具有Clifford斷面的正則純正半群

孫京鋒,邵勇

(西北大學(xué)數(shù)學(xué)系,陜西西安 710127)

具有Clifford斷面的正則純正半群

孫京鋒,邵勇

(西北大學(xué)數(shù)學(xué)系,陜西西安 710127)

給出了具有Clifford斷面的右正規(guī)純正半群的等價(jià)刻畫(huà),得到了具有Clifford斷面的正則純正半群的次直積分解,證明了具有Clifford斷面的正則純正半群一定是正則純正群.

同余；正則純正半群；Clifford斷面；次直積

1 預(yù)備知識(shí)

設(shè)S為半群,a∈S,如果存在x∈S,滿足axa=a,則稱(chēng)a為正則的.如果對(duì)于任意的a∈S,a都是正則的,則稱(chēng)S為正則半群[1].

定義1[1]稱(chēng)正則半群S為純正半群,若S的冪等元之集E(S)是S的子半群.

設(shè)S為純正半群,E(S)為S的冪等元之集.若E(S)為左正則帶,即對(duì)任意的x,y∈E(S) 有xyx=xy,則稱(chēng)S為左逆半群[6].若E(S)為右正則帶,即對(duì)任意的x,y∈E(S)有xyx= yx,則稱(chēng)S為右逆半群[6].若E(S)為正則帶,即對(duì)任意的x,y,z∈E(S)有xyxzx=xyzx,則稱(chēng)S為正則純正半群.

設(shè)S為半群.a∈S,如果有aa'a=a和a'aa'=a'成立,則稱(chēng)a'為a的逆元.用V(a)來(lái)表示a的逆元的全體.若對(duì)于任意的a∈S,|V(a)|=1,則稱(chēng)S為逆半群.

定義2[3]設(shè)S為正則半群.S的逆子半群S?稱(chēng)為S的逆斷面,如果S?含且只含有S的每個(gè)元的一個(gè)逆元,即對(duì)于任意的x∈S,|V(x)∩S?|=1.

x在逆斷面S?中的唯一逆元記作x?,并且用x??來(lái)表示(x?)?.

命題1[4,5]設(shè)S是具有逆斷面S?的正則半群,則下列結(jié)論成立:

(η)S是純正的當(dāng)且僅當(dāng)對(duì)于任意的a,b∈S,(ab)?=b?a?.

(ε)若S是(左,右)正規(guī)純正半群,則對(duì)于任意的a,b,c∈S有a?bc?=a?b??c?.

2 主要結(jié)論

設(shè)S為具有逆斷面S?的正則半群.若S?為Clifford半群,即對(duì)任意的a∈S?有a?a= aa?,則稱(chēng)S?為Clifford斷面.這樣,就有

引理1設(shè)S是具有逆斷面S?的純正半群.若V(x)=V(x?x2),則S?為Clifford斷面.

證明我們只需證明x?x??=x??x?.

對(duì)于任意的x∈S有

于是就有x??x?=x?x??.故S?為Clifford斷面.

設(shè)S是純正半群,E(S)為S的冪等元之集.若E(S)為右正規(guī)帶[2],即對(duì)任意的e,f,g∈E(S)有ef g=f eg,則稱(chēng)S為右正規(guī)純正半群.若E(S)為左正規(guī)帶[2],即對(duì)任意的e,f,g∈E(S)有ef g=egf,則稱(chēng)S為左正規(guī)純正半群.

設(shè)S為半群.稱(chēng)S為完全正則半群[2],是指S的每一個(gè)H-類(lèi)都是S的子群.純正的完全正則半群稱(chēng)為純正群[2].設(shè)S為純正群,E(S)為其冪等元之集.若E(S)為右正規(guī)帶,則稱(chēng)S為右正規(guī)純正群.若E(S)為左正規(guī)帶,則稱(chēng)S為左正規(guī)純正群.這樣,就有下面引理.

引理2若S為具有逆斷面S?的右正規(guī)純正半群,則下列命題等價(jià):

這樣,我們得到本節(jié)的主要結(jié)果.

定理3設(shè)S為具有Clifford斷面的正則純正半群,則S為正則純正群.

證明由定理2知S/σ為左逆半群,且S?/σ為S/σ的Clifford斷面.由推論1知S/σ為左正則純正群.由推論2知S/η為右逆半群,且S?/η為S/η逆斷面.由定理1知S/η為右正則純正群.于是,由引理4可得S同構(gòu)于左正則純正群S/σ與右正則純正群S/η的次直積.從而,S為完全正則的.故S為正則純正群.

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Onregular orthodox sem igroups with Clifford transversals

SUN Jing-feng,SHAO Yong

(Department of Mathematics,Northwest University,Xi’an 710127,China)

We givean equivalent characterization of right norm alorthodox sem igroup with a Clifford transversal and obtain the decom position of subdirect product of regular orthodox sem igroups with Clifford transversals. We also prove that a regu lar orthodox sem igroup with a Clifford transversal is a regular orthogroup.

congruence,regular orthodox sem igroup,Clifford transversal,subdirect product

O152.7

A

1008-5513(2009)02-0310-05

2008-05-21.

國(guó)家自然科學(xué)基金(10471112),陜西省教育廳自然科學(xué)專(zhuān)項(xiàng)基金(07JK 413).

孫京鋒(1982-),碩士,研究方向：半群代數(shù)理論.

2000M SC:20M 10,06B10