $G$ cannot have another
centralizer of index $p$ and consequently $C(y)\lhd G$. Thus $G$ has an
abelian normal subgroup of prime index.

Conversely, if $G$ has an abelian normal subgroup of prime index, then by
Proposition 3.7, $G$ is a CG-group. ∎

As an application of Proposition 3.15, we also have the following result:

######  Proposition 3.26.

Let $G$ be a finite non-abelian solvable group. If $N_{G}(H)=H$ for all non-
abelian $H\leq G$, then $G$ is a CG-group.

###### Proof.

If $G$ is a nilpotent group, then in view of [19, Proposition 2.6], $G$ is a
minimal non-abelian $p$-group for some prime $p$ and therefore by Proposition
3.8, $G$ is a CG-group.

Next, suppose $G$ is non-nilpotent. Then by [19, Theorem 2.13], we have
$\frac{G}{Z(G)}$ is a Frobenius group with complement of prime order $p$.
Moreover, by [19, Proposition 2.2], we have $G^{\prime}$ is abelian. Therefore
by Corollary 3.16, $G$ is a CG-group. ∎

It may be mentioned here that all the examples of CG-groups given in the
earlier section are CA-groups. However, there exists CG-groups which are not
CA-groups. For example, it can be verify that $S_{4}$ is a CG-group but not a
CA-group. In this connection, we give the following result on CG-groups whose
central quotient is $S_{4}$.

######  Proposition 3.27.

Let $G$ be a finite group such that $\frac{G}{Z(G)}\cong S_{4}$. Then $G$ is a
CG-group if and only if $G^{\prime}\cong A_{4}$.

###### Proof.

If $G^{\prime}\cong A_{4}$, then we have $\mid G^{\prime}\cap Z(G)\mid=1$
noting that
$\mid(\frac{G}{Z(G)})^{\prime}\mid=\mid\frac{G^{\prime}}{G^{\prime}\cap
Z(G)}\mid=\mid A_{4}\mid$. Therefore by [9, Lemma 3.1], $\mid Cent(G)\mid=\mid
Cent(\frac{G}{Z(G)})\mid=\mid G^{\prime}\mid+2$.

Conversely, suppose $\mid Cent(G)\mid=\mid G^{\prime}\mid+2$. In view of [8,
Theorem 1.3], we have $\mid Cent(G)\mid=\mid Cent(\frac{G}{Z(G)})\mid$ and
consequently, $\mid
G^{\prime}\mid=\mid(\frac{G}{Z(G)})^{\prime}\mid=\mid\frac{G^{\prime}}{G^{\prime}\cap
Z(G)}\mid$, forcing $\mid G^{\prime}\cap Z(G)\mid=1$. Therefore we have
$A_{4}\cong(\frac{G}{Z(G)})^{\prime}=G^{\prime}$. ∎

######  Remark 3.28.

Recall that a minimal non-nilpotent group is a non-nilpotent group all of
whose proper subgroups are nilpotent. If $G$ is a finite minimal non-nilpotent
group then $G$ is a solvable group of order $p^{m}q^{n}$ ($p,q$ are distinct
primes) with a unique Sylow $p$-subgroup $P$ and a cyclic Sylow $q$-subgroup
$Q$. Moreover, we have $G^{\prime}=P$ (see [37]).

######  Proposition 3.29.

Let $G$ be a finite minimal non-nilpotent group.

  1. (a)

If $G^{\prime}$ is abelian, then $G$ is a CG-group.

  2. (b)

$\frac{G}{Z(G)}$ is a CG-group.

.

###### Proof.

a) By Remark 3.28, we have $G^{\prime}=P$. Again, by [7, Lemma 2.3], we have
$\frac{G}{Z(G)}=\frac{PZ(G)}{Z(G)}\rtimes\frac{QZ(G)}{Z(G)}$ is a Frobenius
group with cyclic complement. Now, if $G^{\prime}$ is abelian, then $PZ(G)$
will be abelian and hence $G$ is a CG-group by Proposition 3.15.

b) Let $G$ be a minimal non-nilpotent group. By [7, Lemma 2.3],
$\frac{G}{Z(G)}$ is a Frobenius group with elementary abelian kernel
$\frac{K}{Z(G)}$ and complement of prime order. Again, by [25, Theorem 2.2],
$(\frac{G}{Z(G)})^{\prime}=\frac{K}{Z(G)}$. Therefore from the definition of
Frobenius group, $\frac{G}{Z(G)}$ is a CG-group. ∎

######  Proposition 3.30.

Let $G$ be a finite group such that $\frac{G}{Z(G)}$ is minimal non-nilpotent.
If $G^{\prime}$ is abelian, then $G$ is a CG-group.

###### Proof.

In view of Remark 3.28, we have
$\frac{G}{Z(G)}=\frac{G^{\prime}Z(G)}{Z(G)}\rtimes\frac{Q}{Z(G)}$, where $Q$
is a subgroup of $G$ such that $\frac{Q}{Z(G)}$ is cyclic. Now, if
$G^{\prime}$ is abelian, then $G^{\prime}Z(G)$ will be abelian and
consequently, by third isomprphic theorem, we have
$\frac{G}{G^{\prime}Z(G)}\cong\frac{Q}{Z(G)}$. Therefore in view of
Proposition 3.14, we have $\mid G^{\prime}Z(G)\mid=\mid G^{\prime}\mid\mid
G^{\prime}Z(G)\cap Z(G)\mid=\mid G^{\prime}\mid\mid Z(G)\mid$ and hence $\mid
G^{\prime}\cap Z(G)\mid=1$. Therefore using [9, Lemma 3.1] and Proposition
3.29, we have $\mid Cent(G)\mid=\mid
Cent(\frac{G}{Z(G)})\mid=\mid(\frac{G}{Z(G)})^{\prime}\mid+2=\mid
G^{\prime}\mid+2$. ∎

######  Proposition 3.31.

Let $G$ be a finite group such that $\frac{G}{Z(G)}$ is minimal non-abelian
which is not a $p$-group, $p$ being prime. If $G^{\prime}$ is abelian, then
$G$ is a CG-group.

###### Proof.

In the present scenario, in view of [8, Theorem 1.1], $\frac{G}{Z(G)}$ is a
Frobenius group with cyclic complement. Now, the result follows using
Corollary 3.16. ∎

Finally, we conclude the paper with a counterexample to the following
conjecture [29, Conjecture 2.3] which is also a counterexample to [29,
Question 1.2] :

######  Conjecture 3.32.

Let $G$ and $S$ be finite groups. Is it true that if
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(S)\mid$ and $\mid
G^{\prime}\mid=\mid S^{\prime}\mid$, then $G$ is isoclinic to $S$?

Consider any non-abelian group $G$ of order $27$ and $S_{3}$. In view of [13,
Corollary 2.5], we have
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(S_{3})\mid=5$ and
$\mid G^{\prime}\mid=\mid{S_{3}}^{\prime}\mid=3$. But $G$ is not isoclinic to
$S_{3}$.

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