# A Characterization of 4-Centralizer Groups

Authors: Jutirekha Dutta
Journal: Chinese Journal of Mathematics
 (2013)

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## Abstract

A finite or infinite group is called ann-centralizer group if it has n numbers of distinct centralizers. In this paper, we prove that a finite or infinite group G is a 4-centralizer group if and only if G/Z(G) is isomorphic to C2×C2. This extends a result of Belcastro and Sherman.


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## Body

## 1. Introduction

Given a finite or infinite groupG and x∈G, the set C(x)={y∈G∣xy=yx} is called the centralizer of x in G. The set of all centralizers in G is denoted by Cent(G). A group G is called an n-centralizer group if |Cent(G)|=n. It is easy to see that one-centralizer groups are precisely the abelian groups. Characterization of finite groups in terms of the number of distinct centralizers has been an interesting topic of research in recent years (see [1–8]). In [7], Belcastro and Sherman have proved the nonexistence of finite n-centralizer group for n=2,3. However, their proof also shows the nonexistence of infinite 2,3-centralizer groups. Belcastro and Sherman [7] also characterize all finite 4-centralizer groups. More precisely, they proved that a finite group G is a 4-centralizer group if and only if G/Z(G) is isomorphic to C2×C2, where Z(G) is the center of G and C2 is the cyclic group having two elements. In this paper, we extend the same characterization for infinite groups using elementary techniques of group theory.Throughout this paperG will denote a finite or infinite group. Recall that for any group G, its center Z(G) is the intersection of all centralizers in G. Also G is the union of all the centralizers of noncentral elements of G. It may be mentioned here that a finite or infinite group can not be written as union of two of its proper subgroups. These facts have important role in proving the main theorem of this paper.


## 2. Main Result

In this section, we proof the following main theorem of this paper.Theorem 1.
A finite or infinite groupG is a 4-centralizer group if and only if G/Z(G)≅C2×C2.Proof.
LetG be a 4-centralizer finite or infinite group and Cent(G)={G,C(p),C(q),C(r)}, where p, q, and r are noncentral elements of G. Then G=C(p)∪C(q)∪C(r), since G is the union of its proper centralizers.
Let us consider the centralizerC(pq). Then C(pq) will be one of G, C(p), C(q), or C(r).
IfC(pq)=G, then pq∈Z(G). This implies that p=zq-1 for some z∈Z(G). Therefore, we get C(p)=C(q), a contradiction. If C(pq)=C(p), then g∈C(p)=C(pq) gives
(1)gpq=pqg⟹pgq=pqg⟹g∈C(q).
Therefore, C(p)⊆C(q). Hence, G=C(q)∪C(r), a contradiction, as a group can not be written as union of two of its proper subgroups. Similarly, it can be seen that C(pq)≠C(q).
Thus,C(pq)=C(r) and so G=C(p)∪C(q)∪C(pq). In a similar way it can be seen that C(qp)=C(r) and so C(pq)=C(qp).
We will now show thatC(p)∩C(q)=Z(G). Clearly, Z(G)⊆C(p)∩C(q). Let g∈C(p)∩C(q). Then
(2)gp=pg⟹(gp)q=(pg)q⟹gpq=pqg⟹g∈C(pq).
Therefore, g∈C(p)∩C(q)∩C(pq)=Z(G) and so C(p)∩C(q)⊆Z(G). Thus,
(3)C(p)∩C(q)=Z(G).
In a similar way, it can be seen that C(q)∩C(pq)=Z(G), and C(p)∩C(pq)=Z(G).
Let us consider the right cosetsZ, Zp, Zq, and Zpq, where Z:=Z(G). As p is a noncentral element of G it follows that Z≠Zp, Zq, and Zpq.
IfZp=Zq, then we have p=zq for some z∈Z. Therefore, we get C(p)=C(q), a contradiction. Again, if Zp=Zpq, then (pq)-1p∈Z which gives q∈Z, a contradiction. Similarly, it can be seen that Zq≠Zpq. Thus the cosets Z, Zp, Zq, and Zpq are mutually disjoint. Clearly, Z∪Zp∪Zq∪Zpq⊆G.
Letg∈Z then g∈Z∪Zp∪Zq∪Zpq. Suppose that g∈G-Z then without any loss of generality we may assume that g∈C(p). Let us consider the centralizer C(gq). Then C(gq) is one of G, C(p), C(q), or C(pq).
Case   1. Let C(gq)=G. In this case, gq∈Z(G) and so gqq=qgq which gives g∈C(q). Therefore, g∈C(p)∩C(q)=Z(G), a contradiction.
Case   2. Let C(gq)=C(p). In this case, gqp=pgq which gives p∈C(q). Therefore, p∈C(p)∩C(q)=Z(G), a contradiction.
Case   3. Let C(gq)=C(q). In this case, gqq=qgq which gives g∈C(q). Therefore, g∈C(p)∩C(q)=Z(G), a contradiction.
Hence,C(gq)=C(pq), since G is a 4-centralizer group. Now, C(gq)=C(pq) implies that gqp=pqg which gives gp-1q=qgp-1 since g∈C(p). Thus, gp-1∈C(q). Also, g∈C(p) implies that gp-1∈C(p). Hence, gp-1∈C(p)∩C(q)=Z(G) and so g=zp for some z∈Z(G). In other words, g∈Zp. Therefore, g∈Z∪Zp∪Zq∪Zpq. Hence, G=Z∪Zp∪Zq∪Zpq. This shows that |G/Z(G)|=4. Finally, the fact that G is nonabelian gives G/Z(G)≅C2×C2.
Conversely, letG/Z(G)≅C2×C2. Therefore, there are four right cosets of Z=Z(G) in G, namely, Z, Zp, Zq, and Zr, where p, q, and r are distinct noncentral elements of G. So, for any element g∈G, either g∈Z or Zp or Zq or Zr. If g∈Z, then C(g)=G. If g∈Zp, then g=zp for some z∈Z, therefore C(g)=C(zp)=C(p). Similarly, g∈Zq gives C(g)=C(q) and g∈Zr gives C(g)=C(r). Hence, G has at most four centralizers, namely, G, C(p), C(q), and C(r). Since |Cent(G)|≥4 we have |Cent(G)|=4. This completes the proof.We conclude this paper by the following remark. In [7], Belcastro and Sherman proved Theorem 1 for finite groups only. Note that their proof can not be extended to infinite groups as they have used the finiteness of the group extensively. Belcastro and Sherman [7] have also obtained the structure of 5-centralizer finite groups as follows.A finite group
G
is a 5-centralizer group if and only if
G
/
Z
(
G
)
≅
C
3
×
C
3
or
G
/
Z
(
G
)
≅
S
3
, where
C
3
is the cyclic group having three elements and
S
3
the symmetric group on three symbols.Note that theif part of the above result also holds for infinite groups. The same proof of Belcastro and Sherman [7] holds for infinite groups. However, the proof of the only if part of the above result is not known for infinite groups. It may be interesting to study the infinite n-centralizer groups for n>4.



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*Source: 871072-2013-12-26.xml*