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        On Non-commuting Sets in a Finite p-group with Derived Subgroup of Prime Order

        2016-10-12 01:46:24WangYuleiandLiuHeguo

        Wang Yu-leiand Liu He-guo

        (1.Department of Mathematics,Henan University of Technology,Zhengzhou,450001) (2.Department of Mathematics,Hubei University,Wuhan,430062)

        Communicated by Du Xian-kun

        ?

        On Non-commuting Sets in a Finite p-group with Derived Subgroup of Prime Order

        Wang Yu-lei1and Liu He-guo2

        (1.Department of Mathematics,Henan University of Technology,Zhengzhou,450001) (2.Department of Mathematics,Hubei University,Wuhan,430062)

        Communicated by Du Xian-kun

        Let G be a finite group.A nonempty subset X of G is said to be noncommuting if xy/=yx for any x,y∈X with x/=y.If|X|≥|Y|for any other non-commuting set Y in G,then X is said to be a maximal non-commuting set.In this paper,we determine upper and lower bounds on the cardinality of a maximal non-commuting set in a finite p-group with derived subgroup of prime order.

        finite p-group,non-commuting set,cardinality

        2010 MR subject classification:20F18

        Document code:A

        Article ID:1674-5647(2016)03-0193-05

        In this paper,p is always a prime.The notation used is standard,see[1].

        Let G be a finite group.A nonempty subset X of G is said to be non-commuting iffor any x,y∈X withAnd further,if|X|≥|Y|for any other noncommuting set Y in G,then X is said to be a maximal non-commuting set in G.We denote by nc(G)the cardinality of a maximal non-commuting set in G,and denote by cc(G)the minimal number of abelian subgroups covering G.Then we can obtain

        Further,Pyber[2]has shown that there exists some constant c such that

        Mason[3]has shown that any finite group G can be covered by at most[|G|/2]+1 abelian subgroups,thus

        For an extraspecial p-group G with order p2n+1,in the case of p=2,Isaacs has obtained that nc(G)=2n+1(see[4]).In the case of p being odd,Chin[5]has determined thefollowing bounds for nc(G),

        Now we determine upper and lower bounds for the cardinality of a maximal noncommuting sets in a finite p-groups with derived subgroup of prime order.

        Lemma 1[6]Let G be a finite p-group with derived subgroup of order p.Then there exist the generators x1,···,x2n,y1,···,ymof G and a generator z of the derived subgroup of G,which satisfy

        Theorem 1Let G be a finite p-group with derived subgroup of order p,andThen

        (1)if p=2,then nc(G)=2n+1;

        (2)if p is odd,then np+1≤nc(G)≤(p(p-1)n-2)/(p-2).

        Proof.There exist the generators of G:x1,···,x2n,y1,···,ym,z,which satisfy the conditions in Lemma 1.Obviously,

        Firstly,we assert

        if and only if

        In fact,since

        and

        we have

        if and only if

        if and only if

        The assertion is true.

        Assuming that n=1,we have

        Then G is the union of four sets as

        Let R be a non-commuting set of G.Since

        we only pick at most one element fromfor R and also at most one element fromfor R.

        For any i∈{1,2,···,p-1},by the assertion above,we have

        Then

        are p-1 mutually disjoint commuting sets.So we only pick at most p-1 elements fromfor R.Hence,the largest possible cardinality of R is

        And we take

        for R to achieve this cardinality.Therefore,

        Suppose that n≥2 in the following section and dn=nc(G).Let S be a non-commuting set of G and{x1,x2}?S.Note that CG(x1)?=CG(x2).Let H:=CG(x1)∩CG(x2).Then

        Obviously,the derived subgroup of H has order p.

        Let dn?1=nc(H).Since

        and

        we have

        Suppose that S′={h1,h2,···,hdn?1}is a maximal non-commuting set of H.

        If p=2,then S{x1,x2}?x1x2H.Hence

        is a maximal non-commuting set of G,thereby dn=2+dn?1.By this recursion formula and d1=3,we have dn=2n+1.

        Suppose that p>2.It is easy to know that

        is a non-commuting set of G,but this set cannot be a maximal non-commuting set of G. Hence dn≥dn?1+p.By this recursion formula and d1=p+1,we have

        Considering the following(p-1)2sets:

        we pick one element from each set above and obtain(p-1)2elements,and these elements are just left coset representatives of left cosetsHence we may pick elements from(p-1)2sets above for S{x1,x2}for convenience.Since the following sets are all commuting sets,

        where i=1,2,···,dn?1,we pick elements from the following sets for S{x1,x2},

        It follows dn≤2+(p-1)dn?1.By the recursion formula and d1=p+1,we have

        Hence,when n≥2,

        Obviously,this result generalizes the main theorem in[5].

        References

        [1]Robinson D J S.A Course in the Theory of Groups(Second edition).New York:Springer-Verlag,1996.

        [2]Pyber L.The number of pairwise non-commuting elements and the index of the center in a finite group.J.London Math.Soc.(2),1987,35(2):287–295.

        [3]Mason D R.On coverings of a finite group by abelian subgroups.Math.Proc.Cambridge Philos. Soc.,1978,83(2):205–209.

        [4]Bertram E A.Some applications of graph theory to finite groups.Discrete Math.,1983,44: 31–43.

        [5]Chin A Y M.On non-commuting sets in an extraspecial p-group.J.Group Theory,2005,8: 189–194.

        [6]Blackburn S R.Groups of prime power order with derived subgroup of prime order.J.Algebra,1999,219:625–657.

        10.13447/j.1674-5647.2016.03.01

        date:Feb.9,2015.

        The NSF(11301150,11371124)of China,the NSF(142300410134)of Henan Province and Plan for Scientific Innovation Talent(11CXRC19)of Henan University of Technology.

        E-mail address:yulwang@163.com(Wang Y L).

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