# On 10-Centralizer Groups of Odd Order

Authors: Z. Foruzanfar; Z. Mostaghim
Journal: ISRN Algebra
 (2014)

---

## Abstract

LetG be a group, and let Cent
(
G
) denote the number of distinct centralizers of its elements. A group G is called n-centralizer if C
e
n
t
(
G
)
=
n. In this paper, we investigate the structure of finite groups of odd order with C
e
n
t
(
G
)
=
10 and prove that there is no finite nonabelian group of odd order with C
e
n
t
(
G
)
=
10.


---

## Body

## 1. Introduction

Throughout this paper all groups mentioned are assumed to be finite, and we will use usual notation; for example,C
n denotes the cyclic group of order n, and C
n
⋊
C
p denotes the semidirect product of C
n and C
p with normal subgroup C
n, where n is a positive integer and p is a prime. For a group G, Z
(
G
) denotes the center of G, and Cent
(
G
)
=
{
C
G
(
x
)
∣
x
∈
G
}, where C
G
(
x
) is the centralizer of the element x in G; that is, C
G
(
x
)
=
{
y
∈
G
|
x
y
=
y
x
}. A group G is a C
A-group if C
G
(
x
) is abelian for every x
∈
G
∖
Z
(
G
). Many authors have studied the influence of |
Cent
(
G
)
| on finite group G (see [1–9]). It is clear that a group G is 1-centralizer if and only if it is abelian. In [6] Belcastro and Sherman proved that there is no n-centralizer group for n
=
2,3. They also proved that G is 4-centralizer if and only if G
/
Z
(
G
)
≅
C
2
×
C
2, and G is 5-centralizer if and only if G
/
Z
(
G
)
≅
C
3
×
C
3 or S
3. In [2] Ashrafi proved that if G is 6-centralizer, then G
/
Z
(
G
)
≅
D
8, A
4, C
2 × C
2 × C
2, or C
2 × C
2 × C
2 × C
2. In [1] Abdollahi et al. proved that G is 7-centralizer if and only if G
/
Z
(
G
)
≅
C
5
×
C
5, D
10 or 〈
x
,
y
∣
x
5
=
y
4
=
1, y
-
1
x
y
=
x
3
〉. They also proved that if G is 8-centralizer, then G
/
Z
(
G
)
≅
C
2
×
C
2
×
C
2, A
4 or D
12.Our main result is as follows.Theorem 1.
There is no finite nonabelian groupG of odd order with |
C
e
n
t
(
G
)
|
=
10.


## 2. Preliminary Results

By [1], a cover Γ for a group G is a collection of proper subgroups whose union is the whole G. We use the term n-cover for a cover with n members. A cover is called irredundant if no proper subcollection is also a cover. A cover is called a partition with kernel k if the intersection of pairwise members of the cover is k. Neumann in [10] obtained a uniform bound for the index of the intersection of an irredundant n-cover in terms of n, and Tomkinson [11] improved this bound. For a natural number n, let f
(
n
) denote the largest index |
G
:
D
|, where G is a group with an irredundant n-cover whose intersection of all of them is D. We know that f
(
3
)
=
4, f
(
4
)
=
9, f
(
5
)
=
16, and f
(
6
)
=
36 (see [12–15], resp.). Now we present some lemmas and propositions that will be used in the proof of Theorem 1.Lemma 2 (Lemma 3.3 of [11]).
LetM be a proper subgroup of the finite group G, and let H
1, H
2
,
…
,
H
k be subgroups of G with |
G
:
H
i
|
=
β
i and β
1
≤
⋯
≤
β
k. If G
=
M
∪
H
1
∪
⋯
∪
H
k, then β
1
≤
k. Furthermore, if β
1
=
k, then β
1
=
β
2
=
⋯
=
β
k
=
k and H
i
∩
H
j
≤
M for any two distinct i and j.Definition 3 (Definition  2.1 of [1]).
A nonempty subsetX
=
{
x
1
,
…
,
x
r
} of a finite group G is called a set of pairwise noncommuting elements if x
i
x
j
≠
x
j
x
i for all distinct i
,
j
∈
{
1
,
…
,
r
}. A set of pairwise noncommuting elements of G is said to have maximal size if its cardinality is the largest one among all such sets.Remark 4.
LetG be a finite group, and let {
x
1
,
…
,
x
r
} be a set of pairwise noncommuting elements of G having maximal size. Then (1)
{
C
G
(
x
i
)
∣
i
=
1
,
…
,
r
} is an irredundant r-cover with the intersection Z
(
G
)
=
∩
i
=
1
r
C
G
(
x
i
) (see Theorem  5.1 of [11]).
(2)
|
G
/
Z
(
G
)
|
≤
f
(
r
) (see Corollary  5.2 of [11]).
(3)
f
(
3
)
=
4, f
(
4
)
=
9, f
(
5
)
=
16, f
(
6
)
=
36, and f
(
7
)
=
81 (see [12–16], resp.).
(4)
LetG be a group such that every proper centralizer in G is abelian. Then for all a
,
b
∈
G
∖
Z
(
G
) either C
G
(
a
)
=
C
G
(
b
) or C
G
(
a
)
∩
C
G
(
b
)
=
Z
(
G
).
Ifz
∈
(
C
G
(
a
)
∩
C
G
(
b
)
)
∖
Z
(
G
), then C
G
(
z
) contains both C
G
(
a
) and C
G
(
b
), since C
G
(
a
) and C
G
(
b
) are abelian. Since z is not in Z
(
G
), C
G
(
z
)
≤
C
G
(
a
) and C
G
(
z
)
≤
C
G
(
b
). Thus, C
G
(
z
)
=
C
G
(
a
)
=
C
G
(
b
). Hence, in such a group G, {
C
G
(
x
)
∣
x
∈
G
∖
Z
(
G
)
} forms a partition with kernel Z
(
G
). It follows that {
C
G
(
x
)
/
Z
(
G
)
∣
x
∈
G
∖
Z
(
G
)
} forms a partition whose kernel is the trivial subgroup (see also Proposition 1.2 of [17]).Lemma 5 (Lemma  2.4 of [1]).
LetG be a finite nonabelian group, and let {
x
1
,
…
,
x
r
} be a set of pairwise noncommuting elements of G with maximal size. Then(1)
r
≥
3;
(2)
r
+
1
≤
|
C
e
n
t
(
G
)
|;
(3)
r
=
3 if and only if |
C
e
n
t
(
G
)
|
=
4;
(4)
r
=
4 if and only if |
C
e
n
t
(
G
)
|
=
5.Proposition 6 (Proposition  2.5 of [1]).
LetG be a finite group, and let X
=
{
x
1
,
…
,
x
r
} be a set of pairwise noncommuting elements of G having maximal size.(a)
If
|
C
e
n
t
(
G
)
|
<
r
+
4
, then
(1)
for each element
x
∈
G
,
C
G
(
x
)
is abelian if and only if
C
G
(
x
)
=
C
G
(
x
i
)
for some
i
∈
{
1
,
…
,
r
}
;
(2)
if
C
G
(
x
i
)
is a maximal subgroup of
G
for some
i
∈
{
1
,
…
,
r
}
, then
Z
(
G
)
=
C
G
(
x
i
)
∩
C
G
(
x
j
)
for all
j
∈
{
1
,
…
,
r
}
∖
{
i
}
. In particular, if
|
G
:
C
G
(
x
1
)
|
≤
|
G
:
C
G
(
x
2
)
|
≤
2
, then
|
C
e
n
t
(
G
)
|
=
4
, and if
|
G
:
C
G
(
x
1
)
|
≤
|
G
:
C
G
(
x
2
)
|
=
3
, then
|
C
e
n
t
(
G
)
|
=
5
.
(b)
If
|
C
e
n
t
(
G
)
|
=
r
+
2
, then there exists a proper nonabelian centralizer
C
G
(
x
)
which contains
C
G
(
x
i
1
)
,
C
G
(
x
i
2
)
, and
C
G
(
x
i
3
)
for three distinct
i
1
,
i
2
,
i
3
∈
{
1
,
…
,
r
}
.
(c)
If
|
C
e
n
t
(
G
)
|
=
r
+
3
, then there exists a proper nonabelian centralizer
C
G
(
x
)
which contains
C
G
(
x
i
1
)
and
C
G
(
x
i
2
)
for two distinct
i
1
,
i
2
∈
{
1
,
…
,
r
}
.Lemma 7 (Lemma  2.6 of [1]).
LetG be a finite nonabelian group. Then every proper centralizer of G is abelian if and only if |
C
e
n
t
(
G
)
|
=
r
+
1, where r is the maximal size of a set of pairwise noncommuting elements of G.Theorem 8 (Theorem  4.2 of [11]).
Suppose thatG is covered by n abelian subgroups A
1
,
A
2
,
…
,
A
n; then(i)
ifG
=
〈
A
1
,
A
2
〉, then |
G
/
Z
(
G
)
|
≤
(
n
-
1
)
2;
(ii)
if〈
A
1
,
A
2
〉
<
G, then |
G
/
Z
(
G
)
|
≤
2
(
n
-
2
)
log
⁡
2
(
n
-
2
).Definition 9 (see [18]).
A groupG is said to be an n-sum group if it can be written as the sum of n of its proper subgroups and of no smaller number. One then writes G
=
∑
r
=
1
n
H
r, where, for each r, H
r is a proper subgroup, which can be assumed to be maximal where convenient.Theorem 10 (Theorem  1 of [18]).
IfG
=
∑
r
=
1
n
H
r, then |
G
|
≤
∑
r
=
2
n
|
H
r
|, with equality if and only if (
a
)
H
1
H
r
=
G; r
≠
1 and (
b
)
H
r
∩
H
s
⊂
H
1; r
≠
s.Remark 11.
IfA and B are subgroups of finite index in a group G, and |
G
:
A
| and |
G
:
B
| are relatively prime, then G
=
A
B.Proposition 12 (Proposition  2.2 of [19]).
Letp be the smallest prime that is dividing |
G
|. If |
G
:
Z
(
G
)
|
=
p
3, then |
C
e
n
t
(
G
)
|
=
p
2
+
p
+
2 or p
2
+
2.Proposition 13 (Proposition  2.9 of [19]).
LetG
/
Z
(
G
) be nonabelian, n an integer, and p a prime. If G
/
Z
(
G
)
≅
C
n
⋊
C
p, then |
C
e
n
t
(
G
)
|
=
n
+
2.Now we are ready to state the main result of this paper.


## 3. The Proof of Theorem1

In this section, we give the proof of the main theorem.Suppose, on the contrary, thatG is a finite nonabelian group of odd order with |
Cent
(
G
)
|
=
10. Let {
x
1
,
…
,
x
r
} be a set of pairwise noncommuting elements of finite nonabelian group G having maximal size. Then X
i
=
C
G
(
x
i
), 1
≤
i
≤
r, is an irredundant r-cover with intersection Z
(
G
). Assume that |
G
:
X
i
|
=
α
i, where α
1
≤
⋯
≤
α
r. Since G is a 10-centralizer group, therefore, by Lemma 5, we have 5
≤
r
≤
9.Supposer
=
5. By Remark 4, |
G
/
Z
(
G
)
|
≤
16 and G
=
X
1
∪
X
2
∪
⋯
∪
X
5; therefore by Lemma 2, |
G
:
X
2
|
≤
4. Since |
G
| is odd, then |
G
:
X
2
|
=
3. On the other hand, 3 is a divisor of |
G
/
Z
(
G
)
|; therefore, we have |
G
/
Z
(
G
)
|
∈
{
9,15
}. If |
G
/
Z
(
G
)
|
=
9, then |
Cent
(
G
)
|
=
5 by [6, Theorem  6], which is a contradiction. Now if |
G
/
Z
(
G
)
|
=
15, then G is abelian which is not possible. It implies that r
≠
5.Supposer
=
6. By Remark 4, |
G
/
Z
(
G
)
|
≤
36 and G
=
X
1
∪
X
2
∪
⋯
∪
X
6; therefore by Lemma 2, |
G
:
X
2
|
≤
5. If |
G
:
X
2
|
=
5, since 5 is a divisor of |
G
/
Z
(
G
)
|, then |
G
/
Z
(
G
)
|
∈
{
15,25,35
}. If |
G
/
Z
(
G
)
|
∈
{
15,35
}, then G is abelian, which is not possible. If |
G
/
Z
(
G
)
|
=
25, then by [6, Theorem  6], |
Cent
(
G
)
|
=
7, a contradiction. Now suppose |
G
:
X
2
|
=
3; since 3 is a divisor of |
G
/
Z
(
G
)
|, therefore |
G
/
Z
(
G
)
|
∈
{
9,15,21,27,33
}. If |
G
/
Z
(
G
)
|
∈
{
15,33
}, then G is abelian which is not possible. If |
G
/
Z
(
G
)
|
=
21, then G
/
Z
(
G
)
≅
C
7
⋊
C
3, and by Proposition 13, |
Cent
(
G
)
|
=
9. If |
G
/
Z
(
G
)
|
=
9, then |
Cent
(
G
)
|
=
5. If |
G
/
Z
(
G
)
|
=
27, then by Proposition 12, |
Cent
(
G
)
|
=
11 or 14. Hence, r
≠
6 and so |
Cent
(
G
)
|
<
r
+
4.Now supposer
=
7. By Remark 4, G
=
X
1
∪
X
2
∪
⋯
∪
X
7, and by Lemma 2, |
G
:
X
2
|
≤
6. Now if |
G
:
X
2
|
=
5, then |
G
:
X
1
|
=
3 or 5. If |
G
:
X
1
|
=
3, then by Remark 11, G
=
X
1
X
2, and by Proposition 6, X
1
∩
X
2
=
Z
(
G
). Hence, |
G
/
Z
(
G
)
|
=
15 which is not possible, since G is nonabelian. If |
G
:
X
1
|
=
5, then |
G
/
Z
(
G
)
|
≤
25, and since 5 is a divisor of |
G
/
Z
(
G
)
|, then |
G
/
Z
(
G
)
|
∈
{
15,25
}, which is again a contradiction. Hence, r
≠
7.Now supposer
=
8. By Remark 4, G
=
X
1
∪
X
2
∪
⋯
∪
X
8, and by Lemma 2, |
G
:
X
2
|
≤
7. Now if |
G
:
X
2
|
=
3, by Proposition 6, |
Cent
(
G
)
|
=
5, a contradiction. If |
G
:
X
2
|
=
5, then |
G
:
X
1
|
=
3 or 5. If |
G
:
X
1
|
=
3, then by Remark 11, G
=
X
1
X
2, and by Proposition 6, X
1
∩
X
2
=
Z
(
G
). Hence, |
G
/
Z
(
G
)
|
=
15 which is not possible, since G is nonabelian. If |
G
:
X
1
|
=
5, then |
G
/
Z
(
G
)
|
≤
25. Now since 5 is a divisor of |
G
/
Z
(
G
)
|, |
G
/
Z
(
G
)
|
∈
{
15,25
}, which is again a contradiction. Finally, suppose that |
G
:
X
2
|
=
7. By Lemma 2, |
G
:
X
2
|
=
|
G
:
X
3
|
=
⋯
=
|
G
:
X
8
|
=
7. Therefore, |
G
|
=
∑
i
=
2
8
‍
|
X
i
|, and by Theorem 10, G
=
X
1
X
2. Again by Proposition 6, X
1
∩
X
2
=
Z
(
G
). Since |
G
| is odd, |
G
:
X
1
|
=
3,5, or 7. If |
G
:
X
1
|
=
3, then |
G
/
Z
(
G
)
|
=
21. So G
/
Z
(
G
)
≅
C
7
⋊
C
3, and by Proposition 13, |
Cent
(
G
)
|
=
9, which is a contradiction. If |
G
:
X
1
|
=
5, then |
G
/
Z
(
G
)
|
=
35. Therefore, G is abelian, which is not possible. If |
G
:
X
1
|
=
7, then |
G
/
Z
(
G
)
|
=
49. Now by [6, Theorem  6], |
Cent
(
G
)
|
=
9, which is a contradiction. Hence, r
≠
8.Thus,r
=
9. By Lemma 7, G is a C
A-group, and by Remark 4, X
i
∩
X
j
=
Z
(
G
) for all distinct i
,
j
∈
{
1,2
,
…
,
9
}. Now G
=
X
1
∪
X
2
∪
⋯
∪
X
9, and by Lemma 2, |
G
:
X
2
|
≤
8. If |
G
:
X
2
|
=
3, then |
Cent
(
G
)
|
=
5, a contradiction. If |
G
:
X
2
|
=
5, then |
G
:
X
1
|
=
3 or 5. If |
G
:
X
1
|
=
3, then by Remark 11, G
=
X
1
X
2. Since X
1
∩
X
2
=
Z
(
G
), |
G
/
Z
(
G
)
|
=
15, which is a contradiction. If |
G
:
X
1
|
=
5, then |
G
/
Z
(
G
)
|
≤
25, and since 5 is a divisor of |
G
/
Z
(
G
)
|, |
G
/
Z
(
G
)
|
∈
{
15,25
}, which is again a contradiction. If |
G
:
X
2
|
=
7, then |
G
:
X
1
|
=
3,5, or 7. If |
G
:
X
1
|
=
3, then by Remark 11, G
=
X
1
X
2, and since X
1
∩
X
2
=
Z
(
G
), therefore |
G
/
Z
(
G
)
|
=
21 and so G
/
Z
(
G
)
≅
C
7
⋊
C
3, and by Proposition 13, |
Cent
(
G
)
|
=
9, which is a contradiction. Similarly, If |
G
:
X
1
|
=
5, then |
G
/
Z
(
G
)
|
=
35, a contradiction. Finally, if |
G
:
X
1
|
=
7, then G
=
〈
X
1
,
X
2
〉. Therefore by Theorem 8, |
G
/
Z
(
G
)
|
≤
64. Since |
G
| is odd and G is nonabelian, |
G
/
Z
(
G
)
|
∈
{
21,35,49,63
}. If |
G
/
Z
(
G
)
|
=
21, then G
/
Z
(
G
)
≅
C
7
⋊
C
3, and by Proposition 13, |
Cent
(
G
)
|
=
9, which is a contradiction. If |
G
/
Z
(
G
)
|
=
35, then G is abelian which is not possible. If |
G
/
Z
(
G
)
|
=
49, then by [6, Theorem  6], |
Cent
(
G
)
|
=
9, which is a contradiction. Now suppose that |
G
/
Z
(
G
)
|
=
63. If X
1 or X
2 is normal in G, then G
=
X
1
X
2. Since X
1
∩
X
2
=
Z
(
G
), |
G
/
Z
(
G
)
|
=
49, and by [6, Theorem  6], |
Cent
(
G
)
|
=
9, a contradiction. It is easy to see that C
G
(
g
)
/
Z
(
G
)
≤
C
G
/
Z
(
G
)
(
g
Z
(
G
)
) for any g
∈
G
∖
Z
(
G
). Therefore, |
G
:
X
i
|
=
7,9, or 21, for i
∈
{
1,2
,
…
,
9
}. Let x
Z
(
G
) be an element of order 7 in G
/
Z
(
G
). Then C
G
/
Z
(
G
)
(
x
Z
(
G
)
) is the normal Sylow 7-subgroup of G
/
Z
(
G
). Since C
G
(
x
)
/
Z
(
G
)
≤
C
G
/
Z
(
G
)
(
x
Z
(
G
)
), C
G
(
x
)
/
Z
(
G
)
=
C
G
/
Z
(
G
)
(
x
Z
(
G
)
). Hence by [19, Lemma  3.1], G has exactly one centralizer of index 9. Suppose |
G
:
X
3
|
=
9. Then |
G
:
X
4
|
=
|
G
:
X
5
|
=
|
G
:
X
6
|
=
|
G
:
X
7
|
=
|
G
:
X
8
|
=
|
G
:
X
9
|
=
21 and |
G
|
>
∑
i
=
2
9
‍
|
X
i
|, which is a contradiction by Theorem 10. Therefore, |
G
:
X
3
|
=
7. Similarly, we can show that |
G
:
X
4
|
=
|
G
:
X
5
|
=
|
G
:
X
6
|
=
|
G
:
X
7
|
=
7. So we have |
G
:
X
1
|
=
⋯
=
|
G
:
X
7
|
=
7, |
G
:
X
8
|
=
9, and |
G
:
X
9
|
=
21. By G
=
X
1
∪
X
2
∪
⋯
∪
X
9 and the property X
i
∩
X
j
=
Z
(
G
) for all distinct i
,
j
∈
{
1,2
,
…
,
9
}, it is easy to see that |
G
|
=
|
X
1
|
+
⋯
+
|
X
9
|
-
8
|
Z
(
G
)
|. So we have |
G
/
Z
(
G
)
|
<
63, a contradiction.Now the proof of Theorem1 is complete.



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*Source: 607984-2014-04-01.xml*