On the direct products of cyclic groups and the quotient group Zm×Zn/N

-, -, -

(School of Mathematics and Information Sciences, Guangzhou University, Guangzhou 510006, China)

Abstract：For the general group, it is not easy to find a simple and concrete group to characterize it.In this paper, we consider the quotient group Zm×Zn/N with normal subgroup N and give its equivalence for the cases gcd(m,n)=1 and gcd(m,n)≠1(m is prime), respectively.

Key words：direct product; normal subgroup; cyclic group; quotient group

0 Introduction

On the other hand, Cayley Theorem proved that any group is isomorphic to a transformation group, especially, any finite group is isomorphic to a permutation group.The meaning of Cayley Theorem tells us that it is convenient to study the structure of the general group if we use the simple and concrete group to express the general group, and thus the above question will be simple.However, for the general groups, it is not easy to give the simple and concrete groups to characterize them.

It is well-known that every cyclic group of ordernis isomorphic toZnand every cyclic group of infinite order is isomorphic toZ[12].If we want to consider the cyclic case of the quotient groupG1×G2/Nfor the general groupsGi, whereNis the normal subgroup ofG1×G2, it will be useful to consider the structure ofZm×Zn/Nfirstly.We know thatZmis cyclic, andZm×Znis cyclic if and only if gcd(m,n)=1, which implies thatZm×Zn?Zmnif and only if gcd(m,n)=1.Following these results, we can characterize the structure ofZm×Zn/N, and give its equivalence.In this paper, the main purpose is to use simple and concrete group to characterize the cyclic case ofZm×Zn/Nwith gcd(m,n)=1 and gcd(m,n)≠1(mis prime), respectively.

The paper is organized as follows.In Section 1, we introduce some fundamental results for the general groups and the direct productZm×Znwith gcd(m,n)=1, and its equivalence.For the quotient group ofZm×Znwith gcd(m,n)=1(resp.with primemand gcd(m,n)≠1), we give an equivalence of the quotient group ofZm×Znin Section 2.

1 Direct products of cyclic groups

In this section, we mainly study the direct productZm×Znunder the operation

([a1],[b1])+([a2],[b2])=([a1+a2],

[b1+b2]),[a1],[a2]∈

Zm,[b1],[b2]∈Zn

(1)

and the direct productsZm1×Zm2×…×Zmrunder the operation

([a1],[a2],…,[ar])+([b1],[b2],…,[br])=

([a1+b1],[a2+b2],…,[ar+br])

(2)

where[ai],[bi]∈Zmi, and consider them in each case to be cyclic groups.In fact, it is clear thatZm×ZnandZm1×Zm2×…×Zmrare groups under the operations(1)and(2), respectively.

Theorem1[12]LetG1,G2,…,Gnbe finite groups, and suppose thatai∈Gihas orderrifor eachi.Then(a1,…,an)∈G1×…×Gnhas orderlcm(r1,…,rn)under the operation(a1,a2,…,an)(b1,b2,…,bn)=(a1b1,a2b2,…,anbn)for anyai,bi∈Gi,i=1,2,…,n.

Theorem2[13]LetGbe a group anda∈Ghas finite orderm.Then

([5],[2])+([5],[2])+([5],[2])+

([5],[2])=([20],[8])=([0],[0]).

In order to consider the structure ofZm×Znin detail, we should first consider the structure of each element inZm×Zn.It is clear that there is no generator inZ2×Z4and([1],[1])is a generator ofZ2×Z3.SinceZm,Znare cyclic groups, so there is a natural question to ask whenZm×Znis a cyclic group?

Theorem3[14]The groupZm×Znunder the operation(1)is cyclic if and only if gcd(m,n)=1.

ProofSuppose thatZm×Znis a cyclic group and gcd(m,n)=d≠1.For any([a],[b])∈Zm×Znand any positive integerk, we have

([ka(modm)],[kb(modn)]).

Conversely, suppose that gcd(m,n)=1 and[a](resp.[b])is the generator ofZm(resp.Zn).Sinceord([a])=m,ord([b])=n,ord(([a],[b]))=lcm(m,n)=mn.This means that([a],[b])is a generator ofZm×Zn, which implies thatZm×Znis a cyclic group.

Following Theorem 3, we can easily obtain the following corollary.

Corollary1Zm×Zn?Zmnif and only if gcd(m,n)=1.

Note that([1],[1])is a generator ofZm×Znif gcd(m,n)=1.

Theorem4 The groupZm1×Zm2×…×Zmrunder the operation(2)is cyclic if and only ifmiandmjare coprime for alli≠j.

ProofAssume that the groupZm1×Zm2×…×Zmris cyclic and gcd(mi,mj)=d≠1 for somei,jwithi≠j.DenoteG=Zm1×Zm2×…×Zmr.

For any([ai],[bj])∈Zmi×Zmjand any positive integerk, we have

([kai(modmi)],[kbj(modmj)]).

([0],[0]).

Conversely, let[ai]be a generator of groupsZmiwith ordermifor eachi, ifmiandmjare relative prime for alli≠j.Then the order of([a1],[a2],…,[ar])ism1m2…mrfollowing Theorem 1.This means that([a1],[a2],…,[ar])is a generator ofZm1×Zm2×…×Zmr.ThusZm1×Zm2×…×Zmris a cyclic group.

Similarly, we have the equivalent withZm1×Zm2×…×Zmrfollowing Corollary 1 and Theorem 4.

Corollary2Zm1×Zm2×…×Zmr?Zm1m2…mrif and only ifmiandmjare relatively prime for alli≠j.

ProofSupposeZm1×Zm2×…×Zmr?Zm1m2…mr.ThenZm1×Zm2×…×Zmris a cyclic group and its order islcm(m1,m2,…,mr)=m1m2…mrby Theorem 1.It follows thatmiandmjare relatively prime for alli≠j.

Conversely, ifmiandmjare relatively prime for alli≠j, thenZm1×Zm2×…×Zmris a cyclic group by Theorem 4, which implies that the order ofZm1×Zm2×…×Zmrism1m2…mr.Therefore,Zm1×Zm2×…×Zmr?Zm1m2…mr.

2 Quotient group Zm×Zn/N

LetG1andG2be two groups, and the left coset ofG1×G2is

G1×G2/H={(a,b)H|(a,b)∈G1×G2}

with the operation

((a1,b1)H)((a2,b2)H)=(a1a2,b1b2)H

(3)

whereHis a subgroup ofG1×G2anda1,a2∈G1,b1,b2∈G2.It is well-known that the operation(3)does not hold in general case.However, if(a,b)H=H(a,b)for any(a,b)∈G1×G2, then the operation(3)is well-defined.Furthermore, ifHis a normal subgroup ofG1×G2, thenG1×G2/His a quotient group.It is well-known that one of the main purpose in group theory is to consider the structure of the quotient groupG1×G2/H.Therefore, there is a natural question to ask whenHis a normal subgroup, and what about the structure ofG1×G2/H? We have known that {e1}×{e2},G1×G2,G1×{e2} and {e1}×G2are normal subgroups ofG1×G2, whereeiis the identity ofGifori=1,2.However, for the general case, it is not easy to determine the normal subgroups ofG1×G2.In this section, we mainly study the left coset(similar for the right coset)

Zm×Zn/N={([a],[b])+N|([a],[b])∈

Zm×Zn}

with the operation

(([a1],[b1])+N)+(([a2],[b2])+N)=

(([a1+a2],[b1+b2])+N)

(4)

whereNis a subgroup ofZm×Znand[a1],[a2]∈Zm,[b1],[b2]∈Zn.SinceZm×Znis Abelian under the operation(1), any subgroups ofZm×Znare normal subgroups.

2.1 The quotient group of Zm×Zn with gcd(m,n)=1

It is well-known thatZm={[0],[1],…,[m-1]} is an Abelian group under the operation[k1]+[k2]=[k1+k2]for any[k1],[k2]∈Zm.Thus, for any element[k]∈Zm,([k])is a normal subgroup ofZm.

Theorem5 For any[k]∈Zm, we haveZm/([k])?Zgcd(m,k).

ProofSetd=gcd(m,k)and let

ψ:Zm→Zd,[n]→[n(modd)].

Assume that[m1]=[m2]for[m1],[m2]∈Zm, thenm|(m1-m2).

It is clear thatd|(m1-m2), which is equivalent to thatm1≡m2(modd), i.e.,[m1(modd)]=[m2(modd)].This means thatψis well-defined.In addition, for any[a],[b]∈Zm, we have

ψ([a]+[b])=ψ([a+b])=

[a+b(modd)]=

[a(modd)+b(modd)]=

[a(modd)]+[b(modd)]=

ψ([a])+ψ([b]).

It follows thatψis a homomorphism.Further, for any[a]∈Zd, we have[td+a]∈Zmandψ([td+a])=[td+a(modd)]=[a]for somet∈Z.It implies thatψis surjective.Also,([k])is the kernel ofψsinceψ([k])=[0].Therefore, following the group homomorphism theorem, we haveZm/([k])?Zgcd(m,k).

Example1 Letm=9 andk=3.ThenZ9/([3])?Z3.

ProofFollowing Theorem 5, we can easily obtain thatZ9/([3])?Z3.In fact, since the subgroup([3])=([0],[3],[6]), all the left cosets for subgroup([3])inZ9are as follows.

[0]+([3])=[3]+([3])=[6]+([3])=([0],[3],[6]),

[1]+([3])=[4]+([3])=[7]+([3])=([1],[4],[7]),

[2]+([3])=[5]+([3])=[8]+([3])=([2],[5],[8]).

Thus, the order ofZ9/([3])is 3 which is equal to the number of all the left cosets.In addition, we know thatZ9/([3])is generated by[1]+([3])with order 3.Therefore,Z9/([3])?Z3.

Let([a],[b])∈Zm×Zn, from the above, we know that 〈([a],[b])〉is a normal subgroup ofZm×Znand thenZm×Zn/〈([a],[b])〉 is a quotient group.

Theorem6 Suppose thatm,nare coprime and[k]∈Zmn.Then

Zm×Zn/〈([k(modm)],[k(modn)])〉?

Zgcd(mn,k).

ProofSuppose thatm,nare coprime.ThenZmn?Zm×Znby Corollary 1.Let

f:Zmn→Zm×Zn,[k]→([k(modm)],[k(modn)]).

For any[k1],[k2]∈Zmn, if[k1]=[k2], i.e.,mn|(k1-k2), thenm|(k1-k2),n|(k1-k2), it follows thatk1(modm)=k2(modm),k1(modn)=k2(modn), which implies that

([k1(modm)],[k1(modn)])=

([k2(modm)],[k2(modn)]).

This means thatfis well-defined.In addition, for any[k1],[k2]∈Zmn, we have

f([k1]+[k2])=f([k1+k2])=

([k1+k2(modm)],[k1+k2(modn)])=

([k1(modm)+k2(modm)],

[k1(modn)+k2(modn)])=

([k1(modm)],[k1(modn)])+

([k2(modm)],[k2(modn)])=

f([k1])+f([k2]).

It follows thatfis a homomorphism.Obviously, we have thatfis surjective.Therefore,fis epimorphism.Sincem,nare coprime, it is clear to know that ker(f)={[0]}.SinceZmn,Zm×Znare Abelian,([k])and 〈([k(modm)],[k(modn)])〉 are normal subgroups ofZmnandZm×Zn, respectively.Furthermore, we have 〈([k(modm)],[k(modn)])〉=f(([k])).Thus, following Lemma 1, we obtain thatZm×Zn/〈([k(modm)],[k(modn)])〉?Zmn/([k]).

On the other hand, from Theorem 5, we haveZmn/([k])?Zgcd(mn,k).Therefore,

Zm×Zn/〈([k(modm)],[k(modn)])〉?

Zgcd(mn,k).

Example2 Letm= 4,n= 9 andk= 24.Then(m,n)= 1 and[24]∈Z36, and

Z4×Z9/〈([24(mod 4)],[24(mod 9)])〉?

Z12.

ProofIt is clear that[24(mod 4)]=[0],[24(mod 9)]=[6].Thus

〈([24(mod 4)],[24(mod 9)])〉=

〈([0],[6])〉=

{([0],[6]),([0],[3]),([0],[0])}.

By computation, we obtain that all the left cosets inZ4×Z9/NwhereN=〈([0],[6])〉 are as follows.

([0],[0])+N=([0],[3])+N=([0],[6])+N={([0],[0]),([0],[3]),([0],[6])},

([0],[1])+N=([0],[4])+N=([0],[7])+N={([0],[1]),([0],[4]),([0],[7])},

([0],[2])+N=([0],[5])+N=([0],[8])+N={([0],[2]),([0],[5]),([0],[8])},

([1],[0])+N=([1],[3])+N=([1],[6])+N={([1],[0]),([1],[3]),([1],[6])},

([1],[1])+N=([1],[4])+N=([1],[7])+N={([1],[1]),([1],[4]),([1],[7])},

([1],[2])+N=([1],[5])+N=([1],[8])+N={([1],[2]),([1],[5]),([1],[8])},

([2],[0])+N=([2],[3])+N=([2],[6])+N={([2],[0]),([2],[3]),([2],[6])},

([2],[1])+N=([2],[4])+N=([2],[7])+N={([2],[1]),([2],[4]),([2],[7])},

([2],[2])+N=([2],[5])+N=([2],[8])+N={([2],[2]),([2],[5]),([2],[8])},

([3],[0])+N=([3],[3])+N=([3],[6])+N={([3],[0]),([3],[3]),([3],[6])},

([3],[1])+N=([3],[4])+N=([3],[7])+N={([3],[1]),([3],[4]),([3],[7])},

([3],[2])+N=([3],[5])+N=([3],[8])+N={([3],[2]),([3],[5]),([3],[8])}.

Thus, the order ofZ4×Z9/〈([0],[6])〉 is 12 which is equal to the number of the left cosets.Furthermore, we know thatZ4×Z9/〈([0],[6])〉 is generated by([1],[2])+Nwith order 12.Therefore, we haveZ4×Z9/〈([24(mod 4)],[24(mod 9)])〉?Z12.In fact, since gcd(4,9)=1, by Theorem 6, we can easily obtain that

Z4×Z9/〈([24(mod 4)],[24(mod 9)])〉?

Zgcd(36,24)?Z12.

Example3 Letm=4,n=6 andk=8.Then(m,n)=2≠1,[8]∈Z24, but

Z4×Z6/〈([8(mod 4)],[8(mod 6)])〉

Zgcd(24,8)?Z8.

ProofIt is clear that

〈([8(mod 4)],[8(mod 6)])〉=

〈([0],[2])〉=

{([0],[2]),([0],[4]),([0],[0])}.

([0],[0])+N=([0],[2])+N=([0],[4])+N={([0],[2]),([0],[4]),([0],[0])},

([0],[1])+N=([0],[3])+N=([0],[5])+N={([0],[3]),([0],[5]),([0],[1])},

([1],[0])+N=([1],[2])+N=([1],[4])+N={([1],[2]),([1],[4]),([1],[0])},

([1],[1])+N=([1],[3])+N=([1],[5])+N={([1],[3]),([1],[5]),([1],[1])},

([2],[0])+N=([2],[2])+N=([2],[4])+N={([2],[2]),([2],[4]),([2],[0])},

([2],[1])+N=([2],[3])+N=([2],[5])+N={([2],[3]),([2],[5]),([2],[1])},

([3],[0])+N=([3],[2])+N=([3],[4])+N={([3],[2]),([3],[4]),([3],[0])},

([3],[1])+N=([3],[3])+N=([3],[5])+N={([3],[3]),([3],[5]),([3],[1])}.

It follows that the order ofZ4×Z6/〈([0],[2])〉 is 8.However, there is no generator inZ4×Z6/〈([0],[2])〉.From the above, we haveZ4×Z6/〈([0],[2])〉?Z4×Z2.

SinceZ4×Z2is not cyclic, we haveZ4×Z2Z8.Therefore,

Z4×Z6/〈([8(mod 4)],[8(mod 6)])〉

Zgcd(24,8).

Similar as the proof of Theorem 6, we can obtain the following corollary easily.

Corollary3 Suppose thatmiandmjare relative prime for alli≠j,i,j=1,2,…,r, and[k]∈Zm1m2…mr.Then

Zm1×Zm2×…×Zmr/N?Zgcd(m1m2…mr,k),

whereN=〈([k(modm1)],[k(modm2)],…,[k(modmr)])〉.

2.2 The quotient group of Zm×Zn with gcd(m,n)≠1 and prime m

From Theorem 6, we know whenm,nare coprime, then the quotient group ofZm×Znis isomorphic to a cyclic groupZr, whereris the order of this quotient group.However, whenm,nare not coprime, do we still have the similar result for the quotient group ofZm×Zn? For example, whenm=n=pis prime, thenZp×Zpis not cyclic.In fact, the orderZp×Zpisp2, and each[a]∈Zpis a generator ofZpfor[a]≠[0].However, for any([a],[b])∈Zp×Zpwith([a],[b])≠([0],[0]), the order of 〈([a],[b])〉 isp(≠p2).This means thatZp×Zpis not cyclic.Hence, we should ask what is the equivalent of the quotient group ofZm×Znwith prime numbermand gcd(m,n)≠1?

It is well-known that every subgroup of a cyclic group is still cyclic, and is also a normal subgroup.Thus, ifG=〈a〉 is a finite cyclic group andHa subgroup ofG.ThenaHgeneratesG/Hwith the operation(xH)(yH)=(xyH)for anyx,y∈G.Indeed, note that any element ofG/His of the formgH={gh|h∈H} forg∈G.SinceG=〈a〉 is a finite cyclic group, we haveg=akfor some positive integerkfor anyg∈G.Thus,gH=(ak)H=(aH)k.This shows thataHgeneratesG/H.

Lemma2[15]LetGbe a group, if |G| is prime, thenGis a cyclic group.

Theorem7 Suppose thatG=〈a〉 is a finite cyclic group with prime orderp>1.ThenG×G/〈(as,at)〉 for 1≤s,t

ProofSuppose thatG=〈a〉 is a finite cyclic group with prime orderp>1, then for any non-identity elementak∈G, the order ofakispby Theorem 2.It follows from Theorem 1 that the order of 〈(as,at)〉 is alsopfor 1≤s,t

Theorem8 Suppose thatG1=〈a〉 andG2=〈b〉 are finite cyclic groups with orderspandkp, respectively, wherep>1 is prime andkis a positive integer.ThenG1×G2/〈(as,at)〉 is cyclic for 1≤s≤pandt=kpif the order of 〈(as,at)〉 iskp.

ProofSuppose the order of 〈(as,at)〉 iskp, from Theorem 1 and Theorem 2, we know that 1≤s≤pandt=kp.According to Lagrange Theorem, the order ofG1×G2/〈(as,at)〉 is a primep.Therefore,G1×G2/〈(as,at)〉 is a cyclic group by the Lemma 2.

Theorem9 Suppose thatp>1 is prime.ThenZp×Zp/〈([a],[b])〉?Zpfor any[a],[b]∈Zpsatisfying([a],[b])≠([0],[0]).

ProofSuppose thatpis prime, then each[a]∈Zpis a generator ofZpfor[a]≠[0].Thus, for any([a],[b])∈Zp×Zpwith([a],[b])≠([0],[0]), the order of 〈([a],[b])〉 ispand 〈([a],[b])〉 is a normal subgroup ofZp×ZpsinceZp×Zpis Abelian.However,Zp×Zpis not cyclic since the order ofZp×Zpisp2.

Case1 If[a]=[0]and[b]≠[0](similar for the case[a]≠[0]and[b]=[0]), then

〈([0],[b])〉=

{([0],[k])|k=0,1,2,…,p-1}

and the left cosets of 〈([0],[b])〉 are

([s],[0])+〈([0],[b])〉=

([s],[1])+〈([0],[b])〉=…=

([s],[p-1])+〈([0],[b])〉=

{([s],[k])|k=0,1,2,…,p-1},

wheres=0, 1, 2,…,p-1.It means that the number of the left cosets of 〈([0],[b])〉 isp.Further,Zp×Zp/〈([0],[b])〉 is generated by([1],[0])+〈([0],[b])〉 with orderp, which means thatZp×Zp/〈([0],[b])〉 is cyclic.Thus,Zp×Zp/〈([0],[b])〉?Zp.

Case2 If[a]≠[0]and[b]≠[0], then

〈([a],[b])〉={([0],[0]),([1],[t1]),

([p-1],[p-t1]),

([2],[t2]),([p-2],[p-t2]),…,

([k],[tk]),([p-k],[p-tk]),…,

([p-1],[tp-1]),([1],[p-tp-1])},

wherek∈{1,2,…,p-1} andt1,t2,…,tp-1∈{1,2,…,p-1} andti≠tjfori≠j, and the left cosets of 〈([a],[b])〉 are

([0],[0])+〈([a],[b])〉=

([1],[t1])+〈([a],[b])〉=

([p-1],[p-t1])+〈([a],[b])〉=

([2],[t2])+〈([a],[b])〉=

([p-2],[p-t2])+〈([a],[b])〉=…=

([p-1],[tp-1])+〈([a],[b])〉=

([1],[p-tp-1])+〈([a],[b])〉,

([0],[1])+〈([a],[b])〉=

([1],[t1+1])+〈([a],[b])〉=

([p-1],[p-t1+1])+〈([a],[b])〉=

([2],[t2+1])+〈([a],[b])〉=

([p-2],[p-t2+1])+〈([a],[b])〉=…=

([p-1],[tp-1+1])+〈([a],[b])〉=

([1],[p-tp-1+1])+〈([a],[b])〉,

?

([0],[p-1])+〈([a],[b])〉=

([1],[t1+p-1])+〈([a],[b])〉=

([p-1],[p-t1+p-1])+〈([a],[b])〉=

([2],[t2+p-1])+〈([a],[b])〉=

([p-2],[p-t2+p-1])+〈([a],[b])〉=…=

([p-1],[tp-1+p-1])+〈([a],[b])〉=

([1],[p-tp-1+p-1])+〈([a],[b])〉,

wheret1,t2,…,tp-1∈{1,2,…,p-1} andti≠tjfori≠j.It follows that the number of left cosets ofZp×Zpmodulo 〈([a],[b])〉 isp.From Theorem 7, we know thatZp×Zp/〈([a],[b])〉 is cyclic and finite.Therefore,Zp×Zp/〈([a],[b])〉?Zp.

Example4Z5×Z5/〈([1],[1])〉?Z5.

ProofLetN=〈([1],[1])〉.The left cosets ofZ5×Z5modulo〈([1],[1])〉 are

([0],[0])+N=([1],[1])+N=([2],[2])+N=([3],[3])+N=([4],[4])+N={([1],[1]),([2],[2]),([3],[3]),([4],[4]),([0],[0])},

([0],[1])+N=([1],[2])+N=([2],[3])+N=([3],[4])+N=([4],[0])+N={([1],[2]),([2],[3]),([3],[4]),([4],[0]),([0],[1])},

([0],[2])+N=([1],[3])+N=([2],[4])+N=([3],[0])+N=([4],[1])+N={([1],[3]),([2],[4]),([3],[0]),([4],[1]),([0],[2])},

([0],[3])+N=([1],[4])+N=([2],[0])+N=([3],[1])+N=([4],[2])+N={([1],[4]),([2],[0]),([3],[1]),([4],[2]),([0],[3])},

([0],[4])+N=([1],[0])+N=([2],[1])+N=([3],[2])+N=([4],[3])+N={([1],[0]),([2],[1]),([3],[2]),([4],[3]),([0],[4])}.

It follows that the number of left cosets ofZ5×Z5modulo 〈([1],[1])〉 is 5.By Theorem 7, we obtain thatZ5×Z5/〈([1],[1])〉 is cyclic and finite.In fact,Z5×Z5/〈([1],[1])〉 is generated by([0],[1])+N.Therefore,Z5×Z5/〈([1],[1])〉?Z5.

In Theorem 9, we consider the case whenm=n=pis prime and obtain an equivalent withZp×Zp/〈([a],[b])〉.In generalization, what is the case whenmis prime,m≠nand gcd(m,n)≠1? That is whenm=pis prime andn=kpfor some positive integerk≥2, whether we still have some similar result for the quotient group ofZp×Zkp? We know that the orderZp×Zkpiskp2, however, the order of 〈([a],[b])〉∈Zp×Zkpfor[a]∈Zp,[b]∈Zkpis less thankp2obviously.That means thatZp×Zkpis not cyclic.For this question, we obtain the following theorem.

Theorem10 Suppose thatpis prime andk≥2 is a positive integer.For any[a]∈Zp,[b]∈Zkpwith([a],[b])≠([0],[0]), if the order of 〈([a],[b])〉 iskp, then

Zp×Zkp/〈([a],[b])〉?Zp.

ProofSuppose that the order of 〈([a],[b])〉 iskp, then the order of[b]inZkpiskpby Theorem 1.It means that[b]≠[0].

Case3 If[a]=[0], then

〈([0],[b])〉=

{([0],[t])|t=0,1,2,…,kp-1}

and the left cosets of 〈([0],[b])〉 are

([s],[0])+〈([0],[b])〉=([s],[1])+〈([0],[b])〉=…=([s],[kp-1])+〈([0],[b])〉={([s],[t])|t=0,1,2,…,kp-1},

wheres= 0, 1, 2,…,p-1.It shows that the number of the left cosets of 〈([0],[b])〉 isp.Furthermore, we know thatZp×Zkp/〈([0],[b])〉 is generated by([1],[0])+〈([0],[b])〉 with orderp,which implies thatZp×Zkp/〈([0],[b])〉 is cyclic.Thus,Zp×Zkp/〈([0],[b])〉?Zp.

Case4 If[a]≠[0], then

〈([a],[b])〉={([0],[0]),([0],[p]),…,([0],[(k-1)p]),

([1],[t1]),([1],[t1+p]),…,([1],[t1+(k-1)p]),

…

([p-1],[tp-1]),([p-1],[tp-1+p]),…,([p-1],[tp-1+(k-1)p])},

wheret1,t2,…,tp-1∈{1,2,…,kp-1}andti≠tjfori≠j, and the left cosets of 〈([a],[b])〉 are

([0],[0])+〈([a],[b])〉=

([0],[p])+〈([a],[b])〉=…=

([0],[(k-1)p])+〈([a],[b])〉=

([1],[t1])+〈([a],[b])〉=

([1],[t1+p])+〈([a],[b])〉=…=

([1],[t1+(k-1)p])+〈([a],[b])〉=…=

([p-1],[tp-1+p])+〈([a],[b])〉=…=

([p-1],[tp-1+(k-1)p])+〈([a],[b])〉;

([0],[1])+〈([a],[b])〉=

([0],[p+1])+〈([a],[b])〉=…=

([0],[(k-1)p+1])+〈([a],[b])〉=

([1],[t1+1])+〈([a],[b])〉=

([1],[t1+p+1])+〈([a],[b])〉=…=

([1],[t1+(k-1)p+1])+〈([a],[b])〉=…=

([p-1],[tp-1+p+1])+〈([a],[b])〉=…=

([p-1],[tp-1+(k-1)p+1])+〈([a],[b])〉;

?

([0],[p-1])+〈([a],[b])〉=

([0],[p+p-1])+〈([a],[b])〉=…=

([0],[(k-1)p+p-1])+〈([a],[b])〉=

([1],[t1+p-1])+〈([a],[b])〉=

([1],[t1+p+p-1])+〈([a],[b])〉=…=

([1],[t1+(k-1)p+p-1])+

〈([a],[b])〉=…=

([p-1],[tp-1+p-1])+〈([a],[b])〉=

([p-1],[tp-1+p+p-1])+

〈([a],[b])〉=…=

([p-1],[tp-1+(k-1)p+p-1])+

〈([a],[b])〉,

wheret1,t2,…,tp-1∈{1,2,…,kp-1} andti≠tjfori≠j.It follows that the left cosets ofZp×Zkpmodulo 〈([a],[b])〉 isp.From Theorem 8, we know thatZp×Zkp/〈([a],[b])〉 is cyclic and finite.Therefore,Zp×Zkp/〈([a],[b])〉?Zp.Thus, we obtain the assertion.

Example5 Letm=p=3,n=3p=9, and([a],[b])=([1],[4]).Then

Z3×Z9/〈([1],[4])〉?Z3.

ProofIt is easy to obtain that

〈([1],[4])〉= {([1],[4]),([2],[8]),([0],[3]),([1],[7]),

([2],[2]),([0],[6]),([1],[1]),([2],[5]),([0],[0])},

and the order of 〈([1],[4])〉 is 9.LetN=〈([1],[4])〉.Then the left cosets ofZ3×Z9modulo 〈([1],[4])〉 are as follows.

([0],[0])+N=([1],[4])+N=([2],[8])+N=([0],[3])+N=([1],[7])+N=([2],[2])+N=([0],[6])+N=([1],[1])+N=([2],[5])+N={([1],[4]),([2],[8]),([0],[3]),([1],[7]),([2],[2]),([0],[6]),([1],[1]),([2],[5]),([0],[0])},

([0],[1])+N=([1],[5])+N=([2],[0])+N=([0],[4])+N=([1],[8])+N=([2],[3])+N=([0],[7])+N=([1],[2])+N=([2],[6])+N={([1],[5]),([2],[0]),([0],[4]),([1],[8]),([2],[3]),([0],[7]),([1],[2]),([2],[6]),([0],[1])},

([0],[2])+N=([1],[6])+N=([2],[1])+N=([0],[5])+N=([1],[0])+N=([2],[4])+N=([0],[8])+N=([1],[3])+N=([2],[7])+N={([1],[6]),([2],[1]),([0],[5]),([1],[0]),([2],[4]),([0],[8]),([1],[3]),([2],[7]),([0],[2])}.

It follows that the number of the left cosets ofZ3×Z9modulo 〈([1],[4])〉 is 3.By Theorem 8, we obtain thatZ3×Z9/〈([1],[4])〉 is cyclic and finite.In fact,Z3×Z9/〈([1],[4])〉 is generated by([0],[1])+Nwith order 3.Therefore,

Z3×Z9/〈([1],[4])〉?Z3.