Each cyclic group G=⟨a⟩ is Abelian. Indeed, aman=am+n=an+m=anam. Therefore, gH=Hg for all g∈G, and H is a normal subgroup of G.
Let [g]=gH be arbitrary element of G/H. Since G=⟨a⟩ is cyclic, g=ak for some k∈Z. By defenition of product of elements of G/H , [g]=[ak]=[a]k. Therefore, G/H=⟨[a]⟩, and G/H
is cyclic.
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