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Cyclic Group $Z_{n}$

How do I find all the Subgroups of $Z_{200}$ under additive modulo 200? And how many subgroups are there? I know the definition of subgroups but to find all the subgroups of such a large set is a bit difficult for me. Is there a smart way to find it?

The answer to this question is 12 and I not sure if this is correct.

Manzoor Haider
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    Denote $[m]$ the equivalence class of $m$ in $\mathbb{Z}_n$. Since $\mathbb{Z}_n$ is cyclic, all its subgroups are cyclic. Denote $([m])$ the subgroup generated by $[m]$. Recall that $([m])=([gcd(m,n)])$ where $gcd(m,n)$ is the greatest common divisor of $m$ and $n$. From these facts, you can determine all subgroups of $\mathbb{Z}_n$. – Frank Lu Mar 02 '18 at 20:26
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    Hint : for every divisor $d$ of $n$, there is exactly one subgroup of order $d$ in $\mathbb{Z}_{n}$ and this gives all the subgroups. Edit : Sorry, Frank Lu, we posted at the same moment. – Panurge Mar 02 '18 at 20:26
  • Please ask one question at a time. – Shaun Mar 02 '18 at 20:27
  • Thanks you all! – Manzoor Haider Mar 03 '18 at 09:03
  • Those who have marked the Question as Duplicate let me tell you that I have joined this website yesterday and don't know whether the question exits. Sorry about that! – Manzoor Haider Mar 03 '18 at 09:07

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Denote by $d(n)$ the number of positive divisors of $n$. Then $\mathbb{Z}_n$ has $d(n)$ subgroups, because a cyclic group of order $n$ has a unique subgroup for each divisor $d\mid n$, see here:

Subgroups of a cyclic group and their order.

Since $d(200)=12$, we are done.

Dietrich Burde
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