Prove that all subgroup of a cyclic group is cyclic.

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- Nov 9th 2008, 10:15 AMroporteCyclic group
Prove that all subgroup of a cyclic group is cyclic.

Thanks! - Nov 9th 2008, 10:39 AMclic-clac
Let $\displaystyle (G,.)$ be a cyclic group, $\displaystyle H$ a subgroup of $\displaystyle G$, and $\displaystyle \alpha \in G$ such that $\displaystyle <\alpha >=G$.

Let $\displaystyle \phi : \mathbb{Z} \rightarrow G : n \mapsto \alpha^{n}$ be the only morphism of groups such that $\displaystyle \phi(1)=\alpha$.

Then $\displaystyle \phi^{-1}(H)$ is a subgroup of $\displaystyle (\mathbb{Z},+)$, so it's a cyclic group.

So $\displaystyle \phi(\phi^{-1}(H))=H$ (because of surjectivity) is a cyclic group.

Why, if $\displaystyle K$ is a cyclic group, $\displaystyle L$ another group, and $\displaystyle \psi$ a morphism of groups between $\displaystyle K$ and $\displaystyle L$, then $\displaystyle \psi(K)$ is cyclic?

Let $\displaystyle a$ be a generator of $\displaystyle K$, and $\displaystyle b$ an element of $\displaystyle \psi(K)$. There is a $\displaystyle n \in \mathbb{N}$ such that $\displaystyle b=\psi(a^{n})=\psi(a)^{n}$. Thus $\displaystyle <\psi(a)>=\psi(K)$ - Nov 9th 2008, 07:17 PMThePerfectHacker