4.如果G为循环群,则有gk1g^{k1}gk1,gk2g^{k2}gk2,gk3g^{k3}gk3∈G,且gk3g^{k3}gk3=gk1g^{k1}gk1*gk2g^{k2}gk2,因为G−>->−>H为群同构,故 ϕ(gk3)ϕ(g^{k3})ϕ(gk3) = ϕ(gk1)ϕ(g^{k1})ϕ(gk1) ◦ ϕ(gk2)ϕ(g^{k2})ϕ(gk2) = ϕ(gk1⋅gk2)ϕ(g^{k1}·g^{k2})ϕ(gk1⋅gk2),得证ϕ(G)ϕ(G)ϕ(G)也为循环群。
如果G为交换群,则有 ∀a,b∈G∀a,b∈G∀a,b∈G,ab=baab=baab=ba,故有ϕ(ab)ϕ(ab)ϕ(ab)=ϕ(ba)ϕ(ba)ϕ(ba),所以ϕ(ab)ϕ(ab)ϕ(ab)=ϕ(a)ϕ(a)ϕ(a) ◦ ϕ(b)ϕ(b)ϕ(b)=ϕ(b)ϕ(b)ϕ(b) ◦ ϕ(a)ϕ(a)ϕ(a)=ϕ(ba)ϕ(ba)ϕ(ba),得证ϕ(G)ϕ(G)ϕ(G)也为交换群。
5.由题意可得群G有两个不同的陪集,对 ∀g∈G∀g∈G∀g∈G:当ggg落在HHH上,则显然gH=Hg=HgH=Hg=HgH=Hg=H;当ggg不落在H上时,根据同一或不相交性,gH=Hg=G−HgH=Hg=G-HgH=Hg=G−H,故对 ∀g∈G∀g∈G∀g∈G,均有gH=HggH=HggH=Hg,故得证HHH是GGG的正规子群。
6.对 ∀aH∀aH∀aH,bH∈G/HbH∈G/HbH∈G/H,存在H1,H2H_1,H_2H1,H2使得aH1aH_1aH1,bH2bH_2bH2落在aaa,bbb上,其中aaa,b∈Gb∈Gb∈G,又因为GGG为阿尔贝群,则有ab=baab=baab=ba,故aH1bH2=bH2aH1aH_1bH_2=bH_2aH_1aH1bH2=bH2aH1,aH1bH2HH=bH2aH1HHaH_1bH_2HH=bH_2aH_1HHaH1bH2HH=bH2aH1HH,得aHbH=bHaHaHbH=bHaHaHbH=bHaH,故得证G/H也为阿尔贝群。
7.因为G是循环群,故有 ∀gk1,gk2∈G∀g^{k1},g^{k2}∈G∀gk1,gk2∈G,使得gk3=gk1∗gk2g^{k3} = g^{k1} * g^{k2}gk3=gk1∗gk2,对于商群G/HG/HG/H有(gk1H)(gk2H)=gk3H(g^{k1}H)(g^{k2}H) = g^{k3}H(gk1H)(gk2H)=gk3H,得证G/HG/HG/H也为循环群。