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Find all subgroups of $\mathbb{Z}_{1001}$, where $1001=7\cdot 11\cdot 13$.

By Lagrange, we know the order of every subgroup must divide 1001, but I'm not sure how the subgroups should be listed/how to find them..

Mikasa
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Bernt
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    This is answered by this: http://math.stackexchange.com/questions/410389/subgroups-of-a-cyclic-group-and-their-order – Najib Idrissi Jun 05 '13 at 10:30
  • Please note that this is not a duplicate question. An answer can be obtained by a statement made in the linked question, but the linked question is not about how to prove that statement, but only about how to prove a small part of it, which is not necessarily relevant to this question. – Tom Oldfield Jun 05 '13 at 11:19
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    I agree with @TomOldfield - essentially, the "duplicate" question is a hint. My personal feeling is that the OP should try to use the hint to answer the question, and then write up the solution as the answer. If the hint does not help, then the OP should say so and help will be provided... – user1729 Jun 05 '13 at 11:26
  • @TomOldfield: I disagree. If you admit Lagrange's theorem (which the OP mentioned), then the linked question answers completely this question: there is exactly one subgroup for every divisor of 1001, and no other. So the question is "how many divisors does 1001 have", which I hope is not a problem for anyone. – Najib Idrissi Jun 07 '13 at 08:03
  • I would also argue that it's a case of "abstract duplicates": http://meta.math.stackexchange.com/questions/1756/coping-with-abstract-duplicate-questions – Najib Idrissi Jun 07 '13 at 08:04
  • @nik If you look at confused reply of the OP tothe answer below, you will hopefully understand why it is important to engage with the OP rather than just closing a questions because it is an "abstract duplicate". Examples are important, and the question you link to gives no examples. – user1729 Jun 07 '13 at 08:22

1 Answers1

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Hints , since $\mathbb{Z_{1001}}$ is finite cyclic group. So for every divisor of $1001$ it has unique subgroup of that order ..

Mikasa
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Andy
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  • Unique as in there is only three subgroups, generated by each of the prime factors? – Bernt Jun 05 '13 at 10:38
  • @Benet, not ont only prime factor ,thre is also other factors ,all factors are 1,7,11,13,711,1113,713,711*13 ,each of the factor produce all subgroups. – Andy Jun 05 '13 at 10:43