• <tr id="yyy80"></tr>
  • <sup id="yyy80"></sup>
  • <tfoot id="yyy80"><noscript id="yyy80"></noscript></tfoot>
  • 99热精品在线国产_美女午夜性视频免费_国产精品国产高清国产av_av欧美777_自拍偷自拍亚洲精品老妇_亚洲熟女精品中文字幕_www日本黄色视频网_国产精品野战在线观看 ?

    The Group of Automorphisms of Total Orthogonal Graphs of Odd Characteristic

    2019-03-30 08:20:48GENGTianzhen耿天真MAXiaobin馬曉玢
    應(yīng)用數(shù)學(xué) 2019年2期
    關(guān)鍵詞:天真

    GENG Tianzhen(耿天真),MA Xiaobin(馬曉玢)

    (School of Mathematics and Big Data,Anhui University of Science and Technology,Huainan 232000,China)

    Abstract: Let Fnq be the n-dimensional row vector space over a finite field Fq of odd characteristic,and let Sn be a non-singular n×n symmetric matrix over Fq.The total orthogonal graph O(Sn,q) with respect to Sn is defined to be the graph with vertices of all 1-dimensional subspace of Fnq ,two vertices [α] and [β] are adjacent if and only if αSnβT≠ 0.The three types of isotropic orthogonal graphs studied in [3] are induced subgraphs of O(Sn;q) with vertices of all isotropic lines.In this paper,the automorphism group of O(Sn,q) is determined.Not like the case of isotopic orthogonal graph studied in[3] (where the graph is vertex transitive),it turns out that the vertex set of O(Sn,q) has three orbits if n is odd,and it has only two orbits if n is even.

    Key words: Orthogonal graph;Graph automorphism;Transitivity

    1.Introduction

    Recently,some graphs based on geometry of classical groups have attracted more attention.We now list some papers concerning with such graphs.TANG and WAN[9]studied the symplectic graph over an arbitrary finite field and determined the full automorphism group.LI and WANG[6]further showed that the subconstituents of the symplectic graph are strictly Deza graphs except the trivial case whenυ= 2.Meemark and Prinyasart[8]introduced the symplectic graphSp(2υ,Zpn) and showed that it is strongly regular whenυ= 1 and it is arc transitive whenpis an odd prime.LI et al.[5]continued the research onSp(2υ,Zpn),showing thatSp(2υ,Zpn) is arc transitive for any primep,and it is a strictly Deza graph ifυ ≥2 andn ≥2.GU and WAN[2]studied the subconstituents of the symplectic graphs modulopn.WAN and ZHOU[11?12]studied the(isotropic)unitary graphs and the(isotropic)orthogonal graphs of Characteristic 2.We here particularly mention [3] due to GU,which seems much related to our topic.Forn=2υor 2υ+1 or 2υ+2,GU[3]defined the(isotropic)orthogonal graphO(n,q) with respect to a symmetric matrixoverFq,where ?=?or ?=1 or ?=diag{1,?z},in such a way: the vertex set of the graph is all isotropic lines ofFnq,two isotropic lines[α]and[β]are adjacent if and only ifαHβT≠0.GU[3]showed thatO(n,q) is strongly regular forυ ≥2,and determined the group of automorphisms ofO(n,q).By the automorphism group of the graph it is easy to see thatO(n,q) is vertex transitive and edge transitive.For more results about the graphs based on geometry of classical groups the reader may refer to [4,7,13].

    One might find that the vertex set of the graphs introduced in [2-9,11-13] are taken to be isotropic lines,or sometimes totally isotropic subspaces of the space in question.In this paper,we shall use all 1-dimensional subspace (isotropic or not) of a orthogonal space to define a new graph,which includes the isotropic orthogonal graph as an induced subgraph.The vertex set of the graph chosen in such a way seems much natural.Indeed,the vertex set of the earliest graph studied by TANG and WAN,i.e.,the symplectic graph,is taken to be all 1-dimensional subspace (although,it can be also viewed consisting of all isotropic lines).

    LetFqbe a finite field withqelements,whereqis a power of an odd primepandn ≥2 be an integer.LetFnq={(a1,a2,...,an) :ai ∈Fq}be then-dimensional row vector space overFq.Denote byei(1≤i ≤n)the row vector inFnqwhoseith coordinate is 1 and all others are 0.For a nonzero vectorα=(a1,...,an)∈Fnq,let [α] denote the 1-dimensional subspace ofFnqspanned byα.Sometimes,[α] = [(a1,...,an)] is directly written as [a1,...,an] for simplicity.Since,[α] = [kα] for anyk ∈F?q=Fq {0},the representation of a vertex in the form [α] is not unique.However if we write the vertex in the form [a1,a2,...,an] such that the first (from left) nonzero coordinate is 1,then the representation is unique.A vertex[a1,a2,...,an] written in such way is said with standard form.

    For ann×nmatrixM,we denote byMTand|M|the transpose and the determinant ofM,respectively.LetGbe the subgroup ofF?qof all square elements.The index ofGinF?qis 2.ThusF?qis exactly a disjoint union ofGandzG,wherezis an arbitrarily given non-square element inFq.Let (α,β)=αSnβTbe a non-degenerate symmetric bilinear form defined onFnqwith respect to a non-singularn×nsymmetric matrixSnoverFq.A vectorαis called isotropic if (α,α) = 0,otherwise,it is called non-isotropic.If a subspace ofFnqconsists of isotopic vectors,it is said to be total isotropic.

    We now define the so-called total orthogonal graphO(Sn,q),with respect to a nonsingularn×nsymmetric matrixSnoverFqin such a way: the vertex set is taken to be all 1-dimensional subspaces (or lines) ofFnq,two lines [α] and [β] are adjacent if and only if (α,β)≠0.Obviously,O(Sn,q) is not a simple graph since for a non-isotropic line [α],[α]~[α].By Theorem 1.23 in[10]we know that ann×nnonsingular symmetric matrix overFqis congruent to diag(In?1,θ),whereIn?1refers to the identity matrix of ordern ?1,andθ= 1 orθ=z.If two non-singular symmetric matricesSnandHnare congruent,that is,Sn=PHnPTfor an invertible matrixP,thenO(Sn,q) is isomorphic toO(Hn,q) under the mapping [α][αP],?[α]∈V(O(Sn,q)).In view of this point,we will directly assume thatSn=InorSn=diag(In?1,z),withza fixed non-square element ofFq.

    A similar discussion as Theorem 2.1 in [9] shows thatO(Sn,q) is also a strongly regular graph with parameters

    In this paper,the automorphism group ofO(Sn,q) is determined,and the partition ofV(O(Sn,q)) as the union of different orbits is done.

    Remark 1.1(i)The three types of isotropic orthogonal graphs studied in[3]are induced subgraphs ofO(Sn,q) with vertices of all isotropic lines.

    (ii) For certain unknown reasons,the case when ?is a non-square element ofFqis left by [3] without consideration.However,the orthogonal graphs studied in this paper cover all cases,especially the case whenSnis congruent towhich is left without consideration in [3].

    2.Automorphisms of O(Sn,q)

    In this section,we determine the automorphism group,denoted by Aut(O(Sn,q)),ofO(Sn,q).Some standard automorphisms are constructed firstly.Based on them every automorphism ofO(Sn,q) is described explicitly.Alln×ninvertible matrices overFqforms a group with respect to matrix multiplication,which is called the general linear group of degreenoverFqand is denoted by GLn(Fq).

    LetP ∈GLn(Fq).We defineσP:V(O(Sn,q))→V(O(Sn,q)) by [α][αP].Then using a similar proof as that of Proposition 3.1 in [3] we obtain:

    Lemma 2.11) LetP ∈GLn(Fq).ThenσPis an automorphism ofO(Sn,q) if and only if PSPT=kSfor somek ∈F?q.

    2) ForP1,P2∈GLn(Fq),σP1=σP2if and only ifP1=kP2for somek ∈F?q.

    IfP ∈GLn(Fq) satisfiesPSnPT=kSnfor somek ∈F?qwe callPa generalized orthogonal matrix.All generalized orthogonal matrices of ordernwith respect toSnforms a subgroup of GLn(Fq),called the generalized orthogonal group of degreenoverFqand denoted by GOn(Fq).The center of GOn(Fq) consists of allkEnwithk ∈F?q.The factor group of GOn(Fq) with respect to its center is called the projective generalized orthogonal group of degreenoverFqand is denoted by PGOn(Fq).According to Lemma 2.1,PGOn(Fq) can be viewed as a subgroup of Aut(O(Sn,q)).

    Letn= 2,and supposeπis a bijective mapping onFqsatisfyingπ(?x?1θ?1) =?θ?1π(x)?1for anyx ∈F?qandπ(0) = 0.DefineσπfromV(O(S2,q)) to itself by sending each vertex [a1,a2],with the standard form,to [a1,π(a2)].

    Lemma 2.2σπis an automorphism ofO(S2,q).

    ProofEach vertex [α] is written in the standard form.Obviously,σπis a bijective mapping onV(O(S2,q)).Suppose that [α] = [a1,a2][b1,b2] = [β].Equivalently,a1b1+θa2b2= 0.Ifa1= 0,thena2=b1= 1 andb2= 0.Clearly,σπ([α])σπ([β]).Ifb1= 0,the result is similar.Now suppose that botha1andb1are 1.Thusb2=?θ?1a?12.By the hypothesis forπ,we haveπ(b2) =π(?θ?1a?12) =?θ?1π(a2)?1,which follows that 1+π(a2)π(b2)θ= 0,showing thatσπ([α])σπ([β]).Consequently,[α][β]?σπ([α])σπ([β]).It follows thatd(σπ([α]))≤d([α])for any vertex[α],whered([α])denotes the degree of [α].By

    we haved(σπ([α]))=d([α]) for any vertex [α].Then we conclude that [α]~[β]?σπ([α])~σπ([β]).Thus,σπis an automorphism.

    Remark 2.1The bijective mappingπonFq,for whichπ(?x?1θ?1)=?θ?1π(x)?1andπ(0)=0,needs not to be an automorphism ofFq.For example,letFqbe the field Z/3Z,and letθ=.Defineπon Z/3Z by fixing,permutingand.Thenπ(?x?1) =?π(x)?1forx=1 and 2.But clearly,πis not an automorphism of Z/3Z.

    Forn ≥3,letπbe an automorphism of the fieldFqand letSn= diag(In?1,θ) withθ= 1 orz.Ifθ=z,thenπ(z)/∈F2qandz?1π(z)∈F2q.Supposer ∈F?qsuch thatπ(z) =zr2.Lety=rifθ=z,and lety= 1 ifθ= 1.Then,π(θ) =θy2.Defineσπ,y:V(O(Sn,q))→V(O(Sn,q)) by

    where [a1,a2,...,an] is written in the standard form.

    Lemma 2.3σπ,yis an automorphism ofO(Sn,q).

    ProofObviously,σπ,yis a bijective mapping onV(O(Sn,q)).The following derivation shows thatσπ,ypreserves adjacency relation of vertices in both directions.

    Hence,σπ,yis an automorphism ofO(Sn,q).

    Letω= (w1,w2,...,wn)∈ Fnq,wherew2i= 1 for eachiandw1= 1.We defineσωfromV(O(Sn,q)) to itself by sending any [a1,a2,...,an],with standard form,to[w1a1,w2a2,...,wnan].

    Lemma 2.4σωis an automorphism ofO(Sn,q).

    ProofObviously,σωis a bijective mapping onV(O(Sn,q)).The procedure

    shows thatσωpreserves adjacency relation of vertices in both directions.Thus the result follows.

    With the known automorphisms in hands,we now describe every automorphism ofO(Sn,q).Letn ≥3 and let a ={α1,α2,...,αn}be a non-isotropic orthogonal basis ofFnq,and suppose that (αi,αi) =a≠ 0 fori= 1,2,...,n ?1 and (αn,αn) =b≠ 0.Letσbe an automorphism ofO(Sn,q).Assumeσ([αi]) = [βi],and assume (βi,βi) =ci.Thenci≠0 fori= 1,2,...,nand (βi,βj) = 0 fori≠j.Namely,{β1,β2,...,βn}also forms a non-isotropic orthogonal basis ofFnq.For a nonzero vectorsuppose thatIt is easy to see thatbi≠0 if and only ifai≠0.Indeed,ifai≠0,then [α]~[αi].Applyingσ,we havewhich leads tobici≠0.Thusbi≠0.Similarly,bi≠0 impliesai≠0.Sinceσis a bijection fromV(O(Sn,q)) to itself,we define permutationsπi,i=2,3,...,nonFqsuch that

    It seems more accurate if one likes to use the symbolπi,σ,ainstead ofπi,as the definition ofπidepends onσand a.

    Next,we useπito characterizeσand give some properties forπiandci.

    Lemma 2.5(i){β1,β2,...,βn}forms a non-isotropic orthogonal basis ofFnq;

    Proof(i) was proved just before this lemma.Assume thatσ([α1+n i=2aiαi]) =Denotebyα,andTo complete the proof of (ii) we need to show thatbi=πi(ai) for eachi ≥2.The case whenai= 0 is trivial.Assume thatai≠0.For the case when 2≤i ≤n ?1,bywe havewhich follows thatThus

    On the other hand,by applyingσtowe havewhich leads to

    Comparing (2.1) and (2.2),we havebi=πi(ai).Next,we consider the case wheni=n.Applyingσto [α][α1?ab?1a?1n αn],we obtain [β][β1+πn(?ab?1a?1n)βn],which follows thatc1+bnπn(?ab?1a?1n)cn=0.Thus

    On the other hand,by applyingσto[α1+anαn][α1?ab?1a?1n αn]we have[β1+πn(an)βn][β1+πn(?ab?1a?1n)βn],which follows that

    Comparing (2.3) and (2.4),we also havebn=πn(an).The discussions above say that for anya2,a3,...,an ∈Fq,

    This completes the proof of (ii).

    Let 2≤i ≤n ?1 and denoteαi+ai+1αi+1+...+anαnalso byα.Assume that

    To complete the proof of (iii) we need to show thatb′j=c?11ciπi(1)πj(aj) fori+1≤j ≤n.Denoteβi+b′i+1βi+1+...+b′nβnalso byβ.For the case wheni+1≤j ≤n ?1,applyingσto [α][α1+αi ?a?1j αj] we have

    which leads to

    Following from (2.2) and (2.6),we obtainb′j=c?11ciπi(1)πj(aj).For the case whenj=n,applyingσto [α][α1+αi ?ab?1a?1n αj] we have

    which leads to

    From (2.4) and (2.7) we obtainb′n=c?11ciπi(1)πn(an).Thus,

    withqi=c?11ciπi(1),which completes the proof of (iii).

    Now we are in a position to complete the proof of (iv).Considering the action ofσon[α2+αn][α2?ab?1αn],we have

    which follows that

    Applyingσto [α1+α2+αn][α2?ab?1αn] we have

    leading to

    Comparing (2.9) and (2.10),we have that

    Applyingσto [α1+α2][α1?α2] we have

    which implies that

    From (2.11) and (2.12) we have

    Ifn=3,then the proof of(iv)is completed.Ifn ≥4,applyingσto[α1?α2+αj][α2+αj]for 3≤j ≤n ?1,we have

    which shows that

    Recalling thatπ2(?1)=?π2(1),we obtain

    On the other hand,the action ofσon [α1+α2+αj][α2?αj] leads to

    showing that

    (2.15) and (2.16) imply that

    The first assertion of (iv) is confirmed by (2.13) together with (2.17),and the second one is confirmed by (2.11) together with (2.15).

    For the proof of (v),considering the action ofσon [α1?α2+αn][α2+ab?1αn],we have

    which implies that

    Comparing (2.10) with (2.19),we have

    The proof of (v) is completed.

    Lemma 2.6Letσ,a,b,πibe defined as in Lemma 2.3.Ifσ([αi])=[αi] for all values ofi,then the following assertions hold.

    ProofBy (iv) of Lemma 2.3,assertion (i) of this lemma is obvious.

    Forx ∈Fqand 2≤j ≤n ?1,applyingσto [α1+xα2+xαj][α2?αj],we have[α1+π2(x)α2+πj(x)αj][α2+π2(1)πj(?1)αj],which leads toπ2(x) =π2(1)πj(1)πj(x).Sinceπj(1)2=1,we have

    For the case whenj=n,applyingσto [α1+xα2+xαn][α2?ab?1αn],we have [α1+π2(x)α2+πn(x)αn][α2+π2(1)πn(?ab?1)αn],which leads to

    By takingx=1,(2.22) becomes

    Substituting this equation into (2.22),we have that

    (2.21) and (2.24) together confirm (ii).

    Now we prove (iii).Ifxy=0,the result is obvious.Suppose thatxy≠0.By [α1+α2+ab?1xαn][α2?x?1αn] we have [α1+π2(1)α2+πn(ab?1x)αn][α2+π2(1)πn(?x?1)αn],which follows that

    Applyingσto [α1+xyα2+yαn][α2?ab?1xαn] we have that

    which implies that

    Substituting (2.25) into (2.26),we obtain

    Multiplying two sides of (2.27) byπ2(1) and using the result of (ii),we have

    Replacingywithxy,andxwithx?1,then we have

    Multiplying two sides of (2.29) byπj(1)2and using the result of (ii),we have

    By (2.30),we easily obtain

    The proof of (iii) is completed.

    For the proof of (iv),the case wheny=0 is trivial.Suppose thaty≠0.Applyingσto[α1?(x+y)α2+xαn][α1+y?1α2+ab?1y?1αn] we have

    which leads toa ?π2(x+y)π2(y?1)a+πn(x)πn(ab?1y?1)b=0.Or equivalently,

    Multiplying two sides of (2.32) byπn(1),we have

    By(iii)of this lemma and(v)of Lemma 2.5,πn(y)πn(ab?1y?1)a?1b=πn(ab?1)πn(1)a?1b=1.Thus (2.33) becomes

    By multiplying two sides of (2.34) byπ2(1),it follows that

    Finally,multiplying two sides of (2.35) byπj(1),we have

    The proof of (iv) is completed.

    It follows from (v) of Lemma 2.5 thatab?1=πn(1)πn(ab?1).(iii) of this lemma shows thatπn(ab?1)=πn(a?1b)?1πn(1)2.Consequently,πn(a?1b)=a?1b·πn(1)3.

    Now,we are in position to announce the main result of this paper.

    Theorem 2.1(i) Ifn= 2,thenσis an automorphism ofO(S2,q) if and only if there exists someK ∈GOn(Fq) and a bijective mappingπonFq,fixing 0 and satisfyingπ(?x?1θ?1)=?θ?1π(x)?1,such thatσ=σK ·σπ.

    (ii) Ifn ≥3,thenσis an automorphism ofO(Sn,q) if and only ifσ=σK ·σω ·σπ,y,whereπis an automorphism of the fieldFqsuch thatπ(θ)=y2θ,ω=(w1,w2,...,wn)∈Fnqwithw2i=1,K ∈GOn(Fq).

    ProofLetn= 2.Lemma 2.1 and 2.2 together confirm the sufficiency of (i).For necessity,letσbe an automorphism ofO(S2,q),and assumeσ([ei])=[fi] fori=1,2.Let

    ThenKS2KT= diag(c1,c2),whereci= (fi,fi) =fiS2fTi.Ifθ=z,we havez|K|2=c1c2.Thus only one ofciis a square element,sayc1.By replacingfiwith suitable multiple we may assume thatc1=1 andc2=z.Thus,KS2KT=S2,showing thatK,K?1∈GOn(Fq).ThusσKandσK?1are automorphisms ofO(S2,q).FromKK?1=I2,it follows thatfiK?1=ei.Thus,(σK?1·σ)([ei]) =σK?1([fi]) = [fiK?1] = [ei] fori= 1,2.Ifc2is a square element,similarly,by replacingfiwith suitable multiple we may assume thatc1=zandc2=z2.ByKS2KT=zS2,we also haveK,K?1∈GO2(Fq).Similarly,we know thatσK?1·σfixes [e1]and [e2].Ifθ= 1,by similar discussions as above,we can choose someK ∈GOn(Fq) such that (σK?1·σ)([ei])=[ei] fori=1,2.The procedure is omitted.

    Now,letσ1=σK?1·σ.For [α] = [1,x]∈V(O(S2,q)),the first coordinate ofσ1([α])cannot be zero sinceσ1fixes[e1].Thus,we can define a mappingπonFqsuch thatσ1([1,x])=[1,π(x)].Obviously,πis bijective andπ(0) = 0.Forx≠ 0,by applyingσ1to [1,x][1,?x?1θ?1],we have [1,π(x)][1,π(?x?1θ?1)],which leads to 1+θπ(x)π(?x?1θ?1) = 0,showing thatπ(?x?1θ?1) =?θ?1π(x)?1.Thusπcan be used to define the automorphismσπofO(S2,q).Clearly,σ1([α])=σπ([α]) for each [α]∈V(O(S2,q)),with the standard form.Finally,we haveσ1=σπandσ=σK ·σπ.

    For the proof of (ii),sufficiency is obvious.Now we consider necessity.Letσbe an automorphism ofO(Sn,q).Obviously,(ei,ei) =eiSeTi= 1 fori= 1,2,...,n ?1,(en,en) =enSeTn=θand (ei,ej) = 0 fori≠j.Thus,a ={e1,e2,...,en}forms a non-isotropic orthogonal basis ofFnq.Assume thatσ([ei])=[fi] and (fi,fi)=ci.Let

    ThenKSKT= diag(c1,c2,...,cn).As what we did just before Lemma 2.3,we define the permutationsπi(i=2,3,...,n),with respect toσand a,onFqsuch that

    By Lemma 2.3,we havecj=c1πj(1)?2forj=2,3,...,n ?1,andcn=c1πn(1)?1πn(θ?1)?1.

    Setπ1(1)=1 andπ1(0)=0.Letfi′=πi(1)fifori=1,2,...,n,and let

    Then we also haveσ([ei])=[f′i],however

    For simplity,f′iis also written asfiandK1is also written asK.

    Now we consider the case whenθ= 1.ThenS=InandKKT=c1In.ThusK,K?1∈GOn(Fq),andσK,σK?1∈Aut(O(Sn,q)).It follows fromKK?1=InthatfiK?1=eifori=1,2,...,n.Thus,(σK?1·σ)([ei])=σK?1([fi])=[fiK?1]=[ei]for eachi.DenoteσK?1·σbyσ1.Thenσ1fixes each [ei].Define permutationsπi,with respect toσ1and a,onFqsuch that

    Let 0≠α ∈Fnq.We write [α] in the standard form [α]=[a1,a2,...,an].Then by using (ii)and (iii) of Lemma 2.3 and recalling thatπi(1)2=1 for 1≤i ≤n,we have

    DefineπonFqbyπ(x) =π2(1)?1π2(x) forx ∈Fq.Sinceπi(1)?1πi(x) =π2(1)?1π2(x) fori= 3,4,...,n,we knowπ=πi(1)?1πifor eachi(≥2).By (iii) and (iv) of Lemma 2.6,it is easy to see thatπ(xy)=π(x)π(y) andπ(x+y)=π(x)+π(y) for allx,y ∈Fq,showing thatπis an automorphism ofFq.Now,(2.37) becomes

    Letω=(π1(1),π2(1),...,πn(1)).Sinceπi(1)2=1 for eachi(thanks to(iv)and(v)of Lemma 2.3),we can useωto define the automorphismσωofO(Sn,q).Now,

    This equation shows thatσω·σ1=σπ,1.Finally,we haveσ=σK ·σω·σπ,1.

    For the left caseθ=z(a fixed non-square element inFq),KSKT=diag(c1,c1,...,c1θ1)withθ1=πn(1)πn(z?1)?1.We claim thatθ1cannot be a square element.Otherwise,sayθ1=r2.By replacingfnwithr?1fn,we haveKSKT=diag(c1,c1,...,c1).Thus(fi,fi)=c1for alli.Sinceσ?1([fi])=[ei] for eachi,applying (v) of Lemma 2.3 toσ?1(viewingfiasαiand viewingeiasβi),we have that 1=(e1,e1)=(en,en)πn(1)2=zπn(1)2,absurd.So we may assume thatθ1=zs2.Replacingfnbys?1fn,we have thatKSKT=diag(c1,c1,...,c1z)=c1S.ThusK?1is a generalized orthogonal matrix.UsingK?1we define the automorphismσK?1onO(Sn,q) and denoteσK?1· σbyσ1.Similar as the case whenθ= 1,we haveσ1([ei])=σK?1([fi])=[fiK?1]=[ei] for eachi.

    Define permutationsπi,with respect toσ1and a,onFqsuch that

    Then by (ii) and (iii) of Lemma 2.3,

    for [α]∈V(O(Sn,q)) with standard form,whereπ1fixes 1 and 0.DefineπonFqbyπ(x)=π2(1)?1π2(x)forx ∈Fq.Also,πis an automorphism ofFq,andπ=πi(1)?1πifor eachi(≥2).Now,(2.39) becomes

    Letω=(π1(1),π2(1),...,πn?1(1),1).By(i)of Lemma 2.4,πi(1)2=1 for each 1≤i ≤n?1.So we can useωto construct the automorphismσωofO(Sn,q).Then

    Lety=πn(1).(v) of Lemma 2.4 shows thatπ(z) =πn(1)?1πn(z) =y2z.Thus we can useπandyto define the automorphismσπ,yofO(Sn,q).It is easy to see thatσω ·σ1=σπ,y.Finally,σ=σK ·σω·σπ,y,as desired.

    3.Orbits Partition of V(O(Sn,q)) Under the Automorphisms

    The study of orbits partition ofV(O(Sn,q)) needs an elementary lemma on GOn(Fq).

    Lemma 3.1LetSn= diag(In?1,θ) withθ= 1 orz.Ifnis odd,then for eachK ∈GOn(Fq) withKSnKT=kSn,kmust be a square element.Ifnis even,then there existsK ∈GOn(Fq) such thatKSnKT=zSn.

    ProofLetnbe odd,and supposeK ∈GOn(Fq) such thatKSnKT=kSn.Thenθ|K|2=knθ,showing thatkn=|K|2,andkis a square element.Now,assume thatnis even.A well known result about finite fields is that the equationX2+Y2=zhas solutions inFq.Suppose thata,b ∈Fqsuch thata2+b2=z.Ifθ=1,for 1≤2k+1<2k+2≤n,set

    Ifθ=z,we set

    for 1≤2k+1<2k+2≤n ?2,and setfn?1=en,fn=zen?1.Let

    Then,we haveQSnQT=zSn.LetV0be the subset ofV(O(Sn,q)) of all isotropic lines and let

    where (F?q)2refers the set of all non-zero square element ofFq.

    Theorem 3.1Ifnis odd,thenV(O(Sn,q)) has three orbits,they areV0,V1andV2.Ifnis even,thenV(O(Sn,q)) has only two orbits,they areV0andV1∪V2.

    ProofFor [α],[β]∈V(O(Sn,q)),if they belong to the sameVi,then by replacingαwith its suitable multiple we may assume that (α,α) = (β,β).Using Lemma 6.8 in [10],we can find an orthogonal matrixQsuch thatαQ=β.ThusσQ([α]) = [β],showing that all elements lying in the sameVibelong to one orbit.If [α]∈V0,clearly,σ([α])∈V0for allσ ∈Aut(O(Sn,q)).It implies thatV0forms an orbit.

    Ifnis odd,we want to prove thatV2is stable under the action of Aut(O(Sn,q)).Let[α]=[a1,a2,...,an]∈V2(with the standard form),and letσbe an automorphism ofO(Sn,q).We consider the action ofσon[α].By Theorem 2.1,σcan be decomposed asσ=σK·σω·σπ,y,whereK ∈GOn(Fq) withKSnKT=kSn,ω= (w1,w2,...,wn)∈Fnqwithw1= 1 andw2i= 1,andπis an automorphism ofFqsatisfyingπ(θ) =θy2fory ∈Fq.Assume thatσ([α])=[β].Then,

    and

    Since (α,α)∈(F?q)2,it follows thatπ(α,α)∈(F?q)2.By Lemma 3.1,k ∈(F?q)2,thusσ([α])=[β]∈V2.It implies thatV2forms an orbit,which immediately follows thatV1forms an orbit.

    Ifnis even,by Lemma 3.1,we can chooseQ ∈GOn(Fq) such thatQSnQT=zSn.Then for [α]∈V2,we have [αQ]∈V1sinceαQSnQTαT=z(α,α)∈z(F?q)2.Consequently,σQ([α])=[αQ]∈V1.It follows thatV1∪V2forms an orbit.

    猜你喜歡
    天真
    彩墨渾成,天真自然
    天真真好
    天真組詩
    滇池(2022年5期)2022-04-30 21:44:36
    天真熱
    天真童年
    小讀者(2020年4期)2020-06-16 03:34:10
    雪天真快樂
    天真給你最美的夢
    雪天真快樂
    古淡天真之美——倪瓚《淡室詩》
    丹青少年(2017年1期)2018-01-31 02:28:21
    年啊年
    夜夜躁狠狠躁天天躁| av视频免费观看在线观看| 91成年电影在线观看| 色精品久久人妻99蜜桃| 老司机午夜福利在线观看视频| 夜夜躁狠狠躁天天躁| 亚洲熟女毛片儿| 国产成人系列免费观看| 免费观看人在逋| 精品无人区乱码1区二区| 精品久久久久久久久久免费视频 | 国产片内射在线| 无限看片的www在线观看| 欧美+亚洲+日韩+国产| 国产欧美日韩精品亚洲av| 在线观看午夜福利视频| 成人三级做爰电影| 真人做人爱边吃奶动态| 久久婷婷成人综合色麻豆| 在线观看免费视频网站a站| 成在线人永久免费视频| 亚洲欧美激情综合另类| 日日摸夜夜添夜夜添小说| 欧美日韩精品网址| av欧美777| 99国产综合亚洲精品| 日本一区二区免费在线视频| 91老司机精品| 国产亚洲av高清不卡| 两个人免费观看高清视频| 亚洲精品中文字幕一二三四区| 国产区一区二久久| 亚洲专区字幕在线| 欧美av亚洲av综合av国产av| 男女之事视频高清在线观看| 欧美成人免费av一区二区三区 | 国产欧美日韩精品亚洲av| 国产精品偷伦视频观看了| 国内久久婷婷六月综合欲色啪| 色老头精品视频在线观看| 久久国产乱子伦精品免费另类| 国产成人精品在线电影| 免费在线观看亚洲国产| 女人爽到高潮嗷嗷叫在线视频| 欧美日韩亚洲高清精品| 国产一卡二卡三卡精品| 欧美亚洲 丝袜 人妻 在线| 高清在线国产一区| 亚洲三区欧美一区| 久久久精品免费免费高清| 51午夜福利影视在线观看| av片东京热男人的天堂| 色综合婷婷激情| 伊人久久大香线蕉亚洲五| 天堂√8在线中文| 两性夫妻黄色片| 黄片播放在线免费| 亚洲精品中文字幕在线视频| 岛国毛片在线播放| 欧美成狂野欧美在线观看| 亚洲欧洲精品一区二区精品久久久| 欧美乱色亚洲激情| www日本在线高清视频| 一进一出抽搐gif免费好疼 | 欧美日韩亚洲综合一区二区三区_| 热99re8久久精品国产| 黄色毛片三级朝国网站| 人人妻,人人澡人人爽秒播| 亚洲人成77777在线视频| 日本wwww免费看| 村上凉子中文字幕在线| 国产亚洲欧美在线一区二区| 一区在线观看完整版| 黄色怎么调成土黄色| 国产一区二区三区视频了| 天堂动漫精品| 日本一区二区免费在线视频| 亚洲欧美日韩高清在线视频| 精品电影一区二区在线| 19禁男女啪啪无遮挡网站| 国产精品成人在线| 精品国内亚洲2022精品成人 | 久久久久国内视频| 黄色怎么调成土黄色| 少妇粗大呻吟视频| 亚洲人成77777在线视频| 国产精品秋霞免费鲁丝片| 丝袜美足系列| 久久久国产欧美日韩av| 交换朋友夫妻互换小说| 免费av中文字幕在线| 侵犯人妻中文字幕一二三四区| 少妇被粗大的猛进出69影院| 欧美成狂野欧美在线观看| 亚洲欧美日韩高清在线视频| 亚洲全国av大片| 国产精品.久久久| 国产麻豆69| 丝袜美足系列| 村上凉子中文字幕在线| 中亚洲国语对白在线视频| 在线播放国产精品三级| 老司机靠b影院| 国产成人系列免费观看| 国产精品二区激情视频| 亚洲七黄色美女视频| 日本撒尿小便嘘嘘汇集6| 美国免费a级毛片| 极品少妇高潮喷水抽搐| 精品国产乱码久久久久久男人| 欧美一级毛片孕妇| av不卡在线播放| avwww免费| 一边摸一边抽搐一进一小说 | 免费高清在线观看日韩| 久久草成人影院| a级毛片黄视频| 丰满的人妻完整版| 嫁个100分男人电影在线观看| 欧美久久黑人一区二区| www.自偷自拍.com| 久久99一区二区三区| 黄色丝袜av网址大全| 亚洲精品在线美女| 国产男女超爽视频在线观看| 亚洲avbb在线观看| 超碰97精品在线观看| 一二三四社区在线视频社区8| 麻豆av在线久日| 免费久久久久久久精品成人欧美视频| 亚洲专区字幕在线| 午夜福利视频在线观看免费| av有码第一页| 亚洲一区二区三区欧美精品| 国产亚洲欧美精品永久| 一区二区三区激情视频| 成人亚洲精品一区在线观看| 国产男女内射视频| 丝袜在线中文字幕| 天天添夜夜摸| 免费久久久久久久精品成人欧美视频| 午夜视频精品福利| 精品免费久久久久久久清纯 | 老汉色av国产亚洲站长工具| 欧美丝袜亚洲另类 | 亚洲色图综合在线观看| 村上凉子中文字幕在线| 欧美日韩成人在线一区二区| 黑人操中国人逼视频| 搡老岳熟女国产| 人成视频在线观看免费观看| 制服人妻中文乱码| 飞空精品影院首页| 中文字幕另类日韩欧美亚洲嫩草| 黑人欧美特级aaaaaa片| 极品少妇高潮喷水抽搐| 高清毛片免费观看视频网站 | 一本大道久久a久久精品| 天天躁狠狠躁夜夜躁狠狠躁| 午夜福利在线观看吧| 人妻丰满熟妇av一区二区三区 | 一区在线观看完整版| 脱女人内裤的视频| 757午夜福利合集在线观看| 久久婷婷成人综合色麻豆| 亚洲av美国av| 国产精品亚洲av一区麻豆| 国产精品国产高清国产av | 男男h啪啪无遮挡| 国产成人系列免费观看| 一区在线观看完整版| 一本综合久久免费| 精品国产美女av久久久久小说| 国产欧美日韩一区二区精品| 在线十欧美十亚洲十日本专区| 午夜免费观看网址| 国产亚洲欧美98| 亚洲精品一卡2卡三卡4卡5卡| 男女床上黄色一级片免费看| 很黄的视频免费| 97人妻天天添夜夜摸| 亚洲久久久国产精品| 黄色片一级片一级黄色片| 99国产精品一区二区三区| 亚洲精品一二三| 日本一区二区免费在线视频| www日本在线高清视频| 深夜精品福利| 欧美日韩亚洲国产一区二区在线观看 | 亚洲第一av免费看| 午夜精品久久久久久毛片777| 国产精品久久久久成人av| 热99久久久久精品小说推荐| 女人被狂操c到高潮| 女同久久另类99精品国产91| 久久中文字幕人妻熟女| 精品少妇久久久久久888优播| 色在线成人网| 国产av又大| 国产亚洲精品一区二区www | 亚洲专区字幕在线| 在线天堂中文资源库| 久久久久国产一级毛片高清牌| 亚洲欧美激情在线| 99久久99久久久精品蜜桃| 日本欧美视频一区| 久久午夜亚洲精品久久| 手机成人av网站| 十八禁网站免费在线| 少妇的丰满在线观看| 女同久久另类99精品国产91| 亚洲欧美一区二区三区黑人| av国产精品久久久久影院| 啦啦啦免费观看视频1| 如日韩欧美国产精品一区二区三区| 91九色精品人成在线观看| 他把我摸到了高潮在线观看| 最新在线观看一区二区三区| 亚洲欧美日韩另类电影网站| а√天堂www在线а√下载 | 母亲3免费完整高清在线观看| 亚洲色图综合在线观看| 1024香蕉在线观看| 女人精品久久久久毛片| ponron亚洲| 国产成人精品久久二区二区91| 50天的宝宝边吃奶边哭怎么回事| 亚洲精品久久成人aⅴ小说| 欧美日韩一级在线毛片| 麻豆乱淫一区二区| 亚洲性夜色夜夜综合| 精品第一国产精品| 日韩 欧美 亚洲 中文字幕| 久久香蕉激情| 亚洲精品成人av观看孕妇| 人人妻人人澡人人爽人人夜夜| 国产精品欧美亚洲77777| 男人舔女人的私密视频| 超色免费av| 一本大道久久a久久精品| 亚洲精品美女久久久久99蜜臀| 中文字幕色久视频| 嫩草影视91久久| a在线观看视频网站| 又黄又粗又硬又大视频| 五月开心婷婷网| 欧洲精品卡2卡3卡4卡5卡区| 国产在线观看jvid| 日本vs欧美在线观看视频| 精品国产亚洲在线| 亚洲国产欧美一区二区综合| 9191精品国产免费久久| 国产一区二区激情短视频| 午夜久久久在线观看| 黑人巨大精品欧美一区二区蜜桃| 欧美乱妇无乱码| 不卡av一区二区三区| 一二三四社区在线视频社区8| 高清av免费在线| 久久人人97超碰香蕉20202| 最新的欧美精品一区二区| 欧美乱色亚洲激情| 国产精品自产拍在线观看55亚洲 | 国产精品一区二区在线观看99| 久久人妻av系列| 亚洲国产精品sss在线观看 | 欧洲精品卡2卡3卡4卡5卡区| 亚洲第一av免费看| 国产精品亚洲av一区麻豆| 国产高清视频在线播放一区| 麻豆av在线久日| 一本大道久久a久久精品| 日韩三级视频一区二区三区| xxx96com| 午夜福利乱码中文字幕| 中文字幕色久视频| 精品亚洲成国产av| 精品国产一区二区久久| 18禁观看日本| 亚洲人成电影免费在线| 身体一侧抽搐| 午夜福利在线观看吧| 两性午夜刺激爽爽歪歪视频在线观看 | 女性生殖器流出的白浆| xxxhd国产人妻xxx| 亚洲男人天堂网一区| 欧美另类亚洲清纯唯美| 欧美在线黄色| 18禁裸乳无遮挡动漫免费视频| 国产欧美日韩一区二区三区在线| 国产熟女午夜一区二区三区| 999久久久精品免费观看国产| 啪啪无遮挡十八禁网站| 久久精品亚洲熟妇少妇任你| 久久久久久久午夜电影 | 免费日韩欧美在线观看| 国产蜜桃级精品一区二区三区 | 啦啦啦免费观看视频1| 叶爱在线成人免费视频播放| 国产精品成人在线| 两个人看的免费小视频| 变态另类成人亚洲欧美熟女 | 欧美一级毛片孕妇| 一区二区三区激情视频| 久久精品aⅴ一区二区三区四区| 久久久久国内视频| 久久ye,这里只有精品| 淫妇啪啪啪对白视频| 美女午夜性视频免费| 欧美日韩中文字幕国产精品一区二区三区 | 中文字幕av电影在线播放| 99re在线观看精品视频| 国产av又大| 淫妇啪啪啪对白视频| 午夜福利欧美成人| 国产av又大| 美国免费a级毛片| 人人妻人人爽人人添夜夜欢视频| 亚洲成a人片在线一区二区| 欧美日韩瑟瑟在线播放| 一进一出好大好爽视频| 亚洲成a人片在线一区二区| 老司机靠b影院| 老鸭窝网址在线观看| 亚洲av成人一区二区三| 欧美黑人精品巨大| 99精品久久久久人妻精品| 久久久久精品国产欧美久久久| 咕卡用的链子| 免费在线观看完整版高清| 免费观看a级毛片全部| 亚洲综合色网址| 亚洲av片天天在线观看| 精品电影一区二区在线| 欧美日韩乱码在线| 久久久国产一区二区| 国产在视频线精品| 精品视频人人做人人爽| 叶爱在线成人免费视频播放| 国产极品粉嫩免费观看在线| 淫妇啪啪啪对白视频| 涩涩av久久男人的天堂| 国产精品自产拍在线观看55亚洲 | 亚洲成人免费av在线播放| 亚洲久久久国产精品| 久久天躁狠狠躁夜夜2o2o| 国产成人啪精品午夜网站| 飞空精品影院首页| 欧美黄色淫秽网站| 淫妇啪啪啪对白视频| 身体一侧抽搐| 一进一出好大好爽视频| 久久精品国产亚洲av高清一级| 窝窝影院91人妻| 一边摸一边做爽爽视频免费| 村上凉子中文字幕在线| 天天操日日干夜夜撸| 精品欧美一区二区三区在线| 每晚都被弄得嗷嗷叫到高潮| 欧美在线黄色| 久久午夜亚洲精品久久| 亚洲精品久久午夜乱码| 亚洲精品在线美女| 久久天堂一区二区三区四区| 国产色视频综合| 免费av中文字幕在线| 国产高清视频在线播放一区| 精品久久久久久久毛片微露脸| a在线观看视频网站| 亚洲精品av麻豆狂野| 婷婷成人精品国产| 欧美黑人欧美精品刺激| 午夜精品久久久久久毛片777| 日日爽夜夜爽网站| 色在线成人网| 欧美日韩视频精品一区| 亚洲在线自拍视频| 亚洲少妇的诱惑av| 国产成人影院久久av| 国产成人av激情在线播放| 侵犯人妻中文字幕一二三四区| 国产欧美日韩一区二区三| 午夜精品久久久久久毛片777| 精品国内亚洲2022精品成人 | 亚洲av电影在线进入| 国产极品粉嫩免费观看在线| 啦啦啦视频在线资源免费观看| 婷婷成人精品国产| 亚洲第一青青草原| 色综合欧美亚洲国产小说| 婷婷精品国产亚洲av在线 | 国产免费男女视频| 高清视频免费观看一区二区| 久久久国产成人精品二区 | 999精品在线视频| 亚洲精品中文字幕一二三四区| 在线观看免费高清a一片| 久久精品国产亚洲av香蕉五月 | 极品少妇高潮喷水抽搐| 久久午夜亚洲精品久久| 美女国产高潮福利片在线看| 国产三级黄色录像| 色尼玛亚洲综合影院| 女人高潮潮喷娇喘18禁视频| 成年人黄色毛片网站| 欧美激情久久久久久爽电影 | 一区二区三区激情视频| 亚洲av第一区精品v没综合| 美女福利国产在线| 91精品三级在线观看| 国产精品永久免费网站| 国产又色又爽无遮挡免费看| 中文字幕色久视频| av天堂在线播放| a在线观看视频网站| 国产亚洲av高清不卡| 欧美成狂野欧美在线观看| 午夜视频精品福利| 久99久视频精品免费| 欧美+亚洲+日韩+国产| 夜夜夜夜夜久久久久| 亚洲第一av免费看| 欧美精品av麻豆av| 99re6热这里在线精品视频| 日日爽夜夜爽网站| 精品国产乱码久久久久久男人| 女性被躁到高潮视频| 欧美精品人与动牲交sv欧美| 精品国产超薄肉色丝袜足j| 男人操女人黄网站| 十八禁高潮呻吟视频| 大码成人一级视频| 少妇被粗大的猛进出69影院| 免费在线观看日本一区| 99精国产麻豆久久婷婷| 欧美日韩精品网址| 制服诱惑二区| 婷婷精品国产亚洲av在线 | 欧美黄色片欧美黄色片| 丝瓜视频免费看黄片| 国产极品粉嫩免费观看在线| 欧美av亚洲av综合av国产av| 国产片内射在线| 99精品欧美一区二区三区四区| 国产成人免费无遮挡视频| 丝袜美腿诱惑在线| 国产精品一区二区精品视频观看| 男人操女人黄网站| 国产野战对白在线观看| 亚洲国产中文字幕在线视频| 天堂动漫精品| 精品久久久精品久久久| av超薄肉色丝袜交足视频| 午夜两性在线视频| 无人区码免费观看不卡| 国产精品久久久av美女十八| 在线播放国产精品三级| 久久香蕉国产精品| 99国产精品一区二区三区| 久久精品91无色码中文字幕| 少妇粗大呻吟视频| 可以免费在线观看a视频的电影网站| 两个人免费观看高清视频| 黄色a级毛片大全视频| 亚洲av电影在线进入| 老司机亚洲免费影院| 国产精品久久视频播放| 久久久久久人人人人人| 少妇的丰满在线观看| 日韩欧美一区二区三区在线观看 | 人妻 亚洲 视频| 人人妻人人添人人爽欧美一区卜| 亚洲人成77777在线视频| a级毛片黄视频| 成人手机av| 婷婷精品国产亚洲av在线 | 国产欧美日韩精品亚洲av| 欧美日韩国产mv在线观看视频| 免费人成视频x8x8入口观看| 国产成人欧美| 国产97色在线日韩免费| 另类亚洲欧美激情| 国精品久久久久久国模美| 叶爱在线成人免费视频播放| 亚洲精品一卡2卡三卡4卡5卡| 成人黄色视频免费在线看| 涩涩av久久男人的天堂| 正在播放国产对白刺激| 人人妻人人澡人人爽人人夜夜| 色老头精品视频在线观看| 青草久久国产| 人人妻,人人澡人人爽秒播| 亚洲精品国产一区二区精华液| 亚洲久久久国产精品| 女人爽到高潮嗷嗷叫在线视频| 老鸭窝网址在线观看| 悠悠久久av| 国产成人免费观看mmmm| 久久天堂一区二区三区四区| 色播在线永久视频| 人成视频在线观看免费观看| 女人被躁到高潮嗷嗷叫费观| 色综合欧美亚洲国产小说| 男女床上黄色一级片免费看| 久久精品亚洲精品国产色婷小说| 高清毛片免费观看视频网站 | 国产高清激情床上av| 精品国内亚洲2022精品成人 | 精品高清国产在线一区| 久久亚洲真实| 999久久久国产精品视频| 欧美成人免费av一区二区三区 | 欧美日韩精品网址| 久99久视频精品免费| 国产一区在线观看成人免费| 久久青草综合色| 午夜亚洲福利在线播放| 精品一区二区三区视频在线观看免费 | 亚洲人成电影观看| 99国产极品粉嫩在线观看| 午夜成年电影在线免费观看| 国产欧美亚洲国产| 亚洲午夜精品一区,二区,三区| 成年人黄色毛片网站| 久久国产乱子伦精品免费另类| 最近最新中文字幕大全电影3 | 一级毛片女人18水好多| 男女免费视频国产| 飞空精品影院首页| 亚洲avbb在线观看| 女人久久www免费人成看片| 色老头精品视频在线观看| 亚洲一卡2卡3卡4卡5卡精品中文| 欧美精品高潮呻吟av久久| 国产日韩一区二区三区精品不卡| 欧美日韩亚洲综合一区二区三区_| 一边摸一边做爽爽视频免费| 国产区一区二久久| 亚洲av美国av| 久久精品亚洲熟妇少妇任你| 欧美人与性动交α欧美精品济南到| 国产xxxxx性猛交| 淫妇啪啪啪对白视频| 国产伦人伦偷精品视频| 国产成人欧美| 校园春色视频在线观看| 国产欧美亚洲国产| 国产精品99久久99久久久不卡| 女警被强在线播放| 国产免费av片在线观看野外av| 色播在线永久视频| 国产成人影院久久av| 午夜影院日韩av| 91大片在线观看| 香蕉丝袜av| 丰满饥渴人妻一区二区三| 久久久久久亚洲精品国产蜜桃av| a级片在线免费高清观看视频| 精品国产国语对白av| 高清毛片免费观看视频网站 | 亚洲,欧美精品.| av电影中文网址| 欧美日韩成人在线一区二区| 久久久国产欧美日韩av| 黄色怎么调成土黄色| 大型av网站在线播放| 一进一出抽搐gif免费好疼 | 国产精品亚洲一级av第二区| 日本vs欧美在线观看视频| 精品乱码久久久久久99久播| 热99国产精品久久久久久7| 亚洲色图 男人天堂 中文字幕| 亚洲欧美色中文字幕在线| 国产99久久九九免费精品| cao死你这个sao货| 精品电影一区二区在线| 国产高清videossex| 精品国产一区二区三区久久久樱花| 久久香蕉精品热| 国产成人欧美在线观看 | 国产精品乱码一区二三区的特点 | 久久性视频一级片| 欧美精品高潮呻吟av久久| 亚洲精品av麻豆狂野| 麻豆av在线久日| 免费黄频网站在线观看国产| 欧美最黄视频在线播放免费 | av福利片在线| av视频免费观看在线观看| av电影中文网址| 国产熟女午夜一区二区三区| 日韩欧美一区视频在线观看| 亚洲av片天天在线观看| 国产精品成人在线| 国产淫语在线视频| 亚洲av片天天在线观看| 国产色视频综合| 一a级毛片在线观看| 王馨瑶露胸无遮挡在线观看| 久久精品亚洲av国产电影网| 日韩 欧美 亚洲 中文字幕| 在线国产一区二区在线| 高清毛片免费观看视频网站 | 中文字幕人妻丝袜一区二区| 成年版毛片免费区| 女人精品久久久久毛片| 久久久水蜜桃国产精品网| 麻豆成人av在线观看| 少妇的丰满在线观看| 国产精品.久久久| 麻豆乱淫一区二区| 色综合欧美亚洲国产小说| 露出奶头的视频| 亚洲精品中文字幕在线视频| 国产精品 欧美亚洲|