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If G is a cyclic group and N is a subgroup of G, show that G/N (or GmodN) is a cyclic group.

What I have so far: Since N is a subgroup of the cyclic group G, G/N is a cyclic group.

I think I'm missing details. What suggestions do you have?

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The exact statement goes for this as it did for your last post, you haven't actually stated why the result is true, you just said it (sorry if I sound harsh). Moreover, the same question I stated last time, applies here. More generally if $C$ is cyclic and $f:C\to H$ is a surjective homomorphism then $H$ is cyclic. Why does this help us? Why is it true?

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Thank you for responding. I didn't state why the result is true, because I don't know why yet. I'm not sure why that helps us. – Lily Nov 3 '11 at 19:07
Because whenever you have a quotient group $G/N$ you have the canonical projection $\pi:G\to G/N$ (i.e. the map $g\mapsto gN$). So, anything preserved under homomorphisms is preserved when going from $G$ to $G/N$. For example, back my actual comment, why if $C=\langle c\rangle$ why does $H=\langle f(c)\rangle$? – Alex Youcis Nov 3 '11 at 19:09

Explicitly: $G$ is isomorphic to either $\mathbb{Z}$ or $\mathbb{Z}_n$ for some $n\in\mathbb{Z_{\geq 2}}$.

In the first case, any subgroup is of the form $m\mathbb{Z}$ for some $m\in\mathbb{Z_{\geq2}}$ and so the quotient group is then isomorphic to $\mathbb{Z_m}$, which is cyclic.

In the second case, any subgroup of $\mathbb{Z}_n$ is isomorphic to $\mathbb{Z}_k$ where $k|n$. The quotient group $\mathbb{Z}_n/\mathbb{Z}_k$ is isomorphic to $\mathbb{Z}_{\frac{n}{k}}$ which again is cyclic.

So basically: all subgroups of cyclic groups are cyclic, and a cyclic group quotiented by a cyclic subgroup is again cyclic.

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Now $G = \langle g \rangle$ and $N \trianglelefteq G$ since $G$ is abelian. Since $N$ is normal in $G$, the quotient group $G/N$ exists and it makes sense to wonder if it is cyclic. Then you can show that $G/N = \langle gN \rangle$.

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