# On groups with same number of centralizers

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

In this paper, among other results, we give some sufficient conditions for
every non-abelian subgroup of a group to be isoclinic with the group itself.
It is also seen that under certain conditions, two groups have same number of
element centralizers implies they are isoclinic. We prove that if $G$ is any
group having $4,5,7$ or $9$ element centralizers and $H$ is any non-abelian
subgroup of $G$, then
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ and
$G^{\prime}\cong H^{\prime}\cong C_{2},C_{3},C_{5}$ or $C_{7}$ respectively.
Furthermore, it is proved that if $G$ is any group having
$n\in\\{4,5,6,7,9\\}$ element centralizers, then $\mid G^{\prime}\mid=n-2$.

###### Key words and phrases:

Finite group, Centralizer, Isoclinic groups

###### 2010 Mathematics Subject Classification:

20D60, 20D99

##  1\. Introduction

Given any group $G$, let $\operatorname{Cent}(G)$ and $nacent(G)$ denotes,
respectively, the set of centralizers and the set of non-abelian centralizers
of elements of $G$. A group $G$ is said to be $n$-centralizer if
$\mid\operatorname{Cent}(G)\mid=n$. In 1994 Belcastro and Sherman [13]
introduced the notion of $n$-centralizer groups and since then the influence
of $\operatorname{Cent}(G)$ on the structure of group have been studied
extensively. See [19, 7, 17, 23, 11, 12] for recent advances on this and
related areas. Perhaps motivated by the impact of
$\mid\operatorname{Cent}(G)\mid$ on the group, Ashrafi and Taeri [5] in 2005
asked the following question which was disproved by Zarrin [30]: Let $G$ and
$H$ be finite simple groups. Is it true that if
$|\operatorname{Cent}(G)|=|\operatorname{Cent}(H)|$, then $G$ is isomorphic to
$H$? Amiri and Rostami [3] in 2015 put forward the following analogue question
which was also disproved by Khoramshahi and Zarrin [23]: Let $G$ and $H$ be
finite simple groups. Is it true that if $|nacent(G)|=|nacent(H)|$, then $G$
is isomorphic to $H$? In this context we have the following natural question:

######  Question 1.1.

What can be said about the relationship between two groups if they have the
same number of element centralizers.

It may be mentioned here that if an $n$-centralizer group $G$ is isoclinic
with a group $H$, then $|\operatorname{Cent}(G)|=|\operatorname{Cent}(H)|$
(see [31, 24]). However, the converse is not true in general. For example, if
$G$ is a non-abelian group of order $27$, then
$|\operatorname{Cent}(G)|=|\operatorname{Cent}(S_{3})|=5$, but $G$ and $S_{3}$
are not isoclinic. The authors in [23] studied and obtained some conditions
under which the converse of this statement holds. In this paper, we continue
with Question 1.1 and improve some earlier results. We obtain some sufficient
conditions for every non-abelian subgroup of a group to be isoclinic with the
group itself. In particular, it is seen that any non-abelian subgroup of a $4$
or $5$-centralizer group is isoclinic with the group itself, which improves
[23, Theorem 3.5]. It is also proved that any two arbitrary $4$-centralizer
groups are isoclinic and any two arbitrary nilpotent $5,7$ or $9$-centralizer
groups are isoclinic. We obtain that if $H$ is any non-abelian subgroup of an
$n$-centralizer group $G$, where $n=4,5,7$ or $9$, then
$|\operatorname{Cent}(G)|=|\operatorname{Cent}(H)|$ and $G^{\prime}\cong
H^{\prime}\cong C_{2},C_{3},C_{5}$ or $C_{7}$ respectively. For any subgroup
$H$ of an arbitrary $8$-centralizer group $G$, it is observed that
$|\operatorname{Cent}(G)|=|\operatorname{Cent}(H)|$ implies $G$ is isoclinic
with $H$. Given any $n$-centralizer group $G$ with $n\in\\{4,5,6,7,9\\}$, we
see that $\mid G^{\prime}\mid=n-2$. A finite group is said to be of conjugate
type $(m,1)$ if every proper element centralizer is of index $m$. For any two
finite groups $G$ and $H$ of conjugate type $(p,1)$, $p$ a prime, it is proved
that $|\operatorname{Cent}(G)|=|\operatorname{Cent}(H)|$ implies $G$ is
isoclinic with $H$. Among other results, we prove that if $G$ is any finite
$(n+2)$-centralizer group of conjugate type $(n,1)$, then $G$ is a CA-group
(i.e., every proper element centralizer of $G$ is abelian) and
$\frac{G}{Z(G)}$ is elementary abelian of order $n^{2}$, which improves [4,
Theorem 3.3].

Throughout this paper, for a group $G$, $Z(G)$ and $G^{\prime}$ denotes its
center and commutator subgroup respectively, $C_{G}(x)$ denotes the
centralizer of $x\in G$ (however, if there is no confusion in the context then
we simply write $C(x)$ in place of $C_{G}(x)$), $C_{n}$ denotes the cyclic
group of order $n$ and $D_{2n}$ denotes the dihedral group of order $2n$. Some
results of this paper holds for finite groups only and we have specifically
mentioned it whenever necessary.

##  2\. Definitions and basic results

We begin with the notion of isoclinism between two groups introduced by P.
Hall [18] in 1940. Two groups $G$ and $H$ are said to be isoclinic if there
are two isomorphisms $\varphi:G/Z(G)\longrightarrow H/Z(H)$ and
$\phi:G^{\prime}\longrightarrow H^{\prime}$ such that if

$\varphi(g_{1}Z(G))=h_{1}Z(H)\;\;\text{and}\;\;\varphi(g_{2}Z(G))=h_{2}Z(H)$

with $g_{1},g_{2}\in G,h_{1},h_{2}\in H$, then

$\phi([g_{1},g_{2}])=[h_{1},h_{2}].$

Isoclinism is an equivalence relation weaker than isomorphism having many
family invariants. Here we list some of the invariants concerning the element
centralizers of two isoclinic groups.

Recall that a group $G$ is called an F-group if every non-central element
centralizer contains no other element centralizer and a CA-group if all non-
central element centralizers are abelian. Finite groups having exactly two
class sizes are called I-groups which are direct product of an abelian group
and a group of prime power order [22]. Two elements of a group are said to be
$z$-equivalent or in the same $z$-class if their centralizers are conjugate in
the group. Being $z$-equivalent is an equivalence relation which is weaker
than conjugacy relation. A $z$-equivalence class is called a $z$-class. In the
following result, $\omega(G)$ denotes the size of a maximal set of pairwise
non-commuting elements of a group $G$.

######  Proposition 2.1.

If an $n$-centralizer group $G$ is isoclinic with a group $H$, then

  1. (a)

$\omega(G)=\omega(H)$ ([31, Lemma 2.1]).

  2. (b)

$z$-classes in $G$=$z$-classes in $H$ ([25, Theorem 2.2]).

  3. (c)

$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ ([31, Lemma
3.2], [24, Theorem A]).

  4. (d)

$\mid nacent(G)\mid=\mid nacent(H)\mid$.

  5. (e)

$G$ is a CA-group implies $H$ is also a CA-group.

  6. (f)

$G$ is an F-group implies $H$ is also an F-group.

  7. (g)

$G$ is an I-group implies $H$ is also an I-group ([18], [21, Proposition
2.2]).

###### Proof.

d) Let $\varphi:G/Z(G)\longrightarrow H/Z(H)$ be the isomorphism. Then
$\varphi$ induces a bijection between the subgroups of $G$ containing $Z(G)$
and the subgroups of $H$ containing $Z(H)$ and the corresponding subgroups are
isoclinic [18, pp. 134]. For any $x\in G$, consider its centralizer $C_{G}(x)$
which contains $Z(G)$. In view of proof of [24, Theorem A], the corresponding
subgroup of $H$ containing $Z(H)$ is $C_{H}(y)$, where $yZ(H)=\varphi(xZ(G))$.
Therefore $C_{G}(x)$ is isoclinic with $C_{H}(y)$. Hence the result follows.

e) It follows from part (d)

f) Let $\varphi:G/Z(G)\longrightarrow H/Z(H)$ be the isomorphism. Suppose $H$
is not an F-group. Then $C_{H}(a)<C_{H}(b)$ for some $a,b\in H\setminus Z(H)$.
Therefore $\frac{C_{H}(a)}{Z(H)}<\frac{C_{H}(b)}{Z(H)}$ and consequently, in
view of proof of [24, Theorem A], we have
$\frac{C_{G}(x)}{Z(G)}<\frac{C_{G}(y)}{Z(G)}$ for some $x,y\in G\setminus
Z(G)$, where $\varphi(\frac{C_{G}(x)}{Z(G)})=\frac{C_{H}(a)}{Z(H)}$ and
$\varphi(\frac{C_{G}(y)}{Z(G)})=\frac{C_{H}(b)}{Z(H)}$. It now follows that
$C_{G}(x)<C_{G}(y)$, which implies $G$ is not an F-group. ∎

The following theorems will be used to obtain some of our results.

######  Theorem 2.2.

(p.135 [18]) Every group is isoclinic to a group whose center is contained in
the commutator subgroup.

######  Theorem 2.3.

([18], [21, Proposition 2.2]) Let $G$ and $H$ be finite $p$-groups ($p$ a
prime). Suppose $G$ is isoclonic with $H$. Then $G$ and $H$ are groups of the
same conjugate type.

######  Theorem 2.4.

(Theorem 11 [24], Theorem 3.3 [31]) The representatives of the families of
isoclinic groups with $n$-centralizers ($n\neq 2,3$) can be chosen to be
finite groups.

For any subgroup $H$ of $G$, it is easy to see that $C_{H}(x)=C_{G}(x)\cap H$,
for any $x\in H$. This gives the following result:

######  Lemma 2.5.

Let $H$ be a subgroup of $G$ such that $H\cap Z(G)\lneq Z(H)$. Then the number
of centralizers of $G$ produced by the elements of $H$ is atleast
$\mid\operatorname{Cent}(H)\mid+1$.

###### Proof.

Clearly, the number of centralizers of $G$ produced by elements of $H$ is
equal to the number of centralizers of $G$ produced by the elements of $H\cap
Z(G)$\+ the number of centralizers of $G$ produced by the elements of
$H\setminus(H\cap Z(G))\geq 1+\mid\operatorname{Cent}(H)\mid$ (note that
elements of $H$ that have the same centralizers in $H$ may have different
centralizers in $G$). ∎

######  Lemma 2.6.

Let $G$ be a finite group and $p$ be a prime. If $G$ has a non-central element
of order $p$, then $\mid\operatorname{Cent}(G)\mid\geq p+2$.

###### Proof.

Let $x\in G\setminus Z(G)$ be an element of order $p$. Let $a\in G\setminus
C(x)$. Clearly, $ax^{i}\in G\setminus C(x)$ for any $i$. Consider the set
$X=\\{x,a,ax,ax^{2},\dots,ax^{p-1}\\}$. Observe that if
$ax^{i}ax^{j}=ax^{j}ax^{i}$ for some $0\leq i<j\leq p-1$, then $a\in
C(x^{j-i})=C(x)$, a contradiction (noting that $gcd((j-i),o(x))=1$). Therefore
$X$ is a set of pairwise non-commuting elements of $G$ and $\mid X\mid=p+1$.
Hence $\mid\operatorname{Cent}(G)\mid\geq p+2$. ∎

For any finite group $G$, the author in [4, Lemma 3.1] proved that if
$G^{\prime}\cap Z(G)=\\{1\\}$, then
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(\frac{G}{Z(G)})\mid$.
However, for any arbitrary $n$-centralizer group we have the following general
result:

######  Proposition 2.7.

Let $G$ be any $n$-centralizer group and $N\unlhd G$. If $N\cap
G^{\prime}=\\{1\\}$, then
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(\frac{G}{N})\mid$.

###### Proof.

In view of [18, pp. 134] and Proposition 2.1 we have the result. ∎

In response to a question raised by Belcastro and Sherman [13], namely,
whether there exists an $n(\neq 2,3)$-centralizer group, Ashrafi showed [4,
Proposition 2.1] that there exists $n$-centralizer groups for $n\neq 2,3$. In
this connection, we have the following result which implies we can say
something more than Ashrafi’s result. It also improves [1, Proposition 2.2]
and [6, Lemma 2]. Furthermore, it improves [19, Example 16], namely, there
exists a $2^{r}$-centralizer CA-group for every $r>1$. In the following result
$C_{n}{\rtimes}_{\theta}C_{p}$ denotes semidirect product of $C_{n}$ and
$C_{p}$, where $\theta:C_{p}\longrightarrow\operatorname{Aut}(C_{n})$ is a
homomorphism.

######  Proposition 2.8.

Given any group $G$, suppose $\frac{G}{Z(G)}$ be non-abelian and $p$ be a
prime. If $\frac{G}{Z(G)}\cong C_{n}{\rtimes}_{\theta}C_{p}$, then $G$ is an
$(n+2)$-centralizer CA-group.

###### Proof.

In view of Theorem 2.4, $G$ is isoclinic with a finite group. Now, the result
follows using Proposition 2.1 and [8, Proposition 2.9 and Lemma 2.10]. ∎

Recall that the generalized quaternion group $Q_{4m}$ has the presentation
$\langle a,b\mid a^{2m}=1,b^{2}=a^{m},bab^{-1}=a^{-1}\rangle,m\geq 2$.

######  Corollary 2.9.

There exists $n$-centralizer CA-groups for $n\geq 4$.

###### Proof.

In view of Proposition 2.8, $Q_{4(n-2)},n\geq 4$ is an $n$-centralizer CA-
group by noting that $\frac{Q_{4(n-2)}}{Z(Q_{4(n-2)})}\cong D_{2(n-2)}$ for
any $n\geq 4$. ∎

######  Remark 2.10.

Let $p$ be a prime. A finite $p$-group $G$ is said to be a special $p$-group
of rank $k$ if $G^{\prime}=Z(G)$ is elementary abelian of order $p^{k}$ and
$\frac{G}{G^{\prime}}$ is elementary abelian. Furthermore, a finite group $G$
is extraspecial if $G$ is a special $p$-group and $\mid G^{\prime}\mid=\mid
Z(G)\mid=p$. It is well known that every extraspecial $p$-group has order
$p^{2a+1}$ for some positive integer $a$. Furthermore, for every prime $p$ and
every positive integer $a$, there exists, upto isomorphism, exactly two
extraspecial groups of order $p^{2a+1}$. Moreover, any two extraspecial groups
of same order are isoclinic (see [28, pp. 7]). Again, if $G$ is any group and
$A$ is an abelian group, then $G$ and $G\times A$ are isoclinic (see [18, pp.
135]).

In this context we have the following result.

######  Proposition 2.11.

There exists $2^{2n}$-centralizer F-groups which are not CA-groups for $n>1$.

###### Proof.

Let $G$ be an extraspecial group of order $2^{2n+1},n>1$. Then in view of [12,
Proposition 2.26] and [11, Proposition 3.13], $G$ is an $2^{2n}$-centralizer
F-group which is not a CA-group. ∎

##  3\. Main results

The following key result helps in determining whether a given group is CA or
not.

######  Proposition 3.1.

An arbitrary group $G$ is a CA-group if and only if $Z(H)=Z(G)\cap H$ for any
non-abelian subgroup $H$ of $G$. In particular,
$\frac{H}{Z(H)}=\frac{H}{Z(G)\cap H}\cong\frac{HZ(G)}{Z(G)}\leq\frac{G}{Z(G)}$
for any non-abelian subgroup $H$ of a CA-group $G$.

###### Proof.

Let $G$ be a CA-group and $H$ be a non-abelian subgroup of $G$. If $a,b\in H$
be such that $ab\neq ba$, then $a,b\in H\setminus(Z(G)\cup Z(H))$. It is easy
to verify that $Z(H)=C_{H}(a)\cap C_{H}(b)=C_{G}(a)\cap H\cap C_{G}(b)\cap
H=Z(G)\cap H$. Conversely, if $Z(H)=Z(G)\cap H$ for any non-abelian subgroup
$H$ of $G$, then $G$ is a CA-group. For if $C_{G}(x)$ is non-abelian for some
$x\in G\setminus Z(G)$, then $C_{G}(x)\cap Z(G)=Z(G)\subsetneq Z(C_{G}(x))$,
which is a contradiction. Last part is trivial. ∎

######  Corollary 3.2.

If $Z(G^{\prime})=\\{1\\}$ for any CA-group $G$, then $G$ is isoclinic with
$\frac{G}{Z(G)}$. In particular, if $G$ is $n$-centralizer, then
$\frac{G}{Z(G)}$ is also an $n$-centralizer CA-group.

###### Proof.

Using Proposition 3.1, [18, pp. 134] and Proposition 2.1 we have the result. ∎

As an application of Proposition 3.1, we also have the following result.

######  Proposition 3.3.

If $H$ is a non-abelian subgroup of $G$ with $\frac{G}{Z(G)}\cong D_{2n}$,
then

  1. (a)

$G$ is an $(n+2)$-centralizer CA-group.

  2. (b)

$\frac{H}{Z(H)}\cong D_{2n/d}$ for some divisor $d$ of $n$.

  3. (c)

$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ implies $G$ is
isoclinic with $H$.

###### Proof.

a) It follows from Proposition 2.8.

b) By part (a) $G$ is a CA-group and consequently, using Proposition 3.1 we
have $\frac{H}{Z(H)}=\frac{H}{Z(G)\cap
H}\cong\frac{HZ(G)}{Z(G)}\leq\frac{G}{Z(G)}$. Now, using [16, Theorem 3.1] we
have the result.

(c) In view of part (b), $\frac{H}{Z(H)}\cong\frac{HZ(G)}{Z(G)}\cong D_{2n/d}$
for $d\mid n$. Now, if $\mid\operatorname{Cent}(H)\mid=n+2$, then by part (a),
we have $HZ(G)=G$ and consequently $G$ is isoclinic with $H$ by [26, Lemma
2.7]. ∎

Let $H$ be a subgroup of $G$. The author in [26, Lemma 2.7] proved that if
$G=HZ(G)$, then $G$ and $H$ are isoclinic, and if $H$ is finite then the
converse is also true. We have the following general result for an arbitrary
$n$-centralizer group.

######  Proposition 3.4.

Let $G$ be any $n$-centralizer group and $H\leq G$. Then $G=HZ(G)$ iff $G$ is
isoclinic with $H$.

###### Proof.

If $G=HZ(G)$, then by [26, Lemma 2.7], we have $G$ is isoclinic with $H$.
Conversely, if $G$ is isoclinic with $H$, then
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ by Proposition
2.1. Therefore using Lemma 2.5, we have $Z(H)=H\cap Z(G)$, which implies
$\frac{H}{Z(H)}=\frac{H}{Z(G)\cap H}=\frac{HZ(G)}{Z(G)}\cong\frac{G}{Z(G)}$
and thus $G=HZ(G)$. ∎

Using arguments similar to Proposition 3.4, we also have the following result:

######  Proposition 3.5.

Let $G$ be any $n$-centralizer group and $H\leq G$ be such that
$\frac{G}{Z(G)}\cong\frac{H}{Z(H)}$. Then
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ iff $G$ is
isoclinic with $H$.

For groups $G_{1}$ and $G_{2}$ in [1, pp.56], we have
$\frac{G_{1}}{Z(G_{1})}\cong\frac{G_{2}}{Z(G_{2})}\cong C_{2}\times
C_{2}\times C_{2}$, $\mid\operatorname{Cent}(G_{1})\mid=6$ and
$\mid\operatorname{Cent}(G_{2})\mid=8$; which implies $G_{1}$ and $G_{2}$ are
not isoclinic by Proposition 2.1. However, for some special situations we have
the following result:

######  Proposition 3.6.

Let $G$ be any $n$-centralizer CA-group and $H\leq G$. Then
$\frac{G}{Z(G)}\cong\frac{H}{Z(H)}$ iff $G$ is isoclinic with $H$. In
particular, $\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$.

###### Proof.

In view of Proposition 3.1, $Z(H)=Z(G)\cap H$ and so
$\frac{G}{Z(G)}\cong\frac{H}{Z(H)}=\frac{H}{Z(G)\cap H}=\frac{HZ(G)}{Z(G)}$.
Therefore $G=HZ(G)$, and hence $G$ is isoclinic with $H$ by [26, Lemma 2.7].
Converse is trivial. Last part follows from Proposition 2.1. ∎

######  Theorem 3.7.

If $H$ is a non-abelian subgroup of $G$ with $\mid\frac{G}{Z(G)}\mid=pq$
($p\leq q$ are primes), then $G$ is isoclinic with $H$. In particular,
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid=q+2$.

###### Proof.

In view of Proposition 3.1, $\frac{H}{Z(H)}=\frac{H}{Z(G)\cap
H}\cong\frac{HZ(G)}{Z(G)}\leq\frac{G}{Z(G)}$ and so $HZ(G)=G$ by noting that
$G$ is a CA-group. Now, using [26, Lemma 2.7], $G$ is isoclinic with $H$. Last
part follows using Theorem 2.4, Proposition 2.1 and [8, Corollary 2.5]. ∎

The following result will be used in the next theorem.

######  Proposition 3.8.

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 $\mid\operatorname{Cent}(G)\mid=\mid
G^{\prime}\mid+2$.

###### 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 [20, Lemma
12.12], $\mid K\mid=\mid G^{\prime}\mid\mid K\cap Z(G)\mid$ which forces
$\mid\frac{K}{Z(G)}\mid=\mid G^{\prime}\mid$. Therefore by [2, Proposition
3.1], $\mid\operatorname{Cent}(G)\mid=\mid G^{\prime}\mid+2$. ∎

In the following result, which improves [23, Theorem 3.5], $(C_{m},C_{n})$
denotes the Frobenius group with complement $C_{m}$ and kernel $C_{n}$.

######  Theorem 3.9.

Let $H$ be a non-abelian subgroup of an $n$-centralizer group $G$.

  1. (a)

If $n=4$ or $5$, then $G$ is isoclinic with $H$.

  2. (b)

If $n=4,5,7$ or $9$, then
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ and
$G^{\prime}\cong H^{\prime}\cong C_{2},C_{3},C_{5}$ or $C_{7}$ respectively.

  3. (c)

If $n=8$, then $\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$
implies $G$ is isoclinic with $H$.

###### Proof.

a) We have $\mid\frac{G}{Z(G)}\mid=4,6$ or $9$ by [31, Theorem 3.5] and hence
the result follows using Theorem 3.7.

b) Following [31, Theorem 3.5] and applying similar arguments to [9, Theorem
2.6] , we have $\frac{G}{Z(G)}\cong(C_{4},C_{5})$, $(C_{6},C_{7})$ or
$\mid\frac{G}{Z(G)}\mid\in\\{4,6,9,10,14,21,25,49\\}$. In the present
scenario, using [10, Lemma 2.1] and Proposition 3.1, we have $Z(H)=Z(G)\cap H$
and hence $\frac{H}{Z(H)}=\frac{H}{Z(G)\cap
H}\cong\frac{HZ(G)}{Z(G)}\leq\frac{G}{Z(G)}$.

Now, if $\mid\frac{G}{Z(G)}\mid\in\\{4,6,9,10,14,21,25,49\\}$, then using
Theorem 3.7 we have
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$.

Next, suppose $\frac{G}{Z(G)}\cong(C_{4},C_{5})$. If
$\frac{HZ(G)}{Z(G)}<\frac{G}{Z(G)}$, then $\mid\frac{H}{Z(H)}\mid=10$ (by
noting that $2$-Sylow subgroup of $\frac{G}{Z(G)}$ is cyclic) and so
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ using [31,
Theorem 3.5]. On the other hand if $\frac{HZ(G)}{Z(G)}=\frac{G}{Z(G)}$, then
$HZ(G)=G$ and hence
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ using
Proposition 3.4 and Proposition 2.1.

Finally, suppose $\frac{G}{Z(G)}\cong(C_{6},C_{7})$. If
$\frac{HZ(G)}{Z(G)}<\frac{G}{Z(G)}$, then $\mid\frac{H}{Z(H)}\mid=14$ or $21$
by noting that $\frac{G}{Z(G)}$ cannot have a non-abelian subgroup of order
$6$ and consequently, applying arguments of [31, Theorem 3.5] to [9, Theorem
2.6] we have $\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$.
On the other hand if $\frac{HZ(G)}{Z(G)}=\frac{G}{Z(G)}$, then $HZ(G)=G$ and
hence $\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ using
Proposition 3.4 and Proposition 2.1.

Second part follows using Proposition 2.1, Theorem 2.4, [8, Theorem 2.3] and
Proposition 3.8.

c) In view of [31, Theorem 3.5], we have $\mid\frac{G}{Z(G)}\mid=8$ or $12$.
In the present scenario, by [10, Lemma 2.1] and Proposition 3.1, we have
$\frac{H}{Z(H)}=\frac{H}{Z(G)\cap
H}\cong\frac{HZ(G)}{Z(G)}\leq\frac{G}{Z(G)}$. Now using [31, Theorem 3.5]
again, we have $HZ(G)=G$ and so by Proposition 3.4, $G$ is isoclinic with $H$.
∎

Note that the $6$-centralizer group $Q_{16}$ has a $4$-centralizer subgroup,
namely $Q_{8}$, which is not isoclinic with $Q_{16}$. The $7$-centralizer
group $(C_{4},C_{5})$ has a $7$-centralizer subgroup of order $10$ which is
not isoclinic with $(C_{4},C_{5})$. The $8$-centralizer group $Q_{24}$ has a
$4$-centralizer subgroup, namely $Q_{8}$. Again, the $9$-centralizer group
$(C_{6},C_{7})$ has a $9$-centralizer subgroup of order $21$ which is not
isoclinic with $(C_{6},C_{7})$.

A finite $p$-group ($p$ a prime) $G$ is semi-extraspecial if for every maximal
subgroup $N$ in $Z(G)$ the quotient $\frac{G}{N}$ is extraspecial. It is known
that every semi-extraspecial $p$-group is special. Furthermore, a group $G$ is
said to be ultraspecial if $G$ is semi-extraspecial and $\mid
G^{\prime}\mid=\sqrt{\mid G:G^{\prime}\mid}$.

######  Proposition 3.10.

If $G$ is an $n$-centralizer group with $n\in\\{4,5,6,7,9\\}$, then $\mid
G^{\prime}\mid=n-2$.

###### Proof.

In view of Proposition 2.1 and Theorem 2.4 without any loss we may assume that
$G$ is a finite group.

Now, suppose $n=6$. Using [31, Theorem 3.5], we have $\frac{G}{Z(G)}\cong
D_{8},A_{4},C_{2}\times C_{2}\times C_{2}$ or $C_{2}\times C_{2}\times
C_{2}\times C_{2}$. If $\mid\frac{G}{Z(G)}\mid=8$, then in view of [10,
Proposition 2.14], $G$ has an abelian centralizer of index $2$ and
consequently, using [8, Theorem 2.3], we have $\mid G^{\prime}\mid=4$. Again,
if $\frac{G}{Z(G)}\cong A_{4}$, then by [10, Proposition 2.12], $G$ has an
abelian normal centralizer of index $3$ and consequently, using [8, Theorem
2.3], we have $\mid G^{\prime}\mid=4$. Finally, if
$\mid\frac{G}{Z(G)}\mid=16$, then in view of [11, Proposition 3.21], $G$ is
isoclinic with an ultraspecial group of order $64$ and hence $\mid
G^{\prime}\mid=4$. Now, the result follows using Theorem 3.9. ∎

Note that for the group $G_{2}$ in [1, pp.56], we have
$\frac{G_{2}}{Z(G_{2})}\cong C_{2}\times C_{2}\times C_{2}$ and
$\mid\operatorname{Cent}(G_{2})\mid=8$. In view of [25, Lemma 3.1], $G_{2}$ is
isoclinic with a finite $2$-group and hence $\mid G_{2}^{\prime}\mid\neq 6$.
From the above result, we can also see that if $G$ and $H$ are $n$-centralizer
groups with $n\in\\{4,5,7,9\\}$, then $G^{\prime}\cong H^{\prime}$. However,
$D_{16}$ and $A_{4}$ are $6$-centralizer groups with $D_{16}^{\prime}\cong
C_{4}$ and $A_{4}^{\prime}\cong C_{2}\times C_{2}$.

######  Proposition 3.11.

Let $G$ and $H$ be two finite groups of conjugate type $(p,1)$, $p$ a prime.
If $\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$, then $G$
is isoclinic with $H$.

###### Proof.

In view of Ito [22], without any loss we may assume that $G$ is a $p$-group.
Furthermore, using Theorem 2.2, $G$ is isoclinic with a group $G_{1}$ such
that $Z(G_{1})\subseteq G_{1}^{\prime}$. Note that since $Z(G_{1})$ is finite,
therefore $G_{1}$ is finite. It now follows using Theorem 2.3 and [21,
Proposition 3.1] that $G_{1}$ is an extraspecial $p$-group. Similarly, we can
see that $H$ is isoclinic with an extraspecial $p$-group $G_{2}$. Now, suppose
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$. In the
present scenario, using Proposition 2.1, we have
$\mid\operatorname{Cent}(G_{1})\mid=\mid\operatorname{Cent}(G_{2})\mid$ and
consequently, applying [11, Proposition 3.13] if follows that $\mid
G_{1}\mid=\mid G_{2}\mid$. Now the result follows using Remark 2.10. ∎

######  Proposition 3.12.

Let $G$ be an extraspecial $p$-group of order $p^{k}$ for some $k$ and prime
$p$. If $H$ is a subgroup of $G$ such that
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$, then $G=H$.

###### Proof.

Since $\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$,
therefore applying Lemma 2.5, we have $H\cap Z(G)=Z(H)=Z(G)$ and consequently,
$\frac{H}{Z(H)}=\frac{H}{Z(G)\cap
H}\cong\frac{HZ(G)}{Z(G)}\leq\frac{G}{Z(G)}$. It now follows that $H$ is an
extraspecial $p$-group of order $p^{l}$ for some $l$. Therefore by [11,
Proposition 3.13], $k=l$ and hence $G=H$. ∎

######  Proposition 3.13.

Let $G$ be any group such that $\frac{G}{Z(G)}\cong C_{p}\times C_{p}$, $p$ a
prime. Then $G$ is isoclinic with an extraspecial group of order $p^{3}$.

###### Proof.

Using Theorem 2.2 and arguments in the proof of Theorem 2.4, $G$ is isoclinic
with a finite group $N$ of order $p^{n}$ with $Z(N)\subseteq N^{\prime}$.
Moreover, since $\frac{N}{Z(N)}\cong C_{p}\times C_{p}$, therefore
$Z(N)=N^{\prime}$. Also note that any proper centralizer of $N$ is abelian
normal of index $p$ in $N$. Hence using [14, Lemma 4, pp. 303], we have $\mid
N\mid=p.\mid Z(N)\mid.\mid N^{\prime}\mid$ and consequently, $N$ is an
extraspecial group of order $p^{3}$. ∎

######  Corollary 3.14.

Let $G$ and $H$ be any two groups such that
$\frac{G}{Z(G)}\cong\frac{H}{Z(H)}\cong C_{p}\times C_{p}$, $p$ a prime. Then
$G$ is isoclinic with $H$.

###### Proof.

The result follows from Proposition 3.13 and Remark 2.10. ∎

The following result shows that any two $4$-centralizer groups are isoclinic.

######  Proposition 3.15.

Any $4$-centralizer group is isoclinic with $Q_{8}$.

###### Proof.

It follows using [31, Theorem 3.5], Proposition 3.13 and Remark 2.10. ∎

######  Lemma 3.16.

If $M$ is a maximal non-abelian subgroup of a CA-group $G$, then either
$Z(G)=Z(M)$ or $G$ is isoclinic with $M$.

###### Proof.

If $Z(G)\subseteq M$, then using Proposition 3.1 we have $Z(G)=Z(M)$. On the
otherhand, if $Z(G)\nsubseteq M$, then $MZ(G)=G$ and hence $G$ is isoclinic
with $M$ by [26, Lemma 2.7]. ∎

######  Proposition 3.17.

Let $H$ be a non-abelian subgroup of $G$ with $\mid\frac{G}{Z(G)}\mid=p^{3}$
($p$ a prime).Then
$\mid\operatorname{Cent}(G)\mid=\mid\operatorname{Cent}(H)\mid$ implies $G$ is
isoclinic with $H$.

###### Proof.

In view of [10, Lemma 2.1] and Proposition 3.1,
$\frac{H}{Z(H)}\cong\frac{HZ(G)}{Z(G)}\leq\frac{G}{Z(G)}$. In the present
scenario using [10, Proposition 2.14], we have $HZ(G)=G$ by noting that if
$\mid\frac{H}{Z(H)}\mid=p^{2}$, then $\mid\operatorname{Cent}(H)\mid=p+2$.
Hence by [26, Lemma 2.7], $G$ is isoclinic with $H$. ∎

Combining Corollary 3.14 and [31, 9] it is easy to see that any two nilpotent
$n\in\\{5,7,9\\}$-centralizer groups are isoclinic. For $5$-centralizer groups
we have the following result:

######  Proposition 3.18.

Any $5$-centralizer group $G$ is isoclinic with $G_{m}=\langle a,b\mid
a^{3}=b^{2^{m}}=1,bab^{-1}=a^{-1}\rangle$, where $m\geq 1$ or an extraspecial
group of order $27$.

###### Proof.

In view of Theorem 2.4 without any loss we may assume that $G$ is finite.
Moreover, by [31, Theorem 3.5] we have $\frac{G}{Z(G)}\cong S_{3}$ or
$C_{3}\times C_{3}$. Now, if $\frac{G}{Z(G)}\cong S_{3}$, then using [27,
Corollary 2.2], we have $G=G_{m}\times A$, where $m\geq 1$ and $A$ is an
abelian group. Hence by Remark 2.10, $G$ is isoclinic with $G_{m}$. Again, if
$\frac{G}{Z(G)}\cong C_{3}\times C_{3}$, then by Proposition 3.13, we have the
result. ∎

Let $p$ be a prime. The author in [4, Theorem 3.3] proved that if $G$ is a
finite $(p^{2}+2)$-centralizer group of conjugate type $(p^{2},1)$ and two of
the proper element centralizers are normal in $G$, then $\frac{G}{Z(G)}$ is
elementary abelian of order $p^{4}$. We conclude the paper with the following
generalization of this result.

######  Theorem 3.19.

Let $G$ be any finite $(n+2)$-centralizer group of conjugate type $(n,1)$.
Then $G$ is a CA-group and $\frac{G}{Z(G)}$ is elementary abelian of order
$n^{2}$.

###### Proof.

In view of Ito [22], without any loss we may assume that $G$ is a $p$-group
for some prime $p$. Let $X_{i}=C(x_{i}),1\leq i\leq n+1$ where $x_{i}\in
G\setminus Z(G)$. We have $G=\overset{n+1}{\underset{i=1}{\cup}}X_{i}$ and
$\mid G\mid=\overset{n+1}{\underset{i=2}{\sum}}\mid X_{i}\mid$. In the present
scenario, interchanging $X_{i}$’s and applying [15, Cohn’s Theorem], we have
$G=X_{i}X_{j}$ and $X_{i}\cap X_{j}=Z(G)$ for any $1\leq i,j\leq n+1,i\neq j$.
It is easy to verify that $\mid\frac{G}{Z(G)}\mid=n^{2}$ and $G$ is a CA-
group. Moreover, using [29, Proposition 2] we have $\frac{G}{Z(G)}$ is
elementary abelian. ∎

## Acknowledgment

I would like to thank Prof. Mohammad Zarrin for carefully reading the
manuscript and giving his valuable suggestions and comments on it.

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