Question #146439

Assume that (G,*) is a cyclic group and that H is a subgroup of G. Explain why H is a normal subgroup G,and prove that G/H is cyclic.

Expert's answer

Each cyclic group G=⟨a⟩G=\langle a \rangle is Abelian. Indeed, aman=am+n=an+m=anama^ma^n=a^{m+n}=a^{n+m}=a^na^m. Therefore, gH=HggH=Hg for all g∈Gg\in G, and HH is a normal subgroup of GG.


Let [g]=gH[g]=gH be arbitrary element of G/HG/H. Since G=⟨a⟩G=\langle a \rangle is cyclic, g=akg=a^k for some k∈Z.k\in \mathbb Z. By defenition of product of elements of G/HG/H , [g]=[ak]=[a]k.[g]=[a^k]=[a]^k. Therefore, G/H=⟨[a]⟩,G/H=\langle[a]\rangle, and G/HG/H

is cyclic.


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