Heguo LIU Yulei WANG

Abstract Let G be a finite p-group with a cyclic Frattini subgroup.In this paper,the automorphism group of G is determined.

Keywords Finite p-groups,Frattini subgroups,Automorphisms

1 Introduction

In this paper,p always is a prime number,only finite groups will be considered.The terminologies and notations used are standard (cf.[1]).

Let G1and G2be any two groups,Z1and Z2be the centers of G1and G2,respectively.Assume that Z1is isomorphic to Z2,and θ : Z1→Z2is the isomorphic mapping,G1∗G2is called the central product of G1and G2relative to Z1,Z2and θ,that is,G1∗G2is the quotient group of G1×G2on the normal subgroup

In particular,let G be any group,Z ≤ζG,the central product G∗G is constructed by virtue of the identity mapping on Z.For any l >1,G∗lis denoted by G∗(l−1)∗G,and G∗1:=G,G∗0:=1.

A finite p-group G is called extraspecial,if G′=Frat G=ζG and have order p.Winter[2]has given the automorphism group of an extraspecial p-group.When p is odd,Dietz [3]generalized the results of Winter,and determined the automorphism group of a finite p-group which is a central extension of a group with order p by an elementary abelian group.

In [1],a finite p-group G is called generalized extraspecial,if the center ζG of G is cyclic and the derived subgroup G′of G has order p.In [4],we determined the automorphism group of the generalized extraspecial p-group.Further,let G be the below central extension

and |G′|≤p.In [5],we determined the automorphism group of the finite p-group,which generalized the results of Winter and Dietz.

Proposition 1.1(cf.[5])Let p be an odd number,G be a finite p-group given by a central extension of the form

and |G′|=p,where m ≥2.Then G=EA,where E is a generalized extraspecial p-group,A=ζG,E ∩A=ζE.Suppose that |E|=p2n+m,|ζE|=pmand |A|=pm+l.Let AutfG={α ∈Aut G |α acts trivially on Frat G}.Then

(i)If both E and A are of exponent pm,then Aut G/AutfGZ(p−1)pm−2,and AutfG/KSp(2n,p)×(GL(l,p)⋉(Zp)l),where K is of order p2n(l+1)+l+1.

(ii)If E and A are of exponent pmand pm+1,respectively,then Aut G/AutfGZ(p−1)pm−1,and AutfG/KSp(2n,p)×(GL(l −1,p)⋉(Zp)l−1),where K is of order p2nl+l.

(iii)If E and A are of exponent pm+1and pm,respectively,then Aut G/AutfGZ(p−1)pm−1,and AutfG/K(I ⋊Sp(2n −2,p))×GL(l,p),where I is an extraspecial p-group with order p2n−1and K is of order p2n(l+1)+l.

Proposition 1.2(cf.[5])Let G be a finite 2-group given by a central extension of the form

and |G′|=2,where m ≥2.Then G=EA,where E is a generalized extraspecial 2-group,A=ζG,E ∩A=ζE.Suppose that |E|=22n+m,|ζE|=2mand |A|=2m+l.Let AutfG={ α ∈Aut G |α acts trivially on FratG}.Then

(i)If both E and A are of exponent 2m,then Aut G/AutfG1(m=2)or Z2×Z2m−3(m ≥3),and AutfG/KSp(2n,2)×(GL(l,2)⋉(Z2)l),where K is of order 22n(l+1)+l+1.

(ii)If E and A are of exponent 2mand 2m+1,respectively,then Aut G/AutfGZ2×Z2m−2,and AutfG/KSp(2n,2)×(GL(l −1,2)⋉(Z2)l−1),where K is of order 22nl+l.

(iii)If E and A are of exponent 2m+1and 2m,respectively,then Aut G/AutfGZ2×Z2m−2,and AutfG/K(I ⋊Sp(2n −2,2))×GL(l,2),where I is an elementary abelian 2-group with order 22n−1and K is of order 22n(l+1)+l.

In [6],the structure and the automorphism group of a finite p-group with a cyclic Frattini subgroup were studied.In this paper,by means of the results in[5],the automorphism group of a finite p-group with a cyclic Frattini subgroup is further determined.On the hand,if p is odd,or p=2 and Frat G ≤ζG,then G is a finite p-group which is a central extension of a cyclic group Frat G by an elementary abelian group and G′has order p by Lemma 1.2 and Lemma 1.3.According to Proposition 1.1 and Proposition 1.2,the automorphism group of G can be determined,on the other hand,if p=2 and Frat GζG,we can obtain the below results.

In what follows,we are going to suppose that |Frat G|=pmand R is an elementary abelian 2-group with rank r.

Theorem 1.1Let G=R×(∗H),where H=H1,H2or H3,which are defined in Lemma 1.6.Let C :=CG(Frat G)and AutfG :={α ∈Aut G |α acts trivially on Frat C}.Then

(1)Aut G/AutfGZ2(if m=2),or Z2m−2×Z2(if m ≥3).

(2)AutfG/KSp(2n,2)×GL(r,2)⋉(Z2)r,where K is of order 2(2n+2)(r+1)+m(ifH=H1or H3),or 2(2n+2)(r+1)+m−1(if H=H2).

Theorem 1.2Let G=R×(∗H),where H=H4or H5,which are defined in Lemma 1.6.Let C :=CG(Frat G)and AutfG:={α ∈Aut G |α acts trivially on Frat C}.Then

(1)Aut G/AutfGZ2(if m=2),or Z2m−2×Z2(if m ≥3).

(2)AutfG/K(I ⋊Sp(2n,2))×GL(r,2),where I is an elementary abelian 2-group with order 22n+1,K is of order 2(2n+2)(r+1)+m+2r.

Theorem 1.3Let G=R×(∗H),where H=H6or H7,which are defined in Lemma 1.6.Let C :=CG(Frat G)and AutfG:={α ∈Aut G |α acts trivially on Frat C}.Then

(1)Aut G/AutfGZ2(if m=2),or Z2m−2×Z2(if m ≥3).

(2)AutfG/KSp(2n,2)×(GL(r,2)⋉(Z2)2r),K is of order 2(2n+2)(r+2)+m−1.

Theorem 1.4Let G=R×(∗H),where H=H8,which is defined in Lemma 1.6.Let C :=CG(Frat G)and AutfG:={α ∈Aut G |α acts trivially on Frat C}.Then

(1)Aut G/AutfGZ2(if m=2),or Z2m−2×Z2(if m ≥3).

(2)AutfG/K(I ⋊Sp(2n,2))×(GL(r,2)⋉(Z2)r),where I is an elementary abelian 2-group with order 22n+1,and K is of order 2(2n+2)(r+2)+2r+m+1.

According to the above theorems,let r=0,then we can obtain the below conclusion in [6].

We need the following several lemmas in order to obtain the above theorems.

Lemma 1.1(cf.[4])Let G be a generalized extraspecial p-group,then

(i)G/ζG is an elementary abelian p-group.

(ii)Let G′=〈c〉.For any two elements=xζG and=yζG of G/ζG,write [x,y]=cr(0 ≤r < p)and f(,)=r,then G/ζG becomes a nondegenerate symplectic space over GF(p).

(iii)G is a central product of some nonabelian subgroups Giwhich satisfy both ζGi=ζG and |Gi/ζGi|=p2.Furthermore,let |Gi|=pm+2,where m ≥2,then Gionly has two types:

or

Lemma 1.2(cf.[6])Let p be odd and G be a nonabelian p-group.If Frat G is cyclic,then Frat G is a central subgroup.

Lemma 1.3Let G be a nonabelian p-group.If Frat G is a cyclic and central subgroup,then G′is of order p.

ProofSince G is a nonabelian p-group,G′is nontrivial,and is included in the cyclic Frattini subgroup Frat G.Now we only need to prove that G′is of order p.

Since G′≤Frat G ≤ζG,for any x,y ∈G,we have that

Moreover,since xp∈Frat G ≤ζG,[xp,y]=1.Consequently,for any x,y ∈G,we have that[x,y]p=1.The lemma is proved.

Lemma 1.4(cf.[6])Let G be a nonabelian 2-group,Φ(G)be cyclic,Frat GζG and|Frat G|=2m,then m>1,and G is isomorphic to the direct product R×(∗H),where R is an elementary abelian 2-group,n ≥0,H is a nontrivial 2-group which is one of the following isomorphic types:

where

and

Lemma 1.5(cf.[4])If m ≥3,then

Lemma 1.6Let G be a nonabelian 2-group,Φ(G)be a cyclic group,and Frat GζG,|Frat G|=2m,then G is isomorphic to the direct product R × (∗H),where R is an elementary abelian 2-group,n ≥0,H is defined in Lemma 1.4.Further,

ProofAssume that

(1)Let H1:=HD2m+2,and H1=thenFrat G=and

Let H2:=HSD2m+2,and H2=If (yk)x=yk,where 0 ≤k < 2m+1,then y−k+2mk=yk.It follows that 2k −2mk ≡0 (mod 2m+1),which implies that (1 −2m−1)k ≡0 (mod 2m).Also 0 ≤k < 2m+1,thus k=2mand ζH2=Consequently,Frat G=〈y2〉.According to the results of H1,we similarly have that CG(Frat G)Nm+1(2)∗n×R.

Let H3:=HQ2m+2,and H3=Obviously,ζH3=Frat G=According to the results of H1,we similarly have that CG(Frat G)Nm+1(2)∗n×R.

(2)Let H4:=and

Let xiyjzk∈ζH4,where 0 ≤i < 2,0 ≤j < 2,0 ≤k < 2m+1,then (xiyjzk)x=xiyjzk.It follows that z2mk+k=zk,thus 2mk ≡0 (mod 2m+1),that is k ≡0 (mod 2).That(xiyjzk)y=xiyjzkimplies that z−k=zk,thus 2k ≡0 (mod 2m+1),that is k ≡0 (mod 2m).Since (xiyjzk)z=xiyjzk,(xi)z=xiz−2miand (yj)z=yjz(−1)j+1+1,−2mi+(−1)j+1+1 ≡0(mod 2m+1),which implies that −2mi+(−1)j+1+1 ≡0 (mod 2m).It follows that(−1)j+1+1 ≡0 (mod 2m),thus j=0.Consequently,i=0.From the above,we have that ζH4=andFrat H4=〈z2〉=Frat G.It follows that

Let H5:=and

Let xiyjzk∈ζH5,where 0 ≤i < 2,0 ≤j < 4 and 0 ≤k < 2m+1,then (xiyjzk)x=xiyjzk.It follows that z2mk+k=zk,thus 2mk ≡0 (mod 2m+1),that is k ≡0 (mod 2).That (xiyjzk)y=xiyjzkimplies that z−k=zk,thus 2k ≡0 (mod 2m+1),therefore k ≡0(mod 2m).Since (xiyjzk)z=xiyjzk,(xi)z=xiz−2miand (yj)z=yjz(−1)j+1+1,−2mi +(−1)j+1+1 ≡0 (mod 2m+1),which implies that −2mi+(−1)j+1+1 ≡0 (mod 2m).It follows that (−1)j+1+1 ≡0 (mod 2m),thus j=0 or 2.Consequently,i=0.From the above,we have that ζH5=and Frat H5==Frat G.According to the results of H4,similarly,CG(Frat G)Nm(2)∗n∗Mm(2)×R.

(3)Let H6:=HD2m+2∗C4,and

It is easy to verify that ζH6=〈z〉,D∗n8∩H6=〈z2〉and Frat H6=〈y2〉.It follows that

Since

Let H7:=HSD2m+2∗C4,and

Obviously,ζH7=and Frat H7=.According to the results of H6,we similarly have that CG(Frat G)Nm+1(2)∗n×R×Z2.

(4)Let H8:=and

Obviously,ζH8=and Frat H8=.It follows that

Since

2 Proof of Theorem 1.1

Since D8∗nis an extraspecial 2-group,we may suppose that x1,x2,··· ,x2n−1,x2n,y2mare the generators of D8∗n,which satisfy the following relations:

According to (1)in Lemma 1.6,we have that

Let Φ:Aut G →Aut(Frat C)be a restriction homomorphism.Obviously,Ker Φ=AutfGAut G.According to (1)in Lemma 1.6,Frat C=.

Theorem 2.1

ProofIf m=2,then Frat CZ4,thus Aut(Frat C)Z2.Define a mapping:

It is easy to verify that σ1is an automorphism of G,which is of order 2.Since Φ(σ1)(y2)=(y2)3and Φ(σ1)2(y2)=y2,Aut(Frat C)=.It follows that Aut G=AutfG ⋊

If m ≥3,then Z∗2m=×,where υ1=3 and υ2=2m−1.By Lemma 1.5,we have that the orders of v1and v2are 2m−2and 2,respectively.Define a mapping:

It is easy to verify that σ1and σ2are commutative automorphisms each other and their orders are 2m−1and 2,respectively.

Take any α ∈Aut G,then α(y2)=y2s1,where s1∈Z∗2m.Hence there exist 0 ≤t1<2m−2and 0 ≤t2<2 such that υt11υt22≡s−11(mod 2m).Since

We claim that 〈σ1〉∩〈σ2〉=1.In fact,letwhere w1,w2∈Z,then

The theorem is proved.

be the natural induced homomorphisms.From this,we may obtain the below homomorphic mapping

From this,C/ζC can become a nondegenerate symplectic space over GF(2).

Take any α ∈AutfG,then [α(a),α(b)]=α[a,b]=[a,b],thus,for any=aζC,b=bζC ∈C/ζC,we have that

therefore Ψ2(α)∈Sp(2n,2).Consequently,Ψ2(AutfG)≤Sp(2n,2).From the above,Ψ is the homomorphic mapping as follows:

Theorem 2.2Im Ψ2=Sp(2n,2).

ProofTake any T ∈Sp(2n,2),let (aik)be the matrix of T relative to a basis {xiζC,i=1,2,··· ,2n} of C/ζC.Define a mapping

where 0 ≤ai< 2,i=1,2,··· ,2n,0 ≤bj< 2,j=1,2,··· ,r,0 ≤c < 2,0 ≤d < 2m+1,

Note that (aik)is a nonsingular matrix.It is easy to verify φ is a bijection.Therefore,φ is an automorphism of G if and only if φ preserves multiplications.By the definition of φ,we have

(3)φ(x)=x.

(4)φ(zj)=zj,j=1,2,··· ,r.

We call the above φ the induced mapping of G by T.

Claim 2.1If φ(xi)2=1,i=1,2,··· ,2n,then φ ∈AutfG.

In fact,let φ(xi)2=1,where i=1,2,··· ,2n.For any g1,g2∈G,we have

and

where ye=and 0 ≤e<2m+1.

Let c1+c2=c+2c′,ai+a′i=ti+2si,bj+b′j=t′j+2s′j,where 0 ≤c,ti,t′j<2,c′,si,s′j∈Z,i=1,2,··· ,2n,j=1,2,··· ,r,then

Hence φ ∈Aut G.Also since φ(y)=y,φ ∈AutfG.

The claim is proved.

For i=1,2,··· ,2n,we have

By Claim 2.1,the induced mapping φ by T is an automorphism of G,and Ψ1(φ)=T.Consequently,Im Ψ1=Sp(2n,2).

The theorem is proved.

Theorem 2.3Im Ψ3GL(r,2)⋉(Z2)r.

ProofLet

where A11is a r×r matrix,A21is a 1×r matrix.It is easy to verify that A ≤GL(r+1,2).For convenience,we may let zr+1:=y.

Take any α ∈AutfG.Let (ajk)be the matrix of Ψ3(α)relative to a basis {zjFrat C,j=1,2,··· ,r+1} of ζC/Frat C.

Let (ajk)be the partitioned matrix as follows:

where A11,A12,A21and A22are r×r,r×1,1×r and 1×1 matrices,respectively.

Since z2j=1 for j=1,2,··· ,r,

thus aj,r+1+ 2aj≡0 (mod 2m).But m > 1 and 0 ≤aj,r+1< 2,consequently,for j=1,2,··· ,r,we have aj,r+1=0,that is A12=0.

Since

ar+1,r+1+2ar+1≡1 (mod 2m).But m > 1 and 0 ≤ar+1,r+1< 2,thus ar+1,r+1=1,that is A22=1.

Conversely,for define a mapping:

It is easy to verify that δ ∈Aut G.Since

δ ∈AutfG and the matrix of Ψ2(δ)is (bjk)relative to a basis {zjFrat C,j=1,2,··· ,r+1} of ζC/Frat C.Hence Im Ψ2A.Also since AGL(r,2)⋉(Z2)r,we have that Ψ2(AutfG)GL(r,2)⋉(Z2)r.

The theorem is proved.

Theorem 2.4(1)If H=H1or H3,then Ker Ψ is a 2-group with order 2(2n+2)(r+1)+m.

(2)If H=H2,then Ker Ψ is a 2-group with order 2(2n+2)(r+1)+m−1.

ProofSince Ker Ψ acts trivially on all factors of the series G ≥C ≥ζC ≥Frat C ≥1,Ker Ψ is a 2-group.

Take any α ∈Ker Ψ,let α be an automorphism as follows:

where zr+1=y,0 ≤ai< 2,0 ≤bj< 2,0 ≤br+1< 2m+1,0 ≤aij< 2,0 ≤ai,r+1< 2m+1,0 ≤ck<2m,i=1,2,··· ,2n,j=1,2,··· ,r,k=1,2,··· ,r+1.

Since α(xi)2=1,where i=1,2,··· ,2n,1=Hence ai,r+1≡0(mod 2m).Consequently,ai,r+1=0 or 2m.

Since α(x)and α(xi)are commutative each other,

If H=H1or H3,then [x,yai,r+1]=y2ai,r+1=1.If H=H2,then[x,yai,r+1]=y2ai,r+1−2mai,r+1=1.In a word,If i is odd,we can let i=2l −1,where l=1,2,··· ,n,then y2ma2l=1,which implies that a2l=0.If i is even,we can let i=2l,where l=1,2,··· ,n,then y2ma2l−1=1,which implies that a2l−1=0.Consequently,for i=1,2,··· ,2n,we have that ai=0.

Since α(x)and α(zk)are commutative each other,where k=1,2,··· ,r,

If H=H1or H3,then y4ck=1.If H=H2,then 1=[x,y2ck]=y4ck−2m+1ck=y4ck.In a word,ck≡0 (mod 2m−1),which implies that ck=0 or 2m−1.Also since α(y2)=y2,y2=(y1+2cr+1)2=y2+4cr+1,which implies that cr+1≡0 (mod 2m−1),thus cr+1=0 or 2m−1.Consequently,for k=1,2,··· ,r+1,we have that ck=0 or 2m−1.

Since α(zk)2=1,where k=1,2,··· ,r,1=(zky2ck)2=y4ck,which implies that ck≡0(mod 2m−1),thus ck=0 or 2m−1.

If H=H1or H3,then α(x)2=which has no effect on the parameters of α.If H=H2,then α(x)2=thus br+1≡0(mod 2).

It is easy to verify other generated relations have no effect on the parameters of α.

In conclusion,α is an automorphism as follows:

where zr+1=y,0 ≤bj< 2,0 ≤aij< 2,ai,r+1=0 or 2m,ck=0 or 2m−1,i=1,2,··· ,2n,j=1,2,··· ,r,k=1,2,··· ,r+1,0 ≤br+1< 2m+1(if H=H1or H3); br+1≡0 (mod 2)(if H=H2).

Conversely,if α is an automorphism of G,which satisfies the above conditions,then α ∈Ker Ψ.Hence,if H=H1or H3,then |Ker Ψ|=2(2n+2)(r+1)+m; if H=H2,then |Ker Ψ|=

The theorem is proved.

3 Proof of Theorem 1.2

For convenience,we may let x3,x4,··· ,x2n+1,x2n+2,z2mbe the generators of D∗n8,which satisfy the following conditions:

According to (2)in Lemma 1.6,we have that

where x1:=z,x2:=x.

For convenience,we sometimes adopt the notations in Theorem 1.1.

Let Φ:Aut G →Aut(Frat C)be the restriction homomorphism.Clearly,Ker Φ=AutfGAut G.According to (2)in Lemma 1.6,we have that Frat C=〈z2〉=Frat GZ2m.

Theorem 3.1

ProofIf m=2,then Frat CZ4,thus Aut(Frat C)Z2.Define a mapping:

It is easy to verify that σ3is an automorphism of G,which is of order 2.Since Φ(σ3)(z2)=(z2)3and Φ(σ3)2(z2)=z2,Aut(Frat C)=Consequently,Aut G=AutfG ⋊

It is easy to verify that σ3and σ4are commutative automorphisms each other and their orders are 2m−1and 2,respectively.

According to the argument in Theorem 2.1,we similarly have that Aut G=AutfG,a nd∩AutfG=Consequently,Aut G/AutfGZ2m−2×Z2.

The theorem is proved.

Let

be the natural induced homomorphisms.Hence we may define the below homomorphic mapping:

Since ζC=〈z2〉×R,we may define the inner product as follows:

From this,C/ζC can become a nondegenerate symplectic space over GF(2).

For any α ∈AutfG,[α(a),α(b)]=α[a,b]=[a,b],thus,for any=aζC,b=bζC ∈C/ζC,we have

therefore Ψ2(α)∈Sp(2n,2).Consequently,Ψ2(AutfG)≤Sp(2n,2).In a word,Ψ is a homomorphic mapping as follows:

Theorem 3.2Im Ψ2=I ⋊Sp(2n,2),where I is an elementary abelian 2-group with order

ProofLet B:={T ∈Sp(2n+2,2)|the first column and second row of the matrix of T are (1,0,··· ,0)Tand (0,1,0,··· ,0)relative to a basis x1ζC,x2ζC,··· ,x2n+2ζC of C/ζC,respectively}.

Take any T ∈B,let (aik)be the matrix of T relative to a basis {xiζC,i=1,2,··· ,2n+2}of C/ζC.Define a mapping:

where 0 ≤ai< 2,i=1,2,··· ,2n+2,0 ≤bj< 2,j=1,2,··· ,r,0 ≤c < 2,0 ≤d < 2m,(mod 2)

Note that (aik)is a nonsingular matrix.It is easy to verify φ is a bijection.Therefore,φ is an automorphism of G if and only if φ preserves multiplications.By the definition of φ,we have

(1)

(2)

(3)φ(zj)=zj,j=1,2,··· ,r.

(4)φ(y)=yxt.

(5)φ(z2)=z2.

Note that

and for any i=2,3,··· ,2n+2,we have that

For g1,g2∈G,

we have that

Let c1+c2=c+2c′,ai+a′i=ti+2si,bj+b′j=t′j+2s′j,2s1+e ≡e1(mod 2m+1),where 0 ≤c,ti,t′j<2,c′,si,s′j∈Z,0 ≤e1<2m+1,i=1,2,··· ,2n,j=1,2,··· ,r,then

therefore φ ∈Aut G.Also since φ(z2)=z2,φ ∈AutfG and Ψ2(φ)=T.

Conversely,take any ϕ ∈AutfG.Let Ψ2(ϕ)=T ∈Sp(2n+2,2),the matrix of T be(aij)relative to a basis{xiζC,i=1,2,··· ,2n+2}of C/ζC,where 0 ≤bik<2,i=1,2,··· ,2n+2,0 ≤di<2m.

Since

According to the results in [2],Ψ2(ϕ)=T ∈BI ⋊Sp(2n,2),where I is an elementary abelian 2-group with order 22n+1.

The theorem is proved.

Theorem 3.3Im Ψ3GL(r,2).

ProofSince Frat C=〈z2〉,{zjFrat C,j=1,2,··· ,r} is a basis of ζC/Frat C.It follows that ζC/Frat C is a linear space over GF(2)with dimension r,which implies that Im Ψ3can be embedded in GL(r,2).

Conversely,for any (djk)r×r∈GL(r,2),we may define a mapping:

where

It is easy to verify that δ1∈AutfG,and the matrix of Ψ2(δ1)is (bjk)relative to a basis{zjFrat C,j=1,2,··· ,r} of ζC/Frat C.Consequently,Ψ2(AutfG)GL(r,2).

The theorem is proved.

Theorem 3.4Ker Ψ is a 2-group with order 2(2n+2)(r+1)+m+2r.

ProofSince Ker Ψ acts trivially on the factors of the series G ≥C ≥ζC ≥Frat C ≥1,thus Ker Ψ is a 2-group.

Take any α ∈Ker Ψ,let α be an automorphism as follows:

where 0 ≤ai< 2,0 ≤bj< 2,0 ≤a < 2m,0 ≤aij< 2,0 ≤ci< 2m,0 ≤dj< 2m,i=1,2,··· ,2n+2,j=1,2,··· ,r.

Since α(z)2=z2,z2=which implies that c1=0 or 2m−1.

Since α(xi)2=1,where i=2,··· ,2n+2,1=which implies that ci≡0 (mod 2m−1),consequently,ci=0 or 2m−1.

Since α(y)is commutative with α(xi),where i=3,4,··· ,2n+2,

Note that 4ci≡0 (mod 2m+1).If i is odd,we can suppose that i=2l+1,where l=1,2,··· ,n,then z2ma2l+2=z4c2l+1+2ma2l+2=1,which implies that a2l+2=0; if i is even,we can suppose that i=2l,where l=2,··· ,n+1,then z2ma2l−1=1,which implies that a2l−1=0.In a word,for i=3,4,··· ,2n+2,we have that ai=0.

Since α(z)−2=[α(z),α(y)],

which implies that a2=0.

Since α(x)is commutative with α(y),

Also since c2=2m−1or 0,we have that a1=0.

Since α(y)is commutative with α(zj),where j=1,2,··· ,r,

which implies that dj=0 or 2m−1.

Since α(zj)2=(zjz2dj)2=z4dj,where j=1,2,··· ,r,dj=0 or 2m−1.

It is easy to verify generated relations of H4and H5have no effect on the parameters of α.

In conclusion,α is an automorphism as follows:

where 0 ≤bj<2,0 ≤a<2m,0 ≤aij<2,ci=0 or 2m−1,dj=0 or 2m−1,i=1,2,··· ,2n+2,j=1,2,··· ,r.

Conversely,if α is an automorphism of G,which satisfies the above conditions,then α ∈Ker Ψ.It follows that |Ker Ψ|=2(2n+2)(r+1)+m+2r.

The theorem is proved.

4 Proof of Theorem 1.3

According to (3)in Lemma 1.6,

For convenience,we may let zr+1:=zy2m−1,then [zr+1,x]=y2m.Let R1:=

Let Φ : Aut G →Aut(Frat C)be the restriction homomorphism.Obviously,Ker Φ=AutfGAut G.According to (3)in Lemma 1.6,Frat C=.

Theorem 4.1

ProofIf m=2,then Frat CZ4,therefore Aut(Frat C)Z2.Define a mapping:

It is easy to verify that σ5is an automorphism of G with order 2.Since Φ(σ5)(y2)=(y2)3and Φ(σ5)2(y2)=y2,Aut(Frat C)=It follows that Aut G=Autf

If m ≥3,then Z∗2m=where υ1=3 and υ2=2m−1.By Lemma 1.5,the orders of v1and v2are 2m−2and 2,respectively.Define a mapping:

It is easy to verify σ5and σ6are the commutative automorphisms of G each other and their orders are 2m−1and 2,respectively.

According to the argument in Theorem 2.1,we similarly have that Aut G=AutfG,∩AutfG=thus Aut G/AutfGZ2m−2×Z2.

The theorem is proved.

Let Ψ1:AutfG →Aut(G/C),Ψ2:AutfG →Aut(C/ζC)and Ψ3:AutfG →Aut(ζC/Frat C)be the natural induced homomorphisms.Define a homomorphic mapping:

From this,C/ζC can become a nondegenerate symplectic space over GF(2).For α ∈AutfG,[α(a),α(b)]=α[a,b]=[a,b],thus,for any=aζC,=bζC ∈C/ζC,we have that

therefore Ψ2(α)∈Sp(2n,2).Hence Ψ2(AutfG)≤Sp(2n,2).It follows that Ψ is a homomorphism as follows:

Theorem 4.2Im Ψ2=Sp(2n,2).

ProofTake any T ∈Sp(2n,2),let (aik)be the matrix of T relative to a basis {xiζC,i=1,2,··· ,2n} of C/ζC.Define a mapping:

where 0 ≤ai< 2,i=1,2,··· ,2n,0 ≤bj< 2,j=1,2,··· ,r+1,0 ≤c < 2,0 ≤d < 2m+1,

Note that (aik)is a nonsingular matrix.It is easy to verify φ is a bijection.Therefore,φ is an automorphism of G if and only if φ preserves multiplications.

According to the argument in Theorem 2.2,we similarly have that Im Ψ1=Sp(2n,2).

The theorem is proved.

Theorem 4.3Im Ψ3GL(r,2)⋉(Z2)2r.

ProofLet

where A11is a r×r matrix,A21is a 2×r matrix,I2is a 2×2 identity matrix.It is easy to verify that A ≤GL(r+2,2).For convenience,we may let zr+2:=y.

Take any α ∈AutfG.Let (ajk)be the (r+2)×(r+2)matrix of Ψ3(α)relative to a basis{zjFrat C,j=1,2,··· ,r+2} of ζC/Frat C.

Let (ajk)be the partitioned matrix as follows:

where A11,A12,A21and A22are r×r,r×2,2×r and 2×2 matrices,respectively.

For j=1,2,··· ,r+1,z2j=1,thus

Hence aj,r+2+2aj≡0 (mod 2m).But m>1 and 0 ≤aj,r+2<2,then,for j=1,2,··· ,r+1,aj,r+2=0,aj=0 or 2m−1.

Since

ar+2,r+2+2ar+2≡1 (mod 2m).But m>1 and 0 ≤ar+2,r+2<2,thus ar+2,r+2=1,ar+2=0 or 2m−1.

From this,2maj,r+1≡0 (mod 2m+1),thus aj,r+1=0.

Since

2mar+1,r+1≡1 (mod 2m+1).Thus ar+1,r+1=1.

If H=H6,then

y2=α(y2)=[α(x),α(zr+2)]

which implies that 2+2mar+2,r+1≡2 (mod 2m+1),therefore ar+2,r+1=0; if H=H7,then y2−2m=α(y2−2m)=[α(x),α(zr+2)]

which implies that 2mar+2,r+1≡0 (mod 2m+1),therefore ar+2,r+1=0.

It is easy to verify that δ2∈Aut G.Also since

δ2∈AutfG,and the matrix of Ψ2(δ2)is(bjk)relative to a basis{zjFrat C,j=1,2,··· ,r+2}of ζC/Frat C.Thus Im Ψ2C.Also since CGL(r,2)⋉(Z2)2r,Ψ2(AutfG)GL(r,2)⋉(Z2)2r.

The theorem is proved.

Theorem 4.4Ker Ψ is a 2-group with order 2(2n+2)(r+2)+m−1.

ProofSince Ker Ψ acts trivially on the factors of the series G ≥C ≥ζC ≥Frat C ≥1,Ker Ψ is a 2-group.

Take any α ∈Ker Ψ.Let

where zr+2=y,0 ≤ai< 2,0 ≤bj< 2,0 ≤br+2< 2m+1,0 ≤aij< 2,0 ≤ai,r+2< 2m+1,0 ≤ck<2m,i=1,2,··· ,2n,j=1,2,··· ,r+1,k=1,2,··· ,r+2.

Since α(xi)2=1,where i=1,2,··· ,2n,1=which implies that ai,r+2≡0 (mod 2m),that is ai,r+2=0 or 2m.

Since α(x)is commutative with α(xi),

If i is odd,let i=2l −1,where l=1,2,··· ,n,then y2m(a2l−1,r+1+a2l)=1,which implies that a2l−1,r+1+a2l≡0 (mod 2).If i is even,let i=2l,where l=1,2,··· ,n,then y2m(a2l,r+1+a2l−1)=1,which implies that a2l,r+1+a2l−1≡0 (mod 2).

Since α(x)is commutative with α(zk),where k=1,2,··· ,r,

If H=H6or H=H7,then y4ck=1,thus ck=0 or 2m−1.Also since

y2m=y2m+4cr+1,which implies that 4cr+1≡0 (mod 2m+1),that is cr+2=0 or 2m−1.If H=H6,

thus 4cr+2≡0 (mod 2m+1),that is cr+2=0 or 2m−1.If H=H7,

thus 4cr+2≡0 (mod 2m+1),that is cr+2=0 or 2m−1.In conclusion,for k=1,2,··· ,r+2,ck=0 or 2m−1.

If H=H6,

For k=1,2,··· ,r+1,1=α(zk)2=z2ky2ck=y4ck,thus 4ck≡0 (mod 2m+1),which implies that ck=0 or 2m−1.Also since y2=α(y2)=(y1+cr+2)2=y2+4cr+2,4cr+2≡0 (mod 2m+1),which implies that cr+2=0 or 2m−1.

It is easy to verify other generated relations of H6and H7which have no effect on the parameters of α.

In conclusion,α is an automorphism as follows:(mod 2)(if H=H6)or

where zr+2=y,0 ≤bj< 2,0 ≤aij< 2,(mod 2)(if H=H7),0 ≤br+2< 2m+1,a2l−1,r+1+a2l≡0(mod 2),a2l,r+1+a2l−1≡0 (mod 2),ai,r+2=0 or 2m,ck=0 or 2m−1,i=1,2,··· ,2n,j=1,2,··· ,r,k=1,2,··· ,r+2,l=1,2,··· ,n.

Conversely,if α is an automorphism of G,which satisfies the above conditions,then α ∈Ker Ψ.It follows that |Ker Ψ|=2(2n+2)(r+2)+m−1.

The theorem is proved.

5 Proof of Theorem 1.4

For convenience,we may suppose that x3,x4,··· ,x2n+1,x2n+2,z2mare the generators of,which satisfy the following conditions:

According to (4)in Lemma 1.6,

where x1:=z,x2:=x.

Let Φ : Aut G →Aut(Frat C)be the restriction homomorphism.Obviously,Ker Φ=AutfGAut G.According to (4)in Lemma 1.6,Frat C==Frat GZ2m.

Theorem 5.1

ProofIf m=2,then Frat CZ4,thus Aut(Frat C)Z2.Define a mapping:

It is easy to verify that σ7is an automorphism of G,which is of order 2.Since Φ(σ7)(z2)=(z2)7and Φ(σ7)2(z2)=z2,Aut(Frat C)=It follows that Aut G=AutfG ⋊.

It is easy to verify that σ7and σ8are the commutative automorphisms of G each other and their orders are 2m−1and 2,respectively.

By means of the argument in Theorem 2.1,we similarly have that Aut G=AutfG,andIt follows that Aut G/AutfGZ2m−2×Z2.

The theorem is proved.

Let

be the natural induced homomorphisms.From this,we can obtain the below homomorphism:

Hence C/ζC can become a nondegenerated symplectic space over GF(2).For any α ∈AutfG,[α(a),α(b)]=α[a,b]=[a,b],then,for any=aζC,=bζC ∈C/ζC,

therefore Ψ2(α)∈Sp(2n,2).Thus Ψ2(AutfG)≤Sp(2n,2).In a word,Ψ is a homomorphism as follows:

Theorem 5.2Im Ψ2=I ⋊Sp(2n,2),where I is an elementary abelian 2-group with order

ProofLet D :={T ∈Sp(2n+2,2)|the first column and second row of the matrix of T are (1,0,··· ,0)Tand (0,1,0,··· ,0)relative to a basis x1ζC,x2ζC,··· ,x2n+2ζC of C/ζC,respectively}.

Take any T ∈D.Let(aik)be the matrix of T relative to a basis{xiζC,i=1,2,··· ,2n+2}of C/ζC.Define a mapping:

where zr+1:=u,0 ≤ai< 2,i=1,2,··· ,2n+2,0 ≤bj< 2,j=1,2,··· ,r+1,0 ≤c < 2,(mod 2))or t=1

Note that (aik)is a nonsingular matrix.It is easy to verify φ is a bijection.Therefore,φ is an automorphism of G if and only if φ preserves multiplications.

According to the argument in Theorem 3.2,we similarly have that Im Ψ2=D=I ⋊Sp(2n,2),where I is an elementary abelian 2-group with order 22n+1.

The theorem is proved.

Theorem 5.3Im Ψ3GL(r,2)⋉(Z2)r.

ProofFor convenience,let zr+1:=uz2m−1,then

Let

where H11is a r×r matrix,H21is a 1×r matrix.It is easy to verify that H ≤GL(r+1,2).

For any α ∈AutfG,let (hjk)be the matrix of Ψ3(α)relative to a basis {zjFrat C,j=1,2,··· ,r+1} of ζC/Frat C.

Let (hjk)be the partitioned matrix as follows:

where H11,H12,H21and H22are r×r,r×1,1×r and 1×1 matrices,respectively.

For j=1,2,··· ,r+1,1=α(zj)2=z4hj,thus 4hj≡0 (mod 2m+1).

Let α(y)=yy1,where y1∈C.Since α(y)is commutative with α(zj)for j=1,2,··· ,r,

Hence hj,r+1=0,that is H12=0.Since

hr+1,r+1=1,that is H22=1.

It is easy to verify that δ3∈AutfG,and the matrix of Ψ2(δ3)is (bjk)relative to a basis{zjFrat C,j=1,2,··· ,r+1} of ζC/Frat C.Hence Im Ψ2H .Also since HGL(r,2)⋉(Z2)r,Ψ2(AutfG)GL(r,2)⋉(Z2)r.

The theorem is proved.

Theorem 5.4Ker Ψ is a 2-group with order 2(2n+2)(r+2)+2r+m+1.

ProofFor convenience,let zr+1:=uz2m−1.

Since Ker Ψ acts trivially on the factors of the series G ≥C ≥ζC ≥Frat C ≥1,Ker Ψ is a 2-group.

For any α ∈Ker Ψ,let

where 0 ≤ai< 2,0 ≤bj< 2,0 ≤a < 2m,0 ≤aij< 2,0 ≤ci< 2m,0 ≤dj< 2m,i=1,2,··· ,2n+2,j=1,2,··· ,r+1.

Since α(z)2=z2,z2=which implies that c1=0 or 2m−1.which implies that ci≡0 (mod 2m−1),that is ci=0 or 2m−1.

Since α(y)is commutative with α(xi),where i=3,4,··· ,2n+2,

Since α(xi)2=1,where i=2,··· ,2n+2,1=

Note that 4ci≡0 (mod 2m+1).If i is odd,let i=2j −1,where j=2,··· ,n + 1,then z2m(a2j−1,r+1+a2j)=1,which implies that a2j−1,r+1+a2j≡0 (mod 2); if i is even,let i=2j,where j=2,··· ,n+1,then z2m(a2j,r+1+a2j−1)=1,which implies that a2j,r+1+a2j−1≡0(mod 2).

Since α(x)is commutative with α(y),

Also since c2=0 or 2m−1,a1+a2,r+1≡0 (mod 2).

Since α(z)−2=[α(z),α(y)],

which implies that a2+a1,r+1≡0 (mod 2).

Since

Since α(y)is commutative with α(zj),where j=1,2,··· ,r,

which implies that dj=0 or 2m−1.Since

dr+1=0 or 2m−1.

Since 1=α(zj)2=(zjz2dj)2=z4dj,where j=1,2,··· ,r+1,dj=0 or 2m−1.

It is easy to verify other generated relations of H8have effect on the parameters of α.

In conclusion,α is an automorphism as follows:

where a2j−1,r+1+a2j≡0 (mod 2),a2j,r+1+a2j−1≡0 (mod 2),(mod 2),0 ≤bj< 2,0 ≤a < 2m,0 ≤aij< 2,ci=0 or 2m−1,dj=0 or 2m−1,i=1,2,··· ,2n+2,j=1,2,··· ,r+1.

Conversely,if α is an automorphism of G,which satisfies the above conditions,then α ∈Ker Ψ.Hence |Ker Ψ|=2(2n+2)(r+2)+2r+m+1.

The theorem is proved.